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Unformatted text preview: Back Print HOLT® ChemFile ® Problem-Solving Workbook Back Print Name Class Date Skills Worksheet Problem Solving Conversions One of the aims of chemistry is to describe changes—to tell what changed, how it changed, and what it changed into. Another aim of chemistry is to look at matter and its changes and to ask questions such as how much, how big, how hot, how many, how hard, and how long did it take. For example, chemistry asks the following: • • • • • • • How much energy is needed to start a reaction? How much will the volume of a gas increase if you heat it? How long will a reaction take? How much can a reaction produce? How much of the reactant is needed to produce a required amount of product? How much energy does a reaction release? How high will the temperature of the solution get as a reaction occurs? To answer these questions, chemists must make measurements. Measurements in science can never be treated as just numbers; they must always involve both a number and a unit. When you use measurements to calculate any quantity, the unit must always accompany the number in the calculation. Sometimes the unit given is not the most appropriate unit for the situation or calculation. In this case, a conversion can change the impractical unit into a more useful one. For instance, you would not want to measure the distance from New York City to San Francisco in inches. A simple conversion can transform the number of inches between the two cities to the much more practical number of miles. General Plan for Converting Measurements State the relationship between the unit given and the unit sought as an equality. 1 Quantity and unit given 2 Relationship between units given and units sought Use the equality to write possible conversion factors. 4 Quantity and unit sought 3 Apply each conversion factor successively so that the units cancel properly. Conversion factor Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 1 Conversions Back Print Name Class Date Problem Solving continued CONVERTING SIMPLE SI UNITS Sample Problem 1 A small bottle contains 45.5 g of calcium chloride. What is the mass of calcium chloride in milligrams? Solution ANALYZE What is given in the problem? What are you asked to find? the mass of calcium chloride in grams the mass of calcium chloride in milligrams A table showing what you know and what you do not know can help you organize the data. Being organized is a key to developing good problem solving skills. Items Quantity given Units of quantity given Units of quantity sought Relationship between units Conversion factor Quantity sought Data 45.5 g calcium chloride grams milligrams 1g ? ? mg calcium chloride 1000 mg PLAN What steps are needed to convert grams to milligrams? Determine a conversion factor that relates grams and milligrams. Multiply the number of grams by that factor. Arrange the factor so that units cancel to give the units sought. 1 Mass of calcium chloride in g write the relationship between g and mg 2 1g 1000 mg possible conversion factors: 1 g or 1000 mg 1g 1000 mg 4 Mass of calcium chloride in mg multiply by conversion factor 3 1000 mg 1g Relationship between units: 1 g Possible conversion factors: 1000 mg 1g 1000 mg or 1000 mg 1g Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 2 Conversions Back Print Name Class Date Problem Solving continued The correct conversion factor is the one that when multiplied by the given quantity causes the units to cancel. conversion factor g 3 mg given g calcium chloride 1000 mg 1g quantity sought mg calcium chloride COMPUTE 45.5 g calcium chloride EVALUATE Are the units correct? Yes; milligrams are the desired units. Grams cancel to give milligrams. 1000mg 1g 45 500 mg calcium chloride Is the answer reasonable? Yes; the number of milligrams is 1000 times the number of grams. Practice 1. State the following measured quantities in the units indicated: a. 5.2 cm of magnesium ribbon in millimeters ans: 52 mm b. 0.049 kg of sulfur in grams ans: 49 g c. 1.60 mL of ethanol in microliters ans: 1600 L d. 0.0025 g of vitamin A in micrograms ans: 2500 g e. 0.020 kg of tin in milligrams ans: 20 000 mg f. 3 kL of saline solution in liters ans: 3000 L Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 3 Conversions Back Print Name Class Date Problem Solving continued 2. State the following measured quantities in the units indicated: a. 150 mg of aspirin in grams ans: 0.15 g b. 2500 mL of hydrochloric acid in liters ans: 2.5 L c. 0.5 g of sodium in kilograms ans: 0.0005 kg d. 55 L of carbon dioxide gas in kiloliters ans: 0.055 kL e. 35 mm in centimeters ans: 3.5 cm h. 8740 m in kilometers ans: 8.74 km i. 209 nm in millimeters ans: 0.000 209 mm j. 500 000 g in kilograms ans: 0.0005 khhtg 3. The greatest distance between Earth and the sun during Earth’s revolution is 152 million kilometers. What is this distance in megameters? ans: 152 000 Mm Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 4 Conversions Back Print Name Class Date Problem Solving continued Sample Problem 2 A metallurgist is going to make an experimental alloy that requires adding 325 g of bismuth to 2.500 kg of molten lead. What is the total mass of the mixture in kilograms? Solution ANALYZE What is given in the problem? What are you asked to find? Items Quantity given Units of quantity given Units of quantity sought Relationship between units Conversion factor Quantity sought Mass of lead Total mass the mass of bismuth in grams, the mass of lead in kg the total mass of the mixture Data 325 g of bismuth grams kilograms 1000 g ? ? kg of bismuth 2.500 kg ? kg of mixture 1 kg PLAN To be added, the quantities must be expressed in the same units—in this case, kilograms. Therefore, 325 g of bismuth must be converted to kilograms of bismuth. What steps are needed to convert grams to kilograms? Determine a conversion factor that relates grams to kilograms. Apply that conversion factor to obtain the quantity sought. What steps are needed to find the total mass of the mixture in kilograms? Add the mass of the lead in kilograms to the mass of the bismuth in kilograms. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 5 Conversions Back Print Name Class Date Problem Solving continued 1 Mass of bismuth in g write the relationship between g and kg 2 1000 g 1 kg possible conversion factors: 1 kg or 1000 g 1000 mg 1 kg 4 Mass of bismuth in kg add the two masses to obtain the mass of the mixture multiply by conversion factor 3 1 kg 1000 g Mass of mixture in kg Relationship between units: 1000 g Possible conversion factors: conversion factor g 3 kg 1 kg 1000 g 1 kg or 1 kg 1000 g quantity sought given g bismuth calculated above 1 kg 1000 g given kg bismuth kg mixture kg bismuth kg lead COMPUTE 325 g bismuth 0.325 kg bismuth EVALUATE Are the units correct? Yes; kilograms are the units sought. 1 kg 1000 g 0.325 kg bismuth 2.825 kg mixture 2.500 kg lead Is the answer reasonable? Yes; the value, 0.325, is one-thousandth the given value, 325. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 6 Conversions Back Print Name Class Date Problem Solving continued Practice 1. How many milliliters of water will it take to fill a 2 L bottle that already contains 1.87 L of water? ans: 130 mL 2. A piece of copper wire is 150 cm long. How long is the wire in millimeters? How many 50 mm segments of wire can be cut from the length? ans: 1500 mm; 30 pieces 3. The ladle at an iron foundry can hold 8500 kg of molten iron. 646 metric tons of iron are needed to make rails. How many ladlefuls of iron will it take to make 646 metric tons of iron? (1 metric ton 1000 kg) ans: 76 ladlefuls Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 7 Conversions Back Print Name Class Date Problem Solving continued CONVERTING DERIVED SI UNITS Sample Problem 3 A balloon contains 0.5 m3 of neon gas. What is the volume of gas in cubic centimeters? Solution ANALYZE What is given in the problem? What are you asked to find? Items Quantity given Units of quantity given Units of quantity sought Relationship between units Conversion factor Quantity sought the volume of neon in cubic meters volume of neon in cubic centimeters Data 0.5 m3 of neon cubic meters cubic centimeters 1m ? ? cm3 neon 100 cm PLAN What steps are needed to convert cubic meters to cubic centimeters? Rewrite the quantity in simple SI units. Determine the relationship between meters and centimeters. Write a conversion factor for each of the units in the given quantity. Multiply that quantity by the conversion factors. Arrange the factors so that units will cancel to give the units of the quantity sought. 1 Volume of neon in m3 Volume of neon in (m m m) 1m 2 100 cm possible conversion factors: 1 m or 100 cm 100 cm 1m 4 Volume of neon in cm3 apply conversion factor three times to convert from cubic meters to cubic centimeters 3 100 cm 1m Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 8 Conversions Back Print Name Class Date Problem Solving continued Rewrite the given quantity in simple units as follows. m3 m m m 100 cm 1m 100 cm or 100 cm 1m Relationship between units: 1 m Possible conversion factors: conversion factors cm 3 m, applied three times given m3 neon 100 cm 1m 100 cm 1m 100 cm 1m quantity sought cm3 neon COMPUTE 0.50 m3 neon 0.50 (m m m) 100 cm 1m 100 cm 1m 100 cm 1m 500 000 cm3 neon EVALUATE Are the units correct? Yes; cubic centimeters were the units sought. Is the answer reasonable? Yes; 500 000 cm3 is half the number of cm3 in 1 m3. Practice 1. State the following measured quantities in the units indicated. a. 310 000 cm3 of concrete in cubic meters ans: 0.31 m3 b. 6.5 m2 of steel sheet in square centimeters ans: 65 000 cm2 c. 0.035 m3 of chlorine gas in cubic centimeters ans: 35 000 cm3 d. 0.49 cm2 of copper in square millimeters ans: 49 mm2 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 9 Conversions Back Print Name Class Date Problem Solving continued e. 1200 dm3 of acetic acid solution in cubic meters ans: 1.2 m3 f. 87.5 mm3 of actinium in cubic centimeters ans: 0.0875 cm3 g. 250 000 cm2 of polyethylene sheet in square meters ans: 25 m2 2. How many palisade cells from plant leaves would fit in a volume of 1.0 cm3 of cells if the average volume of a palisade cell is 0.0147 mm3? ans: 68 027 cells Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 10 Conversions Back Print Name Class Date Problem Solving continued Additional Problems 1. Convert each of the following quantities to the required unit. a. 12.75 Mm to kilometers b. 277 cm to meters c. 30 560 m2 to hectares (1 ha d. 81.9 cm2 to square meters e. 300 000 km to megameters 2. Convert each of the following quantities to the required unit. a. 0.62 km to meters b. 3857 g to milligrams c. 0.0036 mL to microliters d. 0.342 metric tons to kilograms (1 metric ton e. 68.71 kL to liters 3. Convert each of the following quantities to the required unit: a. 856 mg to kilograms b. 1 210 000 g to kilograms c. 6598 L to cubic centimeters (1 mL d. 80 600 nm to millimeters e. 10.74 cm3 to liters 4. Convert each of the following quantities to the required unit: a. 7.93 L to cubic centimeters b. 0.0059 km to centimeters c. 4.19 L to cubic decimeters d. 7.48 m2 to square centimeters e. 0.197 m3 to liters 5. An automobile uses 0.05 mL of oil for each kilometer it is driven. How much oil in liters is consumed if the automobile is driven 20 000 km? 6. How many microliters are there in a volume of 370 mm3 of cobra venom? 7. A baker uses 1.5 tsp of vanilla extract in each cake. How much vanilla extract in liters should the baker order to make 800 cakes? (1 tsp 5 mL) 8. A person drinks eight glasses of water each day, and each glass contains 300 mL. How many liters of water will that person consume in a year? What is the mass of this volume of water in kilograms? (Assume one year has 365 days and the density of water is 1.00 kg/L.) 10 000 m2) 1000 kg) 1 cm3) Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 11 Conversions Back Print Name Class Date Problem Solving continued 9. At the equator Earth rotates with a velocity of about 465 m/s. a. What is this velocity in kilometers per hour? b. What is this velocity in kilometers per day? 10. A chemistry teacher needs to determine what quantity of sodium hydroxide to order. If each student will use 130 g and there are 60 students, how many kilograms of sodium hydroxide should the teacher order? 11. The teacher in item 10 also needs to order plastic tubing. If each of the 60 students needs 750 mm of tubing, what length of tubing in meters should the teacher order? 12. Convert the following to the required units. a. 550 L/h to milliliters per day b. 9.00 metric tons/h to kilograms per minute c. 3.72 L/h to cubic centimeters per minute d. 6.12 km/h to meters per second 13. Express the following in the units indicated. a. 2.97 kg/L as grams per cubic centimeter b. 4128 g/dm2 as kilograms per square centimeter c. 5.27 g/cm3 as kilograms per cubic decimeter d. 6.91 kg/m3 as milligrams per cubic millimeter 14. A gas has a density of 5.56 g/L. a. What volume in milliliters would 4.17 g of this gas occupy? b. What would be the mass in kilograms of 1 m3 of this gas? 15. The average density of living matter on Earth’s land areas is 0.10 g/cm2. What mass of living matter in kilograms would occupy an area of 0.125 ha? 16. A textbook measures 250. mm long, 224 mm wide, and 50.0 mm thick. It has a mass of 2.94 kg. a. What is the volume of the book in cubic meters? b. What is the density of the book in grams per cubic centimeter? c. What is the area of one cover in square meters? 17. A glass dropper delivers liquid so that 25 drops equal 1.00 mL. a. What is the volume of one drop in milliliters? b. How many milliliters are in 37 drops? c. How many drops would be required to get 0.68 L? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 12 Conversions Back Print Name Class Date Problem Solving continued 18. Express each of the following in kilograms and grams: a. 504 700 mg b. 9 200 000 g c. 122 mg d. 7195 cg 19. Express each of the following in liters and milliliters: a. 582 cm3 b. 0.0025 m3 c. 1.18 dm3 d. 32 900 L 20. Express each of the following in grams per liter and kilograms per cubic meter. a. 1.37 g/cm3 b. 0.692 kg/dm3 c. 5.2 kg/L d. 38 000 g/m3 e. 5.79 mg/mm3 f. 1.1 g/mL 21. An industrial chemical reaction is run for 30.0 h and produces 648.0 kg of product. What is the average rate of product production in the stated units? a. grams per minute b. kilograms per day c. milligrams per millisecond 22. What is the speed of a car in meters per second when it is moving at 100. km/h? 23. A heater gives off energy as heat at a rate of 330 kJ/min. What is the rate of energy output in kilocalories per hour? (1 cal 4.184 J) 24. The instructions on a package of fertilizer tell you to apply it at the rate of 62 g/m2. How much fertilizer in kilograms would you need to apply to 1.0 ha? (1 ha 10 000 m2) 25. A water tank leaks water at the rate of 3.9 mL/h. If the tank is not repaired, what volume of water in liters will it leak in a year? Show your setup for solving this. Hint: Use one conversion factor to convert hours to days and another to convert days to years, and assume that one year has 365 days. 26. A nurse plans to give flu injections of 50 L each from a bottle containing 2.0 mL of vaccine. How many doses are in the bottle? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 13 Conversions Back Print Name Class Date Skills Worksheet Problem Solving Significant Figures A lever balance used to weigh a truckload of stone may be accurate to the nearest 100 kg, giving a reading of 15 200 kg, for instance. The measurement should be written in such a way that a person looking at it will understand that it represents the mass of the truck to the nearest 100 kg, that is, that the mass is somewhere between 15 100 kg and 15 300 kg. Some laboratory balances are sensitive to differences of 0.001 g. Suppose you use such a balance to weigh 0.206 g of aluminum foil. A person looking at your data table should be able to see that the measurement was made on a balance that measures mass to the nearest 0.001 g. You should not state the measurement from the laboratory balance as 0.2060 g instead of 0.206 g because the balance was not sensitive enough to measure 0.0001 g. To convey the accuracy of measurements, all people working in science use significant figures. A significant figure is a digit that represents an actual measurement. The mass of the truck was stated as 15 200 kg. The 1, 5, and 2 are significant figures because the balance was able to measure ten-thousands, thousands, and hundreds of kilograms. The truck balance was not sensitive enough to measure tens of kilograms or single kilograms. Therefore, the two zeros are not significant and the measurement has three significant figures. The mass of the foil was correctly stated as 0.206 g. There are three decimal places in this measurement that are known with some certainty. Therefore, this measurement has three significant figures. Had the mass been stated as 0.2060 g, a fourth significant figure would have been incorrectly implied. Rules for Determining Significant Figures A. All digits that are not zeros are significant. All are nonzero digits. 222 3 2 5 mL of ethanol The measurement has three significant figures. All are nonzero digits. 2222 1.3 2 5 g of zinc The measurement has four significant figures. B. Zeros may or may not be significant. To determine whether a zero is significant, use the following rules: Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 14 Significant Figures Back Print Name Class Date Problem Solving continued 1. Zeros appearing between nonzero digits are significant. Nonzero digits 22 4 0. 7 L of ammonia 1 Zero between nonzero digits The measurement has three significant figures. Nonzero digits 22 2 3 2 0 0 6 m of wire 11 Zero between nonzero digits The measurement has five significant figures. 2. Zeros appearing in front of nonzero digits are not significant. Nonzero digits 222 0.0 5 7 2 m2 of foil The measurement has three significant figures. Nonzero digits 2 0.000 2 g of RNA The measurement has one significant figure. 3. Zeros at the end of a number and to the right of a decimal are significant figures. Zeros between nonzero digits and significant zeros are also significant. This is a restatement of Rule 1. Nonzero digits 2 2 9 7. 0 0 kg of tungsten 11 Zeros to the right of a number and after a decimal point The measurement has four significant figures. Nonzero digits 2 2 1 2 0 0.0 0 cm3 of lead 1111 Zeros to the right of a number and after a decimal point The measurement has six significant figures. 4. Zeros at the end of a number but to the left of a decimal may or may not be significant. If such a zero has been measured or is the first estimated digit, it is significant. On the other hand, if the zero has not been measured or estimated but is just a place holder, it is not significant. A decimal placed after the zeros indicates that they are significant. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 15 Significant Figures Back Print Name Class Date Problem Solving continued Nonzero digits 22 3 4 0 0 g of sulfur The measurement has two significant figures. Nonzero digits 22 4 0 0 0. mL of oxygen 11 Decimal point is present, so these zeros are significant. The measurement has four significant figures. The rules are summarized in the following flowchart: General Plan for Determining Significant Figures Look for nonzero digits— all of them are significant. Look for zeros between nonzero digits—they are significant. Look for a decimal point. There is a decimal point. There is no decimal point. Look for zeros at the end of the number—they are significant. Total number of significant figures Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 16 Significant Figures Back Print Name Class Date Problem Solving continued Sample Problem 1 Determine the number of significant figures in the following measurements: a. 30 040 g b. 0.663 kg c. 20.05 mL d. 1500. mg e. 0.0008 m Solution ANALYZE What is given in the problem? What are you asked to find? five measurements the number of significant figures in each measurement Data a Measured quantity 30 040 g b 0.663 kg c 20.05 L d 1500. mg e 0.0008 g Items PLAN What steps are needed to determine the number of significant figures in each measurement? Apply the steps in the flowchart to determine the number of significant figures. Apply the following steps from the flowchart. Eliminate the steps that are not applicable to the measurement in question. How many nonzero digits are there? How many zeros are there between nonzero digits? Is there a decimal point? How many significant zeros are at the end of the number? Total number of significant figures ? ? ? ? ? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 17 Significant Figures Back Print Name Class Date Problem Solving continued SOLVE a. 30 040 g How many nonzero digits are there? How many zeros are there between nonzero digits? Is there a decimal point? How many significant zeros are at the end of the number? Total number of significant figures 2 2 no NA 4 The final zero is not significant. b. 0.663 kg How many nonzero digits are there? How many zeros are there between nonzero digits? Is there a decimal point? How many significant zeros are at the end of the number? Total number of significant figures 3 NA yes NA 3 The zero only locates the decimal point and is not significant. c. 20.05 L How many nonzero digits are there? How many zeros are there between nonzero digits? Is there a decimal point? How many significant zeros are at the end of the number? Total number of significant figures 2 2 yes NA 4 d. 1500. mg How many nonzero digits are there? How many zeros are there between nonzero digits? Is there a decimal point? How many significant zeros are at the end of the number? Total number of significant figures 2 NA yes 2 4 There is a decimal following the final two zeros, so all digits are significant. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 18 Significant Figures Back Print Name Class Date Problem Solving continued e. 0.0008 g How many nonzero digits are there? How many zeros are there between nonzero digits? Is there a decimal point? How many significant zeros are at the end of the number? Total number of significant figures 1 NA yes NA 1 The zeros are only place holders. They are not significant. EVALUATE Are the answers reasonable? Yes; all answers are in agreement with the rules for determining significant figures. Practice 1. Determine the number of significant figures in the following measurements: ans: 2 a. 640 cm3 _____________________ ans: 4 b. 200.0 mL ____________________ ans: 4 c. 0.5200 g _____________________ ans: 4 d. 1.005 kg _____________________ ans: 1 e. 10 000 L _____________________ ans: 5 f. 20.900 cm ___________________ ans: 2 g. 0.000 000 56 g/L ______________ ans: 4 h. 0.040 02 kg/m3 _______________ ans: 6 i. 790 001 cm2 __________________ ans: 6 j. 665.000 kg m/s2 ______________ DETERMINING SIGNIFICANT FIGURES IN CALCULATIONS Suppose you want to determine the density of an ethanol-water solution. You first measure the volume in a graduated cylinder that is accurate to the nearest 0.1 mL. You then determine the mass of the solution on a balance that can measure mass to the nearest 0.001 g. You have read each measuring device as accurately as you can, and you record the following data: Measurement Mass of solution, m Volume of solution, V Density of solution in g/mL, D Data 11.079 g 12.7 mL ? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 19 Significant Figures Back Print Name Class Date Problem Solving continued You can determine density on your calculator and get the following result: D m V 11.079 g 12.7 mL 0.872 362 204 g/mL Although the numbers divide out to give the result shown, it is not correct to say that this quantity is the density of the solution. Remember that you are dealing with measurements, not just numbers. Consider the fact that you measured the mass of the solution with a balance that gave a reading with five significant figures: 11.079 g. In addition, you measured the volume of the solution with a graduated cylinder that was readable only to three significant figures: 12.7 mL. It seems odd to claim that you now know the density with an accuracy of nine significant figures. You can calculate the density—or any measurement—only as accurately as the least accurate measurement that was used in the calculation. In this case the least accurate measurement was the volume because the measuring device you used was capable of giving you a measurement with only three significant figures. Therefore, you can state the density to only three significant figures. Rules for Calculating with Measured Quantities Operation Multiplication and division Rule • Round off the calculated result to the same number of significant figures as the measurement having the fewest significant figures. • Round off the calculated result to the same number of decimal places as the measurement with the fewest decimal places. If there is no decimal point, round the result back to the digit that is in the same position as the leftmost uncertain digit in the quantities being added or subtracted. Addition and subtraction In the example given above, you must round off your calculator reading to a value that contains three significant figures. In this case, you would say: D m V 11.079 g 12.7 mL 0.872 362 204 g/mL 0.872 g/mL Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 20 Significant Figures Back Print Name Class Date Problem Solving continued Sample Problem 2 In an experiment to identify an unknown gas, it is found that 1.82 L of the gas has a mass of 5.430 g. What is the density of the gas in g/L? Solution ANALYZE What is given in the problem? What are you asked to find? Items Mass of the gas, mgas Volume of the gas, Vgas Density of the gas, Dgas (numerical result) Least number of significant figures in measurements Density of the gas, Dgas (rounded) the measured mass and volume of the gas the density of the gas Data 5.430 g 1.82 L ? g/L 3 (in 1.82 L) ? g/L PLAN What step is needed to calculate the density of the gas? Divide the mass measurement by the volume measurement. What steps are necessary to round the calculated value to the correct number of significant figures? Determine which measurement has the fewest significant figures. Round the calculated result to that number of significant figures. Dgas mgas Vgas numerical result round to correct significant figures rounded result COMPUTE Dgas mgas Vgas four significant figures three significant figures the digit following the 8 is less than 5, so the 8 remains unchanged 5.430 g 1.82 L round to three significant figures 2.983 516 484 2.98 g/L EVALUATE Are the units correct? Yes; density is given in units of mass per unit volume. Are the significant figures correct? Yes; the mass had only three significant figures, so the answer was rounded to three significant figures. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 21 Significant Figures Back Print Name Class Date Problem Solving continued Is the answer reasonable? Yes; the mass/volume ratio is roughly 3/1, so the density is approximately 3 g/L. Practice 1. Perform the following calculations, and express the result in the correct units and number of significant figures. a. 47.0 2.2 s ans: 21 m/s b. 140 cm 35 cm ans: 4900 cm2 c. 5.88 kg 200 m3 ans: 0.03 kg/m3 d. 0.00 50 m2 0.042 m ans: 0.000 21 m3 e. 300.3 L 180. s ans: 1.67 L/s f. 33.00 cm2 2.70 cm ans: 89.1 cm3 g. 35 000 kJ 0.250 min ans: 140 000 kJ/min Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 22 Significant Figures Back Print Name Class Date Problem Solving continued Sample Problem 3 Three students measure volumes of water with three different devices. They report the following volumes: Device Large graduated cylinder Small graduated cylinder Calibrated buret Volume measured 164 mL 39.7 mL 18.16 mL If the students pour all of the water into a single container, what is the total volume of water in the container? Solution ANALYZE What is given in the problem? What are you asked to find? Items First volume of water Second volume of water Third volume of water Total volume of water three measured volumes of water the total volume of water Data 164 mL 39.7 mL 18.16 mL ? PLAN What step is needed to calculate the total volume of the water? Add the separate volumes. What steps are necessary to round the calculated value to the correct number of significant figures? Determine which measurement has the fewest decimal places. Round the calculated result to that number of decimal places. COMPUTE Vtotal V1 V2 V3 164 mL 164 mL 39.7 mL 18.16 mL 221.86 mL Round the sum to the same number of decimal places as the measurement with the fewest decimal places (164 mL). Vtotal 221.86 mL 222 mL 39.7 mL 18.16 mL the digit following the 1 is greater than 5, so the 1 is rounded up to 2 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 23 Significant Figures Back Print Name Class Date Problem Solving continued EVALUATE Are the units correct? Yes; the given values have units of mL. Are the significant figures correct? Yes; three significant figures is correct. Is the answer reasonable? Yes; estimating the values as 160, 40, and 20 gives a sum of 220, which is very near the answer. Practice 1. Perform the following calculations and express the results in the correct units and number of significant figures: a. 22.0 m 5.28 m 15.5 m ans: 42.8 m b. 0.042 kg 1.229 kg 0.502 kg ans: 1.773 kg c. 170 cm2 3.5 cm2 28 cm2 ans: 150 cm2 d. 0.003 L 0.0048 L 0.100 L ans: 0.108 L e. 24.50 dL 4.30 dL 10.2 dL ans: 39.0 dL f. 3200 mg 325 mg 688 mg ans: 2800 mg g. 14 000 kg 8000 kg 590 kg ans: 23 000 kg Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 24 Significant Figures Back Print Name Class Date Problem Solving continued Additional Problems 1. Determine the number of significant figures in the following measurements: a. 0.0120 m b. 100.5 mL c. 101 g d. 350 cm 2 f. 1000 kg g. 180. mm h. 0.4936 L i. 0.020 700 s e. 0.97 km 2. Round the following quantities to the specified number of significant figures: a. 5 487 129 m to three significant figures b. 0.013 479 265 mL to six significant figures c. 31 947.972 cm2 to four significant figures d. 192.6739 m2 to five significant figures e. 786.9164 cm to two significant figures f. 389 277 600 J to six significant figures g. 225 834.762 cm3 to seven significant figures 3. Perform the following calculations, and express the answer in the correct units and number of significant figures. a. 651 cm b. 7.835 kg c. 14.75 L 75 cm 2.5 L 1.20 s d. 360 cm e. 5.18 m f. 34.95 g 51 cm 0.77 m 9.07 cm 10.22 m 11.169 cm3 4. Perform the following calculations, and express the answer in the correct units and number of significant figures. a. 7.945 J b. 0.0012 m c. 500 g e. 312 dL f. 1701 kg d. 31.2 kPa 82.3 J 432 g 0.02 J 0.000 11 m 0.147 kPa 2g 3.12 dL 43 kg 0.000 45 m 0.0035 kPa 31.2 dL 50 kg 5. A rectangle measures 87.59 cm by 35.1 mm. Express its area with the proper number of significant figures in the specified unit: a. in cm2 b. in mm2 c. in m2 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 25 Significant Figures Back Print Name Class Date Problem Solving continued 6. A box measures 900. mm by 31.5 mm by 6.3 cm. State its volume with the proper number of significant figures in the specified unit: a. in cm3 b. in m3 c. in mm3 7. A 125 mL sample of liquid has a mass of 0.16 kg. What is the density of the liquid in the following measurements? a. kg/m3 b. g/mL c. kg/dm3 8. Perform the following calculations, and express the results in the correct units and with the proper number of significant figures. a. 13.75 mm b. 89.4 cm2 c. 14.9 m 3 10.1 mm 4.8 cm 3.0 m2 30 m 0.91 mm d. 6.975 m 21.5 m 9. What is the volume of a region of space that measures 752 m 319 m 110 m? Give your answer in the correct unit and with the proper number of significant figures. 10. Perform the following calculations, and express the results in the correct units and with the proper number of significant figures. a. 7.382 g b. 51.3 mg c. 0.007 L d. 253.05 cm e. 14.77 kg f. 319 mL 1.21 g 83 mg 0.0037 L 2 4.7923 g 34.2 mg 0.012 L 28 cm2 0.391 kg 20. mL 33.9 cm2 0.086 kg 13.75 mL 11. A container measures 30.5 mm 202 mm 153 mm. When it is full of a liquid, it has a mass of 1.33 kg. When it is empty, it has a mass of 0.30 kg. What is the density of the liquid in kilograms per liter? 12. If 7.76 km of wire has a mass of 3.3 kg, what is the mass of the wire in g/m? What length in meters would have a mass of 1.0 g? 13. A container of plant food recommends an application rate of 52 kg/ha. If the container holds 10 kg of plant food, how many square meters will it cover (1 ha 10 000 m2)? 14. A chemical process produces 974 550 kJ of energy as heat in 37.0 min. What is the rate in kilojoules per minute? What is the rate in kilojoules per second? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 26 Significant Figures Back Print Name Class Date Problem Solving continued 15. A water pipe fills a container that measures 189 cm 307 cm 272 cm in 97 s. a. What is the volume of the container in cubic meters? b. What is the rate of flow in the pipe in liters per minute? c. What is the rate of flow in cubic meters per hour? 16. Perform the following calculations, and express the results in the correct units and with the proper number of significant figures. Note, in problems with multiple steps, it is better to perform the entire calculation and then round to significant figures. a. (0.054 kg b. 67.35 cm2 c. 4.198 kg d. 3.14159 m e. 690 000 m f. (6.23 cm 1.33 kg) (1.401 cm (1019 m 2 5.4 m2 0.399 cm) 40 m2) 2.150 m) 4.31 h) 0.05 cm) 14.99 cm (54.2 s 31.3 s) (4.17 m (5.022 h 3.111 cm Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 27 Significant Figures Back Print Name Class Date Skills Worksheet Problem Solving Scientific Notation People who work in scientific fields often have to use very large and very small numbers. Look at some examples in the following table: Measurement Density of air at 27°C and 1 atm pressure Radius of a calcium atom One light-year The mass of a neutron Value 0.001 61 g/cm3 0.000 000 000 197 m 9 460 000 000 000 km 0.000 000 000 000 000 000 000 001 675 g You can see that measurements such as these would be awkward to write out repeatedly. Also, calculating with very long numbers is likely to lead to errors because it’s so easy to miscount zeros and decimal places. To make these numbers easier to handle, scientists express them in a form known as scientific notation, which uses powers of 10 to reduce the number of zeros to a minimum. Look at a simple example of the way that scientific notation works. Following are some powers of 10 and their decimal equivalents. 10 10 10 10 0 1 2 2 1 000.01 000.1 001 010 100 10 Suppose we rewrite the values in the table using scientific notation. The numbers become much less cumbersome. Measurement Density of air at 27°C and 1 atm pressure Radius of a calcium atom One light-year Mass of a neutron Value 1.61 1.97 9.46 1.675 10 10 3 g/cm3 m 10 1012 km 10 24 g Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 28 Scientific Notation Back Print Name Class Date Problem Solving continued CONVERTING QUANTITIES TO SCIENTIFIC NOTATION General Plan for Converting Quantities to Scientific Notation Quantity expressed in long form Move the decimal point right or left until there is only one nonzero digit to the left of it. Use the resulting number as the coefficient M. Count the number of places the decimal point moved, and call that number n. Make the number n negative if the decimal moved to the right. M, the coefficient of 10n n, the exponent of 10 Quantity expressed in the form M 10n (scientific notation) Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 29 Scientific Notation Back Print Name Class Date Problem Solving continued Sample Problem 1 Express the following measurements in scientific notation. a. 310 000 L b. 0.000 49 kg Solution ANALYZE What is given in the problem? What are you asked to find? two measured quantities the measured quantities expressed in scientific notation Data a Measured quantity Quantity expressed in scientific notation 310 000 L ?L ? kg b 0.000 49 kg Items PLAN What steps are needed to rewrite the quantities in scientific notation? Move the decimal point in each value until there is only one nonzero digit to the left of it. This number becomes the coefficient, M. Count the number of places the decimal was moved. If it moved to the left, the count is a positive number. If it moved to the right, the count is a negative number. Make this number, n, the exponent of 10. Quantity written in long form COMPUTE a. Express 310 000 L in scientific notation M 3.1 M 10n 310 000 1 decimal point moves 5 places to the left n 5 310 000 L 3.1 10 L 5 b. Express 0.000 49 kg in scientific notation. M 4.9 0. 0 0 0 4 9 1 decimal point moves 4 places to the right n 4 0.000 49 kg 4.9 10 4 kg Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 30 Scientific Notation Back Print Name Class Date Problem Solving continued EVALUATE Are units correct? Is the quantity correctly expressed? a Yes; the original measurement was in liters. Yes; the decimal was moved to the left five places to give a coefficient of 3.1 and an exponent of 5. b Yes; the original measurement was in kilograms. Yes; the decimal was moved to the right four places to give a coefficient of 4.9 and an exponent of 4. Practice 1. Express the following quantities in scientific notation: a. 8 800 000 000 m ans: 8.8 109 m b. 0.0015 kg ans: 1.5 10 3 kg c. 0.000 000 000 06 kg/m3 ans: 6 10 11 kg/m3 d. 8 002 000 Hz ans: 8.002 106 Hz e. 0.009 003 amp ans: 9.003 10 3 amp f. 70 000 000 000 000 000 km ans: 7 1016 km g. 6028 L ans: 6.028 103 L h. 0.2105 g ans: 2.105 10 1 g i. 600 005 000 kJ/h ans: 6.000 05 108 kJ/h j. 33.8 m2 ans: 3.38 101 m2 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 31 Scientific Notation Back Print Name Class Date Problem Solving continued CALCULATING WITH QUANTITIES IN SCIENTIFIC NOTATION Sample Problem 2 What is the total of the measurements 3.61 8.1 102 mm? 104 mm, 5.88 103 mm, and Solution ANALYZE What is given in the problem? What are you asked to find? Items Measured quantity 3.61 104 mm 5.88 three measured quantities expressed in scientific notation the sum of those quantities Data 103 mm 8.1 102 mm PLAN What steps are needed to add the quantities? Convert each quantity so that each exponent is the same as that on the quantity with the largest exponent. The quantities can then be added together. Make sure the result has the correct number of significant figures. (P 10Q) (R 10S ) (T 10V ) ? (P R T) 10Q ? if the exponents are different, convert the quantities so that they have the same exponent as the term with the largest exponent add the quantities P, R , and T , and multiply them by the factor 10 Q (P 10Q) (R 10Q) (T 10Q) ? COMPUTE 3.61 104 mm 5.88 103 mm 8.1 102 mm 4 ? mm Convert the second and third quantities to multiples of 10 . To convert 5.88 103 mm: 5.88 103 mm M 104 mm Because one was added to the exponent, the decimal point must be moved one place to the left. M To convert 8.1 10 mm: 8.1 102 mm M 104 mm 2 0.588 Because two was added to the exponent, the decimal point must be moved two places to the left. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 32 Scientific Notation Back Print Name Class Date Problem Solving continued M 3.610 0.588 0.081 4.275 0.081 104 mm 104 mm 104 mm 104 mm Now the three quantities can be added, as follows: To express the result in the correct number of significant figures, note that the result should only contain two decimal places. 4.275 104 mm 4.28 104 mm this digit is 5, so round up EVALUATE Are the units correct? Yes; units of all quantities were millimeters. Is the quantity correctly expressed in scientific notation? Yes; there is only one number to the left of the decimal point. Is the quantity expressed in the correct number of significant figures? Yes; the result was rounded to give two decimal places to match the least accurate measurement. Practice 1. Carry out the following calculations. Express the results in scientific notation and with the correct number of significant figures. a. 4.74 104 km 7.71 103 km 1.05 103 km ans: 5.62 104 km b. 2.75 10 4 m 8.03 10 5 m 2.122 10 3 m ans: 2.477 10 3 m Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 33 Scientific Notation Back Print Name Class Date Problem Solving continued c. 4.0 10 5 m3 6.85 10 6 m3 1.05 10 5 m3 ans: 3.6 10 5 m3 d. 3.15 102 mg 3.15 103 mg 3.15 104 mg ans: 3.50 104 mg e. 3.01 1022 atoms 1.19 ans: 1.59 1023 atoms 1023 atoms 9.80 1021 atoms f. 6.85 107 nm 4.0229 108 nm 8.38 106 nm ans: 4.624 108 nm Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 34 Scientific Notation Back Print Name Class Date Problem Solving continued Sample Problem 3 Perform the following calculation, and express the result in scientific notation: 3.03 104 cm2 6.29 102 cm Solution ANALYZE What is given in the problem? What are you asked to find? Items Measured quantity 3.03 10 cm 4 2 two quantities expressed in scientific notation the product of the two quantities Data 6.29 102 cm PLAN What steps are needed to multiply quantities expressed in scientific notation? Multiply the coefficients, and add the exponents. Then transform to the correct scientific notation form with the correct units and number of significant figures. (P 10Q) (R 10S ) Result correctly written in scientific notation, with the correct units and number of significant figures if there is more or less than one nonzero digit to the left of the decimal, move the decimal and change the exponent to account for the move, then round to the correct number of significant figures multiply the coefficients and add the exponents (P exponents R) 10(Q S) (P 10Q ) unit 1 coefficients (R 10S) unit 2 product of coefficients sum of exponents (Q S) product of units (P R) 10 (unit 1 unit 2) COMPUTE 3.03 104 cm2 6.29 102 cm (3.03 6.29) 10(4 2) (cm2 cm) 19.0587 106 cm3 To transform the result to the correct form for scientific notation, move the decimal point left one place and increase the exponent by one. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 35 Scientific Notation Back Print Name Class Date Problem Solving continued 19.0587 106 cm3 1.90587 10(6 1) cm3 1.90587 107 cm3 To express the result to the correct number of significant figures, note that both of the original quantities have three significant figures. Therefore, round off the result to three significant figures. 1.905 87 107 cm3 this digit is 5, so round up 1.91 107 cm3 EVALUATE Are the units correct? Yes; the units cm2 and cm are multiplied to give cm3. Is the quantity expressed to the correct number of significant figures? Yes; the number of significant figures is correct because the data were given to three significant figures. Is the quantity expressed correctly in scientific notation? Yes; moving the decimal point decreases the coefficient by a factor of 10, so the exponent increases by one to compensate. Practice 1. Carry out the following computations, and express the result in scientific notation: a. 7.20 103 cm 8.08 103 cm ans: 5.82 107 cm2 b. 3.7 104 mm 6.6 104 mm 9.89 103 mm ans: 2.4 1013 mm3 c. 8.27 102 m 2.5 10 3 m 3.00 10 4 m ans: 6.2 10 4 m3 d. 4.44 10 35 m 5.55 1019 m 7.69 10 12 kg ans: 1.89 10 26 kg m2 e. 6.55 104 dm 7.89 109 dm 4.01893 105 dm ans: 2.08 1020 dm3 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 36 Scientific Notation Back Print Name Class Date Problem Solving continued Sample Problem 4 Perform the following calculation, and express the result in scientific notation: 3.803 103 g 5.3 106 mL Solution ANALYZE What is given in the problem? What are you asked to find? Items Measured quantity 3.803 10 g 3 two quantities expressed in scientific notation the quotient of the two quantities Data 5.3 106 mL PLAN What steps are needed to divide the quantities expressed in scientific notation? Divide the coefficients, and subtract the exponents. Then transform the result to the correct form for scientific notation with the correct units and number of significant figures. (P 10Q) (R 10S ) Result written correctly in scientific notation, with the correct units and number of significant figures if there is more or less than one nonzero digit to the left of the decimal, move the decimal and change the exponent to account for the move, then round to the correct number of significant figures divide the coefficients and subtract the exponents in the order shown (P R) 10(Q S) quotient of coefficients P R 10Q unit 1 10S unit 2 P R 10 difference quotient of exponents of units (Q S ) unit 1 unit 2 COMPUTE 3.803 103 g 5.3 106 mL 3.803 5.3 10(3 6) g mL 0.717 547 10 3 g/mL The measurement 5.3 result accordingly. 106 mL has the fewest significant figures; round the 10 3 0.717 547 g/mL 0.72 10 3 g/mL this digit is greater than 5, so round up Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 37 Scientific Notation Back Print Name Class Date Problem Solving continued To transform the result to the correct form for scientific notation, move the decimal point to the right one place and decrease the exponent by one. 0.72 10 3 g/mL 7.2 10( 3 1) g/mL 7.2 10 4 g/mL EVALUATE Are the units correct? Yes; grams divided by milliliters gives g/mL, a unit of density. Is the quantity expressed to the correct number of significant figures? Yes; the result was limited to two significant figures by the data given. Is the quantity expressed correctly in scientific notation? Yes; there is only one nonzero digit to the left of the decimal point. Practice 1. Carry out the following computations, and express the result in scientific notation: a. 2.290 107 cm 4.33 103 s ans: 5.29 103 cm/s b. 1.788 10 5 L 7.111 10 3 m2 ans: 2.514 10 3 L/m2 c. 5.515 104 L 6.04 103 km ans: 9.13 L/km d. 3.29 10 4 km 1.48 10 2 min ans: 2.22 10 2 km/min e. 4.73 10 4 g (2.08 10 3 km 5.60 10 4 km) ans: 4.06 102 g/km2 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 38 Scientific Notation Back Print Name Class Date Problem Solving continued Additional Problems 1. Express the following quantities in scientific notation: a. 158 000 km b. 0.000 009 782 L c. 837 100 000 cm e. 0.005 93 g 3 2 f. 0.000 000 006 13 m g. 12 552 000 J h. 0.000 008 004 g/L i. 0.010 995 kg j. 1 050 000 000 Hz d. 6 500 000 000 mm 2. Perform the following calculations, and express the result in scientific notation with the correct number of significant figures: a. 2.48 b. 4.07 c. 1.39 d. 7.70 e. 1.111 f. 9.81 g. 1.36 102 kg 10 4 5 9.17 6.52 3.95 5.82 3.456 103 kg 10 10 m 10 10 J 3.18 106 cm 4 8 2 3 4 7.2 mg 4.8 101 kg 7.1 3 mg 3 3.966 10 3 7 2 mg 10 m 10 10 9 5 10 m 10 6 m m 10 1.88 25 m 2.09 1026 molecules 105 cm 10 J 27 3.01 10 J molecules 107 cm 1.01 5.122 molecules 107 cm 3. Perform the following computations, and express the result in scientific notation with the correct number of significant figures: a. 1.54 b. 3.890 c. 9.571 d. 8.33 e. 9.36 f. 6.377 10 1 4 L 2.36 4.71 3.82 1.97 3.82 7.35 10 10 4 s m2 9.01 10 1 10 mm 103 kg 103 km 102 m 10 J 4 10 mm2 1 2 102 s 103 m 10 3 m s 4. Your electric company charges you for the electric energy you use, measured in kilowatt-hours (kWh). One kWh is equivalent to 3 600 000 J. Express this quantity in scientific notation. 5. The pressure in the deepest part of the ocean is 11 200 000 Pa. Express this pressure in scientific notation. 6. Convert 1.5 km to millimeters, and express the result in scientific notation. 7. Light travels at a speed of about 300 000 km/s. a. Express this value in scientific notation. b. Convert this value to meters per hour. c. What distance in centimeters does light travel in 1 s? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 39 Scientific Notation Back Print Name Class Date Problem Solving continued 8. There are 7.11 1024 molecules in 100.0 cm3 of a certain substance. 104 cm3 of the sub- a. What is the number of molecules in 1.09 cm3 of the substance? b. What would be the number of molecules in 2.24 stance? c. What number of molecules are in 9.01 10 6 cm3 of the substance? 9. The number of transistors on a particular integrated circuit is 3 578 000, and the integrated circuit measures 9.5 mm 8.2 mm. a. What is the area occupied by each transistor? b. Using your answer from (a), how many transistors could be formed on a silicon sheet that measures 353 mm 265 mm? 10. A solution has 0.0501 g of a substance in 1.00 L. Express this concentration in grams per microliter. 11. Cesium atoms are the largest of the naturally occurring elements. They have a diameter of 5.30 10 10 m. Calculate the number of cesium atoms that would have to be lined up to give a row of cesium atoms 2.54 cm (1 in.) long. 12. The neutron has a volume of approximately 1.4 10 44 m3 and a mass of 1.675 10 24 g. Calculate the density of the neutron in g/m3. What is the mass of 1.0 cm3 of neutrons in kilograms? 13. The pits in a compact disc are some of the smallest things ever mass-produced mechanically by humans. These pits represent the 1s and 0s of digital information on a compact disc. These pits are only 1.6 10 8 m deep (1/4 the wavelength of red laser light). How many of these pits would have to be stacked on top of each other to make a hole 0.305 m deep? 14. 22 400 mL of oxygen gas contains 6.022 standard atmospheric pressure. 1023 oxygen molecules at 0°C and a. How many oxygen molecules are in 0.100 mL of gas? b. How many oxygen molecules are in 1.00 L of gas? c. What is the average space in milliters occupied by one oxygen molecule? 15. The mass of the atmosphere is calculated to be 5.136 1018 kg, and there are 6 500 000 000 people living on Earth. Calculate the following values. a. The mass of atmosphere in kilograms per person. b. The mass of atmosphere in metric tons per person. c. If the number of people increases to 9 500 000 000, what is the mass in kilograms per person? 16. The mass of the sun is 1.989 1030 kg, and the mass of Earth is 5.974 1024 kilograms. How many Earths would be needed to equal the mass of the sun? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 40 Scientific Notation Back Print Name Class Date Problem Solving continued 17. A new landfill has dimensions of 2.3 km a. What is the volume in cubic kilometer? b. What is the volume in cubic meters? c. If 250 000 000 objects averaging 0.060 m3 each are placed into the landfill each year, how many years will it take to fill the landfill? 18. A dietary calorie (C) is exactly equal to 1000 cal. If your daily intake of food gives you 2400 C, what is your intake in joules per day? (1 cal 4.184 J) 1.4 km 0.15 km. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 41 Scientific Notation Back Print Name Class Date Skills Worksheet Problem Solving Four Steps for Solving Quantitative Problems Maybe you have noticed that the sample problems in the first three chapters of this book are solved in a four-step process. The steps in the process are as follows. 1. ANALYZE 2. PLAN 3. COMPUTE 4. EVALUATE Now it’s time to examine each step to learn how this process can help you solve more-complex problems, like those you will encounter in your chemistry course. 1. ANALYZE In this first step, you should read the problem carefully and then reread it. You must determine what specific information and data you are given in the problem and what you need to find. Try to visualize the situation the problem describes. Look closely at the words in the problem statement for clues to understanding the problem. Are you working with elements, compounds, or mixtures? Are they solids, liquids, or gases? What change is taking place? Is it a chemical change? What are the reactants? What are the products? It is always a good idea to collect and organize all of your information in a table, where you can see it at a glance. Include the things you want to find in the table, too. Be sure to include units with both the data and the quantities you must find. Scanning the quantities and their units will often provide clues about how to set up the problem. Remember, you may be given information not needed to solve the problem. You must analyze the data to determine what is useful and what is not. 2. PLAN In the planning step, you develop a method to solve the problem. Always keep in mind what you want to find and its units. Chances are good that an approach that gives an answer with the correct units is the correct one to use. In any case, a setup that gives an answer with the wrong units is certain to be wrong. You may find it helpful to diagram your solution method. This process helps you organize your thoughts. As you work out your problem-solving method, write down a trial calculation without numbers but with units. When you complete your setup, see if the trial calculation will give you a quantity with the correct units. As stated above, if the setup gives the needed units, it is probably correct. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 42 Four Steps for Solving Quantitative Problems Back Print Name Class Date Problem Solving continued During this planning process you may discover that you need more information, such as atomic masses from the periodic table, the boiling point of alcohol, or the density of tin. You will need to look up such information in the appropriate tables. 3. COMPUTE In this step, you follow your plan, set up a calculation using the data you have assembled, and compute the result. It is a good strategy to write out and check your calculation setup before you start working with your calculator. First reconfirm that the calculation will give a result with the correct units. Go through your calculation and lightly strike through the units that cancel. Be sure the remaining units are those that you want in your answer. Whenever possible, use your calculator in a way that lets you complete the entire problem without writing down numbers and then re-entering them. 4. EVALUATE Everyone makes errors, but good problem solvers always develop strategies to check their work. Confidence in your problem-solving ability will come from knowing how to determine on your own whether your answers are correct. One evaluation strategy is to estimate the numerical value of the answer. In simple problems, you can probably do this in your head. With calculations having several terms, round off each numerical value to the nearest simple value, and then write and compute the estimation. Suppose you had to make the following calculation. 28.8 g 6.30 cm 18.9 cm 30 6 20 30 120 1 4 ? The numerical calculation can be estimated as follows. 0.25 Once you have done the actual computation, compare your result with the estimate. In this case, the calculation gives 0.242 g/cm2, which is close to the estimated result. Therefore, it is likely that you made no mistakes in the calculation. Next, check that your answer is expressed to the correct number of significant figures. Look at the data values you used in the calculation. Usually, significant figures will be limited by the measurement that has the fewest significant figures. Finally, ask yourself the simple question, does this answer make sense based on what you know? If you are calculating the circumference of Earth, an answer of 50 km is obviously much too small. If you are calculating the density of air, a value of 340 g/cm3 is much too large because air density is usually less than 1 g/cm3. You will detect many errors by asking if the answer makes sense. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 43 Four Steps for Solving Quantitative Problems Back Print Name Class Date Problem Solving continued Sample Problem 1 A 10.0% sodium hydroxide solution has a density of 1.11 g/mL. What volume in liters will 2280 g of the solution have? Solution ANALYZE What is given in the problem? the density of the sodium hydroxide solution in g/mL, and the mass of the solution whose volume is to be determined the volume of the specified mass of solution What are you asked to find? Next, bring together in a table everything you might need in the problem. Include what you want to find in the table. The fact that the solution is 10.0% sodium hydroxide is unimportant in the solution of this problem. This piece of data need not appear in your table. Notice that the problem asks you for a volume in liters and that density is given in grams per milliliter. You will need a factor to convert between these units. Therefore, include in the table any relationships between units that will be helpful. Items Density of solution Mass of solution Volume of solution Relationship between mL and L Data 1.11 g/mL 2280 g ?L 1000 mL 1L PLAN What steps are needed to calculate the volume of 2280 g of the solution? Apply the relationship D convert to liters. Mass of solution in g rearrange m D V to solve for V m . Rearrange to solve for V, substitute data, and V Volume of solution in L convert using the factor 1L 1000 mL Volume of solution in mL Solve the density equation for V. V m D volume in mL mass of solution in g density of solution in g/ml Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 44 Four Steps for Solving Quantitative Problems Back Print Name Class Date Problem Solving continued To change the result to liters, multiply by the conversion factor. mass of solution in g density of solution g/ml COMPUTE 2280 g 1.11 g/mL EVALUATE Are the units correct? Yes; units canceled to give liters. 1L 1000 mL volume in L 1L 1000 mL 2.05 L Is the number of significant figures correct? Yes; the number of significant figures is correct because data were given to three significant figures. Is the answer reasonable? Yes; the calculation can be approximated as 2000/1000 2. Also, considering that 2000 g of water occupies 2 L, you would expect 2 L of a slightly more dense material to have a mass slightly greater than 2000 g. Practice 1. Gasoline has a density of 0.73 g/cm3. How many liters of gasoline would be required to increase the mass of an automobile from 1271 kg to 1305 kg? ans: 47 L 2. A swimming pool measures 9.0 m long by 3.5 m wide by 1.75 m deep. What mass of water in metric tons (1 metric ton 1000 kg) does the pool contain when filled? The density of the water in the pool is 0.997 g/cm3. ans: 55 metric tons 3. A tightly packed box of crackers contains 250 g of crackers and measures 7.0 cm 17.0 cm 19.0 cm. What is the average density in kilograms per liter of the crackers in the package? Assume that the unused volume is negligible. ans: 0.11 kg/L Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 45 Four Steps for Solving Quantitative Problems Back Print Name Class Date Problem Solving continued Additional Problems Solve these problems by using the Four Steps for Solving Quantitative Problems. 1. The aluminum foil on a certain roll has a total area of 18.5 m2 and a mass of 1275 g. Using a density of 2.7 g per cubic centimeter for aluminum, determine the thickness in millimeters of the aluminum foil. 2. If a liquid has a density of 1.17 g/cm3, how many liters of the liquid have a mass of 3.75 kg? 3. A stack of 500 sheets of paper measuring 28 cm 21 cm is 44.5 mm high and has a mass of 2090 g. What is the density of the paper in grams per cubic centimeter? 4. A triangular-shaped piece of a metal has a mass of 6.58 g. The triangle is 0.560 mm thick and measures 36.4 mm on the base and 30.1 mm in height. What is the density of the metal in grams per cubic centimeter? 5. A packing crate measures 0.40 m 0.40 m 0.25 m. You must fill the crate with boxes of cookies that each measure 22.0 cm 12.0 cm 5.0 cm. How many boxes of cookies can fit into the crate? 6. Calculate the unknown quantities in the following table. Use the following relationships for volumes of the various shapes. Volume of a cube l l l Volume of a rectangle l w h Volume of a sphere 4/3 r 3 Volume of a cylinder r2 h D m 3 3 V ?L ? cm 3 3 3 3 Shape cube rectangle sphere cylinder rectangle Dimensions ?m ?m ?m 7.2 mm 33 mm 21 mm a. b. c. d. e. 2.27 g/cm 1.85 g/cm 3.21 g/L ? g/cm 3 3.93 kg ?g ? kg 497 g ? dm ?m 3.30 m diameter 7.5 cm diameter 3.5 m 1.2 m 12 cm 0.65 m 0.92 g/cm 3 ? kg ? cm 7. When a sample of a metal alloy that has a mass of 9.65 g is placed into a graduated cylinder containing water, the volume reading in the cylinder increases from 16.0 mL to 19.5 mL. What is the density of the alloy sample in grams per cubic centimeter? 8. Pure gold can be made into extremely thin sheets called gold leaf. Suppose that 50.0 kg of gold is made into gold leaf having an area of 3620 m2. The density of gold is 19.3 g/cm3. a. How thick in micrometers is the gold leaf? b. A gold atom has a radius of 1.44 gold leaf? 10 10 m. How many atoms thick is the Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 46 Four Steps for Solving Quantitative Problems Back Print Name Class Date Problem Solving continued 9. A chemical plant process requires that a cylindrical reaction tank be filled with a certain liquid in 238 s. The tank is 1.2 m in diameter and 4.6 m high. What flow rate in liters per minute is required to fill the reaction tank in the specified time? 10. The radioactive decay of 2.8 g of plutonium-238 generates 1.0 joule of energy as heat every second. Plutonium has a density of 19.86 g/cm3. How many calories (1 cal 4.184 J) of energy as heat will a rectangular piece of plutonium that is 4.5 cm 3.05 cm 15 cm generate per hour? 11. The mass of Earth is 5.974 1024 kg. Assume that Earth is a sphere of diameter 1.28 104 km and calculate the average density of Earth in grams per cubic centimeter. 12. What volume of magnesium in cubic centimeters would have the same mass as 1.82 dm3 of platinum? The density of magnesium is 1.74 g/cm3, and the density of platinum is 21.45 g/cm3. 13. A roll of transparent tape has 66 m of tape on it. If an average of 5.0 cm of tape is needed each time the tape is used, how many uses can you get from a case of tape containing 24 rolls? 14. An automobile can travel 38 km on 4.0 L of gasoline. If the automobile is driven 75% of the days in a year and the average distance traveled each day is 86 km, how many liters of gasoline will be consumed in one year (assume the year has 365 days)? 15. A hose delivers water to a swimming pool that measures 9.0 m long by 3.5 m wide by 1.75 m deep. It requires 97 h to fill the pool. At what rate in liters per minute will the hose fill the pool? 16. Automobile batteries are filled with a solution of sulfuric acid, which has a density of 1.285 g/cm3. The solution used to fill the battery is 38% (by mass) sulfuric acid. How many grams of sulfuric acid are present in 500 mL of battery acid? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 47 Four Steps for Solving Quantitative Problems Back Print Name Class Date Skills Worksheet Problem Solving Mole Concept Suppose you want to carry out a reaction that requires combining one atom of iron with one atom of sulfur. How much iron should you use? How much sulfur? When you look around the lab, there is no device that can count numbers of atoms. Besides, the merest speck (0.001 g) of iron contains over a billion billion atoms. The same is true of sulfur. Fortunately, you do have a way to relate mass and numbers of atoms. One iron atom has a mass of 55.847 amu, and 55.847 g of iron contains 6.022 137 1023 atoms of iron. Likewise, 32.066 g of sulfur contains 6.022 137 1023 atoms of sulfur. Knowing this, you can measure out 55.847 g of iron and 32.066 g of sulfur and be pretty certain that you have the same number of atoms of each. The number 6.022 137 1023 is called Avogadro’s number. For most purposes it is rounded off to 6.022 1023. Because this is an awkward number to write over and over again, chemists refer to it as a mole (abbreviated mol). 6.022 1023 objects is called a mole, just as you call 12 objects a dozen. Look again at how these quantities are related. 55.847 g of iron 32.066 g of sulfur 6.022 6.022 1023 iron atoms 1023 sulfur atoms 1 mol of iron 1 mol of sulfur General Plan for Converting Mass, Amount, and Numbers of Particles 1 Mass of substance Convert using the molar mass of the substance. 2 Amount of substance in moles 3 Use Avogadro's number for conversion. Number of atoms, molecules, or formula units of substance Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 48 Mole Concept Back Print Name Class Date Problem Solving continued PROBLEMS INVOLVING ATOMS AND ELEMENTS Sample Problem 1 A chemist has a jar containing 388.2 g of iron filings. How many moles of iron does the jar contain? Solution ANALYZE What is given in the problem? What are you asked to find? Items Mass of iron Molar mass of iron* Amount of iron mass of iron in grams amount of iron in moles Data 388.2 g 55.85 g/mol ? mol * determined from the periodic table PLAN What step is needed to convert from grams of Fe to number of moles of Fe? The molar mass of iron can be used to convert mass of iron to amount of iron in moles. 1 Mass of Fe in g multiply by the inverse molar mass of Fe 2 Amount of Fe in mol 1 molar mass Fe given g Fe 1 mol Fe 55.85 g Fe mol Fe COMPUTE 388.2 g Fe EVALUATE Are the units correct? Yes; the answer has the correct units of moles of Fe. 1 mol Fe 55.85 g Fe 6.951 mol Fe Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 49 Mole Concept Back Print Name Class Date Problem Solving continued Is the number of significant figures correct? Yes; the number of significant figures is correct because there are four significant figures in the given value of 388.2 g Fe. Is the answer reasonable? Yes; 388.2 g Fe is about seven times the molar mass. Therefore, the sample contains about 7 mol. Practice 1. Calculate the number of moles in each of the following masses: a. 64.1 g of aluminum ans: 2.38 mol Al b. 28.1 g of silicon ans: 1.00 mol Si c. 0.255 g of sulfur ans: 7.95 10 3 mol S d. 850.5 g of zinc ans: 13.01 mol Zn Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 50 Mole Concept Back Print Name Class Date Problem Solving continued Sample Problem 2 A student needs 0.366 mol of zinc for a reaction. What mass of zinc in grams should the student obtain? Solution ANALYZE What is given in the problem? What are you asked to find? Items Amount of zinc Molar mass of zinc Mass of zinc amount of zinc needed in moles mass of zinc in grams Data 0.366 mol 65.39 g/mol ?g PLAN What step is needed to convert from moles of Zn to grams of Zn? The molar mass of zinc can be used to convert amount of zinc to mass of zinc. 2 Amount of Zn in mol multiply by the molar mass of Zn molar mass Zn given 1 Mass of Zn in mol mol Zn 65.39 g Zn 1 mol Zn g Zn COMPUTE 0.366 mol Zn EVALUATE Are the units correct? Yes; the answer has the correct units of grams of Zn. 65.39 g Zn 1 mol Zn 23.9 g Zn Is the number of significant figures correct? Yes; the number of significant figures is correct because there are three significant figures in the given value of 0.366 mol Zn. Is the answer reasonable? Yes; 0.366 mol is about 1/3 mol. 23.9 g is about 1/3 the molar mass of Zn. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 51 Mole Concept Back Print Name Class Date Problem Solving continued Practice 1. Calculate the mass of each of the following amounts: a. 1.22 mol sodium ans: 28.0 g Na b. 14.5 mol copper ans: 921 g Cu c. 0.275 mol mercury ans: 55.2 g Hg d. 9.37 10 3 mol magnesium ans: 0.228 Mg Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 52 Mole Concept Back Print Name Class Date Problem Solving continued Sample Problem 3 How many moles of lithium are there in 1.204 1024 lithium atoms? Solution ANALYZE What is given in the problem? What are you asked to find? Items Number of lithium atoms Avogadro’s number—the number of atoms per mole Amount of lithium number of lithium atoms amount of lithium in moles Data 1.204 6.022 ? mol 1024 atoms 1023 atoms/mol PLAN What step is needed to convert from number of atoms of Li to moles of Li? Avogadro’s number is the number of atoms per mole of lithium and can be used to calculate the number of moles from the number of atoms. 3 Number of Li atoms multiply by the inverse of Avogadro's number 1 Avogadro's number given 2 Amount of Li in mol atoms Li 1 mol Li 6.022 1023 atoms Li mol Li COMPUTE 1.204 1024 atoms Li 1 mol Li 6.022 1023 atoms Li 1.999 mol Li EVALUATE Are the units correct? Yes; the answer has the correct units of moles of Li. Is the number of significant figures correct? Yes; four significant figures is correct. Is the answer reasonable? Yes; 1.204 1024 is approximately twice Avogadro’s number. Therefore, it is reasonable that this number of atoms would equal about 2 mol. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 53 Mole Concept Back Print Name Class Date Problem Solving continued Practice 1. Calculate the amount in moles in each of the following quantities: a. 3.01 1023 atoms of rubidium ans: 0.500 mol Rb b. 8.08 1022 atoms of krypton ans: 0.134 mol Kr c. 5 700 000 000 atoms of lead ans: 9.5 10 15 mol Pb d. 2.997 1025 atoms of vanadium ans: 49.77 mol V Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 54 Mole Concept Back Print Name Class Date Problem Solving continued CONVERTING THE AMOUNT OF AN ELEMENT IN MOLES TO THE NUMBER OF ATOMS In Sample Problem 3, you were asked to determine the number of moles in 1.204 1024 atoms of lithium. Had you been given the amount in moles and asked to calculate the number of atoms, you would have simply multiplied by Avogadro’s number. Steps 2 and 3 of the plan for solving Sample Problem 3 would have been reversed. Practice 1. Calculate the number of atoms in each of the following amounts: a. 1.004 mol bismuth ans: 6.046 1023 atoms Bi b. 2.5 mol manganese ans: 1.5 1024 atoms Mg c. 0.000 000 2 mol helium ans: 1 1017 atoms He d. 32.6 mol strontium ans: 1.96 1025 atoms Sr Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 55 Mole Concept Back Print Name Class Date Problem Solving continued Sample Problem 4 How many boron atoms are there in 2.00 g of boron? Solution ANALYZE What is given in the problem? What are you asked to find? Items Mass of boron Molar mass of boron Avogadro’s number—the number of boron atoms per mole of boron Number of boron atoms mass of boron in grams number of boron atoms Data 2.00 g 10.81 g/mol 6.022 ? atoms 1023 atoms/mol PLAN What steps are needed to convert from grams of B to number of atoms of B? First, you must convert the mass of boron to moles of boron by using the molar mass of boron. Then you can use Avogadro’s number to convert amount in moles to number of atoms of boron. 1 Mass of B in g multiply by the inverse of the molar mass of boron 3 Number of B atoms multiply by Avogadro's number 2 Amount of B in mol 1 molar mass B Avogadro's number given gB 1 mol B 10.81 g B 6.022 1023 atoms B 1 mol B atoms B COMPUTE 2.00 g B 1 mol B 10.81 g B 6.022 1023 atoms B 1 mol B 1.11 1023 atoms B EVALUATE Are the units correct? Yes; the answer has the correct units of atoms of boron. Is the number of significant figures correct? Yes; the mass of boron was given to three significant figures. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 56 Mole Concept Back Print Name Class Date Problem Solving continued Is the answer reasonable? Yes; 2 g of boron is about 1/5 of the molar mass of boron. Therefore, 2.00 g boron will contain about 1/5 of an Avogadro’s constant of atoms. Practice 1. Calculate the number of atoms in each of the following masses: a. 54.0 g of aluminum ans: 1.21 1024 atoms Al b. 69.45 g of lanthanum ans: 3.011 1023 atoms La c. 0.697 g of gallium ans: 6.02 1021 atoms Ga d. 0.000 000 020 g beryllium ans: 1.3 1015 atoms Be Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 57 Mole Concept Back Print Name Class Date Problem Solving continued CONVERTING NUMBER OF ATOMS OF AN ELEMENT TO MASS Sample Problem 4 uses the progression of steps 1 → 2 → 3 to convert from the mass of an element to the number of atoms. In order to calculate the mass from a given number of atoms, these steps will be reversed. The number of moles in the sample will be calculated. Then this value will be converted to the mass in grams. Practice 1. Calculate the mass of the following numbers of atoms: a. 6.022 1024 atoms of tantalum ans: 1810. g Ta b. 3.01 1021 atoms of cobalt ans: 0.295 g Co c. 1.506 1024 atoms of argon ans: 99.91 g Ar d. 1.20 1025 atoms of helium ans: 79.7 g He Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 58 Mole Concept Back Print Name Class Date Problem Solving continued PROBLEMS INVOLVING MOLECULES, FORMULA UNITS, AND IONS How many water molecules are there in 200.0 g of water? What is the mass of 15.7 mol of nitrogen gas? Both of these substances consist of molecules, not single atoms. Look back at the diagram of the General Plan for Converting Mass, Amount, and Numbers of Particles. You can see that the same conversion methods can be used with molecular compounds and elements, such as CO2 , H2O, H2SO4 , and O2 . For example, 1 mol of water contains 6.022 1023 H2O molecules. The mass of a molecule of water is the sum of the masses of two hydrogen atoms and one oxygen atom, and is equal to 18.02 amu. Therefore, 1 mol of water has a mass of 18.02 g. In the same way, you can relate amount, mass, and number of formula units for ionic compounds, such as NaCl, CaBr2 , and Al2(SO4)3. Sample Problem 5 How many moles of carbon dioxide are in 66.0 g of dry ice, which is solid CO2? Solution ANALYZE What is given in the problem? What are you asked to find? Items Mass of CO2 Molar mass of CO2 Amount of CO2 mass of carbon dioxide amount of carbon dioxide Data 66.0 g 44.0 g/mol ? mol PLAN What step is needed to convert from grams of CO2 to moles of CO2? The molar mass of CO2 can be used to convert mass of CO2 to moles of CO2. 1 Mass of CO2 in g multiply by the inverse of the molar mass of CO2 1 molar mass CO2 2 Amount of CO2 in mol given g CO2 1 mol CO2 44.01 g CO2 mol CO2 COMPUTE 66.0 g CO2 1 mol CO2 44.01 g CO2 1.50 mol CO2 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 59 Mole Concept Back Print Name Class Date Problem Solving continued EVALUATE Are the units correct? Yes; the answer has the correct units of moles CO2. Is the number of significant figures correct? Yes; the number of significant figures is correct because the mass of CO2 was given to three significant figures. Is the answer reasonable? Yes; 66 g is about 3/2 the value of the molar mass of CO2. It is reasonable that the sample contains 3/2 (1.5) mol. Practice 1. Calculate the number of moles in each of the following masses: a. 3.00 g of boron tribromide, BBr3 ans: 0.0120 mol BBr3 b. 0.472 g of sodium fluoride, NaF ans: 0.0112 mol NaF c. 7.50 102 g of methanol, CH3OH ans: 23.4 mol CH3OH d. 50.0 g of calcium chlorate, Ca(ClO3)2 ans: 0.242 mol Ca(ClO3)2 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 60 Mole Concept Back Print Name Class Date Problem Solving continued CONVERTING MOLES OF A COMPOUND TO MASS Perhaps you have noticed that Sample Problems 1 and 5 are very much alike. In each case, you multiplied the mass by the inverse of the molar mass to calculate the number of moles. The only difference in the two problems is that iron is an element and CO2 is a compound containing a carbon atom and two oxygen atoms. In Sample Problem 2, you determined the mass of 1.366 mol of zinc. Suppose that you are now asked to determine the mass of 1.366 mol of the molecular compound ammonia, NH3. You can follow the same plan as you did in Sample Problem 2, but this time use the molar mass of ammonia. Practice 1. Determine the mass of each of the following amounts: a. 1.366 mol of NH3 ans: 23.28 g NH3 b. 0.120 mol of glucose, C6H12O6 ans: 21.6 g C6H12O6 c. 6.94 mol barium chloride, BaCl2 ans: 1.45 103 g or 1.45 kg BaCl2 d. 0.005 mol of propane, C3H8 ans: 0.2 g C3H8 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 61 Mole Concept Back Print Name Class Date Problem Solving continued Sample Problem 6 Determine the number of molecules in 0.0500 mol of hexane, C6H14 . Solution ANALYZE What is given in the problem? What are you asked to find? Items Amount of hexane Avogadro’s number—the number of molecules per mole of hexane Molecules of hexane amount of hexane in moles number of molecules of hexane Data 0.0500 mol 6.022 1023 molecules/mol ? molecules PLAN What step is needed to convert from moles of C6H14 to number of molecules of C6H14? Avogadro’s number is the number of molecules per mole of hexane and can be used to calculate the number of molecules from number of moles. 2 Amount of C6H14 in mol multiply by Avogadro's number Avogadro's number given 3 Number of C6H14 molecules mol C6H14 6.022 1023 molecules C6H14 1 mol C6H14 molecules C6H14 COMPUTE 0.0500 mol C6H14 6.022 1023 molecules C6H14 1 mol C6H14 3.01 1022 molecules C6H14 EVALUATE Are the units correct? Yes; the answer has the correct units of molecules of C6H14. Is the number of significant figures correct? Yes; three significant figures is correct. Is the answer reasonable? Yes; multiplying Avogadro’s number by 0.05 would yield a product that is a factor of 10 less with a value of 3 1022. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 62 Mole Concept Back Print Name Class Date Problem Solving continued Practice 1. Calculate the number of molecules in each of the following amounts: a. 4.99 mol of methane, CH4 ans: 3.00 1024 molecules CH4 b. 0.005 20 mol of nitrogen gas, N2 ans: 3.13 1021 molecules N2 c. 1.05 mol of phosphorus trichloride, PCl3 ans: 6.32 1023 molecules PCl3 d. 3.5 10 5 mol of vitamin C, ascorbic acid, C6H8O6 ans: 2.1 cules C6H8O6 1019 mole- Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 63 Mole Concept Back Print Name Class Date Problem Solving continued USING FORMULA UNITS OF IONIC COMPOUNDS Ionic compounds do not exist as molecules. A crystal of sodium chloride, for example, consists of Na ions and Cl ions in a 1:1 ratio. Chemists refer to a combination of one Na ion and one Cl ion as one formula unit of NaCl. A mole of an ionic compound consists of 6.022 1023 formula units. The mass of one formula unit is called the formula mass. This mass is used in the same way atomic mass or molecular mass is used in calculations. Practice 1. Calculate the number of formula units in the following amounts: a. 1.25 mol of potassium bromide, KBr ans: 7.53 1023 formula units KBr b. 5.00 mol of magnesium chloride, MgCl2 ans: 3.01 MgCl2 1024 formula units c. 0.025 mol of sodium carbonate, Na2CO3 ans: 1.5 Na2CO3 1022 formula units d. 6.82 10 6 mol of lead(II) nitrate, Pb(NO3)2 ans: 4.11 units Pb(NO3)2 1018 formula Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 64 Mole Concept Back Print Name Class Date Problem Solving continued CONVERTING NUMBER OF MOLECULES OR FORMULA UNITS TO AMOUNT IN MOLES In Sample Problem 3, you determined the amount in moles of the element lithium. Suppose that you are asked to determine the amount in moles of copper(II) hydroxide in 3.34 1034 formula units of Cu(OH)2. You can follow the same plan as you did in Sample Problem 3. Practice 1. Calculate the amount in moles of the following numbers of molecules or formula units: a. 3.34 1034 formula units of Cu(OH)2 ans: 5.55 1010 mol Cu(OH)2 b. 1.17 1016 molecules of H2S ans: 1.94 10 8 mol H2S c. 5.47 1021 formula units of nickel(II) sulfate, NiSO4 ans: 9.08 NiSO4 10 3 mol d. 7.66 1019 molecules of hydrogen peroxide, H2O2 ans: 1.27 H2O2 10 4 mol Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 65 Mole Concept Back Print Name Class Date Problem Solving continued Sample Problem 7 What is the mass of a sample consisting of 1.00 MgSO4? 1022 formula units of Solution ANALYZE What is given in the problem? What are you asked to find? Items Number of formula units of magnesium sulfate Avogadro’s number—the number of formula units of magnesium sulfate per mole Molar mass of magnesium sulfate Mass of magnesium sulfate number of magnesium sulfate formula units mass of magnesium sulfate in grams Data 1.00 6.022 1022 formula units 1023 formula units/mol 120.37 g/mol ?g PLAN What steps are needed to convert from formula units of MgSO4 to grams of MgSO4? First, you must convert the number of formula units of MgSO4 to amount of MgSO4 by using Avogadro’s number. Then you can use the molar mass of MgSO4 to convert amount in moles to mass of MgSO4. 3 Number of MgSO4 formula units multiply by the inverse of Avogadro's number 1 Mass of MgSO4 in g multiply by the molar mass of MgSO4 2 Amount of MgSO4 in mol 1 Avogadro's number given formula units MgSO4 6.022 1 mol MgSO4 1023 formula units MgSO4 120.37 g MgSO4 1 mol MgSO4 molar mass MgSO4 g MgSO4 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 66 Mole Concept Back Print Name Class Date Problem Solving continued COMPUTE 1.00 1022 formula units MgSO4 6.022 1 mol MgSO4 1023 formula units MgSO4 120.37 g MgSO4 1 mol MgSO4 EVALUATE Are the units correct? Yes; the answer has the correct units of grams of MgSO4. 2.00 g MgSO4 Is the number of significant figures correct? Yes; the number of significant figures is correct because data were given to three significant figures. Is the answer reasonable? Yes; 2 g of MgSO4 is about 1/60 of the molar mass of MgSO4. Therefore, 2.00 g MgSO4 will contain about 1/60 of an Avogadro’s number of formula units. Practice 1. Calculate the mass of each of the following quantities: a. 2.41 1024 molecules of hydrogen, H2 ans: 8.08 g H2 b. 5.00 1021 formula units of aluminum hydroxide, Al(OH)3 ans: 0.648 g Al(OH)3 c. 8.25 1022 molecules of bromine pentafluoride, BrF5 ans: 24.0 g BrF5 d. 1.20 1023 formula units of sodium oxalate, Na2C2O4 ans: 26.7 g Na2C2O4 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 67 Mole Concept Back Print Name Class Date Problem Solving continued CONVERTING MOLECULES OR FORMULA UNITS OF A COMPOUND TO MASS In Sample Problem 4, you converted a given mass of boron to the number of boron atoms present in the sample. You can now apply the same method to convert mass of an ionic or molecular compound to numbers of molecules or formula units. Practice 1. Calculate the number of molecules or formula units in each of the following masses: a. 22.9 g of sodium sulfide, Na2S ans: 1.77 1023 formula units Na2S b. 0.272 g of nickel(II) nitrate, Ni(NO3)2 ans: 8.96 Ni(NO3)2 1020 formula units c. 260 mg of acrylonitrile, CH2CHCN ans: 3.0 1021 molecules CH2CHCN Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 68 Mole Concept Back Print Name Class Date Problem Solving continued Additional Problems 1. Calculate the number of moles in each of the following masses: a. 0.039 g of palladium b. 8200 g of iron c. 0.0073 kg of tantalum d. 0.006 55 g of antimony e. 5.64 kg of barium f. 3.37 10 6 g of molybdenum 2. Calculate the mass in grams of each of the following amounts: a. 1.002 mol of chromium b. 550 mol of aluminum c. 4.08 10 8 mol of neon d. 7 mol of titanium e. 0.0086 mol of xenon f. 3.29 104 mol of lithium 3. Calculate the number of atoms in each of the following amounts: a. 17.0 mol of germanium b. 0.6144 mol of copper c. 3.02 mol of tin d. 2.0 f. 3.227 a. 6.022 b. 1.06 c. 3.008 e. 4.61 106 mol of carbon 10 10 e. 0.0019 mol of zirconium mol of potassium 4. Calculate the number of moles in each of the following quantities: 1024 atoms of cobalt 1023 atoms of tungsten 1019 atoms of silver 1017 atoms of radon d. 950 000 000 atoms of plutonium f. 8 trillion atoms of cerium 5. Calculate the number of atoms in each of the following masses: a. 0.0082 g of gold b. 812 g of molybdenum c. 2.00 102 mg of americium d. 10.09 kg of neon e. 0.705 mg of bismuth f. 37 g of uranium Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 69 Mole Concept Back Print Name Class Date Problem Solving continued 6. Calculate the mass of each of the following: a. 8.22 c. 9.96 1023 atoms of rubidium 1026 atoms of tellurium b. 4.05 Avogadro’s numbers of manganese atoms d. 0.000 025 Avogadro’s numbers of rhodium atoms e. 88 300 000 000 000 atoms of radium f. 2.94 1017 atoms of hafnium 7. Calculate the number of moles in each of the following masses: a. 45.0 g of acetic acid, CH3COOH b. 7.04 g of lead(II) nitrate, Pb(NO3)2 c. 5000 kg of iron(III) oxide, Fe2O3 d. 12.0 mg of ethylamine, C2H5NH2 e. 0.003 22 g of stearic acid, C17H35COOH f. 50.0 kg of ammonium sulfate, (NH4)2SO4 8. Calculate the mass of each of the following amounts: a. 3.00 mol of selenium oxybromide, SeOBr2 b. 488 mol of calcium carbonate, CaCO3 c. 0.0091 mol of retinoic acid, C20H28O2 d. 6.00 f. 3.50 10 10 8 mol of nicotine, C10H14N2 mol of uranium hexafluoride, UF6 e. 2.50 mol of strontium nitrate, Sr(NO3)2 6 9. Calculate the number of molecules or formula units in each of the following amounts: a. 4.27 mol of tungsten(VI) oxide, WO3 b. 0.003 00 mol of strontium nitrate, Sr(NO3)2 c. 72.5 mol of toluene, C6H5CH3 d. 5.11 10 7 mol of -tocopherol (vitamin E), C29H50O2 e. 1500 mol of hydrazine, N2H4 f. 0.989 mol of nitrobenzene C6H5NO2 10. Calculate the number of molecules or formula units in each of the following masses: a. 285 g of iron(III) phosphate, FePO4 b. 0.0084 g of C5H5N c. 85 mg of 2-methyl-1-propanol, (CH3)2CHCH2OH d. 4.6 10 4 g of mercury(II) acetate, Hg(C2H3O2)2 e. 0.0067 g of lithium carbonate, Li2CO3 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 70 Mole Concept Back Print Name Class Date Problem Solving continued 11. Calculate the mass of each of the following quantities: a. 8.39 b. 6.82 c. 7.004 e. 6.3 f. 8.37 1023 molecules of fluorine, F2 1024 formula units of beryllium sulfate, BeSO4 1026 molecules of chloroform, CHCl3 1018 molecules of nitric acid, HNO3 1025 molecules of freon 114, C2Cl2F4 d. 31 billion formula units of chromium(III) formate, Cr(CHO2)3 12. Precious metals are commonly measured in troy ounces. A troy ounce is equivalent to 31.1 g. How many moles are in a troy ounce of gold? How many moles are in a troy ounce of platinum? of silver? 13. A chemist needs 22.0 g of phenol, C6H5OH, for an experiment. How many moles of phenol is this? 14. A student needs 0.015 mol of iodine crystals, I2, for an experiment. What mass of iodine crystals should the student obtain? 15. The weight of a diamond is given in carats. One carat is equivalent to 200. mg. A pure diamond is made up entirely of carbon atoms. How many carbon atoms make up a 1.00 carat diamond? 16. 8.00 g of calcium chloride, CaCl2, is dissolved in 1.000 kg of water. a. How many moles of CaCl2 are in solution? How many moles of water are present? b. Assume that the ionic compound, CaCl2 , separates completely into Ca2 and Cl ions when it dissolves in water. How many moles of each ion are present in the solution? 17. How many moles are in each of the following masses? a. 453.6 g (1.000 pound) of sucrose (table sugar), C12H22O11 b. 1.000 pound of table salt, NaCl 18. When the ionic compound NH4Cl dissolves in water, it breaks into one ammonium ion, NH 4 , and one chloride ion, Cl . If you dissolved 10.7 g of NH4Cl in water, how many moles of ions would be in solution? 19. What is the total amount in moles of atoms in a jar that contains 2.41 1024 atoms of chromium, 1.51 1023 atoms of nickel, and 3.01 1023 atoms of copper? 20. The density of liquid water is 0.997 g/mL at 25°C. a. Calculate the mass of 250.0 mL (about a cupful) of water. b. How many moles of water are in 250.0 mL of water? Hint: Use the result of (a). c. Calculate the volume that would be occupied by 2.000 mol of water at 25°C. d. What mass of water is 2.000 mol of water? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 71 Mole Concept Back Print Name Class Date Problem Solving continued 21. An Avogadro’s number (1 mol) of sugar molecules has a mass of 342 g, but an Avogadro’s number (1 mol) of water molecules has a mass of only 18 g. Explain why there is such a difference between the mass of 1 mol of sugar and the mass of 1 mol of water. 22. Calculate the mass of aluminum that would have the same number of atoms as 6.35 g of cadmium. 23. A chemist weighs a steel cylinder of compressed oxygen, O2 , and finds that it has a mass of 1027.8 g. After some of the oxygen is used in an experiment, the cylinder has a mass of 1023.2 g. How many moles of oxygen gas are used in the experiment? 24. Suppose that you could decompose 0.250 mol of Ag2S into its elements. a. How many moles of silver would you have? How many moles of sulfur would you have? b. How many moles of Ag2S are there in 38.8 g of Ag2S? How many moles of silver and sulfur would be produced from this amount of Ag2S? c. Calculate the masses of silver and sulfur produced in (b). Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 72 Mole Concept Back Print Name Class Date Skills Worksheet Problem Solving Percentage Composition Suppose you are working in an industrial laboratory. Your supervisor gives you a bottle containing a white crystalline compound and asks you to determine its identity. Several unlabeled drums of this substance have been discovered in a warehouse, and no one knows what it is. You take it into the laboratory and carry out an analysis, which shows that the compound is composed of the elements sodium, carbon, and oxygen. Immediately, you think of the compound sodium carbonate, Na2CO3 , a very common substance found in most laboratories and used in many industrial processes. Before you report your conclusion to your boss, you decide to check a reference book to see if there are any other compounds that contain only the elements sodium, carbon, and oxygen. You discover that there is another compound, sodium oxalate, which has the formula Na2C2O4 . When you read about this compound, you find that it is highly poisonous and can cause serious illness and even death. Mistaking sodium carbonate for sodium oxalate could have very serious consequences. What can you do to determine the identity of your sample? Is it the common industrial substance or the dangerous poison? Fortunately, you can determine not only which elements are in the compound, but also how much of each element is present. As you have learned, every compound has a definite composition. Every water molecule is made up of two hydrogen atoms and one oxygen atom, no matter where the water came from. A formula unit of sodium chloride is composed of one sodium atom and one chlorine atom, no matter whether the salt came from a mine or was obtained by evaporating sea water. Likewise, sodium carbonate always has two sodium atoms, one carbon atom, and three oxygen atoms per formula unit, giving it the formula Na2CO3 ; and a formula unit of sodium oxalate always contains two sodium atoms, two carbon atoms, and four oxygen atoms, giving it the formula Na2C2O4 . Because each atom has a definite mass, each compound will have a distinct composition by mass. This composition is usually expressed as the percentage composition of the compound — the percentage by mass of each element in a compound. To identify a compound, you can compare the percentage composition obtained by laboratory analysis with a calculated percentage composition of each possible compound. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 73 Percentage Composition Back Print Name Class Date Problem Solving continued General Plan for Determining Percentage Composition of a Compound 1 Molar mass of element 2 Convert using the formula of the compound. Mass of element per mole of compound Convert by multiplying by the inverse of the molar mass of the compound. Then convert to a percentage by multiplying by 100. 5 Mass of element in a sample of compound 3 Convert by expressing percentage as a fraction and then multiplying by the mass of the sample. Percentage element in the compound Repeat 1, 2, and 3 for each remaining element in the compound. 4 Percentage composition of the compound Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 74 Percentage Composition Back Print Name Class Date Problem Solving continued Sample Problem 1 Determine the percentage composition of sodium carbonate, Na2CO3 . Solution ANALYZE What is given in the problem? What are you asked to find? the formula of sodium carbonate the percentage of each element in sodium carbonate (the percentage composition) Data Na2CO3 Na 22.99 g/mol C 12.01 g/mol O 16.00 g/mol 105.99 g/mol ?% Items Formula of sodium carbonate Molar mass of each element* Molar mass of sodium carbonate Percentage composition of sodium carbonate * determined from the periodic table PLAN What step is needed to determine the mass of each element per mole of compound? Multiply the molar mass of each element by the ratio of the number of moles of that element in a mole of the compound (the subscript of that element in the compound’s formula). What steps are needed to determine the portion of each element as a percentage of the mass of the compound? Multiply the mass of each element by the inverse of the molar mass of the compound, and then multiply by 100 to convert to a percentage. Step 1 1 Molar mass of Na multiply by the subscript of Na in Na2CO3 ratio of mol Na per mol Na2CO3 from formula 2 Mass Na per mole Na2CO3 molar mass Na 22.99 g Na 1 mol Na 2 mol Na 1 mol Na2CO3 g Na 1 mol Na2CO3 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 75 Percentage Composition Back Print Name Class Date Problem Solving continued Step 2 2 Mass Na per mole Na2CO3 multiply by the inverse of the molar mass of Na2CO3 and multiply by 100 3 Percentage Na in Na2CO3 from Step 1 g Na 1 mol Na2CO3 1 mol Na2CO3 105.99 g Na2CO3 1 molar mass Na2CO3 100 percentage Na in Na2CO3 Now you can combine Step 1 and Step 2 into one calculation. combining Steps 1 and 2 22.99 g Na 1 mol Na 2 mol Na 1 mol Na2CO3 1 mol Na2CO3 105.99 g Na2CO3 100 percentage Na in Na2CO3 Finally, determine the percentage of carbon and oxygen in Na2CO3 by repeating the calculation above with each of those elements. 3 Percentage of each element in Na2CO3 repeat Steps 1 and 2 for each remaining element 4 Percentage composition COMPUTE percentage sodium 22.99 g Na 1 mol Na percentage carbon 2 mol Na 1 mol Na2CO3 1 mol C 1 mol Na2CO3 3 mol O 1 mol Na2CO3 1 mol Na2CO3 105.99 g Na2CO3 1 mol Na2CO3 105.99 g Na2CO3 1 mol Na2CO3 105.99 g Na2CO3 100 43.38% Na 12.01 g C 1 mol C percentage oxygen 100 11.33% C 16.00 g O 1 mol O 100 45.29% O Element sodium carbon oxygen Percentage 43.38% Na 11.33% C 45.29% O EVALUATE Are the units correct? Yes; the composition is given in percentages. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 76 Percentage Composition Back Print Name Class Date Problem Solving continued Is the number of significant figures correct? Yes; four significant figures is correct because the molar masses have four significant figures. Is the answer reasonable? Yes; the percentages add up to 100 percent. Practice 1. Determine the percentage composition of each of the following compounds: a. sodium oxalate, Na2C2O4 ans: 34.31% Na, 17.93% C, 47.76% O b. ethanol, C2H5OH ans: 52.13% C, 13.15% H, 34.72% O c. aluminum oxide, Al2O3 ans: 52.92% Al, 47.08% O d. potassium sulfate, K2SO4 ans: 44.87% K, 18.40% S, 36.72% O 2. Suppose that your laboratory analysis of the white powder discussed at the beginning of this chapter showed 42.59% Na, 12.02% C, and 44.99% oxygen. Would you report that the compound is sodium oxalate or sodium carbonate (use the results of Practice Problem 1 and Sample Problem 1)? ans: sodium carbonate Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 77 Percentage Composition Back Print Name Class Date Problem Solving continued Sample Problem 2 Calculate the mass of zinc in a 30.00 g sample of zinc nitrate, Zn(NO3)2. Solution ANALYZE What is given in the problem? What are you asked to find? Items Mass of zinc nitrate Formula of zinc nitrate Molar mass of zinc nitrate Mass of zinc in the sample the mass in grams of zinc nitrate the mass in grams of zinc in the sample Data 30.00 g Zn(NO3)2 189.41 g/mol ?g PLAN What steps are needed to determine the mass of Zn in a given mass of Zn(NO3)2? The percentage of Zn in Zn(NO3)2 can be calculated and used to find the mass of Zn in the sample. 1 Molar mass of Zn multiply by the mole ratio of Zn to Zn(NO3)2 5 Mass Zn in g in sample express percentage as a fraction and multiply by the mass of the sample 2 Mass Zn per mole Zn(NO3 )2 multiply by the inverse of the molar mass of Zn(NO3)2 , then multiply by 100 3 Percentage Zn in Zn(NO3 )2 molar mass Zn 65.39 g Zn 1 mol Zn ratio of mol Zn per mol Zn(NO3)2 from formula 1 mol Zn 1 mol Zn(NO3)2 1 mol 189.41 g Zn(NO3)2 given 1 molar mass Zn(NO3)2 100 percentage Zn percentage Zn expressed as a fraction g Zn 100 g Zn(NO3)2 g Zn(NO3)2 g Zn in sample COMPUTE 65.39 g Zn 1 mol Zn 1 mol Zn 1 mol Zn(NO3)2 1 mol Zn(NO3)2 189.41 g Zn(NO3)2 100 34.52% Zn Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 78 Percentage Composition Back Print Name Class Date Problem Solving continued Note that mass percentage is the same as grams per 100 g, so 34.52% Zn in Zn(NO3)2 is the same as 34.52 g Zn in 100 g Zn(NO3)2 . 34.52 g Zn 100 g Zn(NO3)2 EVALUATE Are the units correct? Yes; units cancel to give the correct units, grams of zinc. 30.00 g Zn(NO3)2 10.36 g Zn Is the number of significant figures correct? Yes; four significant figures is correct because the data given have four significant figures. Is the answer reasonable? Yes; the molar mass of zinc is about one third of the molar mass of Zn(NO3)2, and 10.36 g Zn is about one third of 30.00 g of Zn(NO3)2. Practice 1. Calculate the mass of the given element in each of the following compounds: a. bromine in 50.0 g potassium bromide, KBr ans: 33.6 g Br b. chromium in 1.00 kg sodium dichromate, Na2Cr2O7 ans: 397 g Cr c. nitrogen in 85.0 mg of the amino acid lysine, C6H14N2O2 ans: 16.3 mg N d. cobalt in 2.84 g cobalt(II) acetate, Co(C2H3O2)2 ans: 0.945 g Co Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 79 Percentage Composition Back Print Name Class Date Problem Solving continued HYDRATES Many compounds, especially ionic compounds, are produced and purified by crystallizing them from water solutions. When this happens, some compounds incorporate water molecules into their crystal structure. These crystalline compounds are called hydrates because they include water molecules. The number of water molecules per formula unit is specific for each type of crystal. When you have to measure a certain quantity of the compound, it is important to know how much the water molecules contribute to the mass. You may have seen blue crystals of copper(II) sulfate in the laboratory. When this compound is crystallized from water solution, the crystals include five water molecules for each formula unit of CuSO4 . The true name of the substance is copper(II) sulfate pentahydrate, and its formula is written correctly as CuSO4 5H2O. Notice that the five water molecules are written separately. They are preceded by a dot, which means they are attached to the copper sulfate molecule. On a molar basis, a mole of CuSO4 5H2O contains 5 mol of water per mole of CuSO4 5H2O. The water molecules contribute to the total mass of CuSO4 5H2O. When you determine the percentage water in a hydrate, the water molecules are treated separately, as if they were another element. Sample Problem 3 Determine the percentage water in copper(II) sulfate pentahydrate, CuSO4 5H2O. Solution ANALYZE What is given in the problem? What are you asked to find? Items Formula of copper(II) sulfate pentahydrate Molar mass of H2O Molar mass of copper(II) sulfate pentahydrate* Percentage H2O in CuSO4 5H2O * molar mass of CuSO4 mass of 5 mol H2O the formula of copper(II) sulfate pentahydrate the percentage water in the hydrate Data CuSO4 5H2O 18.02 g/mol 249.72 g/mol ?% Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 80 Percentage Composition Back Print Name Class Date Problem Solving continued PLAN What steps are needed to determine the percentage of water in CuSO4 5H2O? Find the mass of water per mole of hydrate, multiply by the inverse molar mass of the hydrate, and multiply that by 100 to convert to a percentage. 1 Molar mass of H2O multiply by the mole ratio of H2O to CuSO4 5H2O 3 Percentage H2O in CuSO4 5H2O multiply by the inverse of the molar mass of CuSO4 5H2O; then multiply by 100 2 Mass H2O per mole CuSO4 5H2O 18.01 g H2O 1 mol H2O molar mass H2O ratio of moles H2O per mole CuSO4 5H2O from formula 5 mol H2O 1 mol CuSO4 5H2O 1 mol CuSO4 5H2O 249.72 g CuSO4 5H2O 100 percentage H2O 1 molar mass CuSO4 5H2O COMPUTE 18.01 g H2O 1 mol H2O 5 mol H2O 1 mol CuSO4 5H2O 1 mol CuSO4 5H2O 249.72 g CuSO4 5H2O 100 36.08% H2O EVALUATE Are the units correct? Yes; the percentage of water in copper(II) sulfate pentahydrate was needed. Is the number of significant figures correct? Yes; four significant figures is correct because molar masses were given to at least four significant figures. Is the answer reasonable? Yes; five water molecules have a mass of about 90 g, and 90 g is a little more than 1/3 of 250 g; the calculated percentage is a little more than 1/3. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 81 Percentage Composition Back Print Name Class Date Problem Solving continued Practice 1. Calculate the percentage of water in each of the following hydrates: a. sodium carbonate decahydrate, Na2CO3 10H2O ans: 62.97% H2O in Na2CO3 10H2O b. nickel(II) iodide hexahydrate, NiI2 6H2O ans: 25.71% H2O in NiI2 6H2O c. ammonium hexacyanoferrate(III) trihydrate (commonly called ammonium ferricyanide), (NH4)2Fe(CN)6 3H2O ans: 17.89 % H2O in (NH4)2Fe(CN)6 3H2O d. aluminum bromide hexahydrate ans: 28.85% H2O in AlBr3 6H2O Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 82 Percentage Composition Back Print Name Class Date Problem Solving continued Additional Problems 1. Write formulas for the following compounds and determine the percentage composition of each: a. nitric acid b. ammonia c. mercury(II) sulfate d. antimony(V) fluoride 2. Calculate the percentage composition of the following compounds: a. lithium bromide, LiBr b. anthracene, C14H10 c. ammonium nitrate, NH4NO3 d. nitrous acid, HNO2 e. silver sulfide, Ag2S f. iron(II) thiocyanate, Fe(SCN)2 g. lithium acetate h. nickel(II) formate 3. Calculate the percentage of the given element in each of the following compounds: a. nitrogen in urea, NH2CONH2 b. sulfur in sulfuryl chloride, SO2Cl2 c. thallium in thallium(III) oxide, Tl2O3 d. oxygen in potassium chlorate, KClO3 e. bromine in calcium bromide, CaBr2 f. tin in tin(IV) oxide, SnO2 4. Calculate the mass of the given element in each of the following quantities: a. oxygen in 4.00 g of manganese dioxide, MnO2 b. aluminum in 50.0 metric tons of aluminum oxide, Al2O3 c. silver in 325 g silver cyanide, AgCN d. gold in 0.780 g of gold(III) selenide, Au2Se3 e. selenium in 683 g sodium selenite, Na2SeO3 f. chlorine in 5.0 104 g of 1,1-dichloropropane, CHCl2CH2CH3 5. Calculate the percentage of water in each of the following hydrates: a. strontium chloride hexahydrate, SrCl2 6H2O b. zinc sulfate heptahydrate, ZnSO4 7H2O c. calcium fluorophosphate dihydrate, CaFPO3 2H2O d. beryllium nitrate trihydrate, Be(NO3)2 3H2O Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 83 Percentage Composition Back Print Name Class Date Problem Solving continued 6. Calculate the percentage of the given element in each of the following hydrates. You must first determine the formulas of the hydrates. a. nickel in nickel(II) acetate tetrahydrate b. chromium in sodium chromate tetrahydrate c. cerium in cerium(IV) sulfate tetrahydrate 7. Cinnabar is a mineral that is mined in order to produce mercury. Cinnabar is mercury(II) sulfide, HgS. What mass of mercury can be obtained from 50.0 kg of cinnabar? 8. The minerals malachite, Cu2(OH)2CO3 , and chalcopyrite, CuFeS2 , can be mined to obtain copper metal. How much copper could be obtained from 1.00 103 kg of each? Which of the two has the greater copper content? 9. Calculate the percentage of the given element in each of the following hydrates: a. vanadium in vanadium oxysulfate dihydrate, VOSO4 2H2O b. tin in potassium stannate trihydrate, K2SnO3 3H2O c. chlorine in calcium chlorate dihydrate, CaClO3 2H2O 10. Heating copper sulfate pentahydrate will evaporate the water from the crystals, leaving anhydrous copper sulfate, a white powder. Anhydrous means “without water.” What mass of anhydrous CuSO4 would be produced by heating 500.0 g of CuSO4 5H2O? 11. Silver metal may be precipitated from a solution of silver nitrate by placing a copper strip into the solution. What mass of AgNO3 would you dissolve in water in order to get 1.00 g of silver? 12. A sample of Ag2S has a mass of 62.4 g. What mass of each element could be obtained by decomposing this sample? 13. A quantity of epsom salts, magnesium sulfate heptahydrate, MgSO4 7H2O, is heated until all the water is driven off. The sample loses 11.8 g in the process. What was the mass of the original sample? 14. The process of manufacturing sulfuric acid begins with the burning of sulfur. What mass of sulfur would have to be burned in order to produce 1.00 kg of H2SO4 ? Assume that all of the sulfur ends up in the sulfuric acid. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 84 Percentage Composition Back Print Name Class Date Skills Worksheet Problem Solving Empirical Formulas Suppose you analyze an unknown compound that is a white powder and find that it is composed of 36.5% sodium, 38.1% oxygen, and 25.4% sulfur. You can use those percentages to determine the mole ratios among sodium, sulfur, and oxygen and write a formula for the compound. To begin, the mass percentages of each element can be interpreted as “grams of element per 100 grams of compound.” To make things simpler, you can assume you have a 100 g sample of the unknown compound. The unknown compound contains 36.5% sodium by mass. Therefore 100.0 g of the compound would contain 36.5 g of sodium. You already know how to convert mass of a substance into number of moles, so you can calculate the number of moles of sodium in 36.5 g. After you find the number of moles of each element, you can look for a simple ratio among the elements and use this ratio of elements to write a formula for the compound. The chemical formula obtained from the mass percentages is in the simplest form for that compound. The mole ratios for each element, which you determined from the analytical data given, are reduced to the smallest whole numbers. This simplest formula is also called the empirical formula. The actual formula for the compound could be a multiple of the empirical formula. For instance, suppose you analyze a compound and find that it is composed of 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen. If you determine the formula for this compound based only on the analytical data, you will determine the formula to be CH2O. There are, however, other possibilities for the formula. It could be C2H4O2 and still have the same percentage composition. In fact, it could be any multiple of CH2O. It is possible to convert from the empirical formula to the actual chemical formula for the compound as long as the molar mass of the compound is known. Look again at the CH2O example. If the true compound were CH2O, it would have a molar mass of 30.03 g/mol. If you do more tests on the unknown compound and find that its molar mass is 60.06, you know that CH2O cannot be its true identity. The molar mass 60.06 is twice the molar mass of CH2O. Therefore, you know that the true chemical formula must be twice the empirical formula, (CH2O) 2, or C2H4O2. Any correct molecular formula can be determined from an empirical formula and a molar mass in this same way. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 85 Empirical Formulas Back Print Name Class Date Problem Solving continued General Plan for Determining Empirical Formulas and Molecular Formulas 1 Percentage of element expressed as grams of element per 100 g unknown 2 Convert using the molar mass of each element. Amount of each element per 100 g of unknown Use the amount of the least-abundant element to calculate the simplest wholenumber ratio among the elements. 3 4 Empirical formula of the compound The calculated ratio is the simplest Convert using formula. the experimental molar mass of the unknown and the molar mass of the simplest formula. Calculated whole-number ratio among the elements 5 Molecular formula of the compound Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 86 Empirical Formulas Back Print Name Class Date Problem Solving continued Sample Problem 1 Determine the empirical formula for an unknown compound composed of 36.5% sodium, 38.1% oxygen, and 25.4% sulfur by mass. Solution ANALYZE What is given in the problem? What are you asked to find? Items The percentage composition of the unknown subtance the percentage composition of the compound the empirical formula for the compound Data 36.5% sodium 38.1% oxygen 25.4% sulfur 22.99 g Na/mol Na 16.00 g O/mol O 32.07 g S/mol S ? mol ? The molar mass of each element* Amount of each element per 100.0 g of the unknown Simplest mole ratio of elements in the unknown * determined from the periodic table PLAN What steps are needed to calculate the amount in moles of each element per 100.0 g of unknown? State the percentage of the element in grams and multiply by the inverse of the molar mass of the element. What steps are needed to determine the whole-number mole ratio of the elements in the unknown (the simplest formula)? Divide the amount of each element by the amount of the least-abundant element. If necessary, multiply the ratio by a small integer that will produce a whole-number ratio. 1 Mass of Na per 100.0 g unknown multiply by the inverse of the molar mass of Na 2 Amount Na in mol per 100.0 g unknown 1 percent of Na stated as grams Na per 100 g unknown molar mass Na 36.5 g Na 100.0 g unknown 1 mol Na 22.99 g Na mol Na 100.0 g unknown Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 87 Empirical Formulas Back Print Name Class Date Problem Solving continued Repeat this step for the remaining elements. 2 Amount of Na in mol per 100.0 g unknown divide by the amount of the least-abundant element 4 Empirical formula 3 Whole-number ratio among the elements COMPUTE 36.5 g Na 100.0 g unknown 38.1 g O 100.0 g unknown 25.4 g S 100.0 g unknown 1 mol Na 22.99 g Na 1 mol O 16.00 g O 1 mol S 32.07 g S 1.59 mol Na 100.0 g unknown 2.38 mol O 100.0 g unknown 0.792 mol S 100.0 g unknown Divide the amount of each element by the amount of the least-abundant element, which in this example is S. This can be accomplished by multiplying the amount of each element by the inverse of the amount of the least abundant element. 1.59 mol Na 100.0 g unknown 2.38 mol O 100.0 g unknown 0.792 mol S 100.0 g unknown 100.0 g unknown 0.792 mol S 100.0 g unknown 0.792 mol S 100.0 g unknown 0.792 mol S 2.01 mol Na 1 mol S 3.01 mol O 1 mol S 1.00 mol S 1 mol S From the calculations, the simplest mole ratio is 2 mol Na : 3 mol O : 1 mol S. The simplest formula is therefore Na2O3S. Seeing the ratio 3 mol O : 1 mol S, you can use your knowledge of chemistry to suggest that this possibly represents a sulfite group, – SO3 and propose the formula Na2SO3. EVALUATE Are the units correct? Yes; units canceled throughout the calculation, so it is reasonable to assume that the resulting ratio is accurate. Is the number of significant figures correct? Yes; ratios were calculated to three significant figures because percentages were given to three significant figures. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 88 Empirical Formulas Back Print Name Class Date Problem Solving continued Is the answer reasonable? Yes; the formula, Na2SO3 is plausible, given the mole ratios and considering that the sulfite ion has a 2 charge and the sodium ion has a 1 charge. Practice 1. Determine the empirical formula for compounds that have the following analyses: a. 28.4% copper, 71.6% bromine ans: CuBr2 b. 39.0% potassium, 12.0% carbon, 1.01% hydrogen, and 47.9% oxygen ans: KHCO3 c. 77.3% silver, 7.4% phosphorus, 15.3% oxygen ans: Ag3PO4 d. 0.57% hydrogen, 72.1% iodine, 27.3% oxygen ans: HIO3 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 89 Empirical Formulas Back Print Name Class Date Problem Solving continued Sample Problem 2 Determine the empirical formula for an unknown compound composed of 38.4% oxygen, 23.7% carbon, and 1.66% hydrogen. Solution ANALYZE What is given in the problem? What are you asked to find? the percentage composition of the compound the empirical formula for the compound PLAN What steps are needed to calculate the amount in moles of each element per 100.0 g of unknown? State the percentage of the element in grams and multiply by the inverse of the molar mass of the element. What steps are needed to determine the whole-number mole ratio of the elements in the unknown (the simplest formula)? Divide the amount of each element by the amount of the least-abundant element. If necessary, multiply the ratio by a small integer to produce a wholenumber ratio. 1 Mass of K in g per 100.0 g unknown multiply by the inverse of the molar mass of K 2 Amount of K in mol per 100.0 g unknown divide by the amount of the least-abundant element, and multiply by an integer that will produce a wholenumber ratio 4 Empirical formula 3 Whole-number ratio among the elements COMPUTE 38.4 g K 100.0 g unknown 1 mol K 39.10 g K 0.982 mol K 100.0 g unknown Proceed to find the amount in moles per 100.0 g of unknown for the elements carbon, oxygen, and hydrogen, as in Sample Problem 1. When determining the formula of a compound having more than two elements, it is usually advisable to put the data and results in a table. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 90 Empirical Formulas Back Print Name Class Date Problem Solving continued Element Potassium Carbon Oxygen Hydrogen Mass per 100.0 g of unknown 38.4 g K 23.7 g C 36.3 g O 1.66 g H Molar mass 39.10 g/mol 12.01 g/mol 16.00 g/mol 1.01 g/mol Amount in mol per 100.0 g of unknown 0.982 mol K 1.97 mol C 2.27 mol O 1.64 mol H Again, as in Sample Problem 1, divide each result by the amount in moles of the least-abundant element, which in this example is K. You should get the following results: Amount in mol of element per 100.0 g of unknown 0.982 mol K 1.97 mol C 2.27 mol O 1.64 mol H Amount in mol of element per mol of potassium 1.00 mol K 2.01 mol C 2.31 mol O 1.67 mol H Element Potassium Carbon Oxygen Hydrogen In contrast to Sample Problem 1, this calculation does not give a simple whole-number ratio among the elements. To solve this problem, multiply by a small integer that will result in a whole-number ratio. You can pick an integer that you think might work, or you can convert the number of moles to an equivalent fractional number. At this point, you should keep in mind that analytical data is never perfect, so change the number of moles to the fraction that is closest to the decimal number. Then, choose the appropriate integer factor to use. In this case, the fractions are in thirds so a factor of 3 will change the fractions into whole numbers. Amount in mol of element per mole of potassium 1.00 mol K 2.01 mol C 2.31 mol O 1.67 mol H Fraction nearest the decimal value 1 mol K 2 mol C 2 1/3 mol O 1 2/3 mol H Integer factor 3 3 3 3 Whole-number mole ratio 3 mol K 6 mol C 7 mol O 5 mol H Thus, the simplest formula for the compound is K3C6H5O7 , which happens to be the formula for potassium citrate. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 91 Empirical Formulas Back Print Name Class Date Problem Solving continued EVALUATE Is the answer reasonable? Yes; the formula, K3C6H5O7 is plausible, considering that the potassium ion has a 1 charge and the citrate polyatomic ion has a 3 charge. Practice 1. Determine the simplest formula for compounds that have the following analyses. The data may not be exact. a. 36.2% aluminum and 63.8% sulfur ans: Al2S3 b. 93.5% niobium and 6.50% oxygen ans: Nb5O2 c. 57.6% strontium, 13.8% phosphorus, and 28.6% oxygen ans: Sr3P2O8 or Sr3(PO4)2 d. 28.5% iron, 48.6% oxygen, and 22.9% sulfur ans: Fe2S3O12 or Fe2(SO4)3 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 92 Empirical Formulas Back Print Name Class Date Problem Solving continued Sample Problem 3 A compound is analyzed and found to have the empirical formula CH2O. The molar mass of the compound is found to be 153 g/mol. What is the compound’s molecular formula? Solution ANALYZE What is given in the problem? What are you asked to find? Items Empirical formula of unknown Experimental molar mass of unknown Molar mass of empirical formula Molecular formula of the compound the empirical formula, and the experimental molar mass the molecular formula of the compound Data CH2O 153 g/mol 30.03 g/mol ? PLAN What steps are needed to determine the molecular formula of the unknown compound? Multiply the experimental molar mass by the inverse of the molar mass of the empirical formula. The subscripts of the empirical formula are multiplied by the whole-number factor obtained. 4 Empirical formula of unknown multiply the experimental molar mass by the inverse of the molar mass of the empirical formula, and multiply each subscript in the empirical formula by the resulting factor 5 Molecular formula of unknown given factor that shows the number of times the empirical formula 1 molar mass of must be multiplied to get the empirical formula molecular formula 153 g 1 mol unknown 1 mol CH2O 30.03 g mol CH2O 1 mol unknown COMPUTE 153 g 1 mol unknown 1 mol CH2O 30.03 g 5.09 mol CH2O 1 mol unknown Allowing for a little experimental error, the molecular formula must be five times the empirical formula. Molecular formula (CH2O) 5 C5H10O5 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 93 Empirical Formulas Back Print Name Class Date Problem Solving continued EVALUATE Is the answer reasonable? Yes; the calculated molar mass of C5H10O5 is 150.15, which is close to the experimental molar mass of the unknown. Reference books show that there are several different compounds with the formula C5H10O5. Practice 1. Determine the molecular formula of each of the following unknown substances: a. empirical formula CH2, experimental molar mass 28 g/mol ans: C2H4 b. empirical formula B2H5, experimental molar mass 54 g/mol ans: B4H10 c. empirical formula C2HCl, experimental molar mass 179 g/mol ans: C6H3Cl3 d. empirical formula C6H8O, experimental molar mass 290 g/mol ans: C18H24O3 e. empirical formula C3H2O, experimental molar mass 216 g/mol ans: C12H8O4 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 94 Empirical Formulas Back Print Name Class Date Problem Solving continued Additional Problems 1. Determine the empirical formula for compounds that have the following analyses: a. 66.0% barium and 34.0% chlorine b. 80.38% bismuth, 18.46% oxygen, and 1.16% hydrogen c. 12.67% aluminum, 19.73% nitrogen, and 67.60% oxygen d. 35.64% zinc, 26.18% carbon, 34.88% oxygen, and 3.30% hydrogen e. 2.8% hydrogen, 9.8% nitrogen, 20.5% nickel, 44.5% oxygen, and 22.4% sulfur f. 8.09% carbon, 0.34% hydrogen, 10.78% oxygen, and 80.78% bromine 2. Sometimes, instead of percentage composition, you will have the composition of a sample by mass. Use the same method shown in Sample Problem 1, but use the actual mass of the sample instead of assuming a 100 g sample. Determine the empirical formula for compounds that have the following analyses: a. a 0.858 g sample of an unknown substance is composed of 0.537 g of copper and 0.321 g of fluorine b. a 13.07 g sample of an unknown substance is composed of 9.48 g of barium, 1.66 g of carbon, and 1.93 g of nitrogen c. a 0.025 g sample of an unknown substance is composed of 0.0091 g manganese, 0.0106 g oxygen, and 0.0053 g sulfur 3. Determine the empirical formula for compounds that have the following analyses: a. a 0.0082 g sample contains 0.0015 g of nickel and 0.0067 g of iodine b. a 0.470 g sample contains 0.144 g of manganese, 0.074 g of nitrogen, and 0.252 g of oxygen c. a 3.880 g sample contains 0.691 g of magnesium, 1.824 g of sulfur, and 1.365 g of oxygen d. a 46.25 g sample contains 14.77 g of potassium, 9.06 g of oxygen, and 22.42 g of tin 4. Determine the empirical formula for compounds that have the following analyses: a. 60.9% As and 39.1% S b. 76.89% Re and 23.12% O c. 5.04% H, 35.00% N, and 59.96% O d. 24.3% Fe, 33.9% Cr, and 41.8% O e. 54.03% C, 37.81% N, and 8.16% H f. 55.81% C, 3.90% H, 29.43% F, and 10.85% N Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 95 Empirical Formulas Back Print Name Class Date Problem Solving continued 5. Determine the molecular formulas for compounds having the following empirical formulas and molar masses: a. C2H4S; experimental molar mass 179 b. C2H4O; experimental molar mass 176 c. C2H3O2 ; experimental molar mass 119 d. C2H2O, experimental molar mass 254 6. Use the experimental molar mass to determine the molecular formula for compounds having the following analyses: a. 41.39% carbon, 3.47% hydrogen, and 55.14% oxygen; experimental molar mass 116.07 b. 54.53% carbon, 9.15% hydrogen, and 36.32% oxygen; experimental molar mass 88 c. 64.27% carbon, 7.19% hydrogen, and 28.54% oxygen; experimental molar mass 168.19 7. A 0.400 g sample of a white powder contains 0.141 g of potassium, 0.115 g of sulfur, and 0.144 g of oxygen. What is the empirical formula for the compound? 8. A 10.64 g sample of a lead compound is analyzed and found to be made up of 9.65 g of lead and 0.99 g of oxygen. Determine the empirical formula for this compound. 9. A 2.65 g sample of a salmon-colored powder contains 0.70 g of chromium, 0.65 g of sulfur, and 1.30 g of oxygen. The molar mass is 392.2. What is the formula of the compound? 10. Ninhydrin is a compound that reacts with amino acids and proteins to produce a dark-colored complex. It is used by forensic chemists and detectives to see fingerprints that might otherwise be invisible. Ninhydrin’s composition is 60.68% carbon, 3.40% hydrogen, and 35.92% oxygen. What is the empirical formula for ninhydrin? 11. Histamine is a substance that is released by cells in response to injury, infection, stings, and materials that cause allergic responses, such as pollen. Histamine causes dilation of blood vessels and swelling due to accumulation of fluid in the tissues. People sometimes take antihistamine drugs to counteract the effects of histamine. A sample of histamine having a mass of 385 mg is composed of 208 mg of carbon, 31 mg of hydrogen, and 146 mg of nitrogen. The molar mass of histamine is 111 g/mol. What is the molecular formula for histamine? 12. You analyze two substances in the laboratory and discover that each has the empirical formula CH2O. You can easily see that they are different substances because one is a liquid with a sharp, biting odor and the other is an odorless, crystalline solid. How can you account for the fact that both have the same empirical formula? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 96 Empirical Formulas Back Print Name Class Date Skills Worksheet Problem Solving Stoichiometry So far in your chemistry course, you have learned that chemists count quantities of elements and compounds in terms of moles and that they relate moles of a substance to mass by using the molar mass. In addition, you have learned to write chemical equations so that they represent the rearrangements of atoms that take place during chemical reactions, and you have learned to balance these equations. In this chapter you will be able to put these separate skills together to accomplish one of the most important tasks of chemistry—using chemical equations to make predictions about the quantities of substances that react or are given off as products and relating those quantities to one another. This process of relating quantities of reactants and products in a chemical reaction to one another is called stoichiometry. First, look at an analogy. Suppose you need to make several sandwiches to take on a picnic with friends. You decide to make turkey-and-cheese sandwiches using the following “equation:” 2 bread slices 2 turkey slices 1 lettuce leaf 1 cheese slice → 1 turkey-and-cheese sandwich This equation shows that you need those ingredients in a ratio of 2 : 2 : 1 : 1, respectively. You can use this equation to predict that you would need 30 turkey slices to make 15 sandwiches or 6 cheese slices to go with 12 turkey slices. Zinc reacts with oxygen according to the following balanced chemical equation: 2Zn O2 → 2ZnO Like the sandwich recipe, this equation can be viewed as a “recipe” for zinc oxide. It tells you that reacting two zinc atoms with a molecule of oxygen will produce two formula units of zinc oxide. Can you predict how many zinc oxide units could be formed from 500 zinc atoms? Could you determine how many moles of oxygen molecules it would take to react with 4 mol of zinc atoms? What if you had 22 g of zinc and wanted to know how many grams of ZnO could be made from it? Keep in mind that the chemical equation relates amounts, not masses, of products and reactants. The problems in this chapter will show you how to solve problems of this kind. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 97 Stoichiometry Back Print Name Class Date Problem Solving continued General Plan for Solving Stoichiometry Problems 1 Mass of substance A Convert using the molar mass of A. 2 Amount in mol of substance A 3 Convert using the mole ratio A, given in the B balanced chemical equation. Amount in mol of substance B Convert using the molar mass of B. 4 Mass of substance B Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 98 Stoichiometry Back Print Name Class Date Problem Solving continued Sample Problem 1 Ammonia is made industrially by reacting nitrogen and hydrogen under pressure, at high temperature, and in the presence of a catalyst. The equation is N2( g) 3H2( g) → 2NH3( g). If 4.0 mol of H2 react, how many moles of NH3 will be produced? Solution ANALYZE What is given in the problem? What are you asked to find? the balanced equation, and the amount of H2 in moles the amount of NH3 produced in moles Organization of data is extremely important in dealing with stoichiometry problems. You will find that it is most helpful to make data tables such as the following one. Items Substance Coefficient in balanced equation Molar mass Amount Mass of substance * NA means not applicable to the problem Data H2 3 NA* 4.0 mol NA NH3 2 NA ? mol NA PLAN What steps are needed to calculate the amount of NH3 that can be produced from 4.0 mol H2? Multiply by the mole ratio of NH3 to H2 determined from the coefficients of the balanced equation. 2 Amount of H2 in mol given multiply by mole ratio: mole ratio NH3 H2 3 Amount of NH3 in mol mol H2 2 mol NH3 3 mol H2 mol NH3 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 99 Stoichiometry Back Print Name Class Date Problem Solving continued COMPUTE 4.0 mol H2 EVALUATE Are the units correct? Yes; the answer has the correct units of moles NH3. 2 mol NH3 3 mol H2 2.7 mol NH3 Is the number of significant figures correct? Yes; two significant figures is correct because data were given to two significant figures. Is the answer reasonable? Yes; the answer is 2/3 of 4.0. Practice 1. How many moles of sodium will react with water to produce 4.0 mol of hydrogen in the following reaction? 2Na(s) 2H2O(l ) → 2NaOH(aq) H2( g) ans: 8.0 mol Na 2. How many moles of lithium chloride will be formed by the reaction of chlorine with 0.046 mol of lithium bromide in the following reaction? 2LiBr(aq) Cl2( g) → 2LiCl(aq) Br2(l ) ans: 0.046 mol LiCl Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 100 Stoichiometry Back Print Name Class Date Problem Solving continued 3. Aluminum will react with sulfuric acid in the following reaction. 2Al( s) 3H2SO4(l ) → Al2(SO4)3(aq) 3H2( g) a. How many moles of H2SO4 will react with 18 mol Al? ans: 27 mol H2SO4 b. How many moles of each product will be produced? ans: 27 mol H2, 9 mol Al2(SO4)3 4. Propane burns in excess oxygen according to the following reaction. C3H8 5O2 → 3CO2 4H2O a. How many moles each of CO2 and H2O are formed from 3.85 mol of propane? ans: 11.6 mol CO2, 15.4 mol H2O b. If 0.647 mol of oxygen is used in the burning of propane, how many moles each of CO2 and H2O are produced? How many moles of C3H8 are consumed? ans: 0.388 mol CO2, 0.518 mol H2O, 0.129 mol C3H8 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 101 Stoichiometry Back Print Name Class Date Problem Solving continued Sample Problem 2 Potassium chlorate is sometimes decomposed in the laboratory to generate oxygen. The reaction is 2KClO3(s) → 2KCl(s) 3O2(g). What mass of KClO3 do you need to produce 0.50 mol O2? Solution ANALYZE What is given in the problem? What are you asked to find? Items Substance Coefficient in balanced equation Molar mass* Amount Mass * determined from the periodic table the amount of oxygen in moles the mass of potassium chlorate Data KClO3 2 122.55 g/mol ? mol ?g O2 3 NA 0.50 mol NA PLAN What steps are needed to calculate the mass of KClO3 needed to produce 0.50 mol O2? Use the mole ratio to convert amount of O2 to amount of KClO3. Then convert amount of KClO3 to mass of KClO3. 2 Amount of O2 in mol multiply by mole ratio 2 mol KClO3 3 mol O2 multiply by molar mass of KClO3 3 Amount of KClO3 in mol 4 Mass of KClO3 in g mole ratio given molar mass KClO3 mol O2 2 mol KClO3 3 mol O2 122.55 g KClO3 1 mol KClO3 g KClO3 COMPUTE 0.50 mol O2 EVALUATE Are the units correct? Yes; units canceled to give grams of KClO3. Copyright © by Holt, Rinehart and Winston. All rights reserved. 2 mol KClO3 3 mol O2 122.55 g KClO3 1 mol KClO3 41 g KClO3 Holt ChemFile: Problem-Solving Workbook 102 Stoichiometry Back Print Name Class Date Problem Solving continued Is the number of significant figures correct? Yes; two significant figures is correct. Is the answer reasonable? Yes; 41 g is about 1/3 of the molar mass of KClO3, and 0.5 2/3 1/3. Practice 1. Phosphorus burns in air to produce a phosphorus oxide in the following reaction: 4P(s) 5O2( g) → P4O10(s) a. What mass of phosphorus will be needed to produce 3.25 mol of P4O10? ans: 403 g P b. If 0.489 mol of phosphorus burns, what mass of oxygen is used? What mass of P4O10 is produced? ans: 19.6 g O2, 15.4 g P2O4 2. Hydrogen peroxide breaks down, releasing oxygen, in the following reaction: 2H2O2(aq) → 2H2O(l ) O2( g) a. What mass of oxygen is produced when 1.840 mol of H2O2 decomposes? ans: 29.44 g O2 b. What mass of water is produced when 5.0 mol O2 is produced by this reaction? ans: 180 g H2O Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 103 Stoichiometry Back Print Name Class Date Problem Solving continued Sample Problem 3 How many moles of aluminum will be produced from 30.0 kg Al2O3 in the following reaction? 2Al2O3 → 4Al 3O2 Solution ANALYZE What is given in the problem? What are you asked to find? Items Substance Coefficient in balanced equation Molar mass Amount Mass the mass of aluminum oxide the amount of aluminum produced Data Al2O3 2 101.96 g/mol ? mol 30.0 kg Al 4 NA ? mol NA PLAN What steps are needed to calculate the amount of Al produced from 30.0 kg of Al2O3? The molar mass of Al2O3 can be used to convert to moles Al2O3. The mole ratio of Al:Al2O3 from the coefficients in the equation will convert to moles Al from moles Al2O3. 1 Mass of Al2O3 in g multiply by the inverse of the molar mass of Al2O3 multiply by 1000 g 1 kg Mass of Al2O3 in kg 2 Amount of Al2O3 in mol multiply by the mole ratio 4 mol Al 2 mol Al2O3 1 molar mass Al2O3 3 Amount of Al in mol mole ratio given kg Al2 O3 1000 g kg 1 mol Al2O3 101.96 g Al2O3 4 mol Al 2 mol Al2O2 mol Al COMPUTE 30.0 kg Al2O3 1000 g kg 1 mol Al2O3 101.96 g Al2O3 4 mol Al 2 mol Al2O3 588 mol Al Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 104 Stoichiometry Back Print Name Class Date Problem Solving continued EVALUATE Are the units correct? Yes; units canceled to give moles of Al. Is the number of significant figures correct? Yes; three significant figures is correct. Is the answer reasonable? Yes; the molar mass of Al2O3 is about 100, so 30 kg of Al2O3 is about 300 mol. The mole ratio of Al:Al2O3 is 2:1, so the answer should be about 600 mol Al. Practice 1. Sodium carbonate reacts with nitric acid according to the following equation. Na2CO3( s) 2HNO3 → 2NaNO3 CO2 H2O a. How many moles of Na2CO3 are required to produce 100.0 g of NaNO3? ans: 0.5882 mol Na2CO3 b. If 7.50 g of Na2CO3 reacts, how many moles of CO2 are produced? ans: 0.0708 mol CO2 2. Hydrogen is generated by passing hot steam over iron, which oxidizes to form Fe3O4 , in the following equation. 3Fe( s) 4H2O( g) → 4H2( g) Fe3O4( s) a. If 625 g of Fe3O4 is produced in the reaction, how many moles of hydrogen are produced at the same time? ans: 10.8 mol H2 b. How many moles of iron would be needed to generate 27 g of hydrogen? ans: 10. mol Fe Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 105 Stoichiometry Back Print Name Class Date Problem Solving continued Sample Problem 4 Methane burns in air by the following reaction: CH4( g) 2O2( g) → CO2( g) 2H2O( g) What mass of water is produced by burning 500. g of methane? Solution ANALYZE What is given in the problem? What are you asked to find? Items Substance Coefficient in balanced equation Molar mass Amount Mass the mass of methane in grams the mass of water produced Data CH4 1 16.05 g/mol ? mol 500. g H2O 2 18.02 g/mol ? mol ?g PLAN What steps are needed to calculate the mass of H2O produced from the burning of 500. g of CH4? Convert grams of CH4 to moles CH4 by using the molar mass of CH4. Use the mole ratio from the balanced equation to determine moles H2O from moles CH4. Use the molar mass of H2O to calculate grams H2O. 1 Mass of CH4 in g multiply by the inverse of the molar mass of CH4 2 Amount of CH4 in mol multiply by the mole ratio 2 mol H2O 1 mol CH4 multiply by the molar mass of H2O 3 Amount of H2O in mol 4 Mass of H2O in g given 1 molar mass CH4 mole ratio g CH4 1 mol CH4 16.05 g CH4 2 mol H2O 1 mol CH4 18.02 g H2O 1 mol H2O molar mass H2O g H 2O Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 106 Stoichiometry Back Print Name Class Date Problem Solving continued COMPUTE 500. g CH4 1 mol CH4 16.05 g CH4 2 mol H2O 1 mol CH4 18.02 g H2O 1 mol H2O 1.12 103 g H2O EVALUATE Are the units correct? Yes; mass of H2O was required, and units canceled to give grams H2O. Is the number of significant figures correct? Yes; three significant figures is correct because the mass of CH4 was given to three significant figures. Is the answer reasonable? Yes; CH4 and H2O have similar molar masses, and twice as many moles of H2O are produced as moles CH4 burned. So, you would expect to get a little more than 1000 g of H2O. Practice 1. Calculate the mass of silver bromide produced from 22.5 g of silver nitrate in the following reaction: 2AgNO3(aq) MgBr2(aq) → 2AgBr(s) Mg(NO3)2(aq) ans: 24.9 g AgBr 2. What mass of acetylene, C2H2, will be produced from the reaction of 90. g of calcium carbide, CaC2, with water in the following reaction? CaC2( s) 2H2O(l ) → C2H2( g) Ca(OH)2( s) ans: 37 g C2H2 3. Chlorine gas can be produced in the laboratory by adding concentrated hydrochloric acid to manganese(IV) oxide in the following reaction: MnO2(s) 4HCl(aq) → MnCl2(aq) 2H2O(l) Cl2( g) a. Calculate the mass of MnO2 needed to produce 25.0 g of Cl2. ans: 30.7 g MnO2 b. What mass of MnCl2 is produced when 0.091 g of Cl2 is generated? ans: 0.16 g MnCl2 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 107 Stoichiometry Back Print Name Class Date Problem Solving continued Additional Problems 1. How many moles of ammonium sulfate can be made from the reaction of 30.0 mol of NH3 with H2SO4 according to the following equation? 2NH3 H2SO4 → (NH4)2SO4 2. In a very violent reaction called a thermite reaction, aluminum metal reacts with iron(III) oxide to form iron metal and aluminum oxide according to the following equation: Fe2O3 2Al → 2Fe Al2O3 a. What mass of Al will react with 150 g of Fe2O3? b. If 0.905 mol Al2O3 is produced in the reaction, what mass of Fe is produced? c. How many moles of Fe2O3 will react with 99.0 g of Al? 3. As you saw in Sample Problem 1, the reaction N2( g) 3H2( g) → 2NH3( g) is used to produce ammonia commercially. If 1.40 g of N2 are used in the reaction, how many grams of H2 will be needed? 4. What mass of sulfuric acid, H2SO4 , is required to react with 1.27 g of potassium hydroxide, KOH? The products of this reaction are potassium sulfate and water. 5. Ammonium hydrogen phosphate, (NH4)2HPO4 , a common fertilizer, is made from reacting phosphoric acid, H3PO4 , with ammonia. a. Write the equation for this reaction. b. If 10.00 g of ammonia react, how many moles of fertilizer will be produced? c. What mass of ammonia will react with 2800 kg of H3PO4 ? 6. The following reaction shows the synthesis of zinc citrate, a ingredient in toothpaste, from zinc carbonate and citric acid. 3ZnCO3( s) 2C6H8O7(aq) → Zn3(C6H5O7)2(aq) 3H2O(l ) 3CO2( g) a. How many moles of ZnCO3 and C6H8O7 are required to produce 30.0 mol of Zn3(C6H5O7)2? b. What quantities, in kilograms, of H2O and CO2 are produced by the reaction of 500. mol of citric acid? 7. Methyl butanoate, an oily substance with a strong fruity fragrance, can be made by reacting butanoic acid with methanol according to the following equation: C3H7COOH CH3OH → C3H7COOCH3 H2O a. What mass of methyl butanoate is produced from the reaction of 52.5 g of butanoic acid? b. In order to purify methyl butanoate, water must be removed. What mass of water is produced from the reaction of 5800. g of methanol? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 108 Stoichiometry Back Print Name Class Date Problem Solving continued 8. Ammonium nitrate decomposes to yield nitrogen gas, water, and oxygen gas in the following reaction: 2NH4NO3 → 2N2 O2 4H2O a. How many moles of nitrogen gas are produced when 36.0 g of NH4NO3 reacts? b. If 7.35 mol of H2O are produced in this reaction, what mass of NH4NO3 reacted? 9. Lead(II) nitrate reacts with potassium iodide to produce lead(II) iodide and potassium nitrate. If 1.23 mg of lead nitrate are consumed, what is the mass of the potassium nitrate produced? 10. A car battery produces electrical energy with the following chemical reaction: Pb( s) PbO2( s) 2H2SO4(aq) → 2PbSO4( s) 2H2O(l ) If the battery loses 0.34 kg of lead in this reaction, how many moles of lead(II) sulfate are produced? 11. In a space shuttle, the CO2 that the crew exhales is removed from the air by a reaction within canisters of lithium hydroxide. On average, each astronaut exhales about 20.0 mol of CO2 daily. What mass of water will be produced when this amount reacts with LiOH? The other product of the reaction is Li2CO3. 12. Water is sometimes removed from the products of a reaction by placing them in a closed container with excess P4O10 . Water is absorbed by the following reaction: P4O10 6H2O → 4H3PO4 102 g of P4O10 ? a. What mass of water can be absorbed by 1.00 b. If the P4O10 in the container absorbs 0.614 mol of water, what mass of H3PO4 is produced? c. If the mass of the container of P4O10 increases from 56.64 g to 63.70 g, how many moles of water are absorbed? 13. Ethanol, C2H5OH, is considered a clean fuel because it burns in oxygen to produce carbon dioxide and water with few trace pollutants. If 95.0 g of H2O are produced during the combustion of ethanol, how many grams of ethanol were present at the beginning of the reaction? 14. Sulfur dioxide is one of the major contributors to acid rain. Sulfur dioxide can react with oxygen and water in the atmosphere to form sulfuric acid, as shown in the following equation: 2H2O(l ) O2( g) 2SO2( g) → 2H2SO4(aq) If 50.0 g of sulfur dioxide from pollutants reacts with water and oxygen found in the air, how many grams of sulfuric acid can be produced? How many grams of oxygen are used in the process? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 109 Stoichiometry Back Print Name Class Date Problem Solving continued 15. When heated, sodium bicarbonate, NaHCO3 , decomposes into sodium carbonate, Na2CO3 , water, and carbon dioxide. If 5.00 g of NaHCO3 decomposes, what is the mass of the carbon dioxide produced? 16. A reaction between hydrazine, N2H4 , and dinitrogen tetroxide, N2O4 , has been used to launch rockets into space. The reaction produces nitrogen gas and water vapor. a. Write a balanced chemical equation for this reaction. b. What is the mole ratio of N2O4 to N2 ? c. How many moles of N2 will be produced if 20 000 mol of N2H4 are used by a rocket? d. How many grams of H2O are made when 450. kg of N2O4 are consumed? 17. Joseph Priestley is credited with the discovery of oxygen. He produced O2 by heating mercury(II) oxide, HgO, to decompose it into its elements. How many moles of oxygen could Priestley have produced if he had decomposed 517.84 g of mercury oxide? 18. Iron(III) chloride, FeCl3 , can be made by the reaction of iron with chlorine gas. How much iron, in grams, will be needed to completely react with 58.0 g of Cl2? 19. Sodium sulfide and cadmium nitrate undergo a double-displacement reaction, as shown by the following equation: Na2S Cd(NO3)2 → 2NaNO3 CdS What is the mass, in milligrams, of cadmium sulfide that can be made from 5.00 mg of sodium sulfide? 20. Potassium permanganate and glycerin react explosively according to the following equation: 14KMnO4 4C3H5(OH)3 → 7K2CO3 7Mn2O3 5CO2 16H2O a. How many moles of carbon dioxide can be produced from 4.44 mol of KMnO4 ? b. If 5.21 g of H2O are produced, how many moles of glycerin, C3H5(OH)3 , were used? c. If 3.39 mol of potassium carbonate are made, how many grams of manganese(III) oxide are also made? d. How many grams of glycerin will be needed to react with 50.0 g of KMnO4 ? How many grams of CO2 will be produced in the same reaction? 21. Calcium carbonate found in limestone and marble reacts with hydrochloric acid to form calcium chloride, carbon dioxide, and water according to the following equation: CaCO3( s) 2HCl(aq) → CaCl2(aq) CO2( g) 3 H2O(l ) a. What mass of HCl will be needed to produce 5.00 10 kg of CaCl2? b. What mass of CO2 could be produced from the reaction of 750 g of CaCO3? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 110 Stoichiometry Back Print Name Class Date Problem Solving continued 22. The fuel used to power the booster rockets on the space shuttle is a mixture of aluminum metal and ammonium perchlorate. The following balanced equation represents the reaction of these two ingredients: 3Al( s) a. If 1.50 3NH4ClO4( s) → Al2O3( s) 5 AlCl3( g) 3NO( g) 6H2O( g) 10 g of Al react, what mass of NH4ClO4, in grams, is required? b. If aluminum reacts with 620 kg of NH4ClO4 , what mass of nitrogen monoxide is produced? 23. Phosphoric acid is typically produced by the action of sulfuric acid on rock that has a high content of calcium phosphate according to the following equation: 3H2SO4 a. If 2.50 Ca3(PO4)2 6H2O → 3[CaSO4 2H2O] 2H3PO4 105 kg of H2SO4 react, how many moles of H3PO4 can be made? b. What mass of calcium sulfate dihydrate is produced by the reaction of 400. kg of calcium phosphate? c. If the rock being used contains 78.8% Ca3(PO4)2 , how many metric tons of H3PO4 can be produced from 68 metric tons of rock? 24. Rusting of iron occurs in the presence of moisture according to the following equation: 4Fe( s) 3O2( g) → 2Fe2O3( s) Suppose that 3.19% of a heap of steel scrap with a mass of 1650 kg rusts in a year. What mass will the heap have after one year of rusting? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 111 Stoichiometry Back Print Name Class Date Skills Worksheet Problem Solving Limiting Reactants At the beginning of Chapter 8, a comparison was made between solving stoichiometry problems and making turkey sandwiches. Look at the sandwich recipe once more: 2 bread slices 2 turkey slices 1 lettuce leaf 1 cheese slice → 1 turkey-and-cheese sandwich If you have 24 slices of turkey, you can make 12 sandwiches at 2 slices per sandwich if you have enough of all the other ingredients. If, however, you have only 16 slices of bread, you can make only 8 sandwiches, even though you may an ample supply of the other ingredients. The bread is the limiting ingredient that prevents you from making more than 8 sandwiches. The same idea applies to chemical reactions. Look at a reaction used to generate hydrogen gas in the laboratory: Zn(s) H2SO4(aq) → ZnSO4(aq) H2( g) The balanced equation tells you that 1 mol Zn reacts with 1 mol H2SO4 to produce 1 mol ZnSO2 and 1 mol H2 . Suppose you have 1 mol Zn and 5 mol H2SO4 . What will happen, and what will you get? Only 1 mol of H2SO4 will react and only 1 mol of each of the products will be produced because only 1 mol Zn is available to react. In this situation, zinc is the limiting reactant. When it is used up the reaction stops even though more H2SO4 is available. It is difficult to directly observe molar amounts of reactants as they are used up. It is much easier to determine when a certain mass of a reactant has been completely used. Use molar masses to restate the equation in terms of mass, as follows: 65.39 g Zn 98.09 g H2SO4 → 161.46 g ZnSO4 2.02 g H2 This version of the equation tells you that zinc and sulfuric acid will always react in a mass ratio of 65.39 g of Zn : 98.09 g of H2SO4 or 0.667 g of Zn : 1.000 g of H2SO4 . If you have 65.39 g of Zn but only 87.55 g of H2SO4 , you will not be able to make 2.02 g of hydrogen. Sulfuric acid will be the limiting reactant, preventing the zinc from reacting completely. Suppose you place 20 g of zinc and 100 g of sulfuric acid into a flask. Which would be used up first? In other words, is the limiting reactant zinc or sulfuric acid? How much of each product will be produced? The sample problems in this chapter will show you how to answer these questions. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 112 Limiting Reactants Back Print Name Class Date Problem Solving continued General Plan for Solving Limiting Reactant Problems 1A Mass of reactant A available 1B Mass of reactant B available Convert using the molar mass of A. Convert using the molar mass of B. 2A Amount of reactant A in mol available 2B Amount of reactant B in mol available Convert using the mole B ratio, . A If there are more moles of B available than needed, A is the limiting reactant. If there are fewer moles of B than needed, B is the limiting reactant. 3 Amount of reactant B needed to react with A 4 Limiting Reactant Convert using the mole ratio, product . limiting reactant 6 Mass of product Convert using the molar mass of the product. 5 Amount of product in mol Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 113 Limiting Reactants Back Print Name Class Date Problem Solving continued Sample Problem 1 Calcium hydroxide, used to neutralize acid spills, reacts with hydrochloric acid according to the following equation: Ca(OH)2 2HCl → CaCl2 2H2O If you have spilled 6.3 mol of HCl and put 2.8 mol of Ca(OH)2 on it, which substance is the limiting reactant? Solution ANALYZE What is given in the problem? What are you asked to find? Items Reactant Coefficient in balanced equation Molar mass Amount of reactant Mass of reactant Limiting reactant * not applicable to the problem the balanced equation, the amounts of Ca(OH)2 and HCl in moles the limiting reactant Data Ca(OH)2 1 NA* 2.8 mol NA ? HCl 2 NA 6.3 mol NA ? PLAN What steps are needed to determine the limiting reactant? Choose one of the reactants. Use the mole ratio between the two reactants to compute the amount of the other reactant that would be needed to react with it. Compare that amount with the amount available. 2A Amount of Ca(OH)2 in mol multiply by mole ratio HCl Ca(OH)2 compare moles of HCl needed with moles of HCl available 2B Amount of HCl in mol 3 Amount of HCl needed to react with Ca(OH)2 4 Limiting reactant Choose one of the reactants, for instance, Ca(OH)2 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 114 Limiting Reactants Back Print Name Class Date Problem Solving continued mole ratio given mol Ca(OH)2 2 mol HCl 1 mol Ca(OH)2 mol HCl needed COMPUTE 2.8 mol Ca(OH)2 2 mol HCl 1 mol Ca(OH)2 5.6 mol HCl needed The computation shows that more HCl (6.3 mol) is available than is needed (5.6 mol) to react with the 2.8 mol Ca(OH)2 available. Therefore, HCl is present in excess, making Ca(OH)2 the limiting reactant. EVALUATE Is the answer reasonable? Yes; you can see that 6.3 mol HCl is more than is needed to react with 2.8 mol Ca(OH)2. Practice 1. Aluminum oxidizes according to the following equation: 4Al 3O2 → 2Al2O3 Powdered Al (0.048 mol) is placed into a container containing 0.030 mol O2 . What is the limiting reactant? ans: O2 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 115 Limiting Reactants Back Print Name Class Date Problem Solving continued Sample Problem 2 Chlorine can replace bromine in bromide compounds forming a chloride compound and elemental bromine. The following equation is an example of this reaction. 2KBr(aq) Cl2(aq) → 2KCl(aq) Br2(l ) When 0.855 g of Cl2 and 3.205 g of KBr are mixed in solution, which is the limiting reactant? How many grams of Br2 are formed? Solution ANALYZE What is given in the problem? What are you asked to find? the balanced equation, and the masses of Cl2 and KBr available which reactant is limiting, and the mass of Br2 produced Data KBr 2 119.00 g/mol ? mol 3.205 g ? Cl2 1 70.90 g/mol ? mol 0.855 g ? Br2 1 159.80 g/mol ? mol ?g NA Items Substance Coefficient in balanced equation Molar mass* Amount of substance Mass of substance Limiting reactant * determined from the periodic table PLAN What steps are needed to determine the limiting reactant? Convert mass of each reactant to amount in moles. Choose one of the reactants. Compute the amount of the other reactant needed. Compare that with the amount available. What steps are needed to determine the mass of Br2 produced in the reaction? Use amount of the limiting reactant and the mole ratio given in the equation to determine the amount of Br2 . Convert the amount of Br2 to the mass of Br2 using the molar mass. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 116 Limiting Reactants Back Print Name Class Date Problem Solving continued 1a Mass of KBr in g multiply by the inverse molar mass of KBr 1b Mass of Cl2 in g multiply by the inverse molar mass of Cl2 2a Amount of KBr in mol multiply by mole ratio: Cl2 KBr 2b Amount of Cl2 in mol 3 Amount of Cl2 to react with KBr compare moles of reactant needed with moles of reactant available needed 4 Limiting reactant multiply moles of limiting reactant Br2 by mole ratio: limiting reactant 6 Mass of Br2 in g multiply by the molar mass of Br2 1 molar mass KBr given 5 Amount of Br2 in mol g KBr 1 mol KBr 119.00 g KBr 1 molar mass Cl2 mol KBr given g Cl2 1 mol Cl2 70.90 g Cl2 mole ratio mol Cl2 Choose one of the reactants, KBr for instance. calculated above mol KBr 1 mol Cl2 1 mol KBr mole ratio mol Cl2 needed Determine the limiting reactant. calculated above mol limiting reactant mol Br2 mol limiting reactant 159.80 g Br2 1 mol Br2 molar mass Br2 g Br2 COMPUTE 3.205 g KBr 0.855 g Cl2 1 mol KBr 119.00 g KBr 1 mol Cl2 70.90 g Cl2 1 mol Cl2 2 mol KBr 0.02693 mol KBr 0.0121 mol Cl2 Choose one of the reactants, KBr, for instance. 0.02693 mol KBr 0.01346 mol Cl2 needed Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 117 Limiting Reactants Back Print Name Class Date Problem Solving continued Only 0.0121 mol Cl2 is available. For all of the KBr to react, 0.0136 mol Cl2 is needed. Therefore, Cl2 is the limiting reactant. 0.0121 mol Cl2 1 mol Br2 1 mol Cl2 159.80 g Br2 1 mol Br2 1.93 g Br2 EVALUATE Is the determination of limiting reactant reasonable? Yes; the mass of 2 mol KBr is 238 g and the mass of 1 mol Cl2 is about 71 g, so they react in roughly a 3:1 ratio by mass. 3.2 g KBr would require about 1 g of Cl2, but only 0.855 g is available. Are the units and significant figures of the mass of Br2 correct? The number of significant figures is correct because the mass of Cl2 was given to three significant figures. Units cancel to give grams of Br2. Practice 1. A process by which zirconium metal can be produced from the mineral zirconium(IV) orthosilicate, ZrSiO4 , starts by reacting it with chlorine gas to form zirconium(IV) chloride. ZrSiO4 2Cl2 → ZrCl4 SiO2 O2 What mass of ZrCl4 can be produced if 862 g of ZrSiO4 and 950. g of Cl2 are available? You must first determine the limiting reactant. ans: ZrSiO4, 1.10 103 g ZrCl4 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 118 Limiting Reactants Back Print Name Class Date Problem Solving continued Additional Problems 1. Heating zinc sulfide in the presence of oxygen yields the following: ZnS O2 → ZnO SO2 If 1.72 mol of ZnS is heated in the presence of 3.04 mol of O2 , which reactant will be used up? Balance the equation first. 2. Use the following equation for the oxidation of aluminum in the following problems. 4Al 3O2 → 2Al2O3 10 3 a. Which reactant is limiting if 0.32 mol Al and 0.26 mol O2 are available? b. How many moles of Al2O3 are formed from the reaction of 6.38 of O2 and 9.15 10 3 mol of Al? mol c. If 3.17 g of Al and 2.55 g of O2 are available, which reactant is limiting? 3. In the production of copper from ore containing copper(II) sulfide, the ore is first roasted to change it to the oxide according to the following equation: 2CuS 3O2 → 2CuO 2SO2 a. If 100 g of CuS and 56 g of O2 are available, which reactant is limiting? b. What mass of CuO can be formed from the reaction of 18.7 g of CuS and 12.0 g of O2? 4. A reaction such as the one shown here is often used to demonstrate a single replacement reaction. 3CuSO4(aq) 2Fe(s) → 3Cu(s) Fe2(SO4)3(aq) If you place 0.092 mol of iron filings in a solution containing 0.158 mol of CuSO4, what is the limiting reactant? How many moles of Cu will be formed? 5. In the reaction BaCO3 2HNO3 → Ba(NO3)2 CO2 H2O, what mass of Ba(NO3)2 can be formed by combining 55 g BaCO3 and 26 g HNO3? 6. Bromine displaces iodine in magnesium iodide by the following process: MgI2 Br2 → MgBr2 I2 a. Which is the excess reactant when 560 g of MgI2 and 360 g of Br2 react, and what mass remains? b. What mass of I2 is formed in the same process? 7. Nickel displaces silver from silver nitrate in solution according to the following equation: 2AgNO3 Ni → 2Ag Ni(NO3)2 a. If you have 22.9 g of Ni and 112 g of AgNO3 , which reactant is in excess? b. What mass of nickel(II) nitrate would be produced given the quantities above? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 119 Limiting Reactants Back Print Name Class Date Problem Solving continued 8. Carbon disulfide, CS2, is an important industrial substance. Its fumes can burn explosively in air to form sulfur dioxide and carbon dioxide. CS2( g) O2( g) → SO2( g) CO2( g) If 1.60 mol of CS2 burns with 5.60 mol of O2 , how many moles of the excess reactant will still be present when the reaction is over? 9. Although poisonous, mercury compounds were once used to kill bacteria in wounds and on the skin. One was called “ammoniated mercury” and is made from mercury(II) chloride according to the following equation: HgCl2(aq) 2NH3(aq) → Hg(NH2)Cl(s) NH4Cl(aq) a. What mass of Hg(NH2)Cl could be produced from 0.91 g of HgCl2 assuming plenty of ammonia is available? b. What mass of Hg(NH2)Cl could be produced from 0.91 g of HgCl2 and 0.15 g of NH3 in solution? 10. Aluminum chips are sometimes added to sodium hydroxide-based drain cleaners because they react to generate hydrogen gas which bubbles and helps loosen material in the drain. The equation follows. Al(s) NaOH(aq) H2O(l ) → NaAlO2(aq) H2( g) a. Balance the equation. b. How many moles of H2 can be generated from 0.57 mol Al and 0.37 mol NaOH in excess water? c. Which reactant should be limiting in order for the mixture to be most effective as a drain cleaner? Explain your choice. 11. Copper is changed to copper(II) ions by nitric acid according to the following equation: 4HNO3 Cu → Cu(NO3)2 2NO2 2H2O a. How many moles each of HNO3 and Cu must react in order to produce 0.0845 mol of NO2? b. If 5.94 g of Cu and 23.23 g of HNO3 are combined, which reactant is in excess? 12. One industrial process for producing nitric acid begins with the following reaction: 4NH3 5O2 → 4NO 6H2O a. If 2.90 mol NH3 and 3.75 mol O2 are available, how many moles of each product are formed? b. Which reactant is limiting if 4.20 available? 104 g of NH3 and 1.31 105 g of O2 are c. What mass of NO is formed in the reaction of 869 kg of NH3 and 2480 kg O2? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 120 Limiting Reactants Back Print Name Class Date Problem Solving continued 13. Acetaldehyde CH3CHO is manufactured by the reaction of ethanol with copper(II) oxide according to the following equation: CH3CH2OH CuO → CH3CHO H2O Cu What mass of acetaldehyde can be produced by the reaction between 620 g of ethanol and 1020 g of CuO? What mass of which reactant will be left over? 14. Hydrogen bromide can be produced by a reaction among bromine, sulfur dioxide, and water as follows. SO2 Br2 H2O → 2HBr H2SO4 If 250 g of SO2 and 650 g of Br2 react in the presence of excess water, what mass of HBr will be formed? 15. Sulfur dioxide can be produced in the laboratory by the reaction of hydrochloric acid and a sulfite salt such as sodium sulfite. Na2SO3 2HCl → 2NaCl SO2 H2O What mass of SO2 can be made from 25.0 g of Na2SO3 and 22.0 g of HCl? 16. The rare-earth metal terbium is produced from terbium(III) fluoride and calcium metal by the following displacement reaction: 2TbF3 3Ca → 3CaF2 2Tb a. Given 27.5 g of TbF3 and 6.96 g of Ca, how many grams of terbium could be produced? b. How many grams of the excess reactant are left over? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 121 Limiting Reactants Back Print Name Class Date Skills Worksheet Problem Solving Percentage Yield Although we can write perfectly balanced equations to represent perfect reactions, the reactions themselves are often not perfect. A reaction does not always produce the quantity of products that the balanced equation seems to guarantee. This happens not because the equation is wrong but because reactions in the real world seldom produce perfect results. As an example of an imperfect reaction, look again at the equation that shows the industrial production of ammonia. N2( g) 3H2( g) → 2NH3( g) In the manufacture of ammonia, it is nearly impossible to produce 2 mol (34.08 g) of NH3 from the simple reaction of 1 mol (28.02 g) of N2 and 3 mol (6.06 g) of H2 because some ammonia molecules begin breaking down into N2 and H2 molecules as soon as they are formed. There are several reasons that real-world reactions do not produce products at a yield of 100%. Some are simple mechanical reasons, such as: • Reactants or products leak out, especially when they are gases. • The reactants are not 100% pure. • Some product is lost when it is purified. There are also many chemical reasons, including: • The products decompose back into reactants (as with the ammonia process). • The products react to form different substances. • Some of the reactants react in ways other than the one shown in the equation. • These are called side reactions. The reaction occurs very slowly. This is especially true of reactions involving organic substances. Chemists are very concerned with the yields of reactions because they must find ways to carry out reactions economically and on a large scale. If the yield of a reaction is too small, the products may not be competitive in the marketplace. If a reaction has only a 50% yield, it produces only 50% of the amount of product that it theoretically should. In this chapter, you will learn how to solve problems involving real-world reactions and percentage yield. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 122 Percentage Yield Back Print Name Class Date Problem Solving continued General Plan for Solving Percentage-Yield Problems 1 Mass of reactants 2 Convert using the molar mass of the reactants. Amount of reactants Convert using in mol the mole ratio of the limiting reactant to the product. 3 Theoretical amount of product in mol Convert using the molar mass of the product. 4 Theoretical mass of product 6 Percentage yield % yield actual yield theoretical yield 100 5 Actual mass of product Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 123 Percentage Yield Back Print Name Class Date Problem Solving continued Sample Problem 1 Dichlorine monoxide, Cl2O is sometimes used as a powerful chlorinating agent in research. It can be produced by passing chlorine gas over heated mercury(II) oxide according to the following equation: HgO Cl2 → HgCl2 Cl2O What is the percentage yield, if the quantity of reactants is sufficient to produce 0.86 g of Cl2O but only 0.71 g is obtained? Solution ANALYZE What is given in the problem? What are you asked to find? Items Substance Mass available Molar mass Amount of reactant Coefficient in balanced equation Actual yield Theoretical yield (moles) Theoretical yield (grams) Percentage yield the balanced equation, the actual yield of Cl2O, and the theoretical yield of Cl2O the percentage yield of Cl2O Data Cl2O NA* NA NA NA 0.71 g NA 0.86 g ? * Although this table has many Not Applicable entries, you will need much of this information in other kinds of percentage-yield problems. PLAN What steps are needed to calculate the percentage yield of Cl2O? Compute the ratio of the actual yield to the theoretical yield, and multiply by 100 to convert to a percentage. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 124 Percentage Yield Back Print Name Class Date Problem Solving continued 4 Theoretical mass of Cl2O in g % yield actual yield theoretical yield 100 6 Percentage yield of Cl2O 5 Actual mass of Cl2O in g actual mass theoretical mass g Cl2O produced theoretical g Cl2O 100 percentage yield COMPUTE 0.71 g Cl2O 0.86 g Cl2O EVALUATE Are the units correct? Yes; the ratio was converted to a percentage. 100 83% yield Is the number of significant figures correct? Yes; the number of significant figures is correct because the data were given to two significant figures. Is the answer reasonable? Yes; 83% is about 5/6, which appears to be close to the ratio 0.71/0.86. Practice 1. Calculate the percentage yield in each of the following cases: a. theoretical yield is 50.0 g of product; actual yield is 41.9 g ans: 83.8% yield b. theoretical yield is 290 kg of product; actual yield is 270 kg ans: 93% yield Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 125 Percentage Yield Back Print Name Class Date Problem Solving continued c. theoretical yield is 6.05 ans: 69.1% yield 104 kg of product; actual yield is 4.18 104 kg d. theoretical yield is 0.00192 g of product; actual yield is 0.00089 g ans: 46% yield Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 126 Percentage Yield Back Print Name Class Date Problem Solving continued Sample Problem 2 Acetylene, C2H2 , can be used as an industrial starting material for the production of many organic compounds. Sometimes, it is first brominated to form 1,1,2,2-tetrabromoethane, CHBr2CHBr2 , which can then be reacted in many different ways to make other substances. The equation for the bromination of acetylene follows: acetylene C2H2 2Br2 → 1,1,2,2-tetrabromoethane CHBr2CHBR2 If 72.0 g of C2H2 reacts with excess bromine and 729 g of the product is recovered, what is the percentage yield of the reaction? Solution ANALYZE What is given in the problem? the balanced equation, the mass of acetylene that reacts, and the mass of tetrabromoethane produced the percentage yield of tetrabromoethane Data C2H2 72.0 g available 26.04 g/mol ? 1 NA NA NA NA CHBr2CHBr2 NA 345.64 g/mol NA 1 729 g ? ? ? What are you asked to find? Items Substance Mass available Molar mass* Amount of reactant Coefficient in balanced equation Actual yield Theoretical yield (moles) Theoretical yield (grams) Percentage yield * determined from the periodic table PLAN What steps are needed to calculate the theoretical yield of tetrabromoethane? Set up a stoichiometry calculation to find the amount of product that can be formed from the given amount of reactant. What steps are needed to calculate the percentage yield of tetrabromoethane? Compute the ratio of the actual yield to the theoretical yield, and multiply by 100 to convert to a percentage. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 127 Percentage Yield Back Print Name Class Date Problem Solving continued 1 Mass of C2H2 in g multiply by the inverse of the molar mass of C2H2 2 Amount of C2H2 in mol multiply by mole ratio: CHBr2CHBr2 C2H2 3 Theoretical amount of CHBr2CHBr2 in mol multiply by the molar mass of CHBr2CHBr2 4 Theoretical mass of CHBr2CHBr2 Percentage yield CHBr2CHBr2 % yield actual yield theoretical yield 100 5 Actual mass of CHBr2CHBr2 given g C2H2 1 mol C2H2 26.04 g C2H2 actual grams theoretical grams 1 molar mass C2H2 mole ratio 1 mol CHBr2CHBr2 1 mol C2H2 345.64 g CHBr2CHBr2 1 mol CHBr2CHBr2 theoretical g CHBr2CHBr2 molar mass CHBr2CHBr2 g CHBr2CHBr2 produced theoretical g CHBr2CHBr2 100 percentage yield CHBr2CHBr2 COMPUTE 72.0 g C2H2 1 mol C2H2 26.04 g C2H2 1 mol CHBr2CHBr2 1 mol C2H2 100 345.64 g CHBr2CHBr2 1 mol CHBr2CHBr2 956 g CHBr2CHBr2 76.3% yield 729 g CHBr2CHBr2 956 g CHBr2CHBr2 EVALUATE Are the units correct? Yes; units canceled to give grams of CHBr2CHBr2. Also, the ratio was converted to a percentage. Is the number of significant figures correct? Yes; the number of significant figures is correct because the data were given to three significant figures. Is the answer reasonable? Yes; about 3 mol of acetylene were used and the theoretical yield is the mass of about 3 mol tetrabromoethane. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 128 Percentage Yield Back Print Name Class Date Problem Solving continued Practice 1. In the commercial production of the element arsenic, arsenic(III) oxide is heated with carbon, which reduces the oxide to the metal according to the following equation: 2As2O3 3C → 3CO2 4As a. If 8.87 g of As2O3 is used in the reaction and 5.33 g of As is produced, what is the percentage yield? ans: 79.3% yield b. If 67 g of carbon is used up in a different reaction and 425 g of As is produced, calculate the percentage yield of this reaction. ans: 76% yield Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 129 Percentage Yield Back Print Name Class Date Problem Solving continued Additional Problems 1. Ethyl acetate is a sweet-smelling solvent used in varnishes and fingernailpolish remover. It is produced industrially by heating acetic acid and ethanol together in the presence of sulfuric acid, which is added to speed up the reaction. The ethyl acetate is distilled off as it is formed. The equation for the process is as follows. acetic acid ethanol CH3COOH CH3CH2OH H2SO4 ethyl acetate CH3COOCH2CH3 H2O Determine the percentage yield in the following cases: a. 68.3 g of ethyl acetate should be produced but only 43.9 g is recovered. b. 0.0419 mol of ethyl acetate is produced but 0.0722 mol is expected. (Hint: Percentage yield can also be calculated by dividing the actual yield in moles by the theoretical yield in moles.) c. 4.29 mol of ethanol is reacted with excess acetic acid, but only 2.98 mol of ethyl acetate is produced. d. A mixture of 0.58 mol ethanol and 0.82 mol acetic acid is reacted and 0.46 mol ethyl acetate is produced. (Hint: What is the limiting reactant?) 2. Assume the following hypothetical reaction takes place. 2A 7B → 4C 3D Calculate the percentage yield in each of the following cases: a. The reaction of 0.0251 mol of A produces 0.0349 mol of C. b. The reaction of 1.19 mol of A produces 1.41 mol of D. c. The reaction of 189 mol of B produces 39 mol of D. d. The reaction of 3500 mol of B produces 1700 mol of C. 3. Elemental phosphorus can be produced by heating calcium phosphate from rocks with silica sand (SiO2) and carbon in the form of coke. The following reaction takes place. Ca3(PO4)2 3SiO2 5C → 3CaSiO3 2P 5CO a. If 57 mol of Ca3(PO4)2 is used and 101 mol of CaSiO3 is obtained, what is the percentage yield? b. Determine the percentage yield obtained if 1280 mol of carbon is consumed and 622 mol of CaSiO3 is produced. c. The engineer in charge of this process expects a yield of 81.5%. If 1.4 105 mol of Ca3(PO4)2 is used, how many moles of phosphorus will be produced? 4. Tungsten (W) can be produced from its oxide by reacting the oxide with hydrogen at a high temperature according to the following equation: WO3 3H2 → W 3H2O a. What is the percentage yield if 56.9 g of WO3 yields 41.4 g of tungsten? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 130 Percentage Yield Back Print Name Class Date Problem Solving continued b. How many moles of tungsten will be produced from 3.72 g of WO3 if the yield is 92.0%? c. A chemist carries out this reaction and obtains 11.4 g of tungsten. If the percentage yield is 89.4%, what mass of WO3 was used? 5. Carbon tetrachloride, CCl4 , is a solvent that was once used in large quantities in dry cleaning. Because it is a dense liquid that does not burn, it was also used in fire extinguishers. Unfortunately, its use was discontinued because it was found to be a carcinogen. It was manufactured by the following reaction: CS2 3Cl2 → CCl4 S2Cl2 The reaction was economical because the byproduct disulfur dichloride, S2Cl2 , could be used by industry in the manufacture of rubber products and other materials. a. What is the percentage yield of CCl4 if 719 kg is produced from the reaction of 410. kg of CS2 . b. If 67.5 g of Cl2 are used in the reaction and 39.5 g of S2Cl2 is produced, what is the percentage yield? c. If the percentage yield of the industrial process is 83.3%, how many kilograms of CS2 should be reacted to obtain 5.00 104 kg of CCl4 ? How many kilograms of S2Cl2 will be produced, assuming the same yield for that product? 6. Nitrogen dioxide, NO2 , can be converted to dinitrogen pentoxide, N2O5 , by reacting it with ozone, O3 . The reaction of NO2 takes place according to the following equation: 2NO2( g) O3( g) → N2O5(s or g) O2( g) a. Calculate the percentage yield for a reaction in which 0.38 g of NO2 reacts and 0.36 g of N2O5 is recovered. b. What mass of N2O5 will result from the reaction of 6.0 mol of NO2 if there is a 61.1% yield in the reaction? 7. In the past, hydrogen chloride, HCl, was made using the salt-cake method as shown in the following equation: 2NaCl(s) H2SO4(aq) → Na2SO4(s) 2HCl(g) If 30.0 g of NaCl and 0.250 mol of H2SO4 are available, and 14.6 g of HCl is made, what is the percentage yield? 8. Cyanide compounds such as sodium cyanide, NaCN, are especially useful in gold refining because they will react with gold to form a stable compound that can then be separated and broken down to retrieve the gold. Ore containing only small quantities of gold can be used in this form of “chemical mining.” The equation for the reaction follows. 4Au 8NaCN 2H2O O2 → 4NaAu(CN)2 4NaOH a. What percentage yield is obtained if 410 g of gold produces 540 g of NaAu(CN)2? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 131 Percentage Yield Back Print Name Class Date Problem Solving continued b. Assuming a 79.6% yield in the conversion of gold to NaAu(CN)2 , what mass of gold would produce 1.00 kg of NaAu(CN)2 ? c. Given the conditions in (b), what mass of gold ore that is 0.001% gold would be needed to produce 1.00 kg of NaAu(CN)2 ? 9. Diiodine pentoxide is useful in devices such as respirators because it reacts with the dangerous gas carbon monoxide, CO, to produce relatively harmless CO2 according to the following equation: I2O5 5CO → I2 5CO2 a. In testing a respirator, 2.00 g of carbon monoxide gas is passed through diiodine pentoxide. Upon analyzing the results, it is found that 3.17 g of I2 was produced. Calculate the percentage yield of the reaction. b. Assuming that the yield in (a) resulted because some of the CO did not react, calculate the mass of CO that passed through. 10. Sodium hypochlorite, NaClO, the main ingredient in household bleach, is produced by bubbling chlorine gas through a strong lye (sodium hydroxide, NaOH) solution. The following equation shows the reaction that occurs. 2NaOH(aq) Cl2( g) → NaCl(aq) NaClO(aq) H2O(l ) a. What is the percentage yield of the reaction if 1.2 kg of Cl2 reacts to form 0.90 kg of NaClO? b. If a plant operator wants to make 25 metric tons of NaClO per day at a yield of 91.8%, how many metric tons of chlorine gas must be on hand each day? c. What mass of NaCl is formed per mole of chlorine gas at a yield of 81.8%? d. At what rate in kg per hour must NaOH be replenished if the reaction produces 370 kg/h of NaClO at a yield of 79.5%? Assume that all of the NaOH reacts to produce this yield. 11. Magnesium burns in oxygen to form magnesium oxide. However, when magnesium burns in air, which is only about 1/5 oxygen, side reactions form other products, such as magnesium nitride, Mg3N2 . a. Write a balanced equation for the burning of magnesium in oxygen. b. If enough magnesium burns in air to produce 2.04 g of magnesium oxide but only 1.79 g is obtained, what is the percentage yield? c. Magnesium will react with pure nitrogen to form the nitride, Mg3N2 . Write a balanced equation for this reaction. d. If 0.097 mol of Mg react with nitrogen and 0.027 mol of Mg3N2 is produced, what is the percentage yield of the reaction? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 132 Percentage Yield Back Print Name Class Date Problem Solving continued 12. Some alcohols can be converted to organic acids by using sodium dichromate and sulfuric acid. The following equation shows the reaction of 1-propanol to propanoic acid. 3CH3CH2CH2OH 2Na2Cr2O7 8H2SO4 → 3CH3CH2COOH 2Cr2(SO4)3 2Na2SO4 11H2O a. If 0.89 g of 1-propanol reacts and 0.88 g of propanoic acid is produced, what is the percentage yield? b. A chemist uses this reaction to obtain 1.50 mol of propanoic acid. The reaction consumes 136 g of propanol. Calculate the percentage yield. c. Some 1-propanol of uncertain purity is used in the reaction. If 116 g of Na2Cr2O7 are consumed in the reaction and 28.1 g of propanoic acid are produced, what is the percentage yield? 13. Acrylonitrile, C3H3N( g), is an important ingredient in the production of various fibers and plastics. Acrylonitrile is produced from the following reaction: C3H6( g) NH3( g) O2( g) → C3H3N( g) H2O( g) If 850. g of C3H6 is mixed with 300. g of NH3 and unlimited O2, to produce 850. g of acrylonitrile, what is the percentage yield? You must first balance the equation. 14. Methanol, CH3OH, is frequently used in race cars as fuel. It is produced as the sole product of the combination of carbon monoxide gas and hydrogen gas. a. If 430. kg of hydrogen react, what mass of methanol could be produced? b. If 3.12 yield? 103 kg of methanol are actually produced, what is the percentage 15. The compound, C6H16N2 , is one of the starting materials in the production of nylon. It can be prepared from the following reaction involving adipic acid, C6H10O4 : C6H10O4(l ) 2NH3( g) 4H2( g) → C6H16N2(l ) 4H2O(l ) What is the percentage yield if 750. g of adipic acid results in the production of 578 g of C6H16N2 ? 16. Plants convert carbon dioxide to oxygen during photosynthesis according to the following equation: CO2 H2O → C6H12O6 O2 Balance this equation, and calculate how much oxygen would be produced if 1.37 104 g of carbon dioxide reacts with a percentage yield of 63.4%. 17. Lime, CaO, is frequently added to streams and lakes which have been polluted by acid rain. The calcium oxide reacts with the water to form a base that can neutralize the acid as shown in the following reaction: CaO( s) H2O(l ) → Ca(OH)2( s) If 2.67 102 mol of base are needed to neutralize the acid in a lake, and the above reaction has a percentage yield of 54.3%, what is the mass, in kilograms, of lime that must be added to the lake? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 133 Percentage Yield Back Print Name Class Date Skills Worksheet Problem Solving Thermochemistry Thermochemistry deals with the changes in energy that accompany a chemical reaction. Energy is measured in a quantity called enthalpy, represented as H. The change in energy that accompanies a chemical reaction is represented as H. Hess’s law provides a method for calculating the H of a reaction from tabulated data. This law states that if two or more chemical equations are added, the H of the individual equations may also be added to find the H of the final equation. As an example of how this law operates, look at the three reactions below. (1) (2) (3) 2H2( g) O2( g) → 2H2O(l) 2H2O2(l) → 2H2( g) 2O2( g) 2H2O2(l) → 2H2O(l) O2( g) H H H 571.6 kJ/mol 375.6 kJ/mol ? kJ/mol When adding equations 1 and 2, the 2 mol of H2( g) will cancel each other out, while only 1 mol of O2( g) will cancel. 2H2( g) O2( g) 3 2H2O(l) 1 2H2O2(l) 3 2H2( g) 2H2O2(l) → 2H2O(l) 2O2( g) O2( g ) Combining what is left yields the following equation. Notice that this is the same equation as the third equation shown above. Adding the two H values for the reactions 1 and 2 gives the H value for reaction 3. Using Hess’s law to calculate the enthalpy of this reaction, the following answer is obtained. 571.6 kJ/mol 375.6 kJ/mol 196.0 kJ/mol Thus, the H value for the reaction is 196.0 kJ/mol. Equation 1 represents the formation of water from its elemental components. If equation 2 were written in reverse, it would represent the formation of hydrogen peroxide from its elemental components. Therefore, adding equations 1 and 2 is the equivalent of subtracting the equation for the formation of the reactants of equation 3 from the equation for the formation of the products of equation 3. 2H2( g) O2( g) → 2H2O(l) [2H2( g) 2O2( g) → 2H2O2(l)] 2H2O2(l) → 2H2O(l) Hreaction sum of H 0products f O2( g) The enthalpy of the final reaction can be rewritten using the following equation. sum of H 0reactants f The equation states that the enthalpy change of a reaction is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants. This allows Hess’s law to be extended to state that the Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 134 Thermochemistry Back Print Name Class Date Problem Solving continued enthalpy change of any reaction can be calculated by looking up the standard molar enthalpy of formation, H 0 , of each substance involved. Some common f enthalpies of formation may be found in Table 1. The enthalpy change, however, does not account for all of the energy change of a reaction. Changes in the disorder (entropy) of a system can add to or detract from the energy involved in the enthalpy change. This amount of energy is given by the expression T S, where T is the Kelvin temperature and S is the change in entropy during the reaction. A large increase in entropy, such as when a gas is produced from a reaction of liquids or solids, can contribute significantly to the overall energy change. The total amount of energy available from a reaction is called Gibbs energy and is denoted by G. Gibbs energy is given by the following equation. Greaction Hreaction T Sreaction TABLE 1 STANDARD ENTHALPIES OF FORMATION Substance NH3( g) NH4Cl(s) NH4F(s) NH4NO3(s) Br2(l) CaCO3(s) CaO(s) CH4( g) C3H8( g) CO2( g) F2( g) H2( g) HBr( g) HCl( g) 0 H f (kJ/mol) Substance HF( g) H2O( g) H2O(l) H2O2(l) H2SO4(l) FeO(s) Fe2O3(s) MnO2(s) N2O( g) O2( g) Na2O(s) Na2SO3(s) SO2( g) SO3( g) 0 H f (kJ/mol) 45.9 314.4 125 365.56 0.00 1207.6 634.9 74.9 104.7 393.5 0.00 0.00 36.29 92.3 273.3 241.82 285.8 187.8 813.989 825.5 1118.4 520.0 82.1 0.00 414.2 1101 296.8 395.7 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 135 Thermochemistry Back Print Name Class Date Problem Solving continued General Plan for Solving Thermochemistry Problems 1 0 sum of Hfproducts 0 sum of Hfreactants Hreaction Look up H0 for each reactant and f product. Be sure the physical states in the reaction match those given in the reference table. Solve for Hreaction. 2 Hreaction T Sreaction 3 Hreaction T Sreaction Greaction Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 136 Thermochemistry Back Print Name Class Date Problem Solving continued Sample Problem 1 Given the following two reactions and enthalpy data, calculate the enthalpy change for the reaction in which methane and oxygen combine to form ketene, CH2CO, and water. CH2CO( g) 2O2( g) 3 2CO2( g) H2O( g) CH4( g) 2O2( g) 3 CO2( g) 2H2O( g) H H 981.1 kJ 802.3 kJ Solution ANALYZE What is given in the problem? the desired product, chemical equations that can be added to obtain the desired product, and enthalpy changes for these chemical equations enthalpy change for the reaction in which methane and oxygen combine to form ketene and water Data 981.1 kJ 802.3 kJ What are you asked to find? Items H for reaction 1 H for reaction 2 PLAN What steps are needed to calculate H for the reaction between methane and oxygen to form CH2CO? First, the two equations must be added to produce the final reaction. The first equation must be reversed so that ketene is a product, as shown in the final equation. The second equation must be multiplied by 2 so that carbon dioxide cancels out of the final equation. Then the individual enthalpies for the reactions must be added, adjusting for the fact that equation 1 is reversed and equation 2 is doubled. 2 2CO2( g) [CH4( g) 2CH4( g) H2O( g) → CH2CO( g) 2O2( g) 2O2( g) → CO2( g) 2H2O( g)] 2O2( g) → CH2CO( g) (2 Hreaction 1 Hreaction 2) Hfinal reaction 3H2O( g) Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 137 Thermochemistry Back Print Name Class Date Problem Solving continued COMPUTE 2CO2( g) 2CH4( g) 2CO2( g) H2O( g) 2CH4( g) 2CH4( g) H2O( g) 3 CH2CO( g) 2O2( g) 4O2( g) 3 2CO2( g) 4H2O( g) 2 4O2( g) 3 CH2CO( g) ( 981.1 kJ) 802.3 kJ) 623.5 kJ 3 2O2( g) 2CO2( g) 4H2O( g) 2O2( g) 3 CH2CO( g) (2 3H2O( g) EVALUATE Are the units correct? Yes; adding terms in kilojoules gives an answer in kilojoules. Is the number of significant figures correct? Yes; the significant figures are correct. Rules for adding and rounding measurements give a result to four significant figures. Is the answer reasonable? Yes; the result can be approximated as 1600 kJ 1000 kJ 600 kJ. Practice 1. Calculate the reaction enthalpy for the following reaction. 5CO2( g) Si3N4( s) → 3SiO( s) 2N2O( g) 5CO( g) Use the following equations and data. (1) CO( g) SiO2( s) → SiO( g) CO2( g) (2) 8CO2( g) Si3N4( s) → 3SiO2( s) 2N2O( g) Hreaction 1 Hreaction 2 520.9 kJ 461.05 kJ ans: 2024 kJ 8CO( g) Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 138 Thermochemistry Back Print Name Class Date Problem Solving continued Sample Problem 2 Calculate the enthalpy of reaction for the decomposition of hydrogen peroxide to water and oxygen gas according to the following equation. H2O2(l ) 3 H2O(l ) O2( g) Use Table 1 for the necessary enthalpies of formation. Solution ANALYZE What is given in the problem? What are you asked to find? the equation for the decomposition of H2O2, enthalpies of formation given in Table 1 enthalpy of reaction for the decomposition of H2O2 Data ? kJ/mol 187.8 kJ/mol* 0.00 kJ/mol** 285.8 kJ/mol* Items H decomposition of H2O2(l) H 0 H2O2(l) f H 0 O2( g) f H 0 H2O(l) f * from Table 1 ** any pure elemental substance has a H 0 of zero f PLAN What steps are needed to calculate H for the decomposition reaction? First, the equation must be balanced. Add up the enthalpies of formation for the products. From this quantity, subtract the enthalpy of formation for the reactant. Balance the equation for the decomposition of hydrogen peroxide. 2H2O2(l) → 2H2O(l) 1 0 sum of Hfproducts O2( g) sum of H0reactants f look up ∆H for each reactant and product, and solve 0 f Hreaction 2 Hreaction given in Table 1 (2 COMPUTE [2( 285.8 kJ/mol) H0 f H2O Hf0O 2 ) (2 Hf0H O 22 ) Hreaction 0.00 kJ/mol] [2( 187.8 kJ/mol)] 196.0 kJ/mol Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 139 Thermochemistry Back Print Name Class Date Problem Solving continued EVALUATE Are the units correct? Yes; adding terms in kJ/mol gives kJ/mol. Is the number of significant figures correct? Yes; rules for adding and rounding measurements give a result to four significant figures. Is the answer reasonable? Yes; the result can be approximated as 600 kJ/mol 400 kJ/mol 200 kJ/mol. Practice Determine H for each of the following reactions. 1. The following reaction is used to make CaO from limestone. CaCO3(s) → CaO(s) CO2( g) ans: 179.2 kJ/mol 2. The following reaction represents the oxidation of FeO to Fe2O3. 2FeO(s) O2( g) → Fe2O3(s) ans: 533 kJ/mol 3. The following reaction of ammonia and hydrogen fluoride produces ammonium fluoride. NH3( g) HF( g) → NH4F(s) ans: 194 kJ/mol Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 140 Thermochemistry Back Print Name Class Date Problem Solving continued Sample Problem 3 Calculate the Gibbs energy change for the following reaction at 25°C. Ca(s) Use the following data. Hreaction 411.6 kJ/mol; Sreaction 31.8 J/mol K 2H2O(l) 3 Ca(OH)2(s) H2( g) Solution ANALYZE What is given in the problem? What are you asked to find? Items H for the reaction at 25°C S for the reaction at 25°C Temperature G for the reaction at 25°C T, H, and S for the reaction G the Gibbs energy of the reaction, Data 411.6 kJ/mol 31.8 J/mol K 25°C 298 K ? kJ/mol PLAN What steps are needed to calculate G for the given reaction? Apply the relationship G H T S. Sreaction in J/mol K convert using the factor 1 kJ 1000 J 2 Hreaction Kelvin temperature, T Sreaction in kJ/mol K 3 Hreaction given T Sreaction Greaction S J mol K given 1 kJ 1000 given kJ mol K calculated above Hreaction T Sreaction Greaction Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 141 Thermochemistry Back Print Name Class Date Problem Solving continued COMPUTE 31.8 J mol K 411.1 kJ/mol (298 K 1 kJ 1000 J 0.0318 kJ/mol K 411.6 kJ/mol 9.5 kJ/mol 0.0318 kJ/mol K ) Greaction 421.1 kJ/mol Are the units correct? Yes; the kelvin units canceled. Adding terms in kJ/mol gives kJ/mol. Is the number of significant figures correct? Yes; the number of significant figures is correct. Rules for adding and rounding measurements give a result with four significant figures. Is the answer reasonable? Yes; values were given for T, correctly. H, and S and the computation was carried out Practice 1. Calculate the Gibbs energy change, G, for the combustion of hydrogen sulfide according to the following chemical equation. Assume reactants and products are at 25°C. H2S( g) Hreaction Sreaction O2( g) → H2O(l) SO2( g) 562.1 kJ/mol 0.09278 kJ/mol K ans: 534.5 kJ/mol 2. Calculate the Gibbs energy change for the decomposition of sodium chlorate. Assume reactants and products are at 25°C. NaClO3(s) → NaCl(s) Hreaction Sreaction 19.1 kJ/mol 0.1768 kJ/mol K ans: 71.8 kJ/mol O2( g) 3. Calculate the Gibbs energy change for the combustion of 1 mol of ethane. Assume reactants and products are at 25°C. C2H6( g) Hreaction Sreaction O2( g) → 2CO2( g) 1683 kJ/mol 3H2O(l) 1561 kJ/mol 0.4084 kJ/mol K ans: Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 142 Thermochemistry Back Print Name Class Date Problem Solving continued Additional Problems 1. Calculate H for the violent reaction of fluorine with water. F2( g) H2O(l) → 2HF( g) O2( g) 2. Calculate H for the reaction of calcium oxide and sulfur trioxide. CaO(s) H2O(l) SO3( g) → H2SO4(l) H2SO4(l) Ca(s) → CaSO4(s) Ca(s) O2( g) → CaO(s) H2( g) O2( g) → H2O(l) Na2O(s) SO3( g) → CaSO4(s) H H H H 132.5 kJ/mol 602.5 kJ/mol 634.9 kJ/mol 285.8 kJ/mol Use the following equations and data. H2( g) 3. Calculate H for the reaction of sodium oxide with sulfur dioxide. SO2( g) → Na2SO3(s) 4. Use enthalpies of combustion to calculate H for the oxidation of 1-butanol to make butanoic acid. C4H9OH(l) Combustion of butanol: C4H9OH(l) Hc 2675.9 kJ/mol Combustion of butanoic acid: O2( g) → C3H7COOH(l) 6O2( g) → 4CO2( g) H2O(l) 5H2O(l) C3H7COOH(l) Hc 2183.6 kJ/mol 5O2( g) → 4CO2( g) 4H2O(l) 5. Determine the free energy change for the reduction of CuO with hydrogen. Products and reactants are at 25°C. CuO(s) H S 128.5 kJ/mol 70.1 J/mol K H2( g) → Cu(s) H2O(l) 6. Calculate the enthalpy change at 25°C for the reaction of sodium iodide and chlorine. Use only the data given. NaI(s) S G 79.9 J/mol K 98.0 kJ/mol Cl2( g) → NaCl(s) I2(l ) 7. The element bromine can be produced by the reaction of hydrogen bromide and manganese(IV) oxide. 4HBr( g) MnO2(s) → MnBr2(s) 2H2O(l ) Br2(l ) H for the reaction is 291.3 kJ/mol at 25°C. Use this value and values of H 0 from Table 1 to calculate H 0 of MnBr2(s). f f Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbooky 143 Thermochemistry Back Print Name Class Date Problem Solving continued 8. Calculate the change in entropy, S, at 25°C for the reaction of calcium carbide with water to produce acetylene gas. CaC2(s) G H 147.7 kJ/mol 125.6 kJ/mol 2H2O(l) → C2H2( g) Ca(OH)2(s) 9. Calculate the Gibbs energy change for the explosive decomposition of ammonium nitrate at 25°C. Note that H2O is a gas in this reaction. NH4NO3(s) → N2O( g) S 446.4 J/mol K 2H2O( g) 10. In locations where natural gas, which is mostly methane, is not available, many people burn propane, which is delivered by truck and stored in a tank under pressure. a. Write the chemical equations for the complete combustion of 1 mol of methane, CH4, and 1 mol of propane, C3H8. b. Calculate the enthalpy change for each reaction to determine the amount of energy released off by burning 1 mol of each fuel. c. Using the enthapies of combustion you calculated, determine the energy output per kilogram of each fuel. Which fuel yields more energy per unit mass? 11. The hydration of acetylene to form acetaldehyde is shown in the following equation: C2H2( g) H2O(l) → CH3CHO(l) Use enthapies of combustion for acetylene and acetaldehyde to compute the enthalpy of the above reaction. C2H2( g) Hc Hc 1299.6 kJ/mol CH3CHO(l) 1166.9 kJ/mol 2O2( g) → 2CO2( g) 2H2O(l) 2O2( g) → 2CO2( g) H2O(l) 12. Calculate the enthalpy for the combustion of decane. H 0 for liquid decane is f 300.9 kJ/mol. C10H22(l) MgO(s) 15O2( g) → 10CO2( g) 2HCl( g) → MgCl2(s) 11H2O(l) H2O(l) H H H 456.9 kJ/mol 601.6 kJ/mol 285.8 kJ/mol 13. Find the enthalpy of the reaction of magnesium oxide with hydrogen chloride. Use the following equations and data. Mg(s) 2HCl( g) → MgCl2(s) H2( g) Mg(s) O2( g) → MgO(s) H2O(l) → H2( g) O2( g) Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 144 Thermochemistry Back Print Name Class Date Problem Solving continued 14. What is the Gibbs energy change for the following reaction at 25°C? 2NaOH(s) S 10.6 J/mol K H 0NaOH 425.9 kJ/mol f ∆ 2Na(s) O3 2Na2O(s) H2( g) 15. The following equation represents the reaction between gaseous HCl and gaseous ammonia to form solid ammonium chloride. NH3( g) HCl( g) → NH4Cl(s) Calculate the entropy change in J/mol K for the reaction of hydrogen chloride and ammonia at 25°C using the following data and the values fround in Table 1. G 91.2 kJ/mol 16. The production of steel from iron involves the removal of many impurities in the iron ore. The following equations show some of the purifying reactions. Calculate the enthalpy for each reaction. Use Table 1 and the data given. a. 3C(s) Fe2O3(s) → 3CO( g) 110.53 kJ/mol Fe2O3(s) → 3MnO(s) 384.9 kJ/mol 2Fe(s) 2Fe(s) 20Fe(s) 4Fe(s) 4Fe(s) H 0CO( g) f b. 3Mn(s) H 0MnO(s) f c. 12P(s) 10Fe2O3(s) → 3P4O10(s) 10(s) H 0P O f 4 3009.9 kJ/mol 910.9 kJ/mol d. 3Si(s) 2Fe2O3(s) → 3SiO2(s) 2 H 0SiO (s) f e. 3S(s) 2Fe2O3(s) → 3SO2( g) Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 145 Thermochemistry Back Print Name Class Date Skills Worksheet Problem Solving Gas Laws Chemists found that there were relationships among temperature, volume, pressure, and quantity of a gas that could be described mathematically. This chapter deals with Boyle’s law, Charles’s law, Gay-Lussac’s law, the combined gas law, and Dalton’s law of partial pressures. These laws have one condition in common. They all assume that the molar amount of gas does not change. In other words, these laws work correctly only when no additional gas enters a system and when no gas leaks out of it. Remember also that a law describes a fact of nature. Gases do not “obey” laws. The law does not dictate the behavior of the gas. Rather, each gas law describes a certain behavior of gas that occurs if conditions are right. BOYLE’S LAW Robert Boyle, a British chemist who lived from 1627 to 1691 formulated the first gas law, now known as Boyle’s law. This law describes the relationship between the pressure and volume of a sample of gas confined in a container. Boyle found that gases compress, much like a spring, when the pressure on the gas is increased. He also found that they “spring back” when the pressure is lowered. By “springing back” he meant that the volume increases when pressure is lowered. It’s important to note that Boyle’s law is true only if the temperature of the gas does not change and no additional gas is added to the container or leaks out of the container. Boyle’s law states that the volume and pressure of a sample of gas are inversely proportional to each other at constant temperature. This statement can be expressed as follows. Proportionality symbol. It means “is proportional to.” Proportionality constant V 1 P and PV k or V k 1 P According to Boyle’s law, when the pressure on a gas is increased, the volume of the gas decreases. For example, if the pressure is doubled, the volume decreases by half. If the volume quadruples, the pressure decreases to one-fourth of its original value. The expression PV k means that the product of the pressure and volume of any sample of gas is a constant, k. If this is true, then P V under one set of conditions is equal to P V for the same sample of gas under a second set of conditions, as long as the temperature remains constant. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 146 Gas Laws Back Print Name Class Date Problem Solving continued Boyle’s law can be expressed by the following mathematical equation. P1V1 P2V2 Pressure under the first set of conditions Volume under the first set of conditions Pressure under the second set of conditions Volume under the second set of conditions General Plan for Solving Boyle's-Law Problems 1 Given three of the following four quantities: P1, V1, P2, V2 Rearrange the equation P1V1 P2V2 algebraically to solve for the unknown quantity. 2 An equation that can be used to calculate the unknown quantity It will be one of the following four: V2 P1V1 , P2 P2 P1V1 , V1 V2 P2V2 , P1 P1 P2V2 V1 Substitute each of the known quantities, and calculate. 3 Unknown P or V Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 147 Gas Laws Back Print Name Class Date Problem Solving continued Sample Problem 1 A sample of nitrogen collected in the laboratory occupies a volume of 725 mL at a pressure of 0.971 atm. What volume will the gas occupy at a pressure of 1.40 atm, assuming the temperature remains constant? Solution ANALYZE What is given in the problem? What are you asked to find? Items Original pressure, P1 Original pressure, V1 New pressure, P2 New volume, V2 the original volume and pressure of the nitrogen sample, and the new pressure of the sample the volume at the new pressure Data 0.971 atm 725 mL N2 1.40 atm ? mL N2 PLAN What steps are needed to calculate the new volume of the gas? Rearrange the Boyle’s law equation to solve for V2, substitute known quantities, and calculate. to solve for V2, divide both sides of the equation by P2 to isolate V2 insert data and solve for V2 1 P1V1 P2V2 V2 2 P1V1 P2 COMPUTE Substitute data for the terms of the equation, and compute the result. V2 P1V1 P2 0.971 atm 725 mL N2 1.40 atm 503 mL N2 EVALUATE Are the units correct? Yes; units canceled to give mL N2. Is the number of significant figures correct? Yes; the number of significant figures is correct because data were given to three significant figures. Is the answer reasonable? Yes; pressure increased by about 1/3, volume must decrease by about 1/3. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 148 Gas Laws Back Print Name Class Date Problem Solving continued Practice In each of the following problems, assume that the temperature and molar quantity of gas do not change. 1. Calculate the unknown quantity in each of the following measurements of gases. P1 V1 25 mL 550. mL ?L 800. mL ?L P2 6.0 atm ? kPa 3.56 atm 500. kPa 250 atm V2 ? mL 275 mL 20.0 L 160. mL 1.0 10 2 a. b. c. d. e. 3.0 atm 99.97 kPa 0.89 atm ? kPa 0.040 atm ans: 13 mL ans: 200. kPa ans: 80. L ans: 100. kPa L ans: 63 L Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 149 Gas Laws Back Print Name Class Date Problem Solving continued 2. A sample of neon gas occupies a volume of 2.8 L at 1.8 atm. What will its volume be at 1.2 atm? ans: 4.2 L 3. To what pressure would you have to compress 48.0 L of oxygen gas at 99.3 kPa in order to reduce its volume to 16.0 L? ans: 298 kPa 4. A chemist collects 59.0 mL of sulfur dioxide gas on a day when the atmospheric pressure is 0.989 atm. On the next day, the pressure has changed to 0.967 atm. What will the volume of the SO2 gas be on the second day? ans: 60.3 mL 5. 2.2 L of hydrogen at 6.5 atm pressure is used to fill a balloon at a final pressure of 1.15 atm. What is its final volume? ans: 12 L CHARLES’S LAW The French physicist Jacques Charles carried out experiments in 1786 and 1787 that showed a relationship between the temperature and volume of gases at constant pressure. You know that most matter expands as its temperature rises. Gases are no different. When Benjamin Thomson and Lord Kelvin proposed an absolute temperature scale in 1848, it was possible to set up the mathematical expression of Charles’s law. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 150 Gas Laws Back Print Name Class Date Problem Solving continued Charles’s law states that the volume of a sample of gas is directly proportional to the absolute temperature when pressure remains constant. Charles’s law can be expressed as follows. V T and V T k, or V kT According to Charles’s law, when the temperature of a sample of gas increases, the volume of the gas increases by the same factor. Therefore, doubling the Kelvin temperature of a gas will double its volume. Reducing the Kelvin temperature by 25% will reduce the volume by 25%. The expression V/T k means that the result of volume divided by temperature is a constant, k, for any sample of gas. If this is true, then V/T under one set of conditions is equal to V/T for the same sample of gas under another set of conditions, as long as the pressure remains constant. Charles’s law can be expressed by the following mathematical equation. V1 T1 V2 T2 General Plan for Solving Charles’s-Law Problems 1 Given three of the following four quantities: T1, V1, T2, V2 Rearrange the equation V1 V2 T1 T2 algebraically to solve for the unknown quantity. 2 An equation that can be used to calculate the unknown quantity It will be one of the following four: T2V1 T1V2 T1V2 T2V1 V2 , T2 , V1 , T1 T1 V1 T2 V2 Convert Celsius temperatures to Kelvin temperatures if necessary. Substitute each of the known quantities, and compute. 3 Unknown T or V Convert K to C if needed. Unknown t in C Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 151 Gas Laws Back Print Name Class Date Problem Solving continued Sample Problem 2 A container of oxygen has a volume of 349 mL at a temperature of 22°C. What volume will the gas occupy at 50.°C? Solution ANALYZE What is given in the problem? the original volume and temperature of the oxygen sample, and the new temperature of the sample the volume at the new temperature Data 349 mL O2 22°C (22 273) K 295 K What are you asked to find? Items Original volume, V1 Original temperature, t1 Original Kelvin temperature, T1 New volume, V2 New temperature, t2 New Kelvin temperature, T2 ? mL O2 50.°C ? (50. 273) K 323 K PLAN What steps are needed to calculate the new volume of the gas? Rearrange the Charles’s law equation to solve for V2, substitute known quantities, and calculate. 1 V1 T1 V2 T2 to solve for V2 , multiply both sides of the equation by T2 to isolate V2 insert data and solve for V2 2 V1T2 V2 T1 COMPUTE Substitute data for the terms of the equation, and compute the result. V2 EVALUATE Are the units correct? Yes; units canceled to give mL O2. V1T2 T1 349 mL O2 323 K 295 K 382 mL O2 Is the number of significant figures correct? Yes; the number of significant figures is correct because data were given to three significant figures. Is the answer reasonable? Yes; the temperature increased, so the volume must also increase. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 152 Gas Laws Back Print Name Class Date Problem Solving continued Practice In each of the following problems, assume that the pressure and molar quantity of gas do not change. 1. Calculate the unknown quantity in each of the following measurements of gases. V1 T1 280. K 300. K 292 K ?K 22°C V2 ? mL 0.404 L 250. mL 125 mL ?L T2 350. K ?K 365 K 305 K 14°C ans: 50.0 mL ans: 200. K ans: 200. mL ans: 244 K ans: 0.0021 L a. b. c. d. e. 40.0 mL 0.606 L ? mL 100. mL 0.0024 L Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 153 Gas Laws Back Print Name Class Date Problem Solving continued 2. A balloon full of air has a volume of 2.75 L at a temperature of 18°C. What is the balloon’s volume at 45°C? ans: 3.01 L 3. A sample of argon has a volume of 0.43 mL at 24°C. At what temperature in degrees Celsius will it have a volume of 0.57 mL? ans: 121°C GAY-LUSSAC’S LAW You may have noticed a warning on an aerosol spray can that says something similar to Do not incinerate! Do not expose to temperatures greater than 140°F! Warnings such as this appear because the pressure of a confined gas increases with increasing temperature. If the temperature of the can increases enough, the can will explode because of the pressure that builds up inside of it. The relationship between the pressure and temperature of a gas is described by Gay-Lussac’s law. Gay-Lussac’s law states that the pressure of a sample of gas is directly proportional to the absolute temperature when volume remains constant. Gay-Lussac’s law can be expressed as follows. P T and P T k, or P kT According to Gay-Lussac’s law, when the temperature of a sample of gas increases, the pressure of the gas increases by the same factor. Therefore, doubling the temperature of a gas will double its pressure. Reducing the temperature of a gas to 75% of its original value will also reduce the pressure to 75% of its original value. The expression P/T k means that the result of pressure divided by temperature is a constant, k, for any sample of gas. If this is true, then P/T under one set of conditions is equal to P/T for the same sample of gas under another set of conditions, as long as the volume remains constant. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 154 Gas Laws Back Print Name Class Date Problem Solving continued Gay-Lussac’s law can be expressed by the following mathematical equation. P1 T1 P2 T2 General Plan for Solving Gay-Lussac’s-Law Problems 1 Given three of the following four quantities: T1, P1, T2, P2 Rearrange the equation P1 P2 T1 T2 algebraically to solve for the unknown quantity. 2 P2 An equation that can be used to calculate the unknown quantity It will be one of the following four: T2P1 T1P2 T1P2 T2P1 , T2 , P1 , T1 T1 P1 T2 P2 Convert Celsius temperatures to Kelvin temperatures if necessary. Substitute each of the known quantities, and compute. 3 Unknown T or P Convert K to C if needed. Unknown t in C Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 155 Gas Laws Back Print Name Class Date Problem Solving continued Sample Problem 3 A cylinder of gas has a pressure of 4.40 atm at 25°C. At what temperature in °C will it reach a pressure of 6.50 atm? Solution ANALYZE What is given in the problem? the original pressure and temperature of the gas in the cylinder, and the new pressure of the sample the temperature at which the gas reaches the specified pressure Data 4.40 atm 25°C (25 273) K 298 K What are you asked to find? Items Original pressure, P1 Original temperature, t1 Original Kelvin temperature, T1 New pressure, P2 New temperature, t2 New Kelvin temperature, T2 6.50 atm ?°C ?K PLAN What steps are needed to calculate the new temperature of the gas? Rearrange the Gay-Lussac’s law equation to solve for T2, substitute known quantities, and calculate. 1 P1 T1 P2 T2 to solve for T2 , divide both sides of the equation by P2 and invert the equality to isolate T2 insert data and solve for T2 2 T1P2 T2 P1 COMPUTE Substitute data for the terms of the equation, and compute the result. T2 Convert back to °C. 440. K EVALUATE Are the units correct? Yes; units canceled to give Kelvin temperature, which was converted to °C. T1P2 P1 298 K 6.50 atm 4.40 atm (440. 273)°C 440. K 167°C Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 156 Gas Laws Back Print Name Class Date Problem Solving continued Is the number of significant figures correct? Yes; the number of significant figures is correct because the data were given to three significant figures. Is the answer reasonable? Yes; the pressure increases, so the temperature must also increase. Practice In each of the following problems, assume that the volume and molar quantity of gas do not change. 1. Calculate the unknown quantity in each of the following measurements of gases: P1 T1 273 K 300. K 52°C ?°C 4 P2 ? atm 0.156 atm 99.7 kPa 4.16 atm 3.92 10 3 T2 410 K ?K 77°C 13°C atm ?°C ans: 2.25 atm ans: 225 K ans: 92.6 kPa ans: 52°C ans: 616°C a. b. c. d. e. 1.50 atm 0.208 atm ? kPa 5.20 atm 8.33 10 atm 84°C Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 157 Gas Laws Back Print Name Class Date Problem Solving continued 2. A cylinder of compressed gas has a pressure of 4.882 atm on one day. The next day, the same cylinder of gas has a pressure of 4.690 atm, and its temperature is 8°C. What was the temperature on the previous day in °C? ans: 20.°C 3. A mylar balloon is filled with helium gas to a pressure of 107 kPa when the temperature is 22°C. If the temperature changes to 45°C, what will be the pressure of the helium in the balloon? ans: 115 kPa THE COMBINED GAS LAW Look at the relationships among temperature, volume, and pressure of a gas that you have studied so far. Boyle’s law At constant temperature: PV k Charles’s law At constant pressure: V k T Gay-Lussac’s law At constant volume: P k T Notice in these proportions that while P and V are inversely proportional to each other, they are each directly proportional to temperature. These three gas laws can be combined in one combined gas law. This law can be expressed as follows. PV T k If PV/T equals a constant, k, then PV/T for a sample of gas under one set of conditions equals PV/T under another set of conditions, assuming the amount of gas remains the same. Therefore, the combined gas law can be expressed by the following mathematical equation. This equation can be used to solve problems in which pressure, Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 158 Gas Laws Back Print Name Class Date Problem Solving continued volume, and temperature of a gas vary. Only the molar quantity of the gas must be constant. P1V1 T1 P2V2 T2 General Plan for Solving Combined-Gas-Law Problems 1 Given five of the following six quantities: V1,T1, P1, V2,T2, P2 Rearrange the equation P1V1 P2V2 T2 T1 algebraically to solve for the unknown quantity. 2 An equation that can be used to calculate the unknown quantity Convert Celsius temperatures to Kelvin temperatures if necessary. Substitute each of the known quantities, and compute. 3 Unknown V, T, or P Convert K to C if needed. Unknown t in C Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 159 Gas Laws Back Print Name Class Date Problem Solving continued Sample Problem 4 A sample of hydrogen gas has a volume of 65.0 mL at a pressure of 0.992 atm and a temperature of 16°C. What volume will the hydrogen occupy at 0.984 atm and 25°C? Solution ANALYZE What is given in the problem? the original volume, pressure, and temperature of the hydrogen, and the new pressure and temperature of the hydrogen the volume at the new temperature and pressure Data 65.0 mL 0.992 atm 16°C (16 ? mL 0.984 atm 25°C (25 273) K 298 K 273) K 289 K What are you asked to find? Items Original volume, V1 Original pressure, P1 Original temperature, t1 Original Kelvin temperature, T1 New volume, V2 New pressure, P2 New temperature, t2 New Kelvin temperature, T2 PLAN What steps are needed to calculate the new volume of the gas? Rearrange the combined gas law equation to solve for V2, substitute known quantities, and calculate. 1 P1V1 T1 P2V2 T2 to solve for V2 , multiply both sides of the equation by T2 /P2 to isolate V2 insert data and solve for V2 2 T2P1V1 P2T1 V2 COMPUTE Substitute data for the terms of the equation, and compute the result. V2 T2P1V1 P2T1 298 K 0.992 atm 65.0 mL H2 0.984 atm 289 K 67.6 mL H2 EVALUATE Are the units correct? Yes; units canceled to give mL of H2. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 160 Gas Laws Back Print Name Class Date Problem Solving continued Is the number of significant figures correct? Yes; the number of significant figures is correct because data were given to three significant figures. Is the answer reasonable? Yes; the temperature increases, and the pressure decreases, both of which have the effect of making the volume larger. Practice In each of the following problems, it is assumed that the molar quantity of gas does not change. 1. Calculate the unknown quantity in each of the following measurements of gases. P1 V1 225 mL 3.50 L 62 mL 43.2 mL T1 15°C 45°C 373 K 19°C P2 102.8 kPa ? atm 0.0029 atm 101.3 kPa V2 ? mL 3.70 L 64 mL ? mL T2 24°C 37°C ?K 0°C ans: 224 mL ans: 0.884 atm ans: 310 K ans: 39.9 mL a. b. c. d. 99.3 kPa 0.959 atm 0.0036 atm 100. kPa 2. A student collects 450. mL of HCl( g) hydrogen chloride gas at a pressure of 100. kPa and a temperature of 17°C. What is the volume of the HCl at 0°C and 101.3 kPa? ans: 418 mL Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 161 Gas Laws Back Print Name Class Date Problem Solving continued DALTON’S LAW OF PARTIAL PRESSURES Air is a mixture of approximately 78% N2 , 20% O2 , 1% Ar, and 1% other gases by volume, so at any barometric pressure 78% of that pressure is exerted by nitrogen, 20% by oxygen, and so on. This phenomenon is described by Dalton’s law of partial pressures, which says that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. It can be stated mathematically as follows. PTotal PGas 1 PGas 2 PGas 3 PGas 4 … A common method of collecting gas samples in the laboratory is to bubble the gas into a bottle filled with water and allow it to displace the water. When this technique is used, however, the gas collected in the bottle contains a small but significant amount of water vapor. As a result, the pressure of the gas that has displaced the liquid water is the sum of the pressure of the gas plus the vapor pressure of water at that temperature. The vapor pressures of water at various temperatures are given in Table 1 below. TABLE 1 VAPOR PRESSURE OF WATER Temp in °C 10 11 12 13 14 15 16 Vapor pressure in kPa 1.23 1.31 1.40 1.50 1.60 1.71 1.82 Temp in °C 17 18 19 20 21 22 23 Vapor pressure in kPa 1.94 2.06 2.19 2.34 2.49 2.64 2.81 Temp in °C 24 25 26 27 28 29 30 Vapor pressure in kPa 2.98 3.17 3.36 3.57 3.78 4.01 4.25 PTotal PGas PH2O vapor To find the true pressure of the gas alone, the pressure of the water vapor must be subtracted from the total pressure. PGas PTotal PH2O vapor You can use this corrected pressure in gas-law calculations to determine what the volume of the gas alone would be. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 162 Gas Laws Back Print Name Class Date Problem Solving continued Sample Problem 5 A student collects oxygen gas by water displacement at a temperature of 16°C. The total volume is 188 mL at a pressure of 92.3 kPa. What is the pressure of oxygen collected? Solution ANALYZE What is given in the problem? the total pressure, the fact that the gas was collected by water displacement, and the temperature the pressure of oxygen collected Data 92.3 kPa 16°C 1.82 kPa ? kPa What are you asked to find? Items Total pressure, Ptotal Temperature Vapor pressure of water at 16°C, PH2O* Pressure of collected O2, PO2 * determined from Table 1 PLAN What step is needed to calculate the pressure of the oxygen collected? Use Dalton’s law of partial pressures to determine the pressure of the oxygen alone in the container. Use the following equation to determine the pressure of oxygen alone. given from Table 1 PO2 COMPUTE PO2 Ptotal Ptotal PH2 O vapor PH2O vapor 92.3 kPa 1.82 kPa 90.5 kPa EVALUATE Are the units correct? Yes; the units should be kPa. Is the number of significant figures correct? Yes; the number of significant figures is correct because the total pressure was given to one decimal place. Is the answer reasonable? Yes; the pressure of the collected oxygen and the water vapor add up to the total pressure of the system. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 163 Gas Laws Back Print Name Class Date Problem Solving continued Practice 1. A chemist collects a sample of H2S( g) over water at a temperature of 27°C. The total pressure of the gas that has displaced a volume of 15 mL of water is 207.33 kPa. What is the pressure of the H2S gas collected? ans: 203.76 kPa Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 164 Gas Laws Back Print Name Class Date Problem Solving continued Sample Problem 6 Chlorine gas is collected by water displacement at a temperature of 19°C. The total volume is 1.45 L at a pressure of 156.5 kPa. Assume that no chlorine gas dissolves in the water. What is the volume of chlorine corrected to STP? Solution ANALYZE What is given in the problem? the total initial volume, pressure, and temperature; the fact that the gas was collected by water displacement; and the final temperature and pressure of the chlorine the volume of chlorine at STP Data 1.45 L 156.5 kPa 19°C (19 273) K 292 K What are you asked to find? Items Total original volume, V1 Total original pressure, Ptotal Original temperature, t1 Original Kelvin temperature, T1 Vapor pressure of water at 19°C, PH2O Pressure of collected Cl2, P1 New volume, V2 New pressure, P2 New temperature, t2 New Kelvin temperature, T2 2.19 kPa ? kPa ? mL 101.3 kPa 0°C 273 K PLAN What steps are needed to calculate the volume of the chlorine at STP? Use Dalton’s law of partial pressures to determine the pressure of the chlorine alone in the container. Use the combined gas law to solve for the new volume. given from table P1 Ptotal P1V1 T1 PH2O vapor P2V2 T2 Calculate the volume of chlorine at STP using the following equation. Solve for V2 . calculated above given P1 T2V1 P2T1 given Copyright © by Holt, Rinehart and Winston. All rights reserved. V2 Holt ChemFile: Problem-Solving Workbook 165 Gas Laws Back Print Name Class Date Problem Solving continued COMPUTE Determine the pressure of chlorine alone. P1 Ptotal PH2O vapor 156.5 kPa 2.19 kPa 154.3 kPa Calculate the volume of Cl2 at STP. V2 EVALUATE Are the units correct? Yes; the units canceled leaving the correct units, L. 154.3 kPa 273 K 1.45 L 101.3 kPa 292 K 2.06 L Is the number of significant figures correct? Yes; the number of significant figures is correct because the data were given to a minimum of three significant figures. Is the answer reasonable? Yes; the pressure decreased, so the volume had to increase. Practice In each of the following problems, assume that the molar quantity of gas does not change. 1. Some hydrogen is collected over water at 10°C and 105.5 kPa pressure. The total volume of the sample was 1.93 L. Calculate the volume of the hydrogen corrected to STP. ans: 1.92 L 2. One student carries out a reaction that gives off methane gas and obtains a total volume by water displacement of 338 mL at a temperature of 19°C and a pressure of 0.9566 atm. Another student does the identical experiment on another day at a temperature of 26°C and a pressure of 0.989 atm. Which student collected more CH4 ? ans: The second student collected more CH4. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 166 Gas Laws Back Print Name Class Date Problem Solving continued Additional Problems In each of the following problems, assume that the molar quantity of gas does not change. 1. Calculate the unknown quantity in each of the following measurements of gases. P1 V1 796 cm ? mL 1.77 L 2.93 dm 120. mL 3.6 m 3 3 3 P2 ? kPa 9.6 10 1 V2 965 cm3 atm 3.7 2.44 L ? dm3 97.0 mL ? m3 103 mL a. b. c. d. e. f. 127.3 kPa 7.1 ? kPa 114 kPa 1.00 atm 0.77 atm 102 atm 30.79 kPa 4.93 ? atm 1.90 atm 10 kPa 4 2. A gas cylinder contains 0.722 m3 of hydrogen gas at a pressure of 10.6 atm. If the gas is used to fill a balloon at a pressure of 0.96 atm, what is the volume in m3 of the filled balloon? 3. A weather balloon has a maximum volume of 7.50 103 L. The balloon contains 195 L of helium gas at a pressure of 0.993 atm. What will be the pressure when the balloon is at maximum volume? 4. A rubber ball contains 5.70 10 1 dm3 of gas at a pressure of 1.05 atm. What volume will the gas occupy at 7.47 atm? 5. Calculate the unknown quantity in each of the following measurements of gases. V1 T1 ?K 100.°C 104 mm3 10 3 2 3 V2 32.9 mL 0.83 dm 2.59 ?L 819 cm3 ?m 3 3 T2 290. K 9°C ?°C 190. K 409 K 246°C a. b. c. d. e. f. 26.5 mL ? dm 7.44 5.63 ? cm3 679 m 870.°C 132 K 243 K 3°C 102 mm3 L 6. A bubble of carbon dioxide gas in some unbaked bread dough has a volume of 1.15 cm3 at a temperature of 22°C. What volume will the bubble have when the bread is baked and the bubble reaches a temperature of 99°C? 7. A perfectly elastic balloon contains 6.75 dm3 of air at a temperature of 40.°C. What is the temperature if the balloon has a volume of 5.03 dm3? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 167 Gas Laws Back Print Name Class Date Problem Solving continued 8. Calculate the unknown quantity in each of the following measurements of gases. P1 T1 ?°C 302 K 76°C 46°C 37°C 263 K P2 5.6 atm ? kPa 3.97 atm 706 atm 350. atm 0.058 atm T2 192°C 11 K 27°C ?°C 2050°C ?K a. b. c. d. e. f. 0.777 atm 152 kPa ? atm 395 atm ? atm 0.39 atm 9. A 2 L bottle containing only air is sealed at a temperature of 22°C and a pressure of 0.982 atm. The bottle is placed in a freezer and allowed to cool to 3°C. What is the pressure in the bottle? 10. The pressure in a car tire is 2.50 atm at a temperature of 33°C. What would the pressure be if the tire were allowed to cool to 0°C? Assume that the tire does not change volume. 11. A container filled with helium gas has a pressure of 127.5 kPa at a temperature of 290. K. What is the temperature when the pressure is 3.51 kPa? 12. Calculate the unknown quantity in each of the following measurements of gases. P1 V1 1.65 L 3.79 dm 249 mL ? mm 3 3 3 3 T1 19°C 73°C ?K 12°C 22°C 195 K P2 0.920 atm ? kPa 0.098 atm 149 kPa 1.01 atm 2.25 atm V2 ?L 7.58 dm 197 mL 3.18 0.85 m 10 mm 3 3 3 3 3 T2 46°C 217°C 293 K 18°C ?°C 584 K a. b. c. d. e. f. 1.03 atm 107.0 kPa 0.029 atm 113 kPa 1.15 atm ? atm 0.93 m 156 cm 468 cm 13. A scientist has a sample of gas that was collected several days earlier. The sample has a volume of 392 cm3 at a pressure of 0.987 atm and a temperature of 21°C. On the day the gas was collected, the temperature was 13°C and the pressure was 0.992 atm. What volume did the gas have on the day it was collected? 14. Hydrogen gas is collected by water displacement. Total volume collected is 0.461 L at a temperature of 17°C and a pressure of 0.989 atm. What is the pressure of dry hydrogen gas collected? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 168 Gas Laws Back Print Name Class Date Problem Solving continued 15. One container with a volume of 1.00 L contains argon at a pressure of 1.77 atm, and a second container of 1.50 L volume contains argon at a pressure of 0.487 atm. They are then connected to each other so that the pressure can become equal in both containers. What is the equalized pressure? Hint: Each sample of gas now occupies the total space. Dalton’s law of partial pressures applies here. 16. Oxygen gas is collected over water at a temperature of 10.°C and a pressure of 1.02 atm. The volume of gas plus water vapor collected is 293 mL. What volume of oxygen at STP was collected? 17. A 500 mL bottle is partially filled with water so that the total volume of gases (water vapor and air) remaining in the bottle is 325 cm3, measured at 20.°C and 101.3 kPa. The bottle is sealed and taken to a mountaintop where the pressure is 76.24 kPa and the temperature is 10°C. If the bottle is upside down and the seal leaks, how much water will leak out? The key to this problem is to determine the pressure in the 325 cm3 space when the bottle is at the top of the mountain. 18. An air thermometer can be constructed by using a glass bubble attached to a piece of small-diameter glass tubing. The tubing contains a small amount of colored water that rises when the temperature increases and the trapped air expands. You want a 0.20 cm3 change in volume to equal a 1°C change in temperature. What total volume of air at 20.°C should be trapped in the apparatus below the liquid? 19. A sample of nitrogen gas is collected over water, yielding a total volume of 62.25 mL at a temperature of 22°C and a total pressure of 97.7 kPa. At what pressure will the nitrogen alone occupy a volume of 50.00 mL at the same temperature? 20. The theoretical yield of a reaction that gives off nitrogen trifluoride gas is 844 mL at STP. What total volume of NF3 plus water vapor will be collected over water at 25°C and a total pressure of 1.017 atm? 21. A weather balloon is inflated with 2.94 kL of helium at a location where the pressure is 1.06 atm and the temperature is 32°C. What will be the volume of the balloon at an altitude where the pressure is 0.092 atm and the temperature is 35°C? 22. The safety limit for a certain can of aerosol spray is 95°C. If the pressure of the gas in the can is 2.96 atm when it is 17°C, what will the pressure be at the safety limit? 23. A chemistry student collects a sample of ammonia gas at a temperature of 39°C. Later, the student measures the volume of the ammonia as 108 mL, but its temperature is now 21°C. What was the volume of the ammonia when it was collected? 24. A quantity of CO2 gas occupies a volume of 624 L at a pressure of 1.40 atm. If this CO2 is pumped into a gas cylinder that has a volume of 80.0 L, what pressure will the CO2 exert on the cylinder? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 169 Gas Laws Back Print Name Class Date Skills Worksheet Problem Solving The Ideal Gas Law In 1811, the Italian chemist Amedeo Avogadro proposed the principle that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. He determined that at standard temperature and pressure, one mole of gas occupies 22.414 10 L (usually rounded to 22.4 L). At this point, if you know the number of moles of a gas, you can use the molar volume of 22.4 L/mol to calculate the volume that amount of gas would occupy at STP. Then you could use the combined gas law to determine the volume of the gas under any other set of conditions. However, a much simpler way to accomplish the same task is by using the ideal gas law. The ideal gas law is a mathematical relationship that has the conditions of standard temperature (273 K) and pressure (1 atm or 101.3 kPa) plus the molar gas volume (22.4 L/mol) already combined into a single constant. The following equation is the mathematical statement of the ideal gas law. PV in which P V n T R the pressure of a sample of gas the volume of a sample of gas the number of moles of gas present the Kelvin temperature of the gas the ideal gas constant, which combines standard conditions and molar volume into a single constant nRT The value of the ideal gas constant, R, depends on the units of P and V being used in the equation. Temperature is always in kelvins and amount of gas is always in moles. The most common values used for R are shown below. Units of P and V Atmospheres and liters Kilopascals and liters Value of R 0.0821 8.314 L atm mol K L kPa mol K Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt Chemistry 170 The Ideal Gas Law Back Print Name Class Date Problem Solving continued If you have volume units other than liters or pressure units other than atmospheres or kilopascals, it is best to convert volume to liters and pressure to atmospheres or kilopascals. General Plan for Solving Ideal-Gas-Law Problems 1 The equation for the ideal gas law PV nRT Determine from the data which is the unknown quantity. Rearrange the equation algebraically to solve for the unknown quantity. 2 An equation that can be used to calculate the unknown quantity Choose the gas constant, R, that best fits the units of the data. Substitute each of the data values in the equation and calculate. 3 Unknown P, V, n, or T Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 171 The Ideal Gas Law Back Print Name Class Date Problem Solving continued Sample Problem 1 An engineer pumps 5.00 mol of carbon monoxide gas into a cylinder that has a capacity of 20.0 L. What is the pressure in kPa of CO inside the cylinder at 25°C? Solution ANALYZE What is given in the problem? the amount in moles of gas pumped into the cylinder, the volume of the cylinder, and the temperature the pressure of the gas in the cylinder Data 5.00 mol 20.0 L 25°C (25 273) K 298 K What are you asked to find? Items Amount of gas, n Volume of gas in cylinder, V Temperature of gas, t Kelvin temperature of gas, T Ideal gas constant, R Pressure in cylinder, P 0.0821 L atm/mol K or 8.314 L kPa/mol K ? kPa PLAN What steps are needed to calculate the new pressure of the gas? Rearrange the ideal-gas-law equation to solve for P, substitute known quantities, and calculate. 1 Ideal-gas-law equation, PV nRT solve the idealgas-law equation for pressure 2 P nRT V the problem asks for answer in kPa, so choose the appropriate R, substitute known values, and solve 3 Unknown pressure, P Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 172 The Ideal Gas Law Back Print Name Class Date Problem Solving continued PV nRT nRT V Solve the ideal-gas-law equation for P, the unknown quantity. P COMPUTE The problem asks for pressure in kPa, so use R P 5.00 mol 8.314 L kPa/mol K 20.0 L 8.314 L kPa/mol K. 298 K 619 kPa EVALUATE Are the units correct? Yes; the ideal gas constant was selected so that the units canceled to give kPa. Is the number of significant figures correct? Yes; the number of significant figures is correct because data were given to three significant figures. Is the answer reasonable? Yes; the calculation can be approximated as (1/4) equals 600. Thus, 619 kPa is in the right range. (8 300), or 2400/4, which Practice 1. A student collects 425 mL of oxygen at a temperature of 24°C and a pressure of 0.899 atm. How many moles of oxygen did the student collect? ans: 1.57 10 2 mol O2 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 173 The Ideal Gas Law Back Print Name Class Date Problem Solving continued 2. Use the ideal-gas-law equation to calculate the unknown quantity in each of the following sets of measurements. You will need to convert Celsius temperatures to Kelvin temperatures and volume units to liters. P V ?L 0.0350 L 15.7 L 629 mL ?L 39.0 mL n 0.0881 mol ? mol 0.815 mol 0.0337 mol 0.0818 mol ? mol T 302 K 55°C 20.°C ?K 19°C 27°C ans: 2.00 L ans: 1.22 10 3 a. b. c. d. e. f. 1.09 atm 94.9 kPa ? kPa 0.500 atm 0.950 atm 107 kPa mol ans: 109 kPa ans: 114 K ans: 2.06 L ans: 1.67 10 3 mol Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 174 The Ideal Gas Law Back Print Name Class Date Problem Solving continued APPLICATIONS OF THE IDEAL GAS LAW You have seen that you can use the ideal gas law to calculate the moles of gas, n, in a sample when you know the pressure, volume, and temperature of the sample. When you know the amount and identity of the substance, you can use its molar mass to calculate its mass. You did this when you learned how to convert between mass and moles. The relationship is expressed as follows. Mass in grams n Amount in moles m M Molar mass in grams per mole If you substitute the expression m/M for n in the ideal-gas-law equation, you get the following equation. PV m RT M This version of the ideal gas law can be solved for any of the five variables P, V, m, M, or T. It is especially useful in determining the molecular mass of a substance. This equation can also be related to the density of a gas. Density is mass per unit volume, as shown in the following equation. D Solve for m: m DV DV RT M PM RT Then, substitute DV for m in the gas law equation: PV m V The two V terms cancel and the equation is rearranged to give: PM DRT or D This equation can be used to compute the density of a gas under any conditions of temperature and pressure. It can also be used to calculate the molar mass of an unknown gas if its density is known. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 175 The Ideal Gas Law Back Print Name Class Date Problem Solving continued General Plan for Solving Problems Involving Applications of the Ideal Gas Law 1a PV m RT M Determine which equation fits the problem. Rearrange the equation algebraically to solve for the unknown quantity. 1b D PM RT 2 An equation that can be used to calculate the unknown quantity Choose the gas constant, R, that best fits the units of the data. Substitute each of the data values in the equation and calculate. 3 Unknown P, V, m, M, D or T Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 176 The Ideal Gas Law Back Print Name Class Date Problem Solving continued Sample Problem 2 Determine the molar mass of an unknown gas that has a volume of 72.5 mL at a temperature of 68°C, a pressure of 0.980 atm, and a mass of 0.207 g. Solution ANALYZE What is given in the problem? What are you asked to find? Items Volume of gas, V Temperature of gas, t Kelvin temperature of gas, T Pressure of gas, P Mass of gas, m Ideal gas constant, R Molar mass of gas, M the mass, pressure, volume, and temperature of the gas the molar mass of the gas Data 72.5 mL 68°C (68 273) K 341 K 0.980 atm 0.207 g 8.314 L kPa/mol K or 0.0821 L atm/mol K ? g/mol PLAN What steps are needed to calculate the new volume of the gas? Select the equation that will give the desired result. Solve the equation for the unknown quantity. Substitute data values into the solved equation, and calculate. 1a PV m RT M solve this equation for molar mass 2 M mRT PV the problem gives pressure in atm, so choose the appropriate R, substitute known values, and solve 3 Unknown molar mass, M Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 177 The Ideal Gas Law Back Print Name Class Date Problem Solving continued Use the equation that includes m and M. PV m RT M mRT PV Solve the equation for M, the unknown quantity. M COMPUTE Convert the volume in milliliters to liters 72.5 mL 1L 1000 mL 0.0725 L The data give pressure in atm, so use R M mRT PV 0.207 g 0.0821 L atm/mol K. 341 K 81.6 g/mol 0.0821 L atm/mol K 0.980 atm 0.0725 L EVALUATE Are the units correct? Yes; units canceled to give g/mol, the correct units for molar mass. Is the number of significant figures correct? Yes; the number of significant figures is correct because data were given to three significant figures. Is the answer reasonable? Yes; 81.6 g/mol is a reasonable molar mass. The calculation can be approximated as 0.2 341 (8/7), which is roughly 80. Practice 1. A sample of an unknown gas has a mass of 0.116 g. It occupies a volume of 25.0 mL at a temperature of 127°C and has a pressure of 155.3 kPa. Calculate the molar mass of the gas. ans: 99.4 g/mol 2. Determine the mass of CO2 gas that has a volume of 7.10 L at a pressure of 1.11 atm and a temperature of 31°C. Hint: Solve the equation for m, and calculate the molar mass using the chemical formula and the periodic table. ans: 13.9 g Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 178 The Ideal Gas Law Back Print Name Class Date Problem Solving continued Sample Problem 3 Determine the density of hydrogen bromide gas at 3.10 atm and 5°C. Solution ANALYZE What is given in the problem? What are you asked to find? Items Temperature of HBr, t Kelvin temperature of HBr, T Pressure of HBr, P Molar mass of HBr, M* Ideal gas constant, R Density of HBr, D * determined from the periodic table the pressure and temperature of the HBr gas the density of the gas Data 5°C (5 273) K 268 K 3.10 atm 80.91 g/mol 8.314 L kPa/mol K or 0.0821 L atm/mol K ? g/L PLAN What steps are needed to calculate the density of HBr under the conditions given? Select the equation that will give the desired result. Rearrange the equation to solve for the unknown quantity. Substitute data values into the correct equation, and calculate. 1 D PM RT equation is already written correctly to solve for the unknown 2 D PM RT the problem gives pressure in atm, so choose the appropriate R, substitute known values, and solve 3 Unknown density, D Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 179 The Ideal Gas Law Back Print Name Class Date Problem Solving continued Use the equation that includes density. D COMPUTE The data give pressure in atm, so use R D PM RT PM RT 0.0821 L atm/mol K. 11.4 g/L 3.10 atm 80.91 g/mol 0.0821 L atm/mol K 268 K EVALUATE Are the units correct? Yes; units canceled to give g/L, the correct units for gas density. Is the number of significant figures correct? Yes; the number of significant figures is correct because data were given to three significant figures. Is the answer reasonable? Yes; 11.4 g/L is a reasonable density for a heavy gas compressed to 3 atm. The calculation can be approximated as 3 80/(0.08 270) 3 1000/270 11. Practice 1. What is the density of silicon tetrafluoride gas at 72°C and a pressure of 144.5 kPa? ans: 5.24 g/L 2. At what temperature will nitrogen gas have a density of 1.13 g/L at a pressure of 1.09 atm? ans: 329 K or 56°C Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 180 The Ideal Gas Law Back Print Name Class Date Problem Solving continued Additional Problems 1. Use the ideal-gas-law equation to calculate the unknown quantity in each of the following sets of measurements. P V 15 200 L 0.119 mL 250. mL ? n ? mol 0.000 350 mol 0.120 mol 4.7 10 mol 4 t 15°C 0°C ?°C 300.°C a. b. c. d. 0.0477 atm ? kPa 500.0 kPa 19.5 atm 2. Use the ideal-gas-law equation to calculate the unknown quantity in each of the following sets of measurements. P V 3.77 L 50.0 mL ?L 2.00 L 26.1 mL m 8.23 g ?g 3.20 7.19 g 0.414 g 10 3 M ? g/mol 48.02 g/mol g 2.02 g/mol 159.8 g/mol ? g/mol t 25°C 0°C 5°C 185°C 45°C a. b. c. d. e. 0.955 atm 105.0 kPa 0.782 atm ? atm 107.2 kPa 3. Determine the volume of one mole of an ideal gas at 25°C and 0.915 kPa. 4. Calculate the unknown quantity in each of the following sets of measurements. P Molar mass ? g/mol 30.07 g/mol 104.09 g/mol 77.95 g/mol Density 2.40 g/L ? g/L 4.37 g/L 6.27 g/L t 2°C 20.°C ?°C 66°C a. b. c. d. 1.12 atm 7.50 atm 97.4 kPa ? atm 5. What pressure in atmospheres will 1.36 kg of N2O gas exert when it is compressed in a 25.0 L cylinder and is stored in an outdoor shed where the temperature can reach 59°C during the summer? 6. Aluminum chloride sublimes at high temperatures. What density will the vapor have at 225°C and 0.939 atm pressure? 7. An unknown gas has a density of 0.0262 g/mL at a pressure of 0.918 atm and a temperature of 10.°C. What is the molar mass of the gas? 8. A large balloon contains 11.7 g of helium. What volume will the helium occupy at an altitude of 10 000 m, where the atmospheric pressure is 0.262 atm and the temperature is 50.°C? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 181 The Ideal Gas Law Back Print Name Class Date Problem Solving continued 9. A student collects ethane by water displacement at a temperature of 15°C (vapor pressure of water is 1.5988 kPa) and a total pressure of 100.0 kPa. The volume of the collection bottle is 245 mL. How many moles of ethane are in the bottle? 10. A reaction yields 3.75 L of nitrogen monoxide. The volume is measured at 19°C and at a pressure of 1.10 atm. What mass of NO was produced by the reaction? 11. A reaction has a theoretical yield of 8.83 g of ammonia. The reaction gives off 10.24 L of ammonia measured at 52°C and 105.3 kPa. What was the percent yield of the reaction? 12. An unknown gas has a density of 0.405 g/L at a pressure of 0.889 atm and a temperature of 7°C. Calculate its molar mass. 13. A paper label has been lost from an old tank of compressed gas. To help identify the unknown gas, you must calculate its molar mass. It is known that the tank has a capacity of 90.0 L and weighs 39.2 kg when empty. You find its current mass to be 50.5 kg. The gauge shows a pressure of 1780 kPa when the temperature is 18°C. What is the molar mass of the gas in the cylinder? 14. What is the pressure inside a tank that has a volume of 1.20 contains 12.0 kg of HCl gas at a temperature of 18°C? 103 L and 15. What pressure in kPa is exerted at a temperature of 20.°C by compressed neon gas that has a density of 2.70 g/L? 16. A tank with a volume of 658 mL contains 1.50 g of neon gas. The maximum safe pressure that the tank can withstand is 4.50 102 kPa. At what temperature will the tank have that pressure? 17. The atmospheric pressure on Mars is about 6.75 millibars (1 bar 100 kPa 0.9869 atm), and the nighttime temperature can be about 75°C on the same day that the daytime temperature goes up to 8°C. What volume would a bag containing 1.00 g of H2 gas have at both the daytime and nighttime temperatures? 18. What is the pressure in kPa of 3.95 mol of Cl2 gas if it is compressed in a cylinder with a volume of 850. mL at a temperature of 15°C? 19. What volume in mL will 0.00660 mol of hydrogen gas occupy at a pressure of 0.907 atm and a temperature of 9°C? 20. What volume will 8.47 kg of sulfur dioxide gas occupy at a pressure of 89.4 kPa and a temperature of 40.°C? 21. A cylinder contains 908 g of compressed helium. It is to be used to inflate a balloon to a final pressure of 128.3 kPa at a temperature of 2°C. What will the volume of the balloon be under these conditions? 22. The density of dry air at 27°C and 100.0 kPa is 1.162 g/L. Use this information to calculate the molar mass of air (calculate as if air were a pure substance). Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 182 The Ideal Gas Law Back Print Name Class Date Skills Worksheet Problem Solving Stoichiometry of Gases Now that you have worked with relationships among moles, mass, and volumes of gases, you can easily put these to work in stoichiometry calculations. Many reactions have gaseous reactants, gaseous products, or both. Reactants and products that are not gases are usually measured in grams or kilograms. As you know, you must convert these masses to amounts in moles before you can relate the quantities by using a balanced chemical equation. Gaseous products and reactants can be related to solid or liquid products and reactants by using the mole ratio, just as solids and liquids are related to each other. Reactants and products that are gases are usually measured in liters. If the gas is measured at STP, you will need only Avogadro’s law to relate the volume and amount of a gas. One mole of any gas at STP occupies 22.4 L. If the gas is not at STP, you will need to use the ideal gas law to determine the number of moles. Once volume has been converted to amount in moles you can use the mole ratios of products and reactants to solve stoichiometry problems involving multiple phases of products and reactants. n PV RT If the problem which you are trying to solve involves only gases, there is a simpler way of dealing with the stoichiometric amounts. Look again at the expression for the ideal gas law above; the molar amount of a gas is directly related to its volume. Therefore, the mole ratios of gases given by the coefficients in the balanced equation can be used as volume ratios of those gases to solve stoichiometry problems. No conversion from volume to amount is required to determine the volume of one gas from the volume of another gas in a balanced chemical equation. There is one condition that must be observed. Gas volumes can be related by mole ratios only when the volumes are measured under the same conditions of temperature and pressure. If they are not, then the volume of one of the gases must be converted to the conditions of the other gas. Usually you will need to use the combined gas law for this conversion. V2 V1P1T2 T1P2 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt Chemistry 183 Stoichiometry of Gases Back Print Name Class Date Problem Solving continued General Plan for Solving Gas Stoichiometry Problems 1 Mass of substance A — solid, liquid, or gas 6 Mass of substance B — solid, liquid, or gas Convert using the molar mass of B. Convert using the molar mass of A. 2 Amount of substance A in moles — solid, liquid, or gas Convert using the mole ratio between A and B. 5 Amount of substance B in moles — solid, liquid, or gas If A is a gas, convert using gas laws. If B is a gas, convert using gas laws. 3 Volume of substance A if it is a gas 4 The volume ratio is the same as the mole ratio if the two gases are under the same set of conditions. If the two gases exist under different conditions, convert using the gas laws. Volume of substance B if it is a gas Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 184 Stoichiometry of Gases Back Print Name Class Date Problem Solving continued Sample Problem 1 Ammonia can react with oxygen to produce nitrogen and water according to the following equation. 4NH3( g) 3O2( g) 3 2N2( g) 6H2O(l) If 1.78 L of O2 reacts, what volume of nitrogen will be produced? Assume that temperature and pressure remain constant. Solution ANALYZE What is given in the problem? the balanced equation, the volume of oxygen, and the fact that the two gases exist under the same conditions the volume of N2 produced Data O2 3 NA NA NA 1.78 L NA NA N2 2 NA NA NA ?L NA NA What are you asked to find? Items Substance Coefficient in balanced equation Molar mass Moles Mass of substance Volume of substance Temperature conditions Pressure conditions PLAN What steps are needed to calculate the volume of N2 formed from a given volume of O2? The coefficients of the balanced equation indicate the mole ratio of O2 to N2. The volume ratio is the same as the mole ratio when volumes are measured under the same conditions. 3 Volume of O2 in L multiply by the volume ratio, N2 O2 N volume ratio, O2 given 4 Volume of N2 in L 2 L O2 2 L N2 3 L O2 L N2 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 185 Stoichiometry of Gases Back Print Name Class Date Problem Solving continued COMPUTE 1.78 L O2 EVALUATE Are the units correct? Yes; units canceled to give L N2. 2 L N2 3 L O2 1.19 L N2 Is the number of significant figures correct? Yes; the number of significant figures is correct because the data were given to three significant figures. Is the answer reasonable? Yes; the volume of N2 should be 2/3 the volume of O2. Practice 1. In one method of manufacturing nitric acid, ammonia is oxidized to nitrogen monoxide and water. 4NH3( g) 5O2( g) → 4NO( g) 6H2O(l) What volume of oxygen will be used in a reaction of 2800 L of NH3 ? What volume of NO will be produced? All volumes are measured under the same conditions. ans: 3500 L O2, 2800 L NO 2. Fluorine gas reacts violently with water to produce hydrogen fluoride and ozone according to the following equation. 3F2( g) 3H2O(l) → 6HF( g) O3( g) What volumes of O3 and HF gas would be produced by the complete reaction of 3.60 104 mL of fluorine gas? All gases are measured under the same conditions. ans: 1.20 104 mL O3, 7.20 104 mL HF Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 186 Stoichiometry of Gases Back Print Name Class Date Problem Solving continued Sample Problem 2 Ethylene gas burns in air according to the following equation. C2H4( g) 3O2( g) 3 2CO2( g) 2H2O(l) If 13.8 L of C2H4 measured at 21°C and 1.038 atm burns completely with oxygen, calculate the volume of CO2 produced, assuming the CO2 is measured at 44°C and 0.989 atm. Solution ANALYZE What is given in the problem? the balanced equation, the volume of ethylene, the conditions under which the ethylene was measured, and the conditions under which the CO2 is measured the volume of CO2 produced as measured at the specified conditions Data C2H4 1 NA NA NA 13.8 L 21°C 294 K CO2 2 NA NA NA ?L 44°C 317 K What are you asked to find? Items Substance Coefficient in balanced equation Molar mass Moles Mass of substance Volume of substance Temperature conditions Pressure conditions 1.083 atm 0.989 atm PLAN What steps are needed to calculate the volume of CO2 formed from the complete burning of a given volume of C2H4? Use the volume ratio of C2H4 to CO2 to calculate the volume of CO2 at the same conditions as C2H4. Convert to the volume of CO2 for the given conditions using the combined gas law. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 187 Stoichiometry of Gases Back Print Name Class Date Problem Solving continued Volume of C2H4 in L at initial conditions multiply by the volume ratio, CO2 C2H4 Volume of CO2 in L at final conditions use the combined gas law to convert from the initial temperature and pressure to the final temperature and pressure CO Volume of 2 in L at the same conditions as initial C2H4 CO volume ratio, C H2 24 given L C2H4* * at 294 K and 1.083 atm 2 L CO2 1 L C2H4 L CO2* Neither pressure nor temperature is constant; therefore, the combined gas law must be used to calculate the volume of CO2 at the final temperature and pressure. P1V1 T1 given given P2V2 T2 calculated above T2 given P1 P2 V1 given T1 V2 COMPUTE 13.8 L C2H4* 2 L CO2 1 L C2H4 27.6 L CO2* * at 294 K and 1.083 atm Solve the combined-gas-law equation for V2. 317 K 1.083 atm 27.6 L CO2 V2 0.989 atm 294 K EVALUATE Are the units correct? Yes; units canceled to give L CO2. 32.6 L CO2 Is the number of significant figures correct? Yes; the number of significant figures is correct because the data had a minimum of three significant figures. Is the answer reasonable? Yes; the changes in both pressure and temperature increased the volume by small factors. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 188 Stoichiometry of Gases Back Print Name Class Date Problem Solving continued Practice 1. A sample of ethanol burns in O2 to form CO2 and H2O according to the following equation. C2H5OH 3O2 → 2CO2 3H2O If the combustion uses 55.8 mL of oxygen measured at 2.26 atm and 40.°C, what volume of CO2 is produced when measured at STP? ans: 73.3 mL CO2 2. Dinitrogen pentoxide decomposes into nitrogen dioxide and oxygen. If 5.00 L of N2O5 reacts at STP, what volume of NO2 is produced when measured at 64.5°C and 1.76 atm? ans: 7.02 L Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 189 Stoichiometry of Gases Back Print Name Class Date Problem Solving continued Sample Problem 3 When arsenic(III) sulfide is roasted in air, it reacts with oxygen to produce arsenic(III) oxide and sulfur dioxide according to the following equation. 2As2S3(s) 9O2( g) 3 2As2O3(s) 6SO2( g) When 89.5 g of As2S3 is roasted with excess oxygen, what volume of SO2 is produced? The gaseous product is measured at 20°C and 98.0 kPa. Solution ANALYZE What is given in the problem? the balanced equation, the mass of As2S3, and the pressure and temperature conditions under which the SO2 is measured the volume of SO2 produced as measured at the given conditions Data As2S3(s) 2 246.05 g/mol 89.5 g ? mol NA NA NA SO2( g) 6 NA NA ? mol ?L 20°C 293 K What are you asked to find? Items Substance Coefficient in balanced equation Molar mass* Mass of substance Amount Volume of substance Temperature conditions Pressure conditions * determined from the periodic table 98.0 kPa PLAN What steps are needed to calculate the volume of SO2 formed from the reaction of a given mass of As2S3? Use the molar mass of As2S3 to determine the number of moles that react. Use the mole ratio from the balanced chemical equation to determine the amount in moles of SO2 formed. Use the ideal-gas-law equation to determine the volume of SO2 formed from the amount in moles. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 190 Stoichiometry of Gases Back Print Name Class Date Problem Solving continued Mass of As2S3(s) in g multiply by the inverse molar mass of As2S3 Amount of As2S3(s) in mol multiply by the mole ratio, SO2 As2S3 Amount of SO2(g) in mol use the ideal gas law to convert Volume of SO2(g) in L 1 molar mass As2S3 SO2 As2S3 mole ratio, given g As2S3 1 mol As2S3 246.05 g As2S3 6 mol SO2 2 mol As2S3 mol SO2 Rearrange the ideal-gas-law equation to solve for the unknown quantity, V. PV V calculated above nRT nRT P given ideal gas constant mol SO2 8.314 L kPa mol K kPa given K L SO2 COMPUTE 89.5 g As2S3 1.09 mol SO2 1 mol As2S3 246.05 g As2S3 6 mol SO2 2 mol As2S3 293 K 1.09 mol SO2 8.314 L kPa/mol K 98.0 kPa 27.1 L SO2 EVALUATE Are the units correct? Yes; units canceled to give liters of SO2. Is the number of significant figures correct? Yes; the number of significant figures is correct because the data had a minimum of three significant figures. Is the answer reasonable? Yes; computation of the amount of SO2 can be approximated as (9/25) 3 27/25, so you would expect an answer a little greater than 1. At a temperature slightly above standard temperature, you would expect a volume a little greater than 22.4 L. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 191 Stoichiometry of Gases Back Print Name Class Date Problem Solving continued Practice 1. Complete the table below using the following equation, which represents a reaction that produces aluminum chloride. 2Al(s) Mass Al Volume Cl2 ?L ?L ?L 920. L 1.049 mL ? m3 3Cl2( g) → 2AlCl3(s) Mass AlCl3 7.15 g NA NA ?g NA NA ans: 1.80 L Cl2 ans: 24.2 L Cl2 ans: 2.21 ans: 3.65 ans: 3.71 ans: 8.02 103 L Cl2 103 g AlCl3 10 3 Conditions STP STP 20.°C and 0.945 atm STP 37°C and 5.00 atm 15°C and 83.0 kPa a. b. c. d. e. f. excess 19.4 g 1.559 kg excess ?g 500.00 kg g Al 102 m3 Cl2 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 192 Stoichiometry of Gases Back Print Name Class Date Problem Solving continued Additional Problems 1. The industrial production of ammonia proceeds according to the following equation. N2( g) 3H2( g) → 2NH3( g) a. What volume of nitrogen at STP is needed to react with 57.0 mL of hydrogen measured at STP? b. What volume of NH3 at STP can be produced from the complete reaction of 6.39 104 L of hydrogen? c. If 20.0 mol of nitrogen is available, what volume of NH3 at STP can be produced? d. What volume of H2 at STP will be needed to produce 800. L of ammonia, measured at 55°C and 0.900 atm? 2. Propane burns according to the following equation. C3H8( g) 5O2( g) → 3CO2( g) 4H2O( g) a. What volume of water vapor measured at 250.°C and 1.00 atm is produced when 3.0 L of propane at STP is burned? b. What volume of oxygen at 20.°C and 102.6 kPa is used if 640. L of CO2 is produced? The CO2 is also measured at 20.°C and 102.6 kPa. c. If 465 mL of oxygen at STP is used in the reaction, what volume of CO2 , measured at 37°C and 0.973 atm, is produced? d. When 2.50 L of C3H8 at STP burns, what total volume of gaseous products is formed? The volume of the products is measured at 175°C and 1.14 atm. 3. Carbon monoxide will burn in air to produce CO2 according to the following equation. 2CO( g) O2( g) → 2CO2( g) What volume of oxygen at STP will be needed to react with 3500. L of CO measured at 20.°C and a pressure of 0.953 atm? 4. Silicon tetrafluoride gas can be produced by the action of HF on silica according to the following equation. SiO2(s) 4HF( g) → SiF4( g) 2H2O(l) 1.00 L of HF gas under pressure at 3.48 atm and a temperature of 25°C reacts completely with SiO2 to form SiF4 . What volume of SiF4 , measured at 15°C and 0.940 atm, is produced by this reaction? 5. One method used in the eighteenth century to generate hydrogen was to pass steam through red-hot steel tubes. The following reaction takes place. 3Fe(s) 4H2O( g) → Fe3O4(s) 4H2( g) a. What volume of hydrogen at STP can be produced by the reaction of 6.28 g of iron? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 193 Stoichiometry of Gases Back Print Name Class Date Problem Solving continued b. What mass of iron will react with 500. L of steam at 250.°C and 1.00 atm pressure? c. If 285 g of Fe3O4 are formed, what volume of hydrogen, measured at 20.°C and 1.06 atm, is produced? 6. Sodium reacts vigorously with water to produce hydrogen and sodium hydroxide according to the following equation. 2Na(s) 2H2O(l) → 2NaOH(aq) H2( g) If 0.027 g of sodium reacts with excess water, what volume of hydrogen at STP is formed? 7. Diethyl ether burns in air according to the following equation. C4H10O(l) 6O2( g) → 4CO2( g) 5H2O(l) If 7.15 L of CO2 is produced at a temperature of 125°C and a pressure of 1.02 atm, what volume of oxygen, measured at STP, was consumed and what mass of diethyl ether was burned? 8. When nitroglycerin detonates, it produces large volumes of hot gases almost instantly according to the following equation. 4C3H5N3O9(l) → 6N2( g) 12CO2( g) 10H2O( g) O2( g) a. When 0.100 mol of nitroglycerin explodes, what volume of each gas measured at STP is produced? b. What total volume of gases is produced at 300.°C and 1.00 atm when 10.0 g of nitroglycerin explodes? 9. Dinitrogen monoxide can be prepared by heating ammonium nitrate, which decomposes according to the following equation. NH4NO3(s) → N2O( g) 2H2O(l) What mass of ammonium nitrate should be decomposed in order to produce 250. mL of N2O, measured at STP? 10. Phosphine, PH3 , is the phosphorus analogue to ammonia, NH3 . It can be produced by the reaction between calcium phosphide and water according to the following equation. Ca3P2(s) 6H2O(l) → 3Ca(OH)2(s and aq) 2PH3( g) What volume of phosphine, measured at 18°C and 102.4 kPa, is produced by the reaction of 8.46 g of Ca3P2? 11. In one method of producing aluminum chloride, HCl gas is passed over aluminum and the following reaction takes place. 2Al(s) 6HCl( g) → 2AlCl3( g) 3H2( g) What mass of Al should be on hand in order to produce 6.0 103 kg of AlCl3? What volume of compressed HCl at 4.71 atm and a temperature of 43°C should be on hand at the same time? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 194 Stoichiometry of Gases Back Print Name Class Date Problem Solving continued 12. Urea, (NH2)2CO, is an important fertilizer that is manufactured by the following reaction. 2NH3( g) CO2( g) → (NH2)2CO(s) H2O( g) 104 kg of urea if What volume of NH3 at STP will be needed to produce 8.50 there is an 89.5% yield in the process? 13. An obsolete method of generating oxygen in the laboratory involves the decomposition of barium peroxide by the following equation. 2BaO2(s) → 2BaO(s) O2( g) What mass of BaO2 reacted if 265 mL of O2 is collected by water displacement at 0.975 atm and 10.°C? 14. It is possible to generate chlorine gas by dripping concentrated HCl solution onto solid potassium permanganate according to the following equation. 2KMnO4(s) 16HCl(aq) → 2KCl(aq) 2MnCl2(aq) 8H2O(l) 5Cl2( g) If excess HCl is dripped onto 15.0 g of KMnO4 , what volume of Cl2 will be produced? The Cl2 is measured at 15°C and 0.959 atm. 15. Ammonia can be oxidized in the presence of a platinum catalyst according to the following equation. 4NH3( g) 5O2( g) → 4NO( g) 6H2O(l) The NO that is produced reacts almost immediately with additional oxygen according to the following equation. 2NO( g) O2( g) → 2NO2( g) If 35.0 kL of oxygen at STP react in the first reaction, what volume of NH3 at STP reacts with it? What volume of NO2 at STP will be formed in the second reaction, assuming there is excess oxygen that was not used up in the first reaction? 16. Oxygen can be generated in the laboratory by heating potassium chlorate. The reaction is represented by the following equation. 2KClO3(s) → 2KCl(s) 3O2( g) What mass of KClO3 must be used in order to generate 5.00 L of O2, measured at STP? 17. One of the reactions in the Solvay process is used to make sodium hydrogen carbonate. It occurs when carbon dioxide and ammonia are passed through concentrated salt brine. The following equation represents the reaction. NaCl(aq) H2O(l) CO2( g) NH3( g) → NaHCO3(s) NH4Cl(aq) a. What volume of NH3 at 25°C and 1.00 atm pressure will be required if 38 000 L of CO2 , measured under the same conditions, react to form NaHCO3? b. What mass of NaHCO3 can be formed when the gases in (a) react with NaCl? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 195 Stoichiometry of Gases Back Print Name Class Date Problem Solving continued c. If this reaction forms 46.0 kg of NaHCO3 , what volume of NH3 , measured at STP, reacted? d. What volume of CO2 , compressed in a tank at 5.50 atm and a temperature of 42°C, will be needed to produce 100.00 kg of NaHCO3? 18. The combustion of butane is represented in the following equation. 2C4H10( g) 13O2( g) → 8CO2( g) 10H2O(l) a. If 4.74 g of butane react with excess oxygen, what volume of CO2 , measured at 150.°C and 1.14 atm, will be formed? b. What volume of oxygen, measured at 0.980 atm and 75°C, will be consumed by the complete combustion of 0.500 g of butane? c. A butane-fueled torch has a mass of 876.2 g. After burning for some time, the torch has a mass of 859.3 g. What volume of CO2 , at STP, was formed while the torch burned? d. What mass of H2O is produced when butane burns and produces 3720 L of CO2 , measured at 35°C and 0.993 atm pressure? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 196 Stoichiometry of Gases Back Print Name Class Date Skills Worksheet Problem Solving Concentration of Solutions There are three principal ways to express solution concentration in chemistry— percentage by mass, molarity, and molality. The following table compares these three ways of stating solution concentration. Examining the method of preparation of the three types may help you understand the differences among them. Symbol Percentage Molarity % M Meaning Grams solute per 100 g of solution Moles solute per liter of solution How to prepare 5%: Dissolve 5 g of solute in 95 g solvent. 5 M: Dissolve 5 mol of solute in solvent and add solvent to make 1 L of solution. 5 m: Dissolve 5 mol of solute in 1 kg of solvent. Molality m Moles solute per kilogram of solvent PERCENTAGE CONCENTRATION You will find percentages of solutes stated on the labels of many commercial products, such as household cleaners, liquid pesticide solutions, and shampoos. If your sink becomes clogged, you might buy a bottle of drain opener whose label states that it is a 2.4% sodium hydroxide solution. This means that the bottle contains 2.4 g of NaOH for every 100 g of solution. Computing percentage concentration is very much like computing percentage composition. Both involve finding the percentage of a single component of a multicomponent system. In each type of percentage calculation, the mass of the important component (in percentage concentration, the solute) is divided by the total mass of the system and multiplied by 100 to yield a percentage. In percentage concentration, the solute is the important component, and the total mass of the system is the mass of the solute plus the mass of the solvent. General Plan for Solving Percentage Concentration Problems 1 Mass of solvent in g 2 Mass of solute in g Percentage concentration mass of solute mass of solution 3 Mass of solution in g 100 4 Percentage concentration by mass Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 197 Concentration of Solutions Back Print Name Class Date Problem Solving continued Sample Problem 1 What is the percentage by mass of a solution made by dissolving 0.49 g of potassium sulfate in 12.70 g of water? Solution ANALYZE What is given in the problem? What are you asked to find? the mass of solvent, and the mass of solute, K2SO4 the concentration of the solution expressed as a percentage by mass Data 12.70 g H2O 0.49 g K2SO4 ?% Items Mass of solvent Mass of solute Concentration (% by mass) PLAN What step is needed to calculate the concentration of the solution as a percentage by mass? Divide the mass of solute by the mass of the solution and multiply by 100. 1 Mass of water in g 2 Mass of K2SO4 in g percentage concentration solute mass solution mass 3 Mass of K2SO4 solution in g 100 4 Percentage K2SO4 by mass given percentage concentration g K2SO4 g K2SO4 g H2O given given 100 COMPUTE percentage concentration EVALUATE Are the units correct? Yes; percentage K2SO4 was required. 0.49 g K2SO4 0.49 g K2SO4 12.70 g H2O 100 3.7% K2SO4 Is the number of significant figures correct? Yes; the number of significant figures is correct because the data had a minimum of two significant figures. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 198 Concentration of Solutions Back Print Name Class Date Problem Solving continued Is the answer reasonable? Yes; the computation can be approximated as 0.5/13 100 3.8%. Practice 1. What is the percentage concentration of 75.0 g of ethanol dissolved in 500.0 g of water? ans: 13.0% ethanol 2. A chemist dissolves 3.50 g of potassium iodate and 6.23 g of potassium hydroxide in 805.05 g of water. What is the percentage concentration of each solute in the solution? ans: 0.430% KIO3, 0.765% KOH 3. A student wants to make a 5.00% solution of rubidium chloride using 0.377 g of the substance. What mass of water will be needed to make the solution? ans: 7.16 g H2O 4. What mass of lithium nitrate would have to be dissolved in 30.0 g of water in order to make an 18.0% solution? ans: 6.59 g LiNO3 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 199 Concentration of Solutions Back Print Name Class Date Problem Solving continued MOLARITY Molarity is the most common way to express concentration in chemistry. Molarity is the number of moles of solute per liter of solution and is given as a number followed by a capital M. A 2 M solution of nitric acid contains 2 mol of HNO3 per liter of solution. As you know, substances react in mole ratios. Knowing the molar concentration of a solution allows you to measure a number of moles of a dissolved substance by measuring the volume of solution. General Plan for Solving Molarity Problems 1 Mass of solute in g Convert using the molar mass of the solute. 2 Amount of solute in mol 3 moles solute liter solution Molar concentration, M M Volume of solution in L 4 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 200 Concentration of Solutions Back Print Name Class Date Problem Solving continued Sample Problem 2 What is the molarity of a solution prepared by dissolving 37.94 g of potassium hydroxide in some water and then diluting the solution to a volume of 500.00 mL? Solution ANALYZE What is given in the problem? What are you asked to find? the mass of the solute, KOH, and the final volume of the solution the concentration of the solution expressed as molarity Data 37.94 g KOH ? mol KOH 56.11 g/mol 500.00 mL ?M Items Mass of solute Moles of solute Molar mass of solute* Volume of solution Concentration (molarity) * determined from the periodic table PLAN What steps are needed to calculate the concentration of the solution as molarity? Determine the amount in moles of solute; calculate the moles per liter of solution. 1 Mass of KOH in g multiply by the inverted molar mass of KOH Volume of KOH solution in mL multiply by the conversion factor 1L 1000 mL 2 Amount of KOH in mol M moles solute liter solution 3 Volume of KOH solution in L 4 Molarity of KOH solution 1 molar mass of KOH given g KOH 1 mol KOH 56.11 g KOH mol KOH Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 201 Concentration of Solutions Back Print Name Class Date Problem Solving continued given mL solution 1L 1000 mL L solution calculated above calculated above mol KOH L solution M solution COMPUTE 37.94 g KOH 1 mol KOH 56.11 g KOH 1L 1000 mL 0.6762 mol KOH 0.500 00 L solution 1.352 M 500.00 mL solution 0.6762 mol KOH 0.500 00 L solution EVALUATE Are the units correct? Yes; units canceled to give moles KOH per liter of solution. Is the number of significant figures correct? Yes; the number of significant figures is correct because the data had a minimum of four significant figures. Is the answer reasonable? Yes; note that 0.6762 mol is approximately 2/3 mol and 0.500 00 L is 1/2 L. Thus, the calculation can be estimated as (2/3)/(1/2) 4/3, which is very close to the result. Practice 1. Determine the molarity of a solution prepared by dissolving 141.6 g of citric acid, C3H5O(COOH)3, in water and then diluting the resulting solution to 3500.0 mL. ans: 0.2106 M Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 202 Concentration of Solutions Back Print Name Class Date Problem Solving continued 2. What is the molarity of a salt solution made by dissolving 280.0 mg of NaCl in 2.00 mL of water? Assume the final volume is the same as the volume of the water. ans: 2.40 M 3. What is the molarity of a solution that contains 390.0 g of acetic acid, CH3COOH, dissolved in enough acetone to make 1000.0 mL of solution? ans: 6.494 M Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 203 Concentration of Solutions Back Print Name Class Date Problem Solving continued Sample Problem 3 An analytical chemist wants to make 750.0 mL of a 6.00 M solution of sodium hydroxide. What mass of NaOH will the chemist need to make this solution? Solution ANALYZE What is given in the problem? What are you asked to find? Items Mass of solute Molar mass of solute Moles of solute Volume of solution Concentration (molarity) the identity of the solute, the total volume of solution, and the molarity of the solution the mass of solute to dissolve Data ? g NaOH 40.00 g/mol ? mol NaOH 750.0 mL 6.00 M PLAN What steps are needed to calculate the mass of solute needed? Determine the amount in moles needed for the solution required, and convert to grams by multiplying by the molar mass of the solute. COMPUTE Molarity of NaOH solution Volume of NaOH solution in L multiply by the conversion 1L factor 1000 mL Amount of NaOH in mol multiply by the molar mass of NaOH Volume of NaOH solution in mL given Mass of NaOH in g mL solution given 1L 1000 mL L solution mol NaOH L solution calculated above molar mass of NaOH L solution 40.00 g NaOH 1 mol NaOH g NaOH 750.0 mL solution 1L 1000 mL 0.7500 L solution Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 204 Concentration of Solutions Back Print Name Class Date Problem Solving continued 6.00 mol NaOH L solution EVALUATE Are the units correct? Yes; units canceled to give grams of NaOH. 0.7500 L solution 40.00 g NaOH 1 mol NaOH 180. g NaOH Is the number of significant figures correct? Yes; the number of significant figures is correct because the data had a minimum of three significant figures. Is the answer reasonable? Yes; the calculation can be estimated as (3/4) (6)(40) (3/4) 240 180. Practice 1. What mass of glucose, C6H12O6, would be required to prepare 5.000 a 0.215 M solution? ans: 1.94 105 g 103 L of 2. What mass of magnesium bromide would be required to prepare 720. mL of a 0.0939 M aqueous solution? ans: 12.4 g 3. What mass of ammonium chloride is dissolved in 300. mL of a 0.875 M solution? ans: 14.0 g Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 205 Concentration of Solutions Back Print Name Class Date Problem Solving continued MOLALITY Molality is the amount in moles of solute per kilogram of solvent and is given by a number followed by an italic lowercase m. A 5 m aqueous solution of glucose contains 5 mol of C6H12O6 per kilogram of water. Molal concentration is important primarily in working with colligative properties of solutions. General Plan for Solving Molality Problems 1 Mass of solute in g Convert using the molar mass of the solute. 3 Mass of solvent in g Convert using the equation 1 kg 1000 g. 2 Amount of solute in mol 4 mol solute kg solvent m Mass of solvent in kg 5 Molal concentration, m Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 206 Concentration of Solutions Back Print Name Class Date Problem Solving continued Sample Problem 4 Determine the molal concentration of a solution containing 81.3 g of ethylene glycol, HOCH2CH2OH, dissolved in 166 g of water. Solution ANALYZE What is given in the problem? What are you asked to find? Items Mass of solute Molar mass of solute Moles of solute Mass of solvent Concentration (molality) the mass of ethylene glycol dissolved, and the mass of the solvent, water the molal concentration of the solution Data 81.3 g ethylene glycol 62.08 g/mol ethylene glycol ? mol ethylene glycol 166 g H2O ?m PLAN What steps are needed to calculate the molal concentration of the ethylene glycol solution? Determine the amount of solute in moles and the mass of solvent in kilograms; calculate the moles of solute per kilogram of solvent. 1 Mass of C2H6O2 in g multiply by the inverted molar mass of C2H6O2 3 Mass of H2O in g multiply by the conversion factor 1 kg 1000 g moles C2H6O2 kg H2O 2 Amount of C2H6O2 in mol m 4 Mass of H2O in kg 5 Molality of C2H6O2 solution 1 molar mass of C2H6O2 given g C2H6O2 1 mol C2H6O2 62.08 g C2H6O2 given mol C2H6O2 g H2O calculated above 1 kg 1000 g kg H2O mol C2H6O2 kg H2O calculated above m C2H6O2 solution Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 207 Concentration of Solutionss Back Print Name Class Date Problem Solving continued COMPUTE 81.3 g C2H6O2 1 mol C2H6O2 62.08 g C2H6O2 1 kg 1000 g 1.31 mol C2H6O2 166 g H2O 0.166 kg H2O 7.89 m 1.31 mol C2H6O2 0.166 kg H2O EVALUATE Are the units correct? Yes; units canceled to give moles C2H6O2 per kilogram of solvent. Is the number of significant figures correct? Yes; the number of significant figures is correct because the data had a minimum of three significant figures. Is the answer reasonable? Yes; because 1.31 mol is approximately 4/3 mol and 0.166 kg is approximately 1/6 kg, the calculation can be estimated as (4/3)/(1/6) 24/3 8, which is very close to the result. Practice 1. Determine the molality of a solution of 560 g of acetone, CH3COCH3, in 620 g of water. ans: 16 m 2. What is the molality of a solution of 12.9 g of fructose, C6H12O6 , in 31.0 g of water? ans: 2.31 m 3. How many moles of 2-butanol, CH3CHOHCH2CH3 , must be dissolved in 125 g of ethanol in order to produce a 12.0 m 2-butanol solution? What mass of 2butanol is this? ans: 1.50 mol 2-butanol, 111 g 2-butanol Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 208 Concentration of Solutions Back Print Name Class Date Problem Solving continued Additional Problems 1. Complete the table below by determining the missing quantity in each example. All solutions are aqueous. Any quantity that is not applicable to a given solution is marked NA. Mass of solute used ? g KMnO4 ? g BaCl2 ? g glycerol 12.27 g K2Cr2O7 288 g CaCl2 ? g NaCl ? g glucose Quantity of solution made 500.0 g 1.750 L NA 650. mL NA 25.0 mL ? g solution Quantity of solvent used ? g H2O NA 800.0 g H2O NA 2.04 kg H2O NA 1.50 kg H2O Solution made a. b. c. d. e. f. g. 12.0% KMnO4 0.60 M BaCl2 6.20 m glycerol, HOCH2CHOHCH2OH ? M K2Cr2O7 ? m CaCl2 0.160 M NaCl 2.00 m glucose, C6H12O6 2. How many moles of H2SO4 are in 2.50 L of a 4.25 M aqueous solution? 3. Determine the molal concentration of 71.5 g of linoleic acid, C18H32O2 , in 525 g of hexane, C6H14 . 4. You have a solution that is 16.2% sodium thiosulfate, Na2S2O3 , by mass. a. What mass of sodium thiosulfate is in 80.0 g of solution? b. How many moles of sodium thiosulfate are in 80.0 g of solution? c. If 80.0 g of the sodium thiosulfate solution is diluted to 250.0 mL with water, what is the molarity of the resulting solution? 5. What mass of anhydrous cobalt(II) chloride would be needed in order to make 650.00 mL of a 4.00 M cobalt(II) chloride solution? 6. A student wants to make a 0.150 M aqueous solution of silver nitrate, AgNO3 and has a bottle containing 11.27 g of silver nitrate. What should be the final volume of the solution? 7. What mass of urea, NH2CONH2 , must be dissolved in 2250 g of water in order to prepare a 1.50 m solution? 8. What mass of barium nitrate is dissolved in 21.29 mL of a 3.38 M solution? 9. Describe what you would do to prepare 100.0 g of a 3.5% solution of ammonium sulfate in water. 10. What mass of anhydrous calcium chloride should be dissolved in 590.0 g of water in order to produce a 0.82 m solution? 11. How many moles of ammonia are in 0.250 L of a 5.00 M aqueous ammonia solution? If this solution were diluted to 1.000 L, what would be the molarity of the resulting solution? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 209 Concentration of Solutions Back Print Name Class Date Problem Solving continued 12. What is the molar mass of a solute if 62.0 g of the solute in 125 g of water produce a 5.3 m solution? 13. A saline solution is 0.9% NaCl. What masses of NaCl and water would be required to prepare 50. L of this saline solution? Assume that the density of water is 1.000 g/mL and that the NaCl does not add to the volume of the solution. 14. A student weighs an empty beaker on a balance and finds its mass to be 68.60 g. The student weighs the beaker again after adding water and finds the new mass to be 115.12 g. A mass of 4.08 g of glucose is then dissolved in the water. What is the percentage concentration of glucose in the solution? 15. The density of ethyl acetate at 20°C is 0.902 g/mL. What volume of ethyl acetate at 20°C would be required to prepare a 2.0% solution of cellulose nitrate using 25 g of cellulose nitrate? 16. Aqueous cadmium chloride reacts with sodium sulfide to produce brightyellow cadmium sulfide. Write the balanced equation for this reaction and answer the following questions. a. How many moles of CdCl2 are in 50.00 mL of a 3.91 M solution? b. If the solution in (a) reacted with excess sodium sulfide, how many moles of CdS would be formed? c. What mass of CdS would be formed? 17. What mass of H2SO4 is contained in 60.00 mL of a 5.85 M solution of sulfuric acid? 18. A truck carrying 22.5 kL of 6.83 M aqueous hydrochloric acid used to clean brick and masonry has overturned. The authorities plan to neutralize the acid with sodium carbonate. How many moles of HCl will have to be neutralized? 19. A chemist wants to produce 12.00 g of barium sulfate by reacting a 0.600 M BaCl2 solution with excess H2SO4 , as shown in the reaction below. What volume of the BaCl2 solution should be used? BaCl2 H2SO4 → BaSO4 2HCl 20. Many substances are hydrates. Whenever you make a solution, it is important to know whether or not the solute you are using is a hydrate and, if it is a hydrate, how many molecules of water are present per formula unit of the substance. This water must be taken into account when weighing out the solute. Something else to remember when making aqueous solutions from hydrates is that once the hydrate is dissolved, the water of hydration is considered to be part of the solvent. A common hydrate used in the chemistry laboratory is copper sulfate pentahydrate, CuSO4 5H2O. Describe how you would make each of the following solutions using CuSO4 5H2O. Specify masses and volumes as needed. a. 100. g of a 6.00% solution of CuSO4 b. 1.00 L of a 0.800 M solution of CuSO4 c. a 3.5 m solution of CuSO4 in 1.0 kg of water Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 210 Concentration of Solutions Back Print Name Class Date Problem Solving continued 21. What mass of calcium chloride hexahydrate is required in order to make 700.0 mL of a 2.50 M solution? 22. What mass of the amino acid arginine, C6H14N4O2 , would be required to make 1.250 L of a 0.00205 M solution? 23. How much water would you have to add to 2.402 kg of nickel(II) sulfate hexahydrate in order to prepare a 25.00% solution? 24. What mass of potassium aluminum sulfate dodecahydrate, KAl(SO4)2 12H2O, would be needed to prepare 35.00 g of a 15.00% KAl(SO4)2 solution? What mass of water would be added to make this solution? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 211 Concentration of Solutions Back Print Name Class Date Skills Worksheet Problem Solving Dilutions Suppose you work in the laboratory of a paint company where you use 100 mL of a 0.1 M solution of zinc chloride in a quality-control test that you carry out 10 times a day. It would be tedious and time-consuming to continually measure out small amounts of ZnCl2 to make 100 mL of this solution. Of course, you could make many liters of the solution at one time, but that would require several large containers to store the solution. The answer to the problem is to make a much more concentrated solution and then dilute it with water to make the less concentrated solution that you need. The more-concentrated solution is called a stock solution. You could make a 1 M ZnCl2 solution by measuring out 1 mol of zinc chloride, 136.3 g, and dissolving it in enough water to make a liter of solution. This solution is 10 times as concentrated as the solution you need. Every time you need the test solution, you can measure out 10 mL of the 1 M solution and dilute it to 100 mL to yield 100 mL of 0.1 M ZnCl2 solution. To make a solution by dilution, you must determine the volume of stock solution to use and the amount of solvent needed to dilute to the concentration you need. As you have learned, the molarity of a solution is its concentration in moles of solute per liter of solution. Molarity is found by dividing the moles of solute by the number of liters of solution. M moles solute liter solution molarity volume of solution So, for a measured volume of any solution: amount of solute in mol If this measured volume of solution is diluted to a new volume by adding solvent, the new, larger volume still contains the same number of moles of solute. Therefore, where 1 and 2 represent the concentrated and diluted solutions: molarity1 Therefore: molarity1 volume1 molarity2 volume2 This relationship applies whenever solution 2 is made from solution 1 by dilution. volume1 moles solute molarity2 volume2 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 212 Dilutions Back Print Name Class Date Problem Solving continued General Plan for Solving Dilution Problems 1 M1V1 M2V2 Rearrange the equation M1 V1 M2 V2 algebraically to solve for the unknown quantity. 2 The equation used to calculate the unknown quantity will be one of the following four: V2 M1V1 , M2 M2 M1V1 , V1 V2 M2V2 , M1 M1 M2V2 V1 Substitute each of the known quantities for its symbol, and calculate. 3 Unknown molarity or volume Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 213 Dilutions Back Print Name Class Date Problem Solving continued Sample Problem 1 What is the molarity of a solution that is made by diluting 50.00 mL of a 4.74 M solution of HCl to 250.00 mL? Solution ANALYZE What is given in the problem? the molarity of the stock solution, the volume used to dilute, and the volume of the diluted solution the molarity of the diluted solution Data 4.74 M HCl 50.00 mL 250.00 mL ?M What are you asked to find? Items Concentration of the stock solution (M1) Volume of stock solution used (V1) Volume of diluted solution (V2) Concentration of the diluted solution (M2) PLAN What step is needed to calculate the concentration of the diluted solution? Apply the principle that volume1 molarity1 volume2 molarity2. rearrange the equation M1V1 M2V2 algebraically to solve for M2 substitute each of the known quantities for its symbol and calculate given 1 M1V1 M2V2 2 M2 M1V1 V2 M2 M1 V1 V2 COMPUTE Note: Even though molarity is moles per liter, you can use volumes in milliliters along with molarity whenever the units cancel. M2 EVALUATE Are the units correct? Yes; molarity (mol/L) was required. 4.74 M 50.00 mL 250.00 mL 0.948 M Is the number of significant figures correct? Yes; the number of significant figures is correct because the data had a minimum of three significant figures. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 214 Dilutions Back Print Name Class Date Problem Solving continued Is the answer reasonable? Yes; the computation is the same as 4.74/5, which is a little less than 1. Practice 1. Complete the table below by calculating the missing value in each row. Molarity of stock solution Volume of stock solution 20.00 mL ? mL 5.00 mL 0.250 L ? mL Molarity of dilute solution ? M KBr 0.075 M LiOH 0.0493 M HI 1.8 M HCl 0.093 M NH3 Volume of dilute solution 100.00 mL 500.00 mL 100.00 mL ?L 4.00 L ans: 0.100 M ans: 38 mL ans: 0.986 M ans: 1.7 L ans: 50. mL a. b. c. d. e. 0.500 M KBr 1.00 M LiOH ? M HI 12.0 M HCl 7.44 M NH3 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 215 Dilutions Back Print Name Class Date Problem Solving continued Sample Problem 2 What volume of water would you add to 15.00 mL of a 6.77 M solution of nitric acid in order to get a 1.50 M solution? Solution ANALYZE What is given in the problem? the molarity of the stock solution, the volume of stock solution, and the molarity of the diluted solution the volume of water to add to make the dilute solution Data 6.77 M HNO3 15.00 mL 1.50 M HNO3 ? mL ? mL What are you asked to find? Items Concentration of the stock solution (M1) Volume of stock solution used (V1) Molarity of the diluted solution (M2) Volume of diluted solution (V2) Volume of water to add PLAN What steps are needed to calculate the amount of water to add to dilute a solution to the given molarity? Apply the principle that volume1 molarity1 volume2 molarity2. Subtract the stock solution volume from the final volume to determine the amount of water to add. 1 M1V1 M2V2 solve the equation M1V1 M2V2 algebraically for the V2 Volume of water added to dilute subtract the original volume from final diluted volume V2 V1 Vwater added to dilute 2 V2 M1V1 M2 substitute each of the known quantities for its symbol and calculate given 3 Unknown volume V2 calculated above given M1 V1 M2 Vwater added to dilute V2 V1 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 216 Dilutions Back Print Name Class Date Problem Solving continued COMPUTE V2 6.77 M 15.00 mL 1.50 M 15.00 mL 67.7 mL 67.7 mL EVALUATE Are the units correct? 52.7 mL H2O Yes; volume of water was required. Is the number of significant figures correct? Yes; the number of significant figures is correct because the data had a minimum of three significant figures. Is the answer reasonable? Yes; the dilution was to a concentration of less than 1/4 of the original concentration. Thus, the volume of the diluted solution should be more than four times the original volume, 4 15 mL 60 mL. Practice 1. What volume of water would be added to 16.5 mL of a 0.0813 M solution of sodium borate in order to get a 0.0200 M solution? ans: 50.6 mL H2O Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 217 Dilutions Back Print Name Class Date Problem Solving continued Additional Problems 1. What is the molarity of a solution of ammonium chloride prepared by diluting 50.00 mL of a 3.79 M NH4Cl solution to 2.00 L? 2. A student takes a sample of KOH solution and dilutes it with 100.00 mL of water. The student determines that the diluted solution is 0.046 M KOH, but has forgotten to record the volume of the original sample. The concentration of the original solution is 2.09 M. What was the volume of the original sample? 3. A chemist wants to prepare a stock solution of H2SO4 so that samples of 20.00 mL will produce a solution with a concentration of 0.50 M when added to 100.0 mL of water. a. What should the molarity of the stock solution be? b. If the chemist wants to prepare 5.00 L of the stock solution from concentrated H2SO4 , which is 18.0 M, what volume of concentrated acid should be used? c. The density of 18.0 M H2SO4 is 1.84 g/mL. What mass of concentrated H2SO4 should be used to make the stock solution in (b)? 4. To what volume should 1.19 mL of an 8.00 M acetic acid solution be diluted in order to obtain a final solution that is 1.50 M? 5. What volume of a 5.75 M formic acid solution should be used to prepare 2.00 L of a 1.00 M formic acid solution? 6. A 25.00 mL sample of ammonium nitrate solution produces a 0.186 M solution when diluted with 50.00 mL of water. What is the molarity of the stock solution? 7. Given a solution of known percentage concentration by mass, a laboratory worker can often measure out a calculated mass of the solution in order to obtain a certain mass of solute. Sometimes, though, it is impractical to use the mass of a solution, especially with fuming solutions, such as concentrated HCl and concentrated HNO3 . Measuring these solutions by volume is much more practical. In order to determine the volume that should be measured, a worker would need to know the density of the solution. This information usually appears on the label of the solution bottle. a. Concentrated hydrochloric acid is 36% HCl by mass and has a density of 1.18 g/mL. What is the volume of 1.0 kg of this HCl solution? What volume contains 1.0 g of HCl? What volume contains 1.0 mol of HCl? b. The density of concentrated nitric acid is 1.42 g/mL, and its concentration is 71% HNO3 by mass. What volume of concentrated HNO3 would be needed to prepare 10.0 L of a 2.00 M solution of HNO3 ? c. What volume of concentrated HCl solution would be needed to prepare 4.50 L of 3.0 M HCl? See (a) for data. 8. A 3.8 M solution of FeSO4 solution is diluted to eight times its original volume. What is the molarity of the diluted solution? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 218 Dilutions Back Print Name Class Date Problem Solving continued 9. A chemist prepares 480. mL of a 2.50 M solution of K2Cr2O7 in water. A week later, the chemist wants to use the solution, but the stopper has been left off the flask and 39 mL of water has evaporated. What is the new molarity of the solution? 10. You must write out procedures for a group of lab technicians. One test they will perform requires 25.00 mL of a 1.22 M solution of acetic acid. You decide to use a 6.45 M acetic acid solution that you have on hand. What procedure should the technicians use in order to get the solution they need? 11. A chemical test has determined the concentration of a solution of an unknown substance to be 2.41 M. A 100.0 mL volume of the solution is evaporated to dryness, leaving 9.56 g of crystals of the unknown solute. Calculate the molar mass of the unknown substance. 12. Tincture of iodine can be prepared by dissolving 34 g of I2 and 25 g of KI in 25 mL of distilled water and diluting the solution to 500. mL with ethanol. What is the molarity of I2 in the solution? 13. Phosphoric acid is commonly supplied as an 85% solution. What mass of this solution would be required to prepare 600.0 mL of a 2.80 M phosphoric acid solution? 14. Commercially available concentrated sulfuric acid is 18.0 M H2SO4 . What volume of concentrated H2SO4 would you use in order to make 3.00 L of a 4.0 M stock solution? 15. Describe how to prepare 1.00 L of a 0.495 M solution of urea, NH2CONH2 , starting with a 3.07 M stock solution. 16. Honey is a solution consisting almost entirely of a mixture of the hexose sugars fructose and glucose; both sugars have the formula C6H12O6 , but they differ in molecular structure. a. A sample of honey is found to be 76.2% C6H12O6 by mass. What is the molality of the hexose sugars in honey? Consider the sugars to be equivalent. b. The density of the honey sample is 1.42 g/mL. What mass of hexose sugars are in 1.00 L of honey? What is the molarity of the mixed hexose sugars in honey? 17. Industrial chemicals used in manufacturing are almost never pure, and the content of the material may vary from one batch to the next. For these reasons, a sample is taken from each shipment and sent to a laboratory, where its makeup is determined. This procedure is called assaying. Once the content of a material is known, engineers adjust the manufacturing process to account for the degree of purity of the starting chemicals. Suppose you have just received a shipment of sodium carbonate, Na2CO3 . You weigh out 50.00 g of the material, dissolve it in water, and dilute the solution to 1.000 L. You remove 10.00 mL from the solution and dilute it to 50.00 mL. By measuring the amount of a second substance that reacts with Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 219 Dilutions Back Print Name Class Date Problem Solving continued Na2CO3 , you determine that the concentration of sodium carbonate in the diluted solution is 0.0890 M. Calculate the percentage of Na2CO3 in the original batch of material. The molar mass of Na2CO3 is 105.99 g. (Hint: Determine the number of moles in the original solution and convert to mass of Na2CO3 .) 18. A student wants to prepare 0.600 L of a stock solution of copper(II) chloride so that 20.0 mL of the stock solution diluted by adding 130.0 mL of water will yield a 0.250 M solution. What mass of CuCl2 should be used to make the stock solution? 19. You have a bottle containing a 2.15 M BaCl2 solution. You must tell other students how to dilute this solution to get various volumes of a 0.65 M BaCl2 solution. By what factor will you tell them to dilute the stock solution? In other words, when a student removes any volume, V, of the stock solution, how many times V of water should be added to dilute to 0.65 M? 20. You have a bottle containing an 18.2% solution of strontium nitrate (density 1.02 g/mL). a. What mass of strontium nitrate is dissolved in 80.0 mL of this solution? b. How many moles of strontium nitrate are dissolved in 80.0 mL of the solution? c. If 80.0 mL of this solution is diluted with 420.0 mL of water, what is the molarity of the solution? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 220 Dilutions Back Print Name Class Date Skills Worksheet Problem Solving Colligative Properties Colligative properties of solutions are properties that depend solely on the number of particles of solute in solution. In other words, these properties depend only on the concentration of solute particles, not on the identity of those particles. Colligative properties result from the interference of solute particles with the motion of solvent molecules. Solute molecules can be either electrolytes or nonelectrolytes. When a nonelectrolyte dissolves, the molecule remains whole in the solution. Glucose and glycerol are examples of nonelectrolyte solutes. Ionic solutes are electrolytes. When they dissolve, they dissociate into multiple particles, or ions. When magnesium chloride dissolves in water, it dissociates as follows. MgCl2(s) O3 Mg2 (aq) H2O 2Cl (aq) As you can see, when a mole of MgCl2 completely dissociates in solution, it produces 1 mol of Mg2 ions and 2 mol of Cl ions for a total of 3 mol of solute particles. Because colligative properties depend on the number of particles in solution, 1 mol of MgCl2 in solution should have three times the effect of 1 mol of a nonelectrolyte solute, such as glucose. Two important colligative properties are freezing-point depression and boilingpoint elevation. A dissolved solute lowers the freezing point of the solution. The freezing point of a solution differs from that of the pure solvent according to the following equation, in which tf is the change in freezing point. tf Kf m Kf is a constant that differs for each solvent. Because the freezing point of the solution is lower than that of the solvent alone, Kf is a negative number. The symbol m represents the molality (moles of solute per kilogram of solvent) of the solution. Boiling-point elevation works in the same way. The equation to determine the change in boiling point is as follows. tb Kbm Like Kf , Kb is a constant that differs for each solvent. But unlike Kf , Kb is a positive number because the boiling point of the solution is higher than that of the solvent alone. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 221 Colligative Properties Back Print Name Class Date Problem Solving continued General Plan for Solving Problems Involving Freezing-Point Depression and Boiling-Point Elevation 1 Mass of solute Convert using the molar mass of the solute. 2 Amount of solute in moles Determine if solute is an electrolyte or nonelectrolyte. m mol particles kg solvent 3 Amount of particles in solution 4 Molal concentration of particles in solution, m Multiply by the molal boilingpoint constant, Kb. Multiply by the molal freezingpoint constant, Kf. 5a Freezing-point depression, tf 5b Boiling-point elevation, tb Add tf to the normal freezing point of the solvent. Add tb to the normal boiling point of the solvent. 6a Freezing point of solution 6b Boiling point of solution Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 222 Colligative Properties Back Print Name Class Date Problem Solving continued Table 1 lists freezing-point depression and boiling-point elevation constants for common solvents. TABLE 1 Solvent Acetic acid Camphor Ether Naphthalene Phenol Water Normal f.p. 16.6°C 178.8°C 116.3°C 80.2°C 40.9°C 0.00°C Kf 3.90°C/m 39.7°C/m 1.79°C/m 6.94°C/m 7.40°C/m 1.86°C/m Normal b.p. 117.9°C 207.4°C 34.6°C 217.7°C 181.8°C 100.0°C Kb 3.07°C/m 5.61°C/m 2.02°C/m 5.80°C/m 3.60°C/m 0.51°C/m Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 223 Colligative Properties Back Print Name Class Date Problem Solving continued Sample Problem 1 What is the freezing point of a solution of 210.0 g of glycerol, HOCH2CHOHCH2OH, dissolved in 350. g of water? Solution ANALYZE What is given in the problem? What are you asked to find? Items Identity of solute Particles per mole of solute Identity of solvent Freezing point of solvent Mass of solvent Mass of solute Molar mass of solute* Molal concentration of solute particles Molal freezing-point constant for water Freezing-point depression Freezing point of solution * determined from the periodic table the formula and mass of solute, and the mass of water used the freezing point of the solution Data glycerol, HOCH2CHOHCH2OH 1 mol water 0.00°C 350. g 210.0 g 92.11 g/mol ?m 1.86°C/m ?°C ?°C PLAN What steps are needed to calculate the freezing point of the solution? Use the molar mass of the solute to determine the amount of solute. Then apply the mass of solvent to calculate the molality of the solution. From the molality, use the molal freezing-point constant for water to calculate the number of degrees the freezing point is lowered. Add this negative value to the normal freezing point. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 224 Colligative Properties Back Print Name Class Date Problem Solving continued 1 Mass of glycerol in g multiply by the inverse of the molar mass of glycerol Mass of water in g multiply by the conversion factor 1 kg 1000 g 2 Amount of glycerol in mol the solute is a nonelectrolyte, so the amount of solute equals the amount of particles in solution Mass of water in kg divide the amount of the particles in solution by the mass of the solvent in kilograms 3 Amount of particles in solution 4 Molal concentration of glycerol in water, m multiply by the molal freezing-point constant, Kf , for water 5a Freezing-point depression, tf add tf to the normal freezing point of water 6a Freezing point of the glycerol solution given g H2O 1 kg 1000 g kg H2O freezingpoint depression constant 1 molar mass of glycerol given calculated above g glycerol 1 mol glycerol 92.11 g glycerol freezing point of H2O 1 kg H2O calculated above 1.86 C mol/kg tf tf 0.00 C tf COMPUTE 350. g H2O 210.0 g glycerol 1 kg 1000 g 0.350 kg H2O 1.86°C mol/kg 12.1°C 1 mol glycerol 92.11 g glycerol 0.00°C 1 0.350 kg H2O 12.1°C ( 12.1°C) Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 225 Colligative Properties Back Print Name Class Date Problem Solving continued EVALUATE Are the units correct? Yes; units canceled to give Celsius degrees. Is the number of significant figures correct? Yes; three significant figures is correct because the data had a minimum of three significant figures. Is the answer reasonable? Yes; the calculation can be approximated as 200 [90 400/30 13, which is close to the calculated value. 3(350 1000)] 2 Practice 1. Determine the freezing point of a solution of 60.0 g of glucose, C6H12O6 , dissolved in 80.0 g of water. ans: 7.74°C 2. What is the freezing point of a solution of 645 g of urea, H2NCONH2 , dissolved in 980. g of water? ans: 20.4°C Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 226 Colligative Properties Back Print Name Class Date Problem Solving continued Sample Problem 2 What is the boiling point of a solution containing 34.3 g of the ionic compound magnesium nitrate dissolved in 0.107 kg of water? Solution ANALYZE What is given in the problem? What are you asked to find? Items Identity of solute Equation for the dissociation of the solute Amount of ions per mole of solute Identity of solvent Boiling point of solvent Mass of solvent Mass of solute Molar mass of solute Molal concentration of solute particles Molal boiling-point constant for solvent Boiling-point depression Boiling point of solution the formula and mass of solute, and the mass of water used the boiling point of the solution Data magnesium nitrate Mg(NO3)2 → Mg2 3 mol water 100.0°C 0.107 kg H2O 34.3 g 148.32 g/mol ?m 0.51°C/m ?°C ?°C 2NO3 PLAN What steps are needed to calculate the boiling point of the solution? Use the molar mass to calculate the amount of solute in moles. Multiply the amount of solute by the number of moles of ions produced per mole of solute. Use the amount of ions with the mass of solvent to compute the molality of particles in solution. Use this effective molality to determine the boiling-point elevation and the boiling point of the solution. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 227 Colligative Properties Back Print Name Class Date Problem Solving continued 1 Mass of Mg(NO3)2 in g multiply by the inverse of the molar mass of Mg(NO3)2 2 Amount of Mg(NO3)2 in mol the solute is an Mass electrolyte, so multiply the amount of solute by the number of particles divide the amount of the per mole of solute particles in solution by the mass of the solvent in 3 kilograms of water in kg 4 Amount of particles in solution Molal concentration of particles in water, m multiply by the molal boiling-point constant, Kb , for water 5b Boiling-point elevation, tb add tb to the normal boiling point of water 6b Boiling point of the Mg(NO3)2 solution 1 molar mass of Mg(NO3)2 given given g Mg(NO3)2 1 mol Mg(NO3)2 148.32 g Mg(NO3)2 3 mol particles 1 mol Mg(NO3)2 1 kg H2O 0.51 C mol/kg molal boiling-point constant for water tb boiling point of H2O calculated above 100.0 C tb tb COMPUTE 34.3 g Mg(NO3)2 1 mol Mg(NO3)2 148.32 g Mg(NO3)2 3 mol particles 1 mol Mg(NO3)2 1 0.107 kg H2O 0.51°C mol/kg 3.31°C 100.0°C 3.31°C 103.3°C Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 228 Colligative Properties Back Print Name Class Date Problem Solving continued EVALUATE Are the units correct? Yes; units canceled to give Celsius degrees. Is the number of significant figures correct? Yes; the number of significant figures is correct because the boiling point of water was given to one decimal place. Is the answer reasonable? Yes; the calculation can be approximated as [(35 3)/150] 5 (7/10) 3.5, which is close to the calculated value for the boiling-point elevation. 5 Practice 1. What is the expected boiling point of a brine solution containing 30.00 g of KBr dissolved in 100.00 g of water? ans: 102.6°C 2. What is the expected boiling point of a CaCl2 solution containing 385 g of CaCl2 dissolved in 1.230 103 g of water? ans: 104.3°C Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 229 Colligative Properties Back Print Name Class Date Problem Solving continued Sample Problem 3 A solution of 3.39 g of an unknown compound in 10.00 g of water has a freezing point of 7.31°C. The solution does not conduct electricity. What is the molar mass of the compound? Solution ANALYZE What is given in the problem? the freezing point of the solution, the mass of the dissolved compound, the mass of solvent, and the fact that the solution does not conduct electricity the molar mass of the unknown compound Data 3.39 g ? g/mol water 0.00°C 10.00 g 1.86°C/m ?°C 7.31°C ?m What are you asked to find? Items Mass of solute Molar mass of solute Identity of solvent Freezing point of solvent Mass of solvent Molal freezing-point constant for solvent Freezing-point depression Freezing point of solution Molal concentration of solute particles PLAN What steps are needed to calculate the molar mass of the unknown solute? Determine the molality of the solution from the freezing-point depression. Use the molality and the solute and solvent masses to calculate the solute molar mass. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 230 Colligative Properties Back Print Name Class Date Problem Solving continued 6a Freezing point of the solution subtract the normal freezing point of water Molar mass of solute divide the mass of the solute by the amount of solute in moles 5a Freezing-point depression, tf multiply by the inverse of the molal freezing-point constant, Kf , for water 2 Amount of solute in mol the solute is a nonelectrolyte, so the amount of solute equals the amount of particles in solution 4 Molal concentration of particles in water, m 3 Amount of particles in solution multiply the molal concentration by the mass of the water Mass of water in g multiply by the conversion 1 kg factor 1000 g given freezing point of water Mass of water in kg tf given 0.00 C 1 kg 1000 g tf kg H2O g H2O calculated above 1 molal freezing-point constant for water tf mol/kg 1.86 C given calculated above kg H2O mol solute g solute mol solute calculated above molar mass of solute COMPUTE 7.31°C 10.00 g H2O 7.31°C mol/kg 1.86°C 0.00°C 1 kg 1000 g 7.31°C 0.010 00 kg H2O 0.039 30 mol solute 0.010 00 kg H2O 3.39 g solute 0.039 30 mol solute Copyright © by Holt, Rinehart and Winston. All rights reserved. 86.3 g/mol Holt ChemFile: Problem-Solving Workbook 231 Colligative Propertiess Back Print Name Class Date Problem Solving continued EVALUATE Are the units correct? Yes; molar mass has units of g/mol. Is the number of significant figures correct? Yes; the number of significant figures is correct because the data had a minimum of three significant figures. Is the answer reasonable? Yes; the calculation can be approximated as (4/1) (1/100) close to the value of 0.0393 for the amount of solute. 0.04, which is Practice 1. A solution of 0.827 g of an unknown non-electrolyte compound in 2.500 g of water has a freezing point of 10.18°C. Calculate the molar mass of the compound. ans: 60.4 g/mol 2. A 0.171 g sample of an unknown organic compound is dissolved in ether. The solution has a total mass of 2.470 g. The boiling point of the solution is found to be 36.43°C. What is the molar mass of the organic compound? ans: 82.1 g/mol Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 232 Colligative Properties Back Print Name Class Date Problem Solving continued Additional Problems In each of the following problems, assume that the solute is a nonelectrolyte unless otherwise stated. 1. Calculate the freezing point and boiling point of a solution of 383 g of glucose dissolved in 400. g of water. 2. Determine the boiling point of a solution of 72.4 g of glycerol dissolved in 122.5 g of water. 3. What is the boiling point of a solution of 30.20 g of ethylene glycol, HOCH2CH2OH, in 88.40 g of phenol? 4. What mass of ethanol, CH3CH2OH, should be dissolved in 450. g of water to obtain a freezing point of 4.5°C? 5. Calculate the molar mass of a nonelectrolyte that lowers the freezing point of 25.00 g of water to 3.9°C when 4.27 g of the substance is dissolved in the water. 6. What is the freezing point of a solution of 1.17 g of 1-naphthol, C10H8O, dissolved in 2.00 mL of benzene at 20°C? The density of benzene at 20°C is 0.876 g/mL. Kf for benzene is 5.12°C/m, and benzene’s normal freezing point is 5.53°C. 7. The boiling point of a solution containing 10.44 g of an unknown nonelectrolyte in 50.00 g of acetic acid is 159.2°C. What is the molar mass of the solute? 8. A 0.0355 g sample of an unknown molecular compound is dissolved in 1.000 g of liquid camphor at 200.0°C. Upon cooling, the camphor freezes at 157.7°C. Calculate the molar mass of the unknown compound. 9. Determine the boiling point of a solution of 22.5 g of fructose, C6H12O6 , in 294 g of phenol. 10. Ethylene glycol, HOCH2CH2OH, is effective as an antifreeze, but it also raises the boiling temperature of automobile coolant, which helps prevent loss of coolant when the weather is hot. a. What is the freezing point of a 50.0% solution of ethylene glycol in water? b. What is the boiling point of the same 50.0% solution? 11. The value of Kf for cyclohexane is 20.0°C/m, and its normal freezing point is 6.6°C. A mass of 1.604 g of a waxy solid dissolved in 10.000 g of cyclohexane results in a freezing point of 4.4°C. Calculate the molar mass of the solid. 12. What is the expected freezing point of an aqueous solution of 2.62 kg of nitric acid, HNO3 , in a solution with a total mass of 5.91 kg? Assume that the nitric acid is completely ionized. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt Chemistry 233 Colligative Properties Back Print Name Class Date Problem Solving continued 13. An unknown organic compound is mixed with 0.5190 g of naphthalene crystals to give a mixture having a total mass of 0.5959 g. The mixture is heated until the naphthalene melts and the unknown substance dissolves. Upon cooling, the solution freezes at a temperature of 74.8°C. What is the molar mass of the unknown compound? 14. What is the boiling point of a solution of 8.69 g of the electrolyte sodium acetate, NaCH3COO, dissolved in 15.00 g of water? 15. What is the expected freezing point of a solution of 110.5 g of H2SO4 in 225 g of water? Assume sulfuric acid completely dissociates in water. 16. A compound called pyrene has the empirical formula C8H5 . When 4.04 g of pyrene is dissolved in 10.00 g of benzene, the boiling point of the solution is 85.1°C. Calculate the molar mass of pyrene and determine its molecular formula. The molal boiling-point constant for benzene is 2.53°C/m. Its normal boiling point is 80.1°C. 17. What mass of CaCl2 , when dissolved in 100.00 g of water, gives an expected freezing point of 5.0°C; CaCl2 is ionic? What mass of glucose would give the same result? 18. A compound has the empirical formula CH2O. When 0.0866 g is dissolved in 1.000 g of ether, the solution’s boiling point is 36.5°C. Determine the molecular formula of this substance. 19. What is the freezing point of a 28.6% (by mass) aqueous solution of HCl? Assume the HCl is 100% ionized. 20. What mass of ethylene glycol, HOCH2CH2OH, must be dissolved in 4.510 kg of water to result in a freezing point of 18.0°C? What is the boiling point of the same solution? 21. A water solution containing 2.00 g of an unknown molecular substance dissolved in 10.00 g of water has a freezing point of 4.0°C. a. Calculate the molality of the solution. b. When 2.00 g of the substance is dissolved in acetone instead of in water, the boiling point of the solution is 58.9°C. The normal boiling point of acetone is 56.00°C, and its Kb is 1.71°C/m. Calculate the molality of the solution from this data. 22. A chemist wants to prepare a solution with a freezing point of 22.0°C and has 100.00 g of glycerol on hand. What mass of water should the chemist mix with the glycerol? 23. An unknown carbohydrate compound has the empirical formula CH2O. A solution consisting of 0.515 g of the carbohydrate dissolved in 1.717 g of acetic acid freezes at 8.8°C. What is the molar mass of the carbohydrate? What is its molecular formula? 24. An unknown organic compound has the empirical formula C2H2O. A solution of 3.775 g of the unknown compound dissolved in 12.00 g of water is cooled until it freezes at a temperature of 4.72°C. Determine the molar mass and the molecular formula of the compound. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 234 Colligative Properties Back Print Name Class Date Skills Worksheet Problem Solving Equilibrium Not all processes in nature proceed to completion. In fact, most changes hover somewhere between the initial state and what would be the final state. Compare a light switch and a dimmer. If the mechanical switch is working properly, it can be stable in only two positions: on or off. Either current flows or it doesn’t. With a dimmer you can regulate the flow of current so that it stays somewhere between fully on and fully off. If you’ve used mechanical balances, you know that to weigh an object accurately you must adjust the masses so that the pointer hovers in the middle of its range. The balance is in a state of equilibrium. Most chemical reactions also reach a state of equilibrium between no reaction at all and the complete reaction to form products. Equilibrium states occur when reactions are reversible, that is, when products react to re-form the original reactants. When the products re-form reactants at the same rate as the reactants form the products, then the equilibrium point of the reaction has been reached. The progress of a reaction is gauged by measuring the concentrations in moles per liter of reactants and products. At the equilibrium point, these concentrations stop changing. A 2B ^ 2C The equation above represents a reaction in which 1 mol of A reacts with 2 mol of B to produce 2 mol of C. As C is formed, it breaks down to re-form reactants. The extent of the reaction at equilibrium is indicated by the equilibrium constant, Keq. Keq [C]2 [A][B] As you can see, the concentration of each reaction component is raised to the power of its coefficient in the balanced equation. These concentration terms are arranged in a fraction with products in the numerator and reactants in the denominator. Pure substances (substances that appear in the chemical equation as solids or pure liquids) are not included in the equilibrium expressions because their concentrations are meaningless. Problems involving chemical equilibrium will ask you to solve for either the equilibrium constant, Keq, given the concentrations of all of the reaction components, or the concentration of one of the reaction components, given Keq. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 235 Equilibrium Back Print Name Class Date Problem Solving continued General Plan for Solving Equilibrium Problems 1 Balanced chemical equation xA yB 3 zC 3 Unknown concentration x y [C] z Keq[A] [B] Rearrange to solve for the unknown quantity, substitute known values, and solve. 2 Expression for the equilibrium constant, Keq z [C] Keq [A]x[B]y Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 236 Equilibrium Back Print Name Class Date Problem Solving continued Sample Problem 1 The following equation represents the reversible decomposition of PCl5. PCl5( g) ^ PCl3( g) [PCl5] [PCl3] [Cl2] 1.271 M 0.229 M 0.229 M Cl2( g) At 250°C, the equilibrium concentrations of the substances are as follows: What is the value of the equilibrium constant, Keq, for this reaction? Solution ANALYZE What is given in the problem? What are you asked to find? the equilibrium concentrations of the products and reactant at 250°C the value of the equilibrium constant for the reaction Data 1.271 0.229 M 0.229 M ? Items Molar concentration of PCl5 at equilibrium Molar concentration of PCl3 at equilibrium Molar concentration of Cl2 at equilibrium Equilibrium constant Keq PLAN What steps are needed to calculate the equilibrium constant for the given reaction? Set up the equilibrium expression for the reaction using the coefficients as exponents. Substitute the concentration values, and calculate Keq. 1 PCl5(g) ª PCl3(g) Cl2(g) Keq 2 [PCl3][Cl2] [PCl5] Note that since all coefficients have the value 1, there is no need to write in the exponent. given given Keq [PCl3][Cl2] [PCl5] given Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 237 Equilibrium Back Print Name Class Date Problem Solving continued COMPUTE Keq EVALUATE Are the units correct? Yes; the equilibrium constant has no units. [0.229][0.229] [1.271] 0.0413 Is the number of significant figures correct? Yes; the number of significant figures is correct because data values were given to a minimum of three significant figures. Is the answer reasonable? Yes; the calculation may be approximated as (0.2 0.2)/1. This approximation gives a result of 0.04, which is close to the calculated value. Practice 1. Calculate the equilibrium constants for the following hypothetical reactions. Assume that all components of the reactions are gaseous. a. A ^ C D 2 At equilibrium, the concentration of A is 2.24 10 tions of both C and D are 6.41 10 3 M. ans: Keq M and the concentra10 3 1.83 b. A B^C D 10 5 At equilibrium, the concentrations of both A and B are 3.23 concentrations of both C and D are 1.27 10 2 M. ans: Keq M and the 105 1.55 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 238 Equilibrium Back Print Name Class Date Problem Solving continued c. A B ^ 2C 10 3 At equilibrium, the concentrations of both A and B are 7.02 concentration of C is 2.16 10 2 M. ans: Keq 9.47 M and the d. 2A ^ 2C D At equilibrium, the concentration of A is 6.59 10 4 M. The concentration of C is 4.06 10 3 M, and the concentration of D is 2.03 10 3 M. ans: Keq 7.71 10 2 e. A B^C D E 4 At equilibrium, the concentrations of both A and B are 3.73 10 concentrations of C, D, and E are 9.35 10 4 M. ans: Keq 5.88 M and the 10 3 f. 2A B ^ 2C At equilibrium, the concentration of A is 5.50 10 3 M, the concentration of B is 2.25 10 3 M, and the concentration of C is 1.02 10 2 M. ans: Keq 1.53 103 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 239 Equilibrium Back Print Name Class Date Problem Solving continued Sample Problem 2 The following equilibrium reaction is used in the manufacture of methanol. The equilibrium constant at 400 K for the reaction is 1.609. CO( g) 2H2( g) ^ CH3OH( g) At equilibrium, the mixture in the reaction vessel has a concentration of 0.818 M of CH3OH and 1.402 M of CO. Calculate the concentration of H2 in the vessel. Solution ANALYZE What is given in the problem? What are you asked to find? Items Molar concentration of CO at equilibrium Molar concentration of H2 at equilibrium Molar concentration of CH3OH at equilibrium Equilibrium constant Keq the equilibrium concentrations of CO and CH3OH, and the equilibrium constant at 400 K the equilibrium concentration of H2 in the vessel Data 1.402 M ?M 0.818 M 0.609 M PLAN What steps are needed to calculate the concentration of H2? Set up the equilibrium expression for the reaction using coefficients as exponents. Rearrange the expression to solve for [H2]. Substitute known values for [CO2], [CH3OH], and Keq, and solve for [H2]. 1 CO(g) 2H2(g) ª CH3OH(g) 3 [H2] 2 [CH3OH] Keq [CO] Keq [CH3OH] [CO][H2]2 rearrange to solve for [H2] , substitute known values, and solve Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 240 Equilibrium Back Print Name Class Date Problem Solving continued COMPUTE [H2] EVALUATE Are the units correct? Yes; concentrations are in moles per liter. [0.818] 1.609 [1.402] 0.602 M Is the number of significant figures correct? Yes; the number of significant figures is correct because data values were given to a minimum of three significant figures. Is the answer reasonable? Yes; the calculation may be approximated as [1 close to the calculated value. (1.5 1.5)]1/2 0.67, which is Practice 1. Calculate the concentration of product D in the following hypothetical reaction: 2A( g) ^ 2C( g) D( g) At equilibrium, the concentration of A is 1.88 10 1 M, the concentration of C is 6.56 M, and the equilibrium constant is 2.403 102. ans: 0.197 M 2. At a temperature of 700 K, the equilibrium constant is 3.164 103 for the following reaction system for the hydrogenation of ethene, C2H4, to ethane, C2H6. C2H4( g) H2( g) ^ C2H6( g) What will be the equilibrium concentration of ethene if the concentration of H2 is 0.0619 M and the concentration of C2H6 is 1.055 M? ans: 5.39 10 3 M Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 241 Equilibrium Back Print Name Class Date Problem Solving continued Additional Problems—Equilibrium 1. Using the reaction A 2B ^ C 2D, determine Keq if the following equilibrium concentrations are found. All components are gases. [A] [B] [C] [D] 0.0567 M 0.1171 M 0.000 3378 M 0.000 6756 M 2. In the reaction 2A ^ 2C 2D, determine Keq when the following equilibrium concentrations are found. All components are gases. [A] [C] [D] 0.1077 M 0.000 4104 M 0.000 4104 M 3. Calculate the equilibrium constant for the following reaction. Note the phases of the components. 2A( g) [A] [C] [D] 0.0922 M 4.11 8.22 10 10 4 4 B(s) ^ C( g) D( g) The equilibrium concentrations of the components are M M 4. The equilibrium constant of the following reaction for the decomposition of phosgene at 25°C is 4.282 10 2. COCl2( g) ^ CO( g) Cl2( g) a. What is the concentration of COCl2 when the concentrations of both CO and Cl2 are 5.90 10 3 M? b. When the equilibrium concentration of COCl2 is 0.003 70 M, what are the concentrations of CO and Cl2? Assume the concentrations are equal. 5. Consider the following hypothetical reaction. A( g) B(s) ^ C( g) D(s) a. If Keq 1 for this reaction at 500 K, what can you say about the concentrations of A and C at equilibrium? b. If raising the temperature of the reaction results in an equilibrium with a higher concentration of C than A, how will the value of Keq change? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 242 Equilibrium Back Print Name Class Date Problem Solving continued 6. The following reaction occurs when steam is passed over hot carbon. The mixture of gases it generates is called water gas and is useful as an industrial fuel and as a source of hydrogen for the production of ammonia C(s) H2O( g) ^ CO( g) H2( g) The equilibrium constant for this reaction is 4.251 10 2 at 800 K. If the equilibrium concentration of H2O( g) is 0.1990 M, what concentrations of CO and H2 would you expect to find? 7. When nitrogen monoxide gas comes in contact with air, it oxidizes to the brown gas nitrogen dioxide according to the following equation: 2NO( g) O2( g) ^ 2NO2( g) a. The equilibrium constant for this reaction at 500 K is 1.671 104. What concentration of NO2 is present at equilibrium if [NO] 6.200 10 2 M and [O2] 8.305 10 3 M? b. At 1000 K, the equilibrium constant, Keq, for the same reaction is 1.315 10 2. What will be the concentration of NO2 at 1000 K given the same concentrations of NO and O2 as were in (a)? 8. Consider the following hypothetical reaction, for which Keq 1 at 300 K: A( g) B( g) ^ 2C( g) a. If the reaction begins with equal concentrations of A and B and a zero concentration of C, what can you say about the relative concentrations of the components at equilibrium? b. Additional C is introduced at equilibrium, and the temperature remains constant. When equilibrium is restored, how will the concentrations of all components have changed? How will Keq have changed? 9. The equilibrium constant for the following reaction of hydrogen gas and bromine gas at 25°C is 5.628 1018. H2( g) Br2( g) ^ 2HBr( g) a. Write the equilibrium expression for this reaction. b. Assume that equimolar amounts of H2 and Br2 were present at the beginning. Calculate the equilibrium concentration of H2 if the concentration of HBr is 0.500 M. c. If equal amounts of H2 and Br2 react, which reaction component will be present in the greatest concentration at equilibrium? Explain your reasoning. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 243 Equilibrium Back Print Name Class Date Problem Solving continued 10. The following reaction reaches an equilibrium state: N2F4( g) ^ 2NF2( g) At equilibrium at 25°C the concentration of N2F4 is found to be 0.9989 M and the concentration of NF2 is 1.131 10 3 M. Calculate the equilibrium constant of the reaction. 11. The equilibrium between dinitrogen tetroxide and nitrogen dioxide is represented by the following equation: N2O4( g) ^ NO2( g) A student places a mixture of the two gases into a closed gas tube and allows the reaction to reach equilibrium at 25°C. At equilibrium, the concentration of N2O4 is found to be 5.95 10 1 M and the concentration of NO2 is found to be 5.24 10 2 M. What is the equilibrium constant of the reaction? 12. Consider the following equilibrium system: NaCN(s) HCl( g) ^ HCN( g) NaCl(s) a. Write a complete expression for the equilibrium constant of this system. b. The Keq for this reaction is 2.405 106. What is the concentration of HCl remaining when the concentration of HCN is 0.8959 M? 13. The following reaction is used in the industrial production of hydrogen gas: CH4( g) H2O( g) ^ CO( g) 3H2( g) 10 27 The equilibrium constant of this reaction at 298 K (25°C) is 3.896 at 1100 K the constant is 3.112 102. , but a. What do these equilibrium constants tell you about the progress of the reaction at the two temperatures? b. Suppose the reaction mixture is sampled at 1100 K and found to contain 1.56 M of hydrogen, 3.70 10 2 M of methane, and 8.27 10 1 M of gaseous H2O. What concentration of carbon monoxide would you expect to find? 14. Dinitrogen tetroxide, N2O4, is soluble in cyclohexane, a common nonpolar solvent. While in solution, N2O4 can break down into NO2 according to the following equation: N2O4(cyclohexane) ^ NO2(cyclohexane) At 20°C, the following concentrations were observed for this equilibrium reaction: [N2O4] [NO2] 2.55 10.4 10 10 3 3 M M What is the value of the equilibrium constant for this reaction? Note, the chemical equation must be balanced first. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 244 Equilibrium Back Print Name Class Date Problem Solving continued 15. The reaction given in item 14 also occurs when the dinitrogen tetroxide and nitrogen dioxide are dissolved in carbon tetrachloride, CCl4, another nonpolar solvent. N2O4(CCl4) ^ NO2(CCl4) The following experimental data were obtained at 20°C: [N2O4] [NO2] 2.67 10.2 10 10 3 3 M M Calculate the value of the equilibrium constant for this reaction occurring in carbon tetrachloride. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 245 Equilibrium Back Print Name Class Date Skills Worksheet Problem Solving Equilibrium of Salts, Ksp When you try to dissolve a solid substance in water, you expect the solid form to disappear, forming ions or molecules in solution. Most substances, however, are only slightly soluble in water. For example, when you stir silver chloride in water, you may think none of the solid dissolves. Does this mean that some of the solid dissolves, forms a saturated solution, and, after that, experiences no further change between the solid and solution phases? It is true to say that there is no further change in the amount of substance in either the solution or the solid phase, but to say, that no further change occurs is inaccurate. An equilibrium exists between the silver chloride and its dissolved ions, and a state of equilibrium is a dynamic state. The following chemical equation shows this equilibrium. AgCl(s) ^ Ag (aq) Cl (aq) Like the other examples of equilibria that you have studied, the extent to which this solubility equilibrium proceeds toward the products (the ions in solution) is indicated by an equilibrium constant. When an equilibrium constant is written for a solubility equilibrium, it is called a solubility product constant and is symbolized as Ksp. In solubility equilibrium problems, the reactants are pure substances, and pure substances are never included in an equilibrium expression. That means that you will not have anything in the denominator in the expression for the solubility product constant. This Ksp expression for the silver chloride example can be written as follows. Ksp [Ag ][Cl ] Note that the coefficients in the balanced equation are understood to be 1 for both silver and chlorine. Therefore, no exponents appear in the Ksp expression. As with any equilibrium expression, the concentration of each component is raised to the power of its coefficient from the balanced chemical equation. The value for this solubility product constant is 1.77 10 10 at 25°C. The very small value of Ksp indicates that silver chloride is only very slightly soluble in aqueous solution at this temperature. The value of Ksp supports the observation that little seems to occur when silver chloride is stirred into a water solution. In this worksheet, you will learn to apply the solubility equilibrium relationship to determine Ksp for substances and to calculate concentrations of ions in saturated solutions using their Ksp values. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 246 Equilibrium of Salts, Ksp Back Print Name Class Date Problem Solving continued General Plan for Solving Solubility Equilibrium Problems 1 Molarity of solution Write the balanced equation for the dissociation of the solute. 2 Balanced chemical equation Use the mole ratios of the original solute to each ion to calculate each concentration. 3 If the solution is saturated, use the balanced chemical equation to write the Ksp expression. Concentrations of each ion in solution Use the balanced chemical equation to write the ion product expression. 4a Ksp expression If Ksp is greater than the ion product, the solution is not saturated and no precipitation occurs. If Ksp is less than the ion product, precipitation occurs. 4b Ion product Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 247 Equilibrium of Salts, Ksp Back Print Name Class Date Problem Solving continued Sample Problem 1 A saturated solution of magnesium fluoride, MgF2, contains 0.00741 g of dissolved MgF2 per 1.00 102 mL at 25°C. What is the Ksp for magnesium fluoride? Solution ANALYZE What is given in the problem? What are you asked to find? Items Mass of dissolved MgF2 Volume of solution Molar mass of MgF2* Molar concentration of MgF2 Molar concentration of Mg Molar concentration of F Ksp of MgF2 *determined from the periodic table 2 the mass of MgF2 dissolved in 1.00 a saturated solution the solubility product constant, Ksp Data 0.00741 g 1.00 102 mL 102 mL of 62.30 g/mol ?M ?M ?M ? PLAN What steps are needed to calculate the solubility product constant, Ksp, of MgF2? Determine the molar concentration of the saturated MgF2 solution. Write the balanced chemical equation for the dissociation of MgF2, and use this equation to determine the concentrations of each ion in solution. Compute Ksp. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 248 Equilibrium of Salts, Ksp Back Print Name Class Date Problem Solving continued 1 Molarity of solution write a balanced chemical equation for the dissociation 2 MgF2(s) ^ Mg (aq) use coefficients from the balanced chemical equation 2 2F (aq) 3 Molarity of Mg2 3 Molarity of F use the balanced equation to write the Ksp expression, substitute, and solve 4a Ksp [Mg2 ][F ]2 Calculate the molarity of the saturated MgF2 solution. given g MgF2 given 1 mol MgF2 62.30 g MgF2 1L 1000 mL 1 molar mass MgF2 mol MgF2 L solution mL solution calculated above mol MgF2 L solution calculated above [MgF2] Write the balanced chemical equation for the dissociation to determine the mole ratios of solute and ions. MgF2(s) a Mg2 (aq) 2F (aq) [MgF2] [Mg2 ] 2[MgF2] [F ] Write the Ksp expression. Ksp Ksp [Mg2 ][F ]2 [MgF2] (2[MgF2])2 calculated above Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 249 Equilibrium of Salts, Ksp Back Print Name Class Date Problem Solving continued COMPUTE 0.007 41 g MgF2 1 mol MgF2 62.30 g MgF2 1L 1000 mL [MgF2] 1.19 2.38 10 ] 32 1.19 10 4 mol MgF2 100. mL solution 0.100 L solution 1.19 10 10 3 3 1.19 10 4 mol MgF2 0.100 L solution [Mg2 ] [F ] Ksp EVALUATE Are the units correct? Yes; Ksp has no units. 10 M 3 M [MgF2] 2[MgF2] 10 3 M 10 9 [1.19 [2.38 6.74 Is the number of significant figures correct? Yes; the number of significant figures is correct because all data were given to three significant figures. Is the answer reasonable? Yes; the calculation can be approximated as (1 10 3)(2.5 10 3)2 which is of the same order of magnitude as the calculated answer. 6 10 9 , Practice 1. Silver bromate, AgBrO3, is slightly soluble in water. A saturated solution is found to contain 0.276 g AgBrO3 dissolved in 150.0 mL of water. Calculate Ksp for silver bromate. ans: Ksp 6.09 10 5 2. 2.50 L of a saturated solution of calcium fluoride leaves a residue of 0.0427 g of CaF2 when evaporated to dryness. Calculate the Ksp of CaF2. ans: Ksp 4.20 10 11 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 250 Equilibrium of Salts, Ksp Back Print Name Class Date Problem Solving continued Sample Problem 2 The Ksp for lead(II) iodide is 7.08 10 9 at 25°C. What is the molar concentration of PbI2 in a saturated solution? Solution ANALYZE What is given in the problem What are you asked to find? Items Ksp of PbI2 Concentration of Pb2 Concentration of I Concentration of PbI2 in solution the solubility product constant, Ksp of PbI2 the concentration of PbI2 in a saturated solution Data 7.08 ? ? ? 10 9 PLAN What steps are needed to calculate the concentration of dissolved PbI2 in a saturated solution? Write the equation for the dissociation of PbI2. Set up the equation for Ksp, and compute the concentrations of the ions. Determine the concentration of dissolved solute. Write the balanced chemical equation for the dissociation of lead(II) iodide, PbI2 in aqueous solution. PbI2(s) ^ Pb2 (aq) Write the Ksp expression. Ksp [Pb2 ][I ]2 [I ]. Substitute x for [Pb2 ]. The balanced equation gives the following relationship: 2[Pb2 ] Therefore, [I ] 2x. Ksp Rearrange, and solve for x. Ksp Ksp given 2I (aq) [x][2x]2 [x][4x 2] 4x 3 x 3 Ksp 4 [Pb2 ] Relate the substituted value to the unknown solution concentration using the mole ratio from the original balanced chemical equation. The mole ratio shows that [Pb2 ] [PbI2]. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 251 Equilibrium of Salts, Ksp Back Print Name Class Date Problem Solving continued COMPUTE x 3 7.08 4 [Pb2 ] 10 9 [Pb2 ] 1.21 1.21 10 3 10 M 3 M [PbI2] EVALUATE Are the units correct? Yes; concentrations are in molarity (mol/L). Is the number of significant figures correct? Yes; the number of significant figures is correct because data were given to three significant figures. Is the answer reasonable? Yes; the best check is to use the result to calculate Ksp and see if it gives (or is very near) the Ksp you started with. In this case, the calculated Ksp is 7.08 10 9, the same value as was given. Practice 1. The Ksp of calcium sulfate, CaSO4, is 9.1 10 tion of CaSO4 in a saturated solution? ans: 3.0 6 . What is the molar concentra10 3 M 2.A salt has the formula X2Y, and its Ksp is 4.25 10 7 . a. What is the molarity of a saturated solution of the salt? ans: [X2Y] 4.74 10 3 M b. What is the molarity of a solution of AZ if its Ksp is the same value? ans: [AZ] 6.52 10 4 M Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 252 Equilibrium of Salts, Ksp Back Print Name Class Date Problem Solving continued Sample Problem 3 Will precipitation of strontium sulfate occur when 50.0 mL of 0.025 M strontium nitrate solution is mixed with 50.0 mL of a 0.014 M copper(II) sulfate solution? The Ksp of strontium nitrate is 3.2 10 7. Solution ANALYZE What is given in the problem? the molar concentrations of the solutions to be mixed, the identities of the solutes, and the volumes of the solutions to be mixed whether a precipitate of strontium sulfate forms when the two solutions are mixed What are you asked to find? Items Concentration of solution 1 Volume of solution 1 Concentration of solution 2 Volume of solution 2 Volume of combined solution Concentration of combined solution Potential precipitate Ksp of SrSO4 Precipitate/no precipitate forms Data 0.025 M SrNO3 50.0 mL 0.014 M CuSO4 50.0 mL 100.0 mL ? M SrSO4 SrSO4 3.2 ? 10 7 PLAN What steps are needed to determine whether a precipitate will form? Calculate the molar concentrations of the ions that can form a precipitate in the new volume of solution. Use these concentrations to calculate the ion product. Compare the ion product with Ksp. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 253 Equilibrium of Salts, Ksp Back Print Name Class Date Problem Solving continued 2 SrSO4(aq) ^ Sr2 (aq) use the molarity of Sr(NO3)2, and calculate the molarity of Sr 2 in the new diluted solution SO42 (aq) use the molarity of CuSO4 , and calculate the molarity of SO42 in the new diluted solution 3 Molarity of Sr2 use coefficients from the balanced chemical equation 3 Molarity of SO42 4a Ksp ion product if Ksp ion product, precipitate forms, if Ksp ion product, no precipitate forms 4b [Sr2 ][SO42 ] Precipitate or no precipitate Write the balanced equation for the dissociation of SrSO4. SrSO4(aq) ^ Sr 2 (aq) SO2 (aq) 4 Calculate the molarities of Sr 2 and SO2 . This is a simple dilution calculation. 4 The subscript 1 in each case represents that individual solution; the subscript 2 represents the combined solution. Sr(NO3)2(aq) ^ Sr2 (aq) 2NO3 (aq) [Sr 2 ]1 [Sr(NO3)2] [Sr2 ]1V1 [Sr2 ]2V2 calculated above given [Sr2 ]2 [Sr2 ]1 V2 V1 sum of volumes of solutions mixed CuSO4(aq) ^ Cu2 (aq) SO2 (aq) 4 [SO2 ]1 [CuSO4] 4 [SO2 ]1V1 [SO2 ]2V2 4 4 [SO2 ]2 4 [SO2 ]1 4 V2 calculated above given V1 sum of volumes of solutions mixed Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 254 Equilibrium of Salts, Ksp Back Print Name Class Date Problem Solving continued Calculate the ion product for SrSO4. ion product [Sr2 ]2 [SO2 ] 2 4 calculated above Compare the ion product to the Ksp value to determine if precipitation occurs. COMPUTE [Sr2 ]2 [SO 2 ]2 4 ion product 0.025 M 50.0 mL 100.0 mL 0.014 M 50.0 mL 100.0 mL [1.2 10 2][7.0 10 Ksp 3.2 10 7 Ksp ion product 1.2 7.0 3 10 10 8.4 2 M M 10 5 3 Precipitation will occur. EVALUATE Are the units correct? Yes; the ion product has no units. Is the number of significant figures correct? Yes; the number of significant figures is correct because data were given to a minimum of two significant figures. Is the answer reasonable? Yes; the calculation can be approximated as 0.01 0.007 0.000 07 which is of the same order of magnitude as the calculated result. 7 10 5 , Practice In each of the following problems, include the calculated ion product with your answer. 1. Will a precipitate of Ca(OH)2 form when 320. mL of a 0.046 M solution of NaOH mixes with 400. mL of a 0.085 M CaCl2 solution? Ksp of Ca(OH)2 is 5.5 10 6. ans: ion product 1.9 10 5, precipitation occurs. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 255 Equilibrium of Salts, Ksp Back Print Name Class Date Problem Solving continued 2. 20.00 mL of a 0.077 M solution of silver nitrate, AgNO3, is mixed with 30.00 mL of a 0.043 M solution of sodium acetate, NaC2H3O2. Does a precipitate form? The Ksp of AgC2H3O2 is 2.5 10 3. ans: ion product 8.1 10 4, no precipitate 3. If you mix 100. mL of 0.036 M Pb(C2H3O2)2 with 50. mL of 0.074 M NaCl, will a precipitate of PbCl2 form? Ksp of PbCl2 is 1.9 10 4. ans: ion product 1.5 10 5, no precipitate 4. If 20.00 mL of a 0.0090 M solution of (NH4)2S is mixed with 120.00 mL of a 0.0082 M solution of Al(NO3)3, does a precipitate form? The Ksp of Al2S3 is 2.00 10 7. ans: ion product 1.1 10 13, no precipitate Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 256 Equilibrium of Salts, Ksp Back Print Name Class Date Problem Solving continued Additional Problems—Equilibrium of Salts, Ksp 1. The molar concentration of a saturated calcium chromate, CaCrO4, solution is 0.010 M at 25°C. What is the Ksp of calcium chromate? 2. A 10.00 mL sample of a saturated lead selenate solution is found to contain 0.00136 g of dissolved PbSeO4 at 25°C. Determine the Ksp of lead selenate. 3. A 22.50 mL sample of a saturated copper(I) thiocyanate, CuSCN, solution at 25°C is found to have a 4.0 10 6 M concentration. a. Determine the Ksp of CuSCN. b. What mass of CuSCN would be dissolved in 1.0 103 L of solution? 4. A saturated solution of silver dichromate, Ag2Cr2O7, has a concentration of 3.684 10 3 M. Calculate the Ksp of silver dichromate. 5. The Ksp of barium sulfite, BaSO3, at 25°C is 8.0 10 7 . a. What is the molar concentration of a saturated solution of BaSO3? b. What mass of BaSO3 would dissolve in 500. mL of water? 6. The Ksp of lead(II) chloride at 25°C is 1.9 tration of a saturated solution at 25°C? 7. The Ksp of barium carbonate at 25°C is 1.2 10 10 4 . What is the molar concen8 . a. What is the molar concentration of a saturated solution of BaCO3 at 25°C? b. What volume of water would be needed to dissolve 0.10 g of barium carbonate? 8. The Ksp of SrSO4 is 3.2 10 7 at 25°C. a. What is the molar concentration of a saturated SrSO4 solution? b. If 20.0 L of a saturated solution of SrSO4 were evaporated to dryness, what mass of SrSO4 would remain? 9. The Ksp of strontium sulfite, SrSO3, is 4.0 10 8 at 25°C. If 1.0000 g of SrSO3 is stirred in 5.0 L of water until the solution is saturated and then filtered, what mass of SrSO3 would remain? 10. The Ksp of manganese(II) arsenate is 1.9 10 11 at 25°C. What is the molar concentration of Mn3(AsO4)2 in a saturated solution? Note that five ions are produced from the dissociation of Mn3(AsO4)2. 11. Suppose that 30.0 mL of a 0.0050 M solution of Sr(NO3)2 is mixed with 20.0 mL of a 0.010 M solution of K2SO4 at 25°C. The Ksp of SrSO4 is 3.2 10 7. a. What is the ion product of the ions that can potentially form a precipitate? b. Does a precipitate form? 12. Lead(II) bromide, PbBr2, is slightly soluble in water. Its Ksp is 6.3 10 6 at 25°C. Suppose that 120. mL of a 0.0035 M solution of MgBr2 is mixed with 180. mL of a 0.0024 M Pb(C2H3O2)2 solution at 25°C. a. What is the ion product of Br and Pb2 in the mixed solution? b. Does a precipitate form? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 257 Equilibrium of Salts, Ksp Back Print Name Class Date Problem Solving continued 13. The Ksp of Mg(OH)2 at 25°C is 1.5 10 11 . a. Write the equilibrium equation for the dissociation of Mg(OH)2. b. What volume of water would be required to dissolve 0.10 g of Mg(OH)2? c. Considering that magnesium hydroxide is essentially insoluble, why is it possible to titrate a suspension of Mg(OH)2 to an equivalence point with a strong acid such as HCl? 14. Lithium carbonate is somewhat soluble in water; its Ksp at 25°C is 2.51 a. What is the molar concentration of a saturated Li2CO3 solution? b. What mass of Li2CO3 would you dissolve in order to make 3440 mL of saturated solution? 15. A 50.00 mL sample of a saturated solution of barium hydroxide, Ba(OH)2, is titrated to the equivalence point by 31.61 mL of a 0.3417 M solution of HCl. Determine the Ksp of Ba(OH)2. 16. Calculate the Ksp for salts represented by QR that dissociate into two ions, Q and R , in each of the following solutions: a. saturated solution of QR is 1.0 M b. saturated solution of QR is 0.50 M c. saturated solution of QR is 0.1 M d. saturated solution of QR is 0.001 M 17. Suppose that salts QR, X2Y, KL2, A3Z, and D2E3 form saturated solutions that are 0.02 M in concentration. Calculate Ksp for each of these salts. 18. The Ksp at 25°C of silver bromide is 5.0 10 13. What is the molar concentration of a saturated AgBr solution? What mass of silver bromide would dissolve in 10.0 L of saturated solution at 25°C? 19. The Ksp at 25°C for calcium hydroxide is 5.5 10 2 . 10 6 . a. Calculate the molarity of a saturated Ca(OH)2 solution. b. What is the OH concentration of this solution? 20. The Ksp of magnesium carbonate is 3.5 10 would dissolve in 4.00 L of water at 25°C? 8 at 25°C. What mass of MgCO3 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 258 Equilibrium of Salts, Ksp Back Print Name Class Date Skills Worksheet Problem Solving pH In 1909, Danish biochemist S. P. L Sørensen introduced a system in which acidity was expressed as the negative logarithm of the H concentration. In this way, the acidity of a solution having H concentration of 10 5 M would have a value of 5. Because the power of 10 was now a part of the number, the system was called pH, meaning power of hydrogen. Taking the negative logarithm of the hydronium ion concentration will give you a solution’s pH as given by the following equation. pH log[H3O ] Likewise, the value pOH equals the negative logarithm of the hydroxide ion concentration. pOH log[OH ] Water molecules interact with each other and ionize. At the same time, the ions in solution reform molecules of water. This process is represented by the following reversible equation. H2O(l) H2O(l) ^ H3O(aq) OH (aq) In pure water the concentrations of hydroxide ions and hydronium ions will always be equal. These two quantities are related by a term called the ion product constant for water, Kw. Kw [H3O ][OH ] The ion product constant for water can be used to convert from pOH to pH. The following equations derive a simple formula for this conversion. Pure water has a pH of 7. Rearranging the equation for pH, you can solve for the hydronium ion concentration of pure water, which is equal to the hydroxide ion concentration. These values can be used to obtain a numerical value for Kw. Kw [H3O ][OH ] [1 10 7 [1 10 7 =1 10 14 Rearrange the Kw expression to solve for [OH ]. [OH ] Kw [H3O ] Kw [H3O ] Substitute this value into the equation for pOH. pOH log The logarithm of a quotient is the difference of the logarithms of the numerator and the denominator. pOH pOH logKw log(1.0 log[H3O ] log[H3O ]. 10 14 Substitute the value for Kw and pH for ) pH Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 259 pH Back Print Name Class Date Problem Solving continued The negative logarithm of 10 14 is 14. Substitute this value, and rearrange. pH 14 pOH General Plan for Solving pH Problems 1 Mass of solute 2 Convert using the molar mass of the solute. Amount of solute in moles Divide the amount of solute by the volume of the solution. 3 Molar concentration of solution Multiply the concentration of the solution by the moles of H3O per mole of acid. Multiply the concentration of the solution by the moles of OH per mole of base. This step works only for strong acids or bases. Convert using the relationship Kw [H3O ][OH ] Kw 1 10 14 log[OH ] 4b Molar concentration of OH 4a Molar concentration of H3O pH log[H3O ] pOH 5b pOH pH pOH 14 5a pH Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 260 pH Back Print Name Class Date Problem Solving continued Sample Problem 1 A HCl solution has a concentration of 0.0050 M. Calculate [OH ] and [H3O ] for this solution. HCl is a strong acid, so assume it is 100% ionized. Solution ANALYZE What is given in the problem? What are you asked to find? Items Identity of solute Concentration of solute Acid or base Kw [H3O ] [OH ] the molarity of the HCl solution, and the fact that HCl is a strong acid [H3O ] and [OH ] Data HCl 0.0050 M acid 1.0 ?M ?M 10 14 PLAN What steps are needed to calculate the concentration of H3O and OH ? Determine [H3O ] from molarity and the fact that the acid is strong. Use Kw to calculate [OH ]. 3 Molar concentration of the HCl solution rearrange the Kw equation to solve for [OH ], substitute known values and solve each HCl molecule dissociates to produce one H3O ion, so [HCl ] [H3O ] 4b Molar concentration of OH given 4a Molar concentration of H3O [HCl] Kw [H3O ] calculated above 1 10 14 [OH ] Kw [H3O ] [OH ] 1 calculated above 10 14 [H3O ] Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 261 pH Back Print Name Class Date Problem Solving continued COMPUTE [H3O ] [OH ] EVALUATE Are the units correct? Yes; molarity, or mol/L, was required. 0.0050 M 14 1.0 10 0.0050 2.0 10 12 M Is the number of significant figures correct? Yes; the number of significant figures is correct because molarity of HCl was given to two significant figures. Is the answer reasonable? Yes; the two concentrations multiply to give 10 1.0 10 14. 10 15 , which is equal to Practice 1. The hydroxide ion concentration of an aqueous solution is 6.4 What is the hydronium ion concentration? ans: 1.6 10 10 M 10 5 M. 2. Calculate the H3O and OH concentrations in a 7.50 10 4 M solution of HNO3, a strong acid. ans: [H3O ] 7.50 10 4 M; [OH ] 1.33 10 11 M Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 262 pH Back Print Name Class Date Problem Solving continued Sample Problem 2 Calculate the pH of a 0.000 287 M solution of H2SO4. Assume 100% ionization. Solution ANALYZE What is given in the problem? What are you asked to find? Items Identity of solute Concentration of solute Acid or base Kw [H3O ] pH the molarity of the H2SO4 solution, and the fact that H2SO4 is completely ionized pH Data H2SO4 0.000 287 M acid 1.0 ?M ? 10 14 PLAN What steps are needed to calculate the concentration of H3O and the pH? Determine [H3O ] from molarity and the fact that the acid is 100% ionized. Determine the pH, negative logarithm of the concentration. 3 Molar concentration of the H2SO4 solution each H2SO4 molecule dissociates to produce two H3O ions, so 2 [H2SO4] [H3O ] 4a Molar concentration of H3O pH log[H3O ] 5a pH given 2 [H2SO4] calculated above [H3O ] pH log[H3O ] Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 263 pH Back Print Name Class Date Problem Solving continued COMPUTE [H3O ] 2 0.000 287 M 10 4 5.74 ] 3.24 10 4 M log[5.74 EVALUATE Are the units correct? Yes; there are no units on a pH value. Is the number of significant figures correct? Yes; the number of significant figures is correct because molarity of H2SO4 was given to three significant figures. Is the answer reasonable? Yes. You would expect the pH of a dilute H2SO4 solution to be below 7. Practice 1. Determine the pH of a 0.001 18 M solution of HBr. ans: 2.93 2. What is the pH of a solution that has a hydronium ion concentration of 1.0 M? ans: 0.00 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 264 pH Back Print Name Class Date Problem Solving continued 3. What is the pH of a 2.0 M solution of HCl, assuming the acid remains 100% ionized? ans: 0.30 4. What is the theoretical pH of a 10.0 M solution of HCl? ans: 1.00 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 265 pH Back Print Name Class Date Problem Solving continued Sample Problem 3 A solution of acetic acid has a pH of 5.86. What are the pOH and [OH ] of the solution? Solution ANALYZE What is given in the problem? What are you asked to find? Items Identity of solute Acid or base pH pOH [OH ] the pH of the acetic acid solution pOH and [OH ] Data acetic acid acid 5.86 ? ?M PLAN What steps are needed to calculate the pOH? The sum of the pH and pOH of any solution is 14.00. Use this relationship to find the pOH of the acetic acid solution. What steps are needed to calculate [OH ]? The pOH of a solution is the negative logarithm of the hydroxide ion concentration. Therefore, calculate [OH ] using the inverse logarithm of the negative pOH. 4b Molar concentration of [OH ] convert using the relationship pOH log [OH ] 5b pOH convert using the relationship pH pOH 14.00 5a pH given pH 14.00 calculated above pOH given 14.00 pOH pH pOH 10 calculated above log[OH ] pOH [OH ] Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 266 pH Back Print Name Class Date Problem Solving continued COMPUTE 14.00 10 EVALUATE Are the units correct? Yes; there are no units on a pOH value, and [OH ] has the correct units of molarity. 8.14 5.86 7.2 8.14 10 9 M Is the number of significant figures correct? Yes; the number of significant figures is correct because the data were given to three significant figures. Is the answer reasonable? Yes; you would expect the pOH of an acid to be above 7, and the hydroxide ion concentration to be small. Practice 1. What is the pH of a solution with the following hydroxide ion concentrations? a. 1 10 5 M ans: 9.0 b. 5 10 8 M ans: 6.7 11 c. 2.90 10 M ans: 3.46 2. What are the pOH and hydroxide ion concentration of a solution with a pH of 8.92? ans: pOH 5.08, [OH ] 8.3 10 6 M 3. What are the pOH values of solutions with the following hydronium ion concentrations? a. 2.51 10 3 13 M ans: 1.40 b. 4.3 10 M ans: 11.6 c. 9.1 10 6 M ans: 8.96 d. 0.070 M ans: 12.8 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 267 pH Back Print Name Class Date Problem Solving continued Sample Problem 4 Determine the pH of a solution made by dissolving 4.50 g NaOH in a 0.400 L aqueous solution. NaOH is a strong base. Solution ANALYZE What is given in the problem? What are you asked to find? Items Identity of solute Mass of solute Molar mass of solute Volume of solution Concentration of solute Acid or base [OH ] pOH pH the mass of NaOH, and the solution volume pH Data NaOH 4.50 g 40.00 g/mol 0.400 L ?M base ?M ? ? PLAN What steps are needed to calculate the pH? First determine the concentration of the solution. Then find the concentration of hydroxide ions. Calculate the pOH of the solution, and use this to find the pH. 1 Mass of NaOH in g multiply the mass by the inverse of the molar mass of NaOH divide the amount of NaOH by the volume of the solution 2 Amount of NaOH in mol 4b Molar concentration of OH pOH NaOH dissociates to produce one OH per NaOH molecule, so [NaOH ] is equal to [OH ] 3 Molar concentration of NaOH(aq) log [OH ] 5b pOH convert using the relationship pH pOH 14 5a pH Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 268 pH Back Print Name Class Date Problem Solving continued 1 molar mass of NaOH given g NaOH 1 mol NaOH 40.00 g NaOH 1 L solution given calculated above 1 mol OH 1 mol NaOH [OH ] pOH pH 14.00 log[OH ] calculated above pOH pOH 14.00 pH calculated above COMPUTE 4.50 g NaOH 1 mol NaOH 40.00 g NaOH 14.00 EVALUATE Are the units correct? Yes; pH has no units. 1 0.400 L solution log[0.281] 0.551 0.551 13.45 1 mol OH 1 mol NaOH 0.281 M Is the number of significant figures correct? Yes; the number of significant figures is correct because the value 14.00 has two decimal places. Is the answer reasonable? Yes; NaOH is a strong base, so you would expect it to have a pH around 14. Practice 1. A solution is prepared by dissolving 3.50 g of sodium hydroxide in water and adding water until the total volume of the solution is 2.50 L. What are the OH and H3O concentrations? ans: [OH ] 0.0350 M, [H3O ] 2.86 10 13 M 2. If 1.00 L of a potassium hydroxide solution with a pH of 12.90 is diluted to 2.00 L, what is the pH of the resulting solution? ans: 13.20 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 269 pH Back Print Name Class Date Problem Solving continued Additional Problems—pH 1. Calculate the H3O and OH concentrations in the following solutions. Each is either a strong acid or a stong base. a. 0.05 M sodium hydroxide b. 0.0025 M sulfuric acid c. 0.013 M lithium hydroxide d. 0.150 M nitric acid e. 0.0200 M calcium hydroxide f. 0.390 M perchloric acid g. What is the pH of each solution in items 1a. to 1f.? 2. Calculate [H3O ] and [OH ] in a 0.160 M solution of potassium hydroxide. Assume that the solute is 100% dissociated at this concentration. 3. The pH of an aqueous solution of sodium hydroxide is 12.9. What is the molarity of the solution? 4. What is the pH of a 0.001 25 M HBr solution? If 175 mL of this solution is diluted to a total volume of 3.00 L, what is the pH of the diluted solution? 5. What is the pH of a 0.0001 M solution of NaOH? What is the pH of a 0.0005 M solution of NaOH? 6. A solution is prepared using 15.0 mL of 1.0 M HCl and 20.0 mL of 0.50 M HNO3. The final volume of the solution is 1.25 L. Answer the following questions. a. What are the [H3O ] and [OH ] in the final solution? b. What is the pH of the final solution? 7. A container is labeled 500.0 mL of 0.001 57 M nitric acid solution. A chemist finds that the container was not sealed and that some evaporation has taken place. The volume of solution is now 447.0 mL. a. What was the original pH of the solution? b. What is the pH of the solution now? 8. Calculate the hydroxide ion concentration in an aqueous solution that has a 0.000 35 M hydronium ion concentration. 9. A solution of sodium hydroxide has a pH of 12.14. If 50.00 mL of the solution is diluted to 2.000 L with water, what is the pH of the diluted solution? 10. An acetic acid solution has a pH of 4.0. What are the [H3O ] and [OH ] in this solution? 11. What is the pH of a 0.000 460 M solution of Ca(OH)2? 12. A solution of strontium hydroxide with a pH of 11.4 is to be prepared. What mass of Sr(OH)2 would be required to make 1.00 L of this solution? 13. A solution of NH3 has a pH of 11.00. What are the concentrations of hydronium and hydroxide ions in this solution? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 270 pH Back Print Name Class Date Problem Solving continued 14. Acetic acid does not completely ionize in solution. Percent ionization of a substance dissolved in water is equal to the moles of ions produced as a percentage of the moles of ions that would be produced if the substance were completely ionized. Calculate the percent ionization of acetic acid the following solutions. a. 1.0 M acetic acid solution with a pH of 2.40 b. 0.10 M acetic acid solution with a pH of 2.90 c. 0.010 M acetic acid solution, with a pH of 3.40 15. Calculate the pH of an aqueous solution that contains 5.00 g of HNO3 in 2.00 L of solution. 16. A solution of HCl has a pH of 1.50. Determine the pH of the solutions made in each of the following ways. a. 1.00 mL of the solution is diluted to 1000. mL with water. b. 25.00 mL is diluted to 200 mL with distilled water. c. 18.83 mL of the solution is diluted to 4.000 L with distilled water. d. 1.50 L is diluted to 20.0 kL with distilled water. 17. An aqueous solution contains 10 000 times more hydronium ions than hydroxide ions. What is the concentration of each ion? 18. A potassium hydroxide solution has a pH of 12.90. Enough acid is added to react with half of the OH ions present. What is the pH of the resulting solution? Assume that the products of the neutralization have no effect on pH and that the amount of additional water produced is negligible. 19. A hydrochloric acid solution has a pH of 1.70. What is the [H3O ] in this solution? Considering that HCl is a strong acid, what is the HCl concentration of the solution? 20. What is the molarity of a solution of the strong base Ca(OH)2 in a solution that has a pH of 10.80? 21. You have a 1.00 M solution of the strong acid, HCl. What is the pH of this solution? You need a solution of pH 4.00. To what volume would you dilute 1.00 L of the HCl solution to get this pH? To what volume would you dilute 1.00 L of the pH 4.00 solution to get a solution of pH 6.00? To what volume would you dilute 1.00 L of the pH 4.00 solution to get a solution of pH 8.00? 22. A solution of perchloric acid, HClO3, a strong acid, has a pH of 1.28. How many moles of NaOH would be required to react completely with the HClO3 in 1.00 L of the solution? What mass of NaOH is required? 23. A solution of the weak base NH3 has a pH of 11.90. How many moles of HCl would have to be added to 1.00 L of the ammonia to react with all of the OH ions present at pH 11.90? 24. The pH of a citric acid solution is 3.15. What are the [H3O ] and [OH ] in this solution? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 271 pH Back Print Name Class Date Skills Worksheet Problem Solving Titrations Chemists have many methods for determining the quantity of a substance present in a solution or other mixture. One common method is titration, in which a solution of known concentration reacts with a sample containing the substance of unknown quantity. There are two main requirements for making titration possible. Both substances must react quickly and completely with each other, and there must be a way of knowing when the substances have reacted in precise stoichiometric quantities. The most common titrations are acid-base titrations. These reactions are easily monitored by keeping track of pH changes with a pH meter or by choosing an indicator that changes color when the acid and base have reacted in stoichiometric quantities. This point is referred to as the equivalence point. Look at the following equation for the neutralization of KOH with HCl. KOH(aq) HCl(aq) → KCl(aq) H2O(l) Suppose you have a solution that contains 1.000 mol of KOH. All of the KOH will have reacted when 1.000 mol of HCl has been added. This is the equivalence point of this reaction. Titration calculations rely on the relationship between volume, concentration, and amount. volume of solution molarity of solution amount of solute in moles If a titration were carried out between KOH and HCl, according the reaction above, the amount in moles of KOH and HCl would be equal at the equivalence point. The following relationship applies to this system: molarityKOH volumeKOH amount of KOH in moles amount of HCl in moles molarityHCl volumeHCl amount of KOH in moles amount of HCl in moles Therefore: molarityKOH volumeKOH molarityHCl volumeHCl The following plan for solving titration problems may be applied to any acidbase titration, regardless of whether the equivalence point occurs at equivalent volumes. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 272 Titrations Back Print Name Class Date Problem Solving continued General Plan for Solving Titration Problems 1a Molarity of known acid 2a Volume of known acid 1b Molarity of known base 2b Volume of known base The product of molarity and volume in liters is the amount in moles. The product of molarity and volume in liters is the amount in moles. 3a Amount of acid in moles Convert using the mole ratio of acid to base. 3b Amount of base in moles 4a Volume of acid used in titration Divide the amount in moles by the volume in liters to compute molarity. 4b Volume of base used in titration 5a Molarity of unknown acid 5b Molarity of unknown base Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 273 Titrations Back Print Name Class Date Problem Solving continued Sample Problem 1 A titration of a 25.00 mL sample of a hydrochloric acid solution of unknown molarity reaches the equivalence point when 38.28 mL of 0.4370 M NaOH solution has been added. What is the molarity of the HCl solution? HCl(aq) NaOH(aq) 3 NaCl(aq) H2O(l) Solution ANALYZE What is given in the problem? the volume of the HCl solution titrated, and the molarity and volume of NaOH solution used in the titration figures. the molarity of the HCl solution Data 25.00 mL ?M 1 mol base: 1 mol acid 38.28 mL 0.4370 M What are you asked to find? Items Volume of acid solution Molarity of acid solution Mole ratio of base to acid in titration reaction Volume of base solution Molarity of base solution PLAN What steps are needed to calculate the molarity of the HCl solution? Use the volume and molarity of the NaOH to calculate the number of moles of NaOH that reacted. Use the mole ratio between base and acid to determine the moles of HCl that reacted. Use the volume of the acid to calculate molarity. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 274 Titrations Back Print Name Class Date Problem Solving continued Volume of NaOH in mL multiply by the conversion factor 1L 1000 mL Volume of HCl in mL multiply by the conversion factor 1L 1000 mL 1b Molarity of NaOH the product of molarity and volume is the amount of NaOH in moles 2b Volume of NaOH in L 4a Volume of HCl in L 3a Amount of HCl in mol multiply by the mole ratio mol HCl mol NaOH 3b Amount of NaOH in mol divide amount of HCl by volume to yield molarity 5a Molarity of HCl given mL NaOH given 1L 1000 mL 1L 1000 mL L NaOH L HCl mL HCl calculated above given given in balanced calculated chemical equation above L NaOH COMPUTE mol NaOH L NaOH mol HCl 1 mol NaOH 1 L HCl M HCl 38.28 mL NaOH 25.00 mL HCl 0.03828 L NaOH 0.4370 mol NaOH L NaOH 1L 1000 mL 1L 1000 mL 0.03828 L NaOH 0.02500 L HCl 1 mol HCl 1 mol NaOH 1 0.02500 L HCl 0.6691 M HCl Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 275 Titrations Back Print Name Class Date Problem Solving continued EVALUATE Are the units correct? Yes; molarity, or mol/L, was required. Is the number of significant figures correct? Yes; the number of significant figures is correct because all data were given to four significant figures. Is the answer reasonable? Yes; a larger volume of base was required than the volume of acid used. Therefore, the HCl must be more concentrated than the NaOH. Practice In each of the following problems, the acids and bases react in a mole ratio of 1mol base : 1 mol acid. 1. A student titrates a 20.00 mL sample of a solution of HBr with unknown molarity. The titration requires 20.05 mL of a 0.1819 M solution of NaOH. What is the molarity of the HBr solution? ans: 0.1824 M HBr 2. Vinegar can be assayed to determine its acetic acid content. Determine the molarity of acetic acid in a 15.00 mL sample of vinegar that requires 22.70 mL of a 0.550 M solution of NaOH to reach the equivalence point. ans: 0.832 M Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 276 Titrations Back Print Name Class Date Problem Solving continued Sample Problem 2 A 50.00 mL sample of a sodium hydroxide solution is titrated with a 1.605 M solution of sulfuric acid. The titration requires 24.09 mL of the acid solution to reach the equivalence point. What is the molarity of the base solution? H2SO4(aq) 2NaOH(aq) 3 Na2SO4(aq) 2H2O(l) Solution ANALYZE What is given in the problem? the balanced chemical equation for the acidbase reaction, the volume of the base solution, and the molarity and volume of the acid used in the titration the molarity of the sodium hydroxide solution Data 24.09 mL 1.605 M 2 mol base: 1 mol acid 50.00 mL ?M What are you asked to find? Items Volume of acid solution Molarity of acid solution Mole ratio of base to acid in titration reaction Volume of base solution Molarity of base solution PLAN What steps are needed to calculate the molarity of the NaOH solution? Use the volume and molarity of the acid to calculate the number of moles of acid that reacted. Use the mole ratio between base and acid to determine the moles of base that reacted. Use the volume of the base to calculate molarity. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 277 Titrations Back Print Name Class Date Problem Solving continued Volume of H2SO4 in mL multiply by the conversion factor 1L 1000 mL 1a Molarity of H2SO4 2a Volume of H2SO4 in L the product of molarity and volume is the amount of H2SO4 in moles Volume of NaOH in mL multiply by the conversion factor 1L 1000 mL 3a Amount of H2SO4 multiply by the in mol mole ratio mol NaOH mol H2SO4 3b Amount of NaOH in mol 4b Volume of NaOH in L divide the amount of NaOH by volume to yield molarity 5b Molarity of NaOH given mL NaOH given 1L 1000 mL 1L 1000 mL 2 mol NaOH 1 mol H2SO4 L NaOH L H2SO4 calculated above mL H2SO4 calculated above given given in balanced chemical equation L H2SO4 COMPUTE mol H2SO4 L H2SO4 1 L NaOH M NaOH 50.00 mL NaOH 24.09 mL H2SO4 0.02409 L H2SO4 1.605 mol H2SO4 L H2SO4 1L 1000 mL 1L 1000 mL 0.05000 L NaOH 0.02409 L H2SO4 2 mol NaOH 1 mol H2SO4 1 0.05000 L NaOH 1.547 M NaOH Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 278 Titrations Back Print Name Class Date Problem Solving continued EVALUATE Are the units correct? Yes; molarity, or mol/L, was required. Is the number of significant figures correct? Yes; the number of significant figures is correct because all data were given to four significant figures. Is the answer reasonable? Yes; the volume of acid required was approximately half the volume of base used. Because of the 1:2 mole ratio, the acid must be about the same as the concentration of the base, which agrees with the result obtained. Practice 1. A 20.00 mL sample of a solution of Sr(OH)2 is titrated to the equivalence point with 43.03 mL of 0.1159 M HCl. What is the molarity of the Sr(OH)2 solution? ans: 0.1247 M Sr(OH)2 2. A 35.00 mL sample of ammonia solution is titrated to the equivalence point with 54.95 mL of a 0.400 M sulfuric acid solution. What is the molarity of the ammonia solution? ans: 1.26 M NH3 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 279 Titrations Back Print Name Class Date Problem Solving continued Sample Problem 3 A supply of NaOH is known to contain the contaminants NaCl and MgCl2. A 4.955 g sample of this material is dissolved and diluted to 500.00 mL with water. A 20.00 mL sample of this solution is titrated with 22.26 mL of a 0.1989 M solution of HCl. What percentage of the original sample is NaOH? Assume that none of the contaminants react with HCl. Solution ANALYZE What is given in the problem? the mass of the original solute sample, the volume of the solution of the sample, the volume of the sample taken for titration, the molarity of the acid solution, and the volume of the acid solution used in the titration the percentage by mass of NaOH in the original sample Data 22.26 mL 0.1989 M ? 20.00 mL ? mol NaOH 500.00 mL ? mol NaOH 4.955 g impure NaOH ? g NaOH ?% NaOH What are you asked to find? Items Volume of acid solution Molarity of acid solution Mole ratio of base to acid in titration reaction Volume of base solution titrated Moles of base in solution titrated Volume of original sample solution Moles of base in original sample Mass of original sample Mass of base in original sample Percentage of NaOH in original sample PLAN What steps are needed to calculate the concentration of NaOH in the sample? Determine the balanced chemical equation for the titration reaction. Use the volume and molarity of the HCl to calculate the moles of HCl that reacted. Use the mole ratio between base and acid to determine the amount of NaOH that reacted. Divide by the volume titrated to obtain the concentration of NaOH. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 280 Titrations Back Print Name Class Date Problem Solving continued What steps are needed to calculate the percentage of NaOH in the sample? Convert the concentration of NaOH to the amount of NaOH in the original sample by multiplying the concentration by the total volume. Convert amount of NaOH to mass NaOH by using the molar mass of NaOH. Use the mass of NaOH and the mass of the sample to calculate the percentage of NaOH. You must first determine the equation for titration reaction. HCl(aq) NaOH(aq) → NaCl(aq) H2O(l) Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 281 Titrations Back Print Name Class Date Problem Solving continued Volume of HCl in mL multiply by the conversion factor 1L 1000 mL 1a Molarity of HCl 2a Volume of HCl in L the product of molarity and volume is the amount of HCl in moles Volume of NaOH in mL multiply by the conversion factor 1L 1000 mL 3a Amount of HCl in mol multiply by the mole ratio mol NaOH mol HCl 3b 4b Amount of Volume of NaOH NaOH used in in mol titration in L divide amount of NaOH by volume to yield molarity 5b Volume of original solution in L Molarity of NaOH Amount of NaOH in the original solution in mol multiply by the molar mass of NaOH Mass of NaOH in the original solution in g divide by the total mass of the solute and multiply by 100 Percentage of NaOH in original solution Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 282 Titrations Back Print Name Class Date Problem Solving continued given mL NaOHtitrated given 1L 1000 mL 1L 1000 mL L NaOHtitrated L HCl mL HCl calculated above given given in balanced chemical equation L HCl mol HCl L HCl given 1 mol NaOH 1 mol HCl 1L 1000 mL 1 L NaOHtitrated M NaOH mL NaOHoriginal calculated above calculated above L NaOHoriginal molar mass of NaOH mol NaOH L NaOH calculated above L NaOHoriginal 40.00 g NaOH mol NaOH g NaOHoriginal g NaOHoriginal g solute given 100 percentage of NaOH in solute COMPUTE 20.00 mL NaOHtitrated 22.26 mL HCl 0.02226 L HCl 0.1989 mol HCl L HCl 1L 1000 mL 1L 1000 mL 1 mol NaOH 1 mol HCl 1 0.02000 L NaOH 500.00 mL NaOHoriginal 0.2214 mol NaOH L NaOH 1L 1000 mL 0.2214 M NaOH 0.02000 L NaOHtitrated 0.02226 L HCl 0.500 00 L NaOHoriginal 40.00 g NaOH mol NaOH 89.35% NaOH 4.428 g NaOH 0.500 00 L NaOH 100 4.428 g NaOH 4.955 g solute EVALUATE Are the units correct? Yes; units canceled to give percentage of NaOH in sample. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 283 Titrations Back Print Name Class Date Problem Solving continued Is the number of significant figures correct? Yes; the number of significant figures is correct because all data were given to four significant figures. Is the answer reasonable? Yes; the calculation can be approximated as (0.02 400/5 80, which is close to the calculated result. 0.2 25 40 100)/5 Practice In the problems below, assume that impurities are not acidic or basic and that they do not react in an acid-base titration. 1. A supply of glacial acetic acid has absorbed water from the air. It must be assayed to determine the actual percentage of acetic acid. 2.000 g of the acid is diluted to 100.00 mL, and 20.00 mL is titrated with a solution of sodium hydroxide. The base solution has a concentration of 0.218 M, and 28.25 mL is used in the titration. Calculate the percentage of acetic acid in the original sample. Write the titration equation to get the mole ratio. ans: 92.5% acetic acid 2. A shipment of crude sodium carbonate must be assayed for its Na2CO3 content. You receive a small jar containing a sample from the shipment and weigh out 9.709 g into a flask, where it is dissolved in water and diluted to 1.0000 L with distilled water. A 10.00 mL sample is taken from the flask and titrated to the equivalence point with 16.90 mL of a 0.1022 M HCl solution. Determine the percentage of Na2CO3 in the sample. Write the titration equation to get the mole ratio. ans: 94.28% Na2CO3 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 284 Titrations Back Print Name Class Date Problem Solving continued Additional Problems—Titrations 1. A 50.00 mL sample of a potassium hydroxide is titrated with a 0.8186 M HCl solution. The titration requires 27.87 mL of the HCl solution to reach the equivalence point. What is the molarity of the KOH solution? 2. A 15.00 mL sample of acetic acid is titrated with 34.13 mL of 0.9940 M NaOH. Determine the molarity of the acetic acid. 3. A 12.00 mL sample of an ammonia solution is titrated with 1.499 M HNO3 solution. A total of 19.48 mL of acid is required to reach the equivalence point. What is the molarity of the ammonia solution? 4. A certain acid and base react in a 1:1 ratio. a. If the acid and base solutions are of equal concentration, what volume of acid will titrate a 20.00 mL sample of the base? b. If the acid is twice as concentrated as the base, what volume of acid will be required to titrate 20.00 mL of the base? c. How much acid will be required if the base is four times as concentrated as the acid, and 20.00 mL of base is used? 5. A 10.00 mL sample of a solution of hydrofluoric acid, HF, is diluted to 500.00 mL. A 20.00 mL sample of the diluted solution requires 13.51 mL of a 0.1500 M NaOH solution to be titrated to the equivalence point. What is the molarity of the original HF solution? 6. A solution of oxalic acid, a diprotic acid, is used to titrate a 16.22 mL sample of a 0.5030 M KOH solution. If the titration requires 18.41 mL of the oxalic acid solution, what is its molarity? 7. A H2SO4 solution of unknown molarity is titrated with a 1.209 M NaOH solution. The titration requires 42.27 mL of the NaOH solution to reach the equivalent point with 25.00 mL of the H2SO4 solution. What is the molarity of the acid solution? 8. Potassium hydrogen phthalate, KHC8H4O4, is a solid acidic substance that reacts in a 1:1 mole ratio with bases that have one hydroxide ion. Suppose that 0.7025 g of KHC8H4O4 is titrated to the equivalence point by 20.18 mL of a KOH solution. What is the molarity of the KOH solution? 9. A solution of citric acid, a triprotic acid, is titrated with a sodium hydroxide solution. A 20.00 mL sample of the citric acid solution requires 17.03 mL of a 2.025 M solution of NaOH to reach the equivalence point. What is the molarity of the acid solution? 10. A flask contains 41.04 mL of a solution of potassium hydroxide. The solution is titrated and reaches an equivalence point when 21.65 mL of a 0.6515 M solution of HNO3 is added. Calculate the molarity of the base solution. 11. A bottle is labeled 2.00 M H2SO4. You decide to titrate a 20.00 mL sample with a 1.85 M NaOH solution. What volume of NaOH solution would you expect to use if the label is correct? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 285 Titrations Back Print Name Class Date Problem Solving continued 12. What volume of a 0.5200 M solution of H2SO4 would be needed to titrate 100.00 mL of a 0.1225 M solution of Sr(OH)2? 13. A sample of a crude grade of KOH is sent to the lab to be tested for KOH content. A 4.005 g sample is dissolved and diluted to 200.00 mL with water. A 25.00 mL sample of the solution is titrated with a 0.4388 M HCl solution and requires 19.93 mL to reach the equivalence point. How many moles of KOH were in the 4.005 g sample? What mass of KOH is this? What is the percent KOH in the crude material? 14. What mass of magnesium hydroxide would be required for the magnesium hydroxide to react to the equivalence point with 558 mL of 3.18 M hydrochloric acid? 15. An ammonia solution of unknown concentration is titrated with a solution of hydrochloric acid. The HCl solution is 1.25 M, and 5.19 mL are required to titrate 12.61 mL of the ammonia solution. What is the molarity of the ammonia solution? 16. What volume of 2.811 M oxalic acid solution is needed to react to the equivalence point with a 5.090 g sample of material that is 92.10% NaOH? Oxalic acid is a diprotic acid. 17. Standard solutions of accurately known concentration are available in most laboratories. These solutions are used to titrate other solutions to determine their concentrations. Once the concentration of the other solutions are accurately known, they may be used to titrate solutions of unknowns. The molarity of a solution of HCl is determined by titrating the solution with an accurately known solution of Ba(OH)2, which has a molar concentration of 0.1529 M. A volume of 43.09 mL of the Ba(OH)2 solution titrates 26.06 mL of the acid solution. The acid solution is in turn used to titrate 15.00 mL of a solution of rubidium hydroxide. The titration requires 27.05 mL of the acid. a. What is the molarity of the HCl solution? b. What is the molarity of the RbOH solution? 18. A truck containing 2800 kg of a 6.0 M hydrochloric acid has been in an accident and is in danger of spilling its load. What mass of Ca(OH)2 should be sent to the scene in order to neutralize all of the acid in case the tank bursts? The density of the 6.0 M HCl solution is 1.10 g/mL. 19. A 1.00 mL sample of a fairly concentrated nitric acid solution is diluted to 200.00 mL. A 10.00 mL sample of the diluted solution requires 23.94 mL of a 0.0177 M solution of Ba(OH)2 to be titrated to the equivalence point. Determine the molarity of the original nitric acid solution. 20. What volume of 4.494 M H2SO4 solution would be required to react to the equivalence point with 7.2280 g of LiOH(s)? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 286 Titrations Back Print Name Class Date Skills Worksheet Problem Solving Equilibrium of Acids and Bases, Ka and Kb The ionization of a strong acid, such as HCl, will proceed to completion in reasonably dilute solutions. This process is written as follows. HCl(g) H2O(l) → H3O (aq) Cl (aq) goes 100% to the right; NOT an equilibrium process. However, when weak acids and weak bases dissolve, they only partially ionize, resulting in an equilibrium between ionic and molecular forms. The following equation shows the equilibrium process that occurs when hydrogen fluoride, a weak acid, dissolves in water. HF(g) H2O(l) ^ H3O (aq) F (aq) does not go 100% to the right; IS an equilibrium process Most weak acids react in this way, that is, by donating a proton to a water molecule to form a H3O ion. The weak base ammonia establishes the following equilibrium in water. NH3(g) H2O(l) ^ NH4 (aq) OH (aq) Most weak bases react in this way, that is, by accepting a proton from a water molecule to leave an OH ion. You learned to write an equilibrium expression to solve for K, the equilibrium constant of a reaction. As with equilibrium reactions, equilibrium constants can be calculated for the ionization and dissociation processes shown above. The equilibrium constants indicate how far the equilibrium goes toward the ionic “products.” Recall that the concentration of each reaction component is raised to the power of its coefficient in the balanced equation. These concentration terms are arranged in a fraction with products in the numerator and reactants in the denominator. The equilibrium constants calculated for acid ionization reactions are called acid ionization constants and have the symbol Ka. The equilibrium constants calculated for base dissociation reactions are called base dissociation constants and have the symbol Kb. From the diagram on the next page and the problems in this chapter, you will see how these constants are calculated and how they relate to concentrations of the reactants and products and to pH. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 287 Equilibrium of Acids and Bases, Ka and Kb Back Print Name Class Date Problem Solving continued General Plan for Solving Ka and Kb Problems 1a pH pH log[H3O 1b pH pOH 14 pOH log[OH pOH 2a Molar concentration of H3O Calculate the remaining concentrations. Convert using the relationship Kw [H3O ][OH Kw 1 10 14 2b Molar concentration of OH Calculate the remaining concentrations. 3a Balanced chemical equation HA H2O º H3O A Set up the equilibrium expression and substitute values. 3b Balanced chemical equation H2O B º OH BH Set up the equilibrium expression and substitute values. 4a Ka [H3O ][A ] [HA] 4b Kb [OH ][BH ] [B] Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 288 Equilibrium of Acids and Bases, Ka and Kb Back Print Name Class Date Problem Solving continued Sample Problem 1 The hydronium ion concentration of a 0.500 M solution of HF at 25°C is found to be 0.0185 M. Calculate the ionization constant for HF at this temperature. Solution ANALYZE What is given in the problem? the molarity of the acid solution, the equilibrium concentration of hydronium ions, and the temperature the acid ionization constant, Ka Data 0.0185 M ?M 0.500 M ?M ? What are you asked to find? Items Molar concentration of H3O at equilibrium Molar concentration of F at equilibrium Initial molar concentration of HF Molar concentration of HF at equilibrium Acid ionization constant, Ka, of HF PLAN What steps are needed to calculate the acid ionization constant for HF? Write the equation for the ionization reaction. Set up the equilibrium expression for the ionization of HF in water. Determine the concentrations of all components at equilbrium, and calculate Ka. 2a [H3O calculate the remaining concentrations 3a HF H2O º H3O F Ka 4a [H3O ][F ] [HF] use the balanced equation to relate the unknown quantities to known quantities, and substitute these values into the Ka expression Ka [H3O ][F ] [HF]initial [H3O ] Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 289 Equilibrium of Acids and Bases, Ka and Kb Back Print Name Class Date Problem Solving continued Write the balanced chemical equation for the ionization of HF in aqueous solution. HF(aq) H2O(l) → H3O (aq) F (aq) Write the acid ionization expression for Ka. Remember that pure substances are not included in equilibrium expressions. For this reason, [H2O] does not appear in the expression for Ka. Ka [H3O ][F ] [HF] From the balanced chemical equation, you can see that one HF molecule ionizes to give one fluoride ion and one hydronium ion. Therefore, at equilibrium, the concentration of F must equal the concentration of H3O . [F ] [H3O ] 0.0185 M When the HF ionizes in solution, the HF concentration decreases from its initial value. The amount by which it decreases is equal to the concentration of either the fluoride ion or the hydronium ion. [HF]equilibrium Set up the equilibrium expression. given calculated above [HF]initial [H3O ]equilibrium Ka [H3O ] [F ] [HF]initial [H3O ] given given COMPUTE Ka EVALUATE Are the units correct? Yes; the acid ionization constant has no units. [0.0185][0.0185] [0.500] [0.0185] 7.11 10 4 Is the number of significant figures correct? Yes; the number of significant figures is correct because data values have as few as three significant figures. Is the answer reasonable? Yes; the calculation can be approximated as (0.02 of the same magnitude as the calculated result. 0.02)/0.5 0.0008, which is PRACTICE 1. At 25°C, a 0.025 M solution of formic acid, HCOOH, is found to have a hydronium ion concentration of 2.03 10 3 M. Calculate the ionization constant of formic acid. ans: Ka 1.8 10 4 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 290 Equilibrium of Acids and Bases, Ka and Kb Back Print Name Class Date Problem Solving continued Sample Problem 2 At 25°C, the pH of a 0.315 M solution of nitrous acid, HNO2, is 1.93. Calculate the Ka of nitrous acid at this temperature. Solution ANALYZE What is given in the problem? What are you asked to find? Items pH of solution Molar concentration of H3O at equilibrium Molar concentration of NO2 at equilibrium Initial molar concentration of HNO2 Molar concentration of HNO2 at equilibrium Acid ionization constant, Ka, of HNO2 the pH of the acid solution and the original concentration of HNO2 the acid ionization constant, Ka Data 1.93 ?M ?M 0.315 M ?M ? PLAN What steps are needed to calculate the acid ionization constant for HNO2? Determine the H3O concentration from the pH. Write the equation for the ionization reaction. Set up the equilibrium expression. Determine all equilibrium concentrations and calculate Ka. 1a pH log[H3O rearrange to solve for [H3O ] 2a [H3O ] 10 pH 3a HNO2 H2O º H3O NO2 Ka 4a [H3O ][NO2 ] [HNO2] use the balanced equation to relate the unknown quantities to known quantities, and substitute these values into the Ka expression Ka Copyright © by Holt, Rinehart and Winston. All rights reserved. [H3O ][NO2 ] [HNO2]initial [H3O ] Holt ChemFile: Problem-Solving Workbook 291 Equilibrium of Acids and Bases, Ka and Kb Back Print Name Class Date Problem Solving continued Calculate the H3O concentration from the pH. pH log [H3O ] 10 given pH [H3O ] Write the balanced chemical equation for the ionization of HNO2 in aqueous solution. HNO2(aq) H2O(l) ^ H3O (aq) [NO2 ][H3O ] [HNO2] NO2 (aq) Write the mathematical equation to compute Ka. Ka Because each HNO2 molecule dissociates into one hydronium ion and one nitrite ion, [NO2 ] [H3O ] the HNO2 concentration at equilibrium will be its initial concentration minus any HNO2 that has ionized. The amount ionized will equal the concentration of H3O or NO2 . [HNO2]equilibrium [HNO2]initial calculated above [H3O ] Substitute known values into the Ka expression. calculated above Ka [H3O ] [NO2 ] [HNO2]initial [H3O ] given calculated above COMPUTE [H3O ] Ka EVALUATE Are the units correct? Yes; the acid ionization constant has no units. 10 1.93 1.2 10 2 [1.2 10 2][1.2 10 2] = 4.4 [0.315] [1.2 10 2] 10 4 Is the number of significant figures correct? Yes; the number of significant figures is correct because the pH was given to two decimal places. Is the answer reasonable? Yes; the calculation can be approximated as (0.01 0.01)/0.3 is of the same order of magnitude as the calculated result. 3 10 4 , which Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 292 Equilibrium of Acids and Bases, Ka and Kb Back Print Name Class Date Problem Solving continued Practice 1. The pH of a 0.400 M solution of iodic acid, HIO3, is 0.726 at 25°C. What is the Ka at this temperature? ans: Ka 0.167 2. The pH of a 0.150 M solution of hypochlorous acid, HClO, is found to be 4.55 at 25°C. Calculate the Ka for HClO at this temperature. ans: Ka 5.2 10 9 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 293 Equilibrium of Acids and Bases, Ka and Kb Back Print Name Class Date Problem Solving continued Sample Problem 3 A 0.450 M ammonia solution has a pH of 11.45 at 25°C. Calculate the [H3O ] and [OH ] of the solution, and determine the base dissociation constant, Kb, of ammonia. Solution ANALYZE What is given in the problem? What are you asked to find? the pH of the base solution, the original concentration of NH3, and the temperature [H3O ], [OH ], and the base dissociation constant, Kb Data 11.45 ?M ?M ?M 0.450 M ?M ? Items pH of solution Molar concentration of H3O at equilibrium Molar concentration of OH at equilibrium Molar concentration of NH4 at equilibrium Initial molar concentration of NH3 Molar concentration of NH3 at equilibrium Base dissociation constant, Kb, of NH3 PLAN What steps are needed to calculate the base dissociation constant for NH3? Determine the H3O concentration from the pH. Calculate the OH concentration. Write the balanced chemical equation for the dissociation reaction. Write the mathematical equation for Kb, substitute values, and calculate. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 294 Equilibrium of Acids and Bases, Ka and Kb Back Print Name Class Date Problem Solving continued 1a pH log[H3O rearrange to solve for [H3O ] 2a 2b 10 pH [H3O ] convert using the relationship Kw [OH ][H3O ] [OH ] Kw H3O [ 3b NH3 H2O º NH4 OH 4b Kb [NH4][OH ] [NH3] use the balanced equation to relate the unknown quantities to known quantities, and substitute these values into the Kb expression Kb [NH4][OH ] NH3]initial [OH [ Determine [H3O ]. [H3O ] Determine [OH ]. Kw [H3O ][OH ] Kw [H3O ] 10 given pH [OH ] calculated above Write the equilibrium expression for the ionization reaction. NH3(aq) H2O(l) ^ NH4 (aq) Kb [OH ][NH4 ] [NH3] OH (aq) Because 1 mol of NH3 reacts with 1 mol of H2O to produce 1 mol of NH4 and 1 mol of OH : [OH ] [NH4 ] Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 295 Equilibrium of Acids and Bases, Ka and Kb Back Print Name Class Date Problem Solving continued The equilibrium concentration of NH3 will be the initial concentration minus any NH3 that has reacted. The amount reacted will equal the concentration of NH4 or OH . [NH3]equilibrium [NH3]initial calculated above [OH ] Substitute known quantities into the Kb expression. [OH ][NH4 ] [NH3]initial [OH ] given Kb calculated above COMPUTE [H3O ] [OH ] Kb EVALUATE Are the units correct? Yes; the equilibrium constant has no units. 10 10 3.5 11.45 14 3.5 12 10 12 3 10 2.9 10 1.9 [2.9 10 3][2.9 10 3] [0.450] [2.9 10 3] 10 5 Is the number of significant figures correct? Yes; the number of significant figures is correct because the pH was given to two decimal places. Is the answer reasonable? Yes. The calculation can be approximated as (0.003 1.8 10 5. 0.003)/0.5 0.00009/5 Practice 1. The compound propylamine, CH3CH2CH2NH2, is a weak base. At equilibrium, a 0.039 M solution of propylamine has an OH concentration of 3.74 10 3 M. Calculate the pH of this solution and Kb for propylamine. ans: pH 11.573; Kb 4.0 10 4 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 296 Equilibrium of Acids and Bases, Ka and Kb Back Print Name Class Date Problem Solving continued Sample Problem 4 A 1.00 M iodic acid, HIO3, solution has an acid ionization constant of 0.169 at 25°C. Calculate the hydronium ion concentration of the solution at this temperature. Solution ANALYZE What is given in the problem? What are you asked to find? Items Molar concentration of H3O at equilibrium Initial molar concentration of HIO3 Molar concentration of HIO3 at equilibrium Acid ionization constant, Ka, of HIO3 the original concentration of HIO3, the temperature, and Ka [H3O ] Data ?M 1.00 M ?M 0.169 PLAN What steps are needed to calculate the acid ionization constant for HIO3? Write the balanced chemical equation for the ionization reaction. Write the equation for Ka. Substitute x for the unknown values. Rearrange the Ka expression so that a quadratic equation remains. Substitute known values, and solve for x. Write the equilibrium equation for the ionization reaction. HIO3(aq) H2O(l) ^ H3O (aq) [H3O ][IO3 ] [HIO3] IO3 (aq) Use the chemical equation to write an expression for Ka. Ka Since 1 mol of HIO3 reacts with 1 mol of H2O to produce 1 mol of H3O and 1 mol of IO3 : [H3O ] [IO3 ] Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 297 Equilibrium of Acids and Bases, Ka and Kb Back Print Name Class Date Problem Solving continued The equilibrium concentration of HIO3 will be the initial concentration minus any HIO3 that has reacted. The amount reacted will equal the concentration of H3O or IO3 . [HIO3]equilibrium [HIO3]equilibrium given [HIO3]initial [HIO3]initial x2 x [H3O ] x Because [H3O ] is an unknown quantity, substitute the variable x for it to solve. Ka [HIO3]initial given COMPUTE 0.169 0.169(1.00 0.169 Rearrange the above equation. x2 0.169x ax2 where a 1, b 0.169, c 0.169 b2 2a bx 0.169 c 0 0 Notice that this equation fits the form for a general quadratic equation. x2 1.00 x x) x2 x2 0.169x Use the formula for solving quadratic equations. x b 4ac Substitute the values given above into the quadratic equation, and solve for x. x 0.169 x EVALUATE Are the units correct? Yes; the value has units of mol/L, or M. (0.169)2 2(1) 0.335 M 4(1)( 0.169) Is the number of significant figures correct? Yes; the number of significant figures is correct because the data values had three significant figures. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 298 Equilibrium of Acids and Bases, Ka and Kb Back Print Name Class Date Problem Solving continued Is the answer reasonable? Yes; substituting the calculated [H3O ] back into the equation for Ka yields the given value. Practice 1. The Ka of nitrous acid is 4.6 10 4 at 25°C. Calculate the [H3O ] of a 0.0450 M nitrous acid solution. ans: 4.4 10 3 M Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 299 Equilibrium of Acids and Bases, Ka and Kb Back Print Name Class Date Problem Solving continued Additional Problems—Equilibrium of Acids and Bases 1. Hydrazoic acid, HN3, is a weak acid. The [H3O ] of a 0.102 M solution of hydrazoic acid is 1.39 10 3 M. Determine the pH of this solution, and calculate Ka at 25°C for HN3. 2. Bromoacetic acid, BrCH2COOH, is a moderately weak acid. A 0.200 M solution of bromoacetic acid has a H3O concentration of 0.0192 M. Determine the pH of this solution and the Ka of bromoacetic acid at 25°C. 3. A base, B, dissociates in water according to the following equation: B H2O ^ BH OH Complete the following table for base solutions with the characteristics given. Initial [B] a. 0.400 M b. 0.005 50 M c. 0.0350 M d. ? M [B] at equilibrium NA ?M ?M 0.006 28 M [OH ] 2.70 8.45 ?M 0.000 92 M 10 10 4 4 Kb M M ? ? ? ? [H3O ] ?M NA ?M NA ? ? pH 11.29 ? 4. The solubility of benzoic acid, C6H5COOH, in water at 25°C is 2.9 g/L. The pH of this saturated solution is 2.92. Determine Ka at 25°C for benzoic acid. (Hint: first calculate the initial concentration of benzoic acid.) 5. A 0.006 50 M solution of ethanolamine, H2NCH2CH2OH, has a pH of 10.64 at 25°C. Calculate the Kb of ethanolamine. What concentration of undissociated ethanolamine remains at equilibrium? 6. The weak acid hydrogen selenide, H2Se, has two hydrogen atoms that can form hydronium ions. The second ionization is so small that the concentration of the resulting H3O is insignificant. If the [H3O ] of a 0.060 M solution of H2Se is 2.72 10 3 M at 25°C, what is the Ka of the first ionization? 7. Pyridine, C5H5N, is a very weak base. Its Kb at 25°C is 1.78 x 10–9. Calculate the [OH ] and pH of a 0.140 M solution. Assume that the concentration of pyridine at equilibrium is equal to its initial concentration because so little pyridine is dissociated. 8. A solution of a monoprotic acid, HA, at equilibrium is found to have a 0.0208 M concentration of nonionized acid. The pH of the acid solution is 2.17. Calculate the initial acid concentration and Ka for this acid. 9. Pyruvic acid, CH3COCOOH, is an important intermediate in the metabolism of carbohydrates in the cells of the body. A solution made by dissolving 438 mg of pyruvic acid in 10.00 mL of water is found to have a pH of 1.34 at 25°C. Calculate Ka for pyruvic acid. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 300 Equilibrium of Acids and Bases, Ka and Kb Back Print Name Class Date Problem Solving continued 10. The [H3O ] of a solution of acetoacetic acid, CH3COCH2COOH, is 4.38 10 3 M at 25°C. The concentration of nonionized acid is 0.0731 M at equilibrium. Calculate Ka for acetoacetic acid at 25°C. 11. The Ka of 2-chloropropanoic acid, CH3CHClCOOH, is 1.48 10 3. Calculate the [H3O ] and the pH of a 0.116 M solution of 2-chloropropionic acid. Let x [H3O ]. The degree of ionization of the acid is too large to ignore. If your set up is correct, you will have a quadratic equation to solve. 12. Sulfuric acid ionizes in two steps in water solution. For the first ionization shown in the following equation, the Ka is so large that in moderately dilute solution the ionization can be considered 100%. H2SO4 HSO4 H2O → H3O H2O ^ H3O HSO4 1.3 SO4 2 The second ionization is fairly strong, and Ka 10 2 . Calculate the total [H3O ] and pH of a 0.0788 M H2SO4 solution. Hint: If the first ionization is 100%, what will [HSO4 ] and [H3O ] be? Remember to account for the already existing concentration of H3O in the second ionization. Let x [SO42 ]. 13. The hydronium ion concentration of a 0.100 M solution of cyanic acid, HOCN, is found to be 5.74 10 3 M at 25°C. Calculate the ionization constant of cyanic acid. What is the pH of this solution? 14. A solution of hydrogen cyanide, HCN, has a 0.025 M concentration. The cyanide ion concentration is found to be 3.16 10 6 M. a. What is the hydronium ion concentration of this solution? b. What is the pH of this solution? c. What is the concentration of nonionized HCN in the solution? Be sure to use the correct number of significant figures. d. Calculate the ionization constant of HCN. e. How would you characterize the strength of HCN as an acid? f. Determine the [H3O ] for a 0.085 M solution of HCN. 15. A 1.20 M solution of dichloroacetic acid, CCl2HCOOH, at 25°C has a hydronium ion concentration of 0.182 M. a. What is the pH of this solution? b. What is the Ka of dichloroacetic acid at 25°C? c. What is the concentration of nonionized dichloroacetic acid in this solution? d. What can you say about the strength of dichloroacetic acid? Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 301 Equilibrium of Acids and Bases, Ka and Kb Back Print Name Class Date Problem Solving continued 16. Phenol, C6H5OH, is a very weak acid. The pH of a 0.215 M solution of phenol at 25°C is found to be 5.61. Calculate the Ka for phenol. 17. A solution of the simplest amino acid, glycine (NH2CH2COOH), is prepared by dissolving 3.75 g in 250.0 mL of water at 25°C. The pH of this solution is found to be 0.890. a. Calculate the molarity of the glycine solution. b. Calculate the Ka for glycine. 18. Trimethylamine, (CH3)3N, dissociates in water the same way that NH3 does— by accepting a proton from a water molecule. The [OH ] of a 0.0750 M solution of trimethylamine at 25°C is 2.32 10 3 M. Calculate the pH of this solution and the Kb of trimethylamine. 19. Dimethylamine, (CH3)2NH, is a weak base similar to the trimethylamine in item 18. A 5.00 10 3 M solution of dimethylamine has a pH of 11.20 at 25°C. Calculate the Kb of dimethylamine. Compare this Kb with the Kb for trimethylamine that you calculated in item 18. Which substance is the stronger base? 20. Hydrazine dissociates in water solution according to the following equations: H2NNH2 H2NNH3 (aq) H2O(l) ^ NH2NNH3 (aq) H2O(l) ^ NH3NNH2 (aq) 3 16 OH (aq) OH (aq) The Kb of this second dissociation is 8.9 10 , so it contributes almost no hydroxide ions in solution and can be ignored here. a. The pH of a 0.120 M solution of hydrazine at 25°C is 10.50. Calculate Kb for the first ionization of hydrazine. Assume that the original concentration of H2NNH2 does not change. b. Make the same assumption as you did in (a) and calculate the [OH ] of a 0.020 M solution. c. Calculate the pH of the solution in (b). Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 302 Equilibrium of Acids and Bases, Ka and Kb Back Print Name Class Date Skills Worksheet Problem Solving Redox Equations The feature that distinguishes redox reactions from other types of reactions is that elements change oxidation state by gaining or losing electrons. Compare the equations for the following two reactions: (1) KBr(aq) AgNO3(aq) → AgBr(s) Cl2( g) → 2KCl(aq) KNO3(aq) Br2(l) (2) 2KBr(aq) Equation 1 represents the combining of the salts potassium bromide solution and a silver nitrate solution to form a precipitate of insoluble silver bromide, leaving potassium nitrate in solution. The only change that occurs is that ions trade places, forming an insoluble compound. This reaction is a typical doubledisplacement reaction. It is driven by the removal of Ag and Br from solution in the form of a precipitate. You will recognize that Equation 2 is a single-displacement reaction in which chlorine atoms replace bromine atoms in the salt KBr. Although it is not complex, Equation 2 differs from Equation 1 in a fundamental way. In order for chlorine to replace bromine, the uncharged atoms of elemental chlorine must change into chloride ions, each having a 1 charge. Also, bromide ions with a 1 charge must change into uncharged bromine atoms. The K is a spectator ion that doesn’t participate in the process. In fact, it could be Na , Ca2 , Fe3 , H , or any other stable cation. The loss of electrons by bromine is oxidation, and the gain of electrons by chlorine is reduction. Formation of the chloride ions and the bromine molecule involves the complete transfer of two electrons. The two chlorine atoms gain two electrons, and two bromide ions lose two electrons. Oxidation and reduction can involve the partial transfer of electrons as well as the complete transfer seen in the preceding example. The oxidation number of an atom is not synonymous with the charge on that atom, so a change in oxidation number does not require a change in actual charge. Take the example of the following half-reaction: H2C2O4(aq) → 2CO2( g) 2H (aq) 2e Carbon changes oxidation state from 3 to 4. Carbon is oxidized even though it is not ionized. The potassium bromide–chlorine reaction is simple, but many redox reactions are not. In this worksheet, you will practice the art of balancing redox equations and try your hand at more-complex ones. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 303 Redox Equations Back Print Name Class Date Problem Solving continued There are seven simple steps to balancing redox equations. You will find these steps in the General Plan for Balancing Redox Equations. General Plan for Balancing Redox Equations 1 Write the unbalanced formula equation if it is not given. List formulas for any ionic substances as their individual ions, and write a total ionic equation. 2 Assign oxidation numbers to each element. Then rewrite the equation, leaving out any ions or molecules whose elements do not change oxidation state during the reaction. 3 Write the half-reaction for reduction. You must decide which element of the ions and molecules left after item 2 is reduced. Once you have written the half-reaction, you must balance it for charge and mass. 4 Write the half-reaction for oxidation. You must decide which element of the ions and molecules left after item 2 is oxidized. Once you have written the half-reaction, you must balance it for charge and mass. 5 Adjust the coefficients of the two half-reactions so that the same number of electrons are gained in reduction as are lost in oxidation. 6 Add the two half-reactions together. Cancel out anything common to both sides of the new equation. Note that the electrons should always cancel out of the total equation. 7 Combine ions to form the compounds shown in the original formula equation. Check to ensure that all other ions and atoms balance. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 304 Redox Equations Back Print Name Class Date Problem Solving continued REACTIONS IN ACIDIC SOLUTION Sample Problem 1 Write a balanced redox equation for the reaction of hydrochloric acid with nitric acid to produce aqueous hypochlorous acid and nitrogen monoxide. Solution ANALYZE What is given in the problem? What are you asked to find? Items Reactants Products Solution type Oxidized species Reduced species Balanced equation the reactants and products of a redox reaction the balanced redox reaction Data HCl, HNO3 HClO, NO acidic ? ? ? PLAN What steps are needed to balance the redox equation? 1. Write the unbalanced formula equation followed by the ionic equation. 2. Assign oxidation numbers to each element. Delete any ion or molecule in which there is no change in oxidation state. 3. Write the half-reaction for reduction, and balance the mass and charge. H and H2O may be added to either side of the equation to balance mass. 4. Repeat step 3 for the oxidation half-reaction. 5. Adjust the coefficients so that the number of electrons lost equals the number of electrons gained. 6. Combine the half-reactions, and cancel anything common to both sides of the equation. 7. Combine ions to change the equation back to its original form, and check the balance of everything. COMPUTE 1. Write the formula equation. HCl(aq) H Cl HNO3(aq) → HClO(aq) H NO3 → H ClO NO( g) NO Write the total ionic equation. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 305 Redox Equations Back Print Name Class Date Problem Solving continued 2. Assign oxidation numbers to each element. 1 1 1 H Cl 1 H 5 NO3 → H 1 52 1 21 22 OCl 2 NO Delete any ion or molecule in which there is no change in oxidation state. Cl NO3 → OCl 5 2 NO 3. Write the half-reaction for reduction. NO3 → NO Balance the mass by adding H and H2O. 4H NO3 → 5 2 O 2H2O Balance the charge by adding electrons to the side with the higher positive charge. 5 4H NO3 1 3e → 1 2 O 2H2O 4. Write the half-reaction for oxidation. Cl → OCl Balance the mass by adding H and H2O. 1 Cl 1 H2O → OCl H2O → OCl 1 1 2H Balance charge by adding electrons to the side with the higher positive charge. Cl 2H 2e 5. Multiply by factors so that the number of electrons lost equals the number of electrons gained. 2e are lost in oxidation; 3e are gained in reduction. Therefore, to get 6e on both sides, calculate as follows: 2 3 2 3 3Cl 3H2O 2NO3 [4H [Cl [4H [Cl 8H 2 NO3 3e → NO 2H 2H2O] 2e ] 2H2O] e] 6e 2NO 4H2O H2O → OCl 6. Combine the half-reactions. NO3 3e → NO H2O → OCl 2H 6e → 3OCl 6e → 3OCl 6H Cancel out anything common to both sides of the equation. 3Cl 3H2O 2NO3 8H 6H 6e 2NO 4H2O 7. Combine ions to change the equation back to its original form. There must be five H ions on the reactant side of the equation to bind with the three chloride ions and two nitrate ions, so three H ions must be added to each side. 3Cl 2NO3 3HCl 5H → 3OCl 2HNO3 → 3HOCl 2NO 2NO H2O H2O 3H Check the balance. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 306 Redox Equations Back Print Name Class Date Problem Solving continued EVALUATE Are the units correct? NA Is the number of significant figures correct? NA Is the answer reasonable? Yes; the reaction has the reactants and products required and is balanced. Practice Balance the following redox equations. Assume that all reactions take place in an acid environment where H and H2O are readily available. 1. Fe SnCl4 → FeCl3 SnCl2 ans: 2Fe 3SnCl4 3 2FeCl3 3SnCl2 2. H2O2 FeSO4 H2SO4 → Fe2(SO4)3 H2O ans: H2O2 2FeSO4 H2SO4 3 Fe2(SO4)3 2H2O 3. CuS HNO3 → Cu(NO3)2 NO 3Cu(NO3)2 2NO 3S 4H2O S H2O ans: 3CuS 8HNO3 3 4. K2Cr2O7 HI → CrI3 KI 2CrI3 2KI 3I2 7H2O I2 H2O ans: K2Cr2O7 14HI 3 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 307 Redox Equations Back Print Name Class Date Problem Solving continued REACTIONS IN BASIC SOLUTION Sample Problem 2 Write a balanced equation for the reaction in a basic solution of NiO2 and Fe to produce Ni(OH)2 and Fe(OH)2. Solution ANALYZE What is given in the problem? What are you asked to find? Items Reactants Products Solution type Oxidized species Reduced species Balanced equation the reactants and products of a redox reaction the balanced redox reaction Data NiO2, Fe Ni(OH)2, Fe(OH)2 basic ? ? ? PLAN What steps are needed to balance the redox equation? 1. Write the formula equation followed by the ionic equation. 2. Assign oxidation numbers to each element. Delete any ion or molecule in which there is no change in oxidation state. 3. Write the half-reaction for reduction, and balance the mass and charge. OH and H2O may be added to either side. 4. Repeat step 3 for the oxidation half-reaction. 5. Adjust the coefficients so that the number of electrons lost equals the number of electrons gained. 6. Combine the half-reactions, and cancel anything common to both sides of the equation. 7. Combine ions to change the equation back to its original form, and check the balance of everything. COMPUTE 1. Write the formula equation. NiO2 NiO2 Fe → Ni(OH)2 2OH Fe(OH)2 Fe2 2OH Write the total ionic equation. Fe → Ni2 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 308 Redox Equations Back Print Name Class Date Problem Solving continued 2. Assign oxidation numbers to each element. 42 NiO2 Fe → Ni 2 4 0 0 2 21 2OH 2 Fe2 Fe2 2 2 21 2OH Delete any ions or molecules in which there is no change in oxidation state. NiO2 Fe → Ni 2 NiO2 → Ni 2 Balance the mass by adding OH and H2O. 4 4 2 3. Write the half-reaction for reduction. NiO2 2H2O → Ni 2 2 4OH Balance the charge by adding electrons to the side with the higher positive charge. 4 NiO2 2H2O 0 2e → Ni 2 2 2 4OH 4. Write the half-reaction for oxidation. Fe → Fe2 The mass is already balanced. Balance the charge by adding electrons to the side with the higher positive charge. Fe → Fe2 6. Combine the half-reactions. 4 0 2 2e 5. The numbers of electrons lost and gained are already the same. NiO2 NiO2 2H2O 0 2e → Ni 2 2 2 4OH Fe → Fe2 2H2O Fe → Ni2 2e Fe2 4OH 7. Combine ions to change the equation back to its original form. The four OH ions combine with the nickel and iron to make nickel(II) hydroxide and iron(II) hydroxide. NiO2 Check the balance. EVALUATE Are the units correct? NA 2H2O Fe → Ni(OH)2 Fe(OH)2 Is the number of significant figures correct? NA Is the answer reasonable? Yes; the reaction has the reactants and products required and is balanced. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 309 Redox Equations Back Print Name Class Date Problem Solving continued Practice Balance the following redox equations. Assume that all reactions take place in a basic environment where OH and H2O are readily available. 1. CO2 NH2OH → CO N2 H2O ans: CO2 2NH2OH 3 CO N2 3H2O 2. Bi(OH)3 K2SnO2 → Bi K2SnO3 (Both of the potassium-tin-oxygen compounds dissociate into potassium ions and tin-oxygen ions.) ans: 2Bi(OH)3 3K2SnO2 3 2Bi 3K2SnO3 3H2O Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 310 Redox Equations Back Print Name Class Date Problem Solving continued Additional Problems Balance each of the following redox equations. Unless stated otherwise, assume that the reaction occurs in acidic solution. 1. Mg 2. SO2 3. H2S 4. PbO2 5. S N2 → Mg3N2 Br2 H2O → HBr HCl Br2 H2O H2O NaICl2 O2 H2O (N2H4 is hydrazine; do not H2O H2SO4 H2SO4 Cl2 → S HBr → PbBr2 HNO3 → NO2 6. NaIO3 N2H4 HCl → N2 separate it into ions.) 7. MnO2 H2O2 HCl → MnCl2 8. AsH3 NaClO3 → H3AsO4 ammonia, NH3.) NaCl (AsH3 is arsine, the arsenic analogue of KCl H2O (H2C2O4 is oxalic 9. K2Cr2O7 H2C2O4 HCl → CrCl3 CO2 C2O2 .) acid; it can be treated as 2H 4 10. Hg(NO3)2 O HgO 3 11. HAuCl4 12. Sb2(SO4)3 13. Mn(NO3)2 14. H3AsO4 15. KClO3 NO2 N2 O2 (The reaction is not in solution.) HCl (HAuCl4 can be considered as H K2SO4 H2O MnSO4 H2SO4 H2O HMnO4 NaNO3 AuCl4 .) N2H4 → Au KMnO4 NaBiO3 Zn H2O → H3SbO4 ZnCl2 KCl HNO3 → Bi(NO3)2 H2O HCl → AsH3 HCl → Cl2 16. The same reactants as in Item 15 can combine in the following way when more KClO3 is present. Balance the equation. KClO3 17. MnCl3 18. NaOH 19. Br2 20. N2O HCl → Cl2 MnO2 HCl ClO2 H2O KCl H2O → MnCl2 H2O NaClO Ca(OH)2 → CaBr2 Al → NaAl(OH)4 NaOH → NaCl H2 in basic solution H2O in basic solution H2O in basic solution NaNO2 Ca(BrO3)2 21. Balance the following reaction, which can be used to prepare bromine in the laboratory: HBr MnO2 → MnBr2 H2O Br2 22. The following reaction occurs when gold is dissolved in aqua regia. Balance the equation. Au HCl HNO3 → HAuCl4 NO H2O Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 311 Redox Equations Back Print Name Class Date Skills Worksheet Problem Solving Electrochemistry The potential in volts has been measured for many different reduction halfreactions. The potential value is measured against the standard hydrogen electrode, which is assigned a value of zero. For consistency, these half-reactions are always written in the direction of the reduction. A half-reaction that has a positive reduction potential proceeds in the direction of the reduction when it is coupled with a hydrogen electrode. A reaction that has a negative reduction potential proceeds in the oxidation direction when it is coupled with a hydrogen electrode. Table 1 gives some common standard reduction potentials. TABLE 1 Standard electrode potential, E 0 (in volts) 5e ^ 1.50 1.50 1.36 6e ^ 1.23 1.22 1.07 0.85 0.80 0.80 0.77 0.56 0.54 0.34 0.14 0.00 Standard electrode potential, E 0 (in volts) 0.04 0.13 0.14 0.26 0.40 0.45 0.48 0.76 1.66 2.37 2.71 2.87 2.91 2.93 3.04 Reduction half-reaction MnO4 Mn2 Au3 Cl2 Cr2O2 7 2Cr3 MnO2 Mn2 Br2 Hg Ag Hg2 2 Fe3 MnO4 I2 Cu 2 2 Reduction half-reaction Fe3 Pb2 Sn 2 8H 4H2O 2e ^ 2Cl 14H 7H2O 3e ^ Fe 2e ^ Pb 2e ^ Sn 2e ^ Ni 2e ^ Cd 2e ^ Fe 2e ^ S 2 3e ^ Au Ni2 Cd2 Fe2 S Zn2 Al 3 4H 2e ^ 2H2O 2e ^ 2Br 2e ^ Hg e ^ Ag 2e ^ 2Hg e ^ Fe2 e^ MnO2 4 2e ^ Zn 3e ^ Al 2e ^ Mg e ^ Na 2e ^ Ca 2e ^ Ba e ^K e ^ Li Mg2 Na Ca2 Ba K Li 2 2e ^ 2I 2e ^ Cu 2e ^ S 2H (aq) H2S(aq) 2H (aq) 2e ^ H2 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 312 Electrochemistry Back Print Name Class Date Problem Solving continued You can use reduction potentials to predict the direction in which any redox reaction will be spontaneous. A spontaneous reaction occurs by itself, without outside influence. The redox reaction will proceed in the direction for which the difference between the two half-reaction potentials is positive. This direction is the same as the direction of the more positive half-reaction. General Plan for Solving Electrochemical Problems 1 Balanced ionic equation for the redox reaction Write the individual halfreactions for both oxidation and reduction. 2a Balanced equation for reduction Reduction occurs at the cathode. 2b Balanced equation for oxidation Oxidation occurs at the anode. 3a Cathode Look up the reduction potential for the halfreaction in Table 1. 3b Anode Look up the reduction potential for the halfreaction in Table 1. 4a E0cathode Substitute and solve. 4b E0anode 5 E0cell E0cathode E0anode Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 313 Electrochemistry Back Print Name Class Date Problem Solving continued Sample Problem 1 Calculate the cell potential to determine whether the following reaction is spontaneous in the direction indicated. Cd2 (aq) 2I (aq) → Cd(s) I2(s) Solution ANALYZE What is given in the problem? What are you asked to find? Items Reactants Products E 0cathode E 0anode E 0cell reactants and products whether the reaction is spontaneous Data Cd2 (aq) Cd(s) ?V ?V ?V 2I (aq) I2(s) PLAN What steps are needed to determine whether the reaction is spontaneous? Separate into oxidation and reduction half-reactions. Find reduction potentials for each. Solve the equation for the cell potential to determine if the reaction is spontaneous. 1 Cd2 (aq) 2I (aq) 3 Cd(s) write the individual half-reactions I2(s) 2a Cd2 (aq) 2e 3 Cd(s) reduction occurs at the cathode 2b 2I (aq) 3 I2(s) oxidation occurs at the anode 2e 3a Cathode look up the reduction potential for the half-reaction in Table 1 3b Anode look up the reduction potential for the half-reaction in Table 1 4a E0cathode substitute and solve 4b E0anode 5 E0cell Holt ChemFile: Problem-Solving Workbook E0cathode E0anode Electrochemistry Copyright © by Holt, Rinehart and Winston. All rights reserved. 314 Back Print Name Class Date Problem Solving continued Write the given equation. Cd2 (aq) 2I (aq) → Cd(s) 2e → Cd(s) 2e I2(s) The oxidation number of cadmium decreases; it is reduced. Cd2 (aq) The oxidation number of iodine increases; it is oxidized. 2I (aq) → I2(s) Cadmium is the cathode, and iodine is the anode. from Table 1 E0cathode E0anode given above 0.40 V from Table 1 0.54 V given above E0cell E0cathode E0anode Determine spontaneity. If the cell potential is positive, the reaction is spontaneous as written. If the cell potential is negative, the reaction is not spontaneous as written, but the reverse reaction is spontaneous. COMPUTE E 0cell 0.40 V 0.54 V 0.94 V The reaction potential is negative. Therefore, the reaction is not spontaneous. The reverse reaction would have a positive potential and would, therefore, be spontaneous. Cd2 (aq) Cd(s) EVALUATE Are the units correct? Yes; cell potentials are in volts. 2I (aq) → Cd(s) I2(s) → Cd2 (aq) I2(s) not spontaneous 2I (aq) spontaneous Is the number of significant figures correct? Yes; the number of significant figures is correct because the half-cell potentials have two significant figures. Is the answer reasonable? Yes; the reduction potential for the half-reaction involving iodine was more positive than the potential for the reaction involving cadmium, which means that I2 has a greater attraction for electrons than Cd2 . Therefore, I2 is more likely to be reduced than Cd2 . The reverse reaction is favored. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 315 Electrochemistry Back Print Name Class Date Problem Solving continued Practice Use the reduction potentials in Table 1 to determine whether the following reactions are spontaneous as written. Report the E 0cell for the reactions. 1. Cu2 Fe → Fe2 Cu ans: 0.79 V; spontaneous 2. Pb2 Fe2 → Fe3 Pb ans: 0.90 V; nonspontaneous 3. Mn2 4H2O Sn2 → MnO4 8H Sn ans: 1.64 V; nonspontaneous 4. MnO2 4 Cl2 → MnO4 2Cl ans: 0.80 V; spontaneous Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 316 Electrochemistry Back Print Name Class Date Problem Solving continued 5. Hg2 2 2MnO2 → 2Hg 4 2MnO4 ans: 0.24 V; spontaneous 6. 2Li Pb → 2Li Pb2 ans: 2.91 V; nonspontaneous 7. Br2 2Cl → 2Br Cl2 ans: 0.29 V; nonspontaneous 8. S 2I → S2 I2 ans: 1.02 V; nonspontaneous Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 317 Electrochemistry Back Print Name Class Date Problem Solving continued Sample Problem 2 A cell is constructed in which the following two half-reactions can occur in either direction. Zn2 Br2 2e ^ Zn 2e ^ 2Br Write the full ionic equation for the cell in the spontaneous direction, identify the reactions occurring at the anode and cathode, and determine the cell’s voltage. Solution ANALYZE What is given in the problem? What are you asked to find? the reversible half-reactions of the cell the equation in the spontaneous direction; the voltage of the cell Data Zn2 Br2 2e ^ Zn 2e ^ 2Br Items Half-reaction 1 Half-reaction 2 Reduction potential of 1 Reduction potential of 2 Full ionic reaction Cell voltage 0.76 V 1.07 V ? ? PLAN What steps are needed to determine the spontaneous reaction of the cell and the cell voltage? Determine which half-reaction has the more positive reduction potential. This will be the reduction half-reaction; it occurs at the cathode. Reverse the other half-reaction so that it becomes an oxidation half-reaction; it occurs at the anode. Adjust the half-reactions so that the same number of electrons are lost as are gained. Add the reactions together. Compute the cell voltage by the formula E 0cell E 0cathode E 0anode, using the reduction potentials for the reaction at each electrode. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 318 Electrochemistry Back Print Name Class Date Problem Solving continued Br2 2e ^ 2Br look up the reduction potential for the half-reaction in Table 1 Zn2 2e ^ Zn look up the reduction potential for the half-reaction in Table 1 E0half-reaction the half-cell with the larger reduction potential is the cathode E0half-reaction the half-cell with the smaller reduction potential is the anode Cathode reduction occurs at the cathode Anode oxidation occurs at the anode Br2 2e 3 2Br combine to write the full ionic equation Zn 3 Zn2 2e Br2 Zn 3 2Br Zn2 substitute and solve E0cell E0cathode E0anode First, look up the reduction potentials for the two half-reactions in Table 1. from Table 1 0 EBr2 0 EZn 1.07 V from Table 1 0.76 V Br2 has the larger reduction potential; therefore, it is the cathode. Zn has the smaller reduction potential; therefore, it is the anode. The cathode half-reaction is Br2 The anode half-reaction is Zn → Zn2 The full-cell equation is Br2 Br2 2e → 2Br Zn → Zn2 Zn → 2Br 2e Zn2 2e 2e → 2Br Substitute the reduction potentials for the anode and cathode into the cell potential equation, and solve the equation. E0cell E0cathode E0 2 Br E0anode E0 Zn Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 319 Electrochemistry Back Print Name Class Date Problem Solving continued COMPUTE Br2 E 0cell EVALUATE Are the units correct? Yes; the cell potential is in volts. Zn → 2Br Zn2 1.83 V 1.07 V ( 0.76 V) Is the number of significant figures correct? Yes; the half-cell potentials were given to two decimal places. Is the answer reasonable? Yes; you would expect the reaction to have a positive cell potential because it should be spontaneous. Practice If a cell is constructed in which the following pairs of reactions are possible, what would be the cathode reaction, the anode reaction, and the overall cell voltage? 1. Ca2 2e ^ Ca 3e ^ Fe Fe3 ans: cathode: Fe3 3e 3 Fe, anode: Ca 3 Ca2 2e , E 0cell 2.83 V 2. Ag e ^ Ag 2e ^ H2S S 2H ans: cathode: Ag e 3 Ag, anode: H2S 3 S 2H 2e , E 0cell 0.66 V 3. Fe3 e ^ Fe2 Sn2 2e ^ Sn ans: cathode: Fe3 e 3 Fe2 , anode: Sn 3 Sn2 2e , E 0cell 0.91 V 4. Cu2 2e ^ Cu 3 Au 3e ^ Au ans: cathode: Au3 3e 3 Au, anode: Cu 3 Cu2 2e , E 0cell 1.16 V Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 320 Electrochemistry Back Print Name Class Date Problem Solving continued Additional Problems Use reduction potentials to determine whether the reactions in the following 10 problems are spontaneous. 1. Ba 2. Ni 3. 2Cr 4. Cl2 5. Al 6. 3 Sn2 → Ba2 Hg 2 Sn Hg 3 → Ni 7H2O 2 6Fe → Cr2O2 7 14H 6Fe2 Sn → 2Cl 3Ag → Al3 S Ca 2 Sn2 3Ag S 2Ag Ca 2 2 Hg2 2 → 2Hg → I2 7. Ba 8. 2I 9. Zn 10. 2Cr3 2Ag → Ba 2 2MnO4 → Zn 3Mg2 2MnO2 4 14H 3Mg 7H2O → Cr2O2 7 In the following problems, you are given a pair of reduction half-reactions. If a cell were constructed in which the pairs of half-reactions were possible, what would be the balanced equation for the overall cell reaction that would occur? Write the half-reactions that occur at the cathode and anode, and calculate the cell voltage. 11. Cl2 Ni2 12. Fe3 Hg2 2e ^ 2Cl 2e ^ Ni 3e ^ Fe 2e ^ Hg 13. MnO4 e ^ MnO2 4 Al3 3e ^ Al 14. MnO4 S 2H 15. Ca2 Li 8H 5e ^ Mn2 2e ^ H2S 4H2O 2e ^ Ca e ^ Li 4H2O 16. Br2 2e ^ 2Br MnO4 8H 5e ^ Mn2 17. Sn Fe3 2 2e ^ Sn e ^ Fe2 2e ^ Zn 14H 6e ^ 2Cr3 2e ^ Ba 2e ^ Ca 2e ^ 2Hg 2e ^ Cd 7H2O 18. Zn2 Cr2O2 7 19. Ba Ca2 20. Hg2 2 Cd2 2 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt ChemFile: Problem-Solving Workbook 321 Electrochemistry Back Print Answer Key The Science of Chemistry CONVERSIONS 1. a. 12 750 km b. 2.77 m c. 3.056 hectares d. 0.008 19 m2 e. 300 Mm 2. a. 620 m b. 3 875 000 mg c. 3.6 L d. 342 kg e. 68 710 L 3. a. 0.000 856 kg b. 0.001 21 kg c. 6.598 cm3 d. 0.0806 mm e. 0.010 74 L 4. a. 7930 cm3 b. 590 cm c. 4.19 dm3 d. 74 800 cm2 e. 197 L 5. 1 L 6. 370 L 7. 6 L 8. 0.876 L of water per year; 876 kg of 9. 10. 11. 12. 18. a. 0.5047 kg; 504.7 g b. 0.0092 kg; 9.2 g c. 0.000 122 kg; 0.122 g d. 0.071 95 kg; 71.95 g 19. a. 0.582 L; 582 mL b. 2.5 L; 2500 mL c. 1.18 L; 1180 mL d. 0.0329 L; 32.9 mL 20. a. 1370 g/L; 1370 kg/m3 b. 692 g/L; 692 kg/m3 c. 5200 g/L; 5200 kg/m3 d. 38 g/L; 38 kg/m3 e. 5790 g/L; 5790 kg/m3 f. 0.0011 g/L; 0.0011 kg/m3 21. a. 360 g/min b. 518.4 kg/day c. 6 mg/ms 22. 27.8 m/s 23. 4732 kcal/h 24. 620 kg 1L 24 h 25. 3.9 mL h 1000 mL 1 day 365 days 34.164 L /year 1 year 26. 40 doses Matter and Energy SIGNIFICANT FIGURES 1. a. b. c. d. e. f. g. h. i. 2. a. b. c. d. e. f. g. 3. a. b. c. d. 13. 14. 15. 16. 17. water a. 1674 km/h b. 40 176 km/day 7.8 kg of sodium hydroxide 45 m of plastic tubing a. 13.2 mL/day b. 150 kg/min c. 62 cm3/min d. 1.7 m/s a. 2.97 g/cm3 b. 0.041 28 kg/cm2 c. 5.27 kg/dm3 d. 0.006 91 mg/mm3 a. 750 mL b. 5.56 kg 1250 kg a. 0.0028m3 b. 1.05 g/cm3 c. 0.056 m2 a. 0.04 mL per drop b. 1.48 mL c. 17 000 drops 3 4 3 2 2 1 3 4 5 5 490 000 m 0.013 479 3 mL 31 950 cm2 192.67 m2 790 cm 389 278 000 J 225 834.8 cm3 49 000 cm2 3.1 kg/L 12.3 L/sec 170 000 cm3 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt Chemistry 322 Answer Key Back Print e. f. a. b. c. d. e. f. a. b. c. a. b. c. a. b. c. a. b. c. d. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 41 m3 3.129 g/cm3 90.2 J 0.0006 m 900 g 31.1 kPa 278 dL 1790 kg 307 cm2 30 700 mm2 0.0307 m2 1800 cm3 0.0018 m3 1 800 000 mm3 1300 kg/m3 1.3 g/mL 1.3 kg/dm3 130 mm3 430 cm3 5.0 m 4000 m3 26 000 000 m3 a. 13.38 g b. 100. mg c. 0.015 L d. 315 cm2 e. 14.47 kg f. 353 mL 1.09 kg/L 0.43 g/m; 2.3 m 2000 m2 26 300 kJ/min; 439 kJ/s a. 15.8 m3 b. 9800 L/min c. 590 m3/h a. 7.5 kg m2 b. 67.22 cm c. 2.4 kg m2/s2 d. 19.9 m2 e. 970 000 m/h f. 139 cm2 1.58 105 km 9.782 10 6 L 8.371 108 cm3 6.5 109 mm2 5.93 10 3 g 6.13 10 9 m 1.2552 107 J 8.004 10 6 g/L 1.0995 10 2 kg 1.05 109 Hz 2. a. 9.49 103 kg b. 7.1 10 2 mg c. 9.8 103 m3 d. 1.56 10 7 m e. 3.18 106 J f. 9.63 1027 molecules g. 7.47 106 cm 3. a. 6.53 102 L/s b. 1.83 107 mm3 c. 2.51 104 kg/m2 d. 4.23 101 km/s e. 3.22 106 m3 f. 8.68 106 J/s 4. 3.6 106 J 5. 1.12 107 Pa 6. 1.5 106 mm 7. a. 3 105 km/s b. 1 1012 m/h c. 3 104 cm 8. a. 7.75 1022 molecules b. 1.59 1027 molecules c. 6.41 1017 molecules 9. a. 2.2 10 5 mm2/transistor b. 4.3 109 transistors 10. 5.01 10 8 g/ L 11. 4.79 107 cesium atoms 12. 1.2 1020 g/m3; 1.2 1014 kg 13. 1.9 107 pits 14. a. 2.69 1018 molecules of oxygen b. 2.69 1022 molecules of oxygen c. 3.72 10 20 mL/molecule 15. a. 7.9 108 kg/person b. 7.9 105 metric ton/person c. 5.4 108 kg/person 16. 3.329 105 Earths 17. a. 4.8 10 1 km3 b. 4.8 108 m3 c. 32 years 18. 1.0 107 J/day SCIENTIFIC NOTATION 1. a. b. c. d. e. f. g. h. i. j. The Mole and Chemical Composition FOUR STEPS FOR SOLVING QUANTITATIVE PROBLEMS 1. 2. 3. 4. 5. 6. 0.026 mm 3.21 L 0.80 g/cm3 21.4 g/cm3 30 boxes a. 1.73 L 0.120 m 0.120 m b. 9.2 g; 5.0 cm3 0.120 m Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt Chemistry 323 Answer Key Back Print c. 60.4 kg; 1.88 104 dm3 d. 0.94 g/cm3; 5.3 10 4 m3 e. 2.5 103 kg; 2.7 106 cm3 7. 2.8 g/cm3 8. a. 0.72 m b. 2.5 103 atoms 9. 1300 L/min 10. 1.3 106 cal/h 11. 5.44 g/ cm3 12. 2.24 104 cm3 13. 32 000 uses 14. 2500 L 15. 9.5 L/min c. d. e. f. 8. a. b. c. d. e. f. 9. a. b. c. d. e. f. 10. a. b. c. MOLE CONCEPT 1. a. b. c. d. e. f. 2. a. b. c. d. e. f. 3. a. b. c. d. e. f. 4. a. b. c. d. e. f. 5. a. b. c. d. e. f. 6. a. b. c. d. e. f. 7. a. b. 3.7 10 4 mol Pd 150 mol Fe 0.040 mol Ta 5.38 10 5 mol Sb 41.1 mol Ba 3.51 10 8 mol Mo 52.10 g Cr 1.5 104 g or 15 kg Al 8.23 10 7 g Ne 3 102 g or 0.3 kg Ti 1.1 g Xe 2.28 105 g or 228 kg Li 1.02 1025 atoms Ge 3.700 1023 atoms Cu 1.82 1024 atoms Sn 1.2 1030 atoms C 1.1 1021 atoms Zr 1.943 1014 atoms K 10.00 mol Co 0.176 mol W 4.995 10 5 mol Ag 1.6 10 15 mol Pu 7.66 10 7 mol Rn 1 10 11 mol Ce 2.5 1019 atoms Au 5.10 1024 atoms Mo 4.96 1020 atoms Am 3.011 1026 atoms Ne 2.03 1018 atoms Bi 9.4 1016 atoms U 117 g Rb 223 g Mn 2.11 105 g Te 2.6 10 3 g Rh 3.31 10 8 g Ra 8.71 10 5 g Hf 0.749 mol CH3COOH 0.0213 mol Pb(NO3)2 11. 12. 14. 15. 16. 17. 18. 19. 20. 21. 22. 3 104 mol Fe2O3 2.66 10 4 mol C2H5NH2 1.13 10 5 mol C17H35COOH 378 mol (NH4)2SO4 764 g SeOBr2 4.88 104 g CaCO3 2.7 g C20H28O2 9.74 10 6 g C10H14N2 529 g Sr(NO3)2 1.23 10 3 g UF6 2.57 1024 formula units WO3 1.81 1021 formula units Sr(NO3)2 4.37 1025 molecules C6H5CH3 3.08 1017 molecules C29H50O2 9.0 1026 molecules N2H4 5.96 1023 molecules C6H5NO2 1.14 1024 formula units FePO4 6.4 1019 molecules C5H5N 6.9 1020 molecules (CH3)2CHCH2OH d. 8.7 1017 formula units Hg(C2H3O2)2 e. 5.5 1019 formula units Li2CO3 a. 52.9 g F2 b. 1.19 103 g or 1.19 kg BeSO4 c. 1.388 105 g or 138.8 kg CHCl3 d. 9.6 10 12 g Cr(CHO2)3 e. 6.6 10 4 g HNO3 f. 2.38 104 g or 23.8 kg C2Cl2F4 0.158 mol Au 0.159 mol Pt 0.288 mol Ag 13.0.234 mol C6H5OH 3.8 g I2 1.00 1022 atoms C a. 0.0721 mol CaCl2 55.49 mol H2O b. 0.0721 mol Ca2 0.144 mol Cl a. 1.325 mol C12H22O11 b. 7.762 mol NaCl 0.400 mol ions 4.75 mol atoms a. 249 g H2O b. 13.8 mol H2O c. 36.1 mL H2O d. 36.0 g H2O The mass of a sugar molecule is much greater than the mass of a water molecule. Therefore, the mass of 1 mol of sugar molecules is much greater than the mass of 1 mol of water molecules. 1.52 g Al Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt Chemistry 324 Answer Key Back Print 23. 0.14 mol O2 24. a. 0.500 mol Ag d. e. f. 4. a. b. c. d. e. f. 5. a. b. c. d. 6. a. 0.250 mol S b. 0.157 mol Ag2S 0.313 mol Ag 0.157 mol S c. 33.8 g Ag 5.03 g S PERCENTAGE COMPOSITION 1. a. HNO3 b. c. d. 2. a. b. c. d. e. f. g. h. 3. a. b. c. 1.60% H 22.23% N 76.17% O NH3 82.22% N 17.78% H HgSO4 67.616% Hg 10.81% S 21.57% O SbF5 56.173% Sb 43.83% F 7.99% Li 92.01% Br 94.33% C 5.67% H 35.00% N 5.05% H 59.96% O 2.15% H 29.80% N 68.06% O 87.059% Ag 12.94% S 32.47% Fe 13.96% C 16.29% N 37.28% S LiC2H3O2 10.52% Li 36.40% C 4.59% H 48.49% O Ni(CHO2)2 39.46% Ni 16.15% C 1.36% H 43.03% O 46.65% N 23.76% S 89.491% Tl 7. 8. 9. 10. 11. 12. 13. 14. 39.17% O 79.95% Br in CaBr2 78.767% Sn in SnO2 1.47 g O 26.5 metric tons Al 262 g Ag 0.487 g Au 312 g Se 3.1 104 g Cl 40.55% H2O 43.86% H2O 20.70% H2O 28.90% H2O Ni(C2H3O2)2 4H2O 23.58% Ni b. Na2CrO4 4H2O 22.22% Cr c. Ce(SO4)2 4H2O 34.65% Ce 43.1 kg Hg malachite: 5.75 102 kg Cu chalcopyrite: 3.46 102 kg Cu malachite has a greater Cu content a. 25.59% V b. 39.71% Sn c. 22.22% Cl 319.6 g anhydrous CuSO4 1.57 g AgNO3 54.3 g Ag 8.08 g S 23.1 g MgSO4 7H2O 3.27 102 g S EMPIRICAL FORMULAS 1. a. b. c. d. e. f. 2. a. b. c. 3. a. b. c. d. 4. a. b. c. d. e. f. BaCl2 BiO3H3 or Bi(OH)3 AlN3O9 or Al(NO3)3 ZnC4H6O4 or Zn(CH3COO)2 NiN2S2H8O8 or Ni(NH4)2SO4 C2HBr3O2 or CBr3COOH CuF2 Ba(CN)2 MnSO4 NiI2 MgN2O6 or Mg(NO3)2 MgS2O3, magnesium thiosulfate K2SnO3, potassium stannate As2S3 Re2O7 N2H4O3 or NH4NO3 Fe2Cr3O12 or Fe2(CrO4)3 C5H9N3 C6H5F2N or C6H3F2NH2 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt Chemistry 325 Answer Key Back Print 5. a. C6H12S3 b. C8H16O4 c. C4H6O4 d. C12H12O6 6. a. C4H4O4 b. C4H8O2 c. C9H12O3 7. K2S2O5, potassium metabisulfite 8. Pb3O4 9. Cr2S3O12 or Cr2(SO4)3, chromium(III) 16. a. 2N2H4 N2O4 3 3N2 4H2O b. 1 mol N2O4 to 3 mol N2 c. 30 000 mol N2 d. 3.52 105 g H2O 17. 2HgO(s) 3 2Hg(l) O2( g); 1.1954 mol O2 18. 2Fe 3Cl2 3 2FeCl3; 30.5 g Fe 19. 9.26 mg CdS 20. a. 1.59 mol CO2 b. 0.0723 mol C3H5(OH)3 c. 535 g Mn2O3 d. 8.33 g C3H5(OH)3; 4.97 g CO2 21. a. 3.29 103 kg of HCl b. 330 g CO2 (s) 22. a. 6.53 105 g NH4ClO4 b. 160 kg NO(g) 23. a. 1.70 106 mol H3PO4 b. 666 kg of CaSO4 2H2O c. 34 metric tons of H3PO4 24. 1670 kg sulfate 10. C9H6O4 11. C5H9N3, the empirical formula and the molecular formula are the same 12. The molecular formulas of the compounds are different multiples of the same empirical formula. (FYI: The first could be acetic acid, C2H4O2, and the second could be glucose, C6H12O6, or some other simple sugar.) Stoichiometry STOICHIOMETRY 1. 15.0 mol (NH4)2SO4 2. a. 51 g Al b. 101 g Fe c. 1.83 mol Fe2O3 3. 0.303 g H2 4. H2SO4 2KOH 3 K2SO4 5. LIMITING REACTANTS 1. 2ZnS 2. 3. 4. 5. 6. 7. 8. 9. 10. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 2H2O; 1.11 g H2SO4 a. H3PO4 2NH3 3 (NH4)2HPO4 b. 0.293 mol (NH4)2HPO4 c. 970 kg NH3 a. 90.0 mol ZnCO3; 60.0 mol C6H8O7 b. 13.5 kg H2O; 33.0 kg CO2 a. 60.9 g methyl butanoate b. 3261 g H2O a. 0.450 mol N2 b. 294 g NH4NO3 Pb(NO3)2 2KI 3 PbI2 2KNO3; 0.751 mg KNO3 3.3 mol PbSO4 2LiOH CO2 3 H2O Li2CO3; 360 g H2O a. 38.1 g H2O b. 40.1 g H3PO4 c. 0.392 mol H2O C2H5OH 3O2 3 2CO2 3H2O; 81.0 g C2H5OH 76.5 g H2SO4; 12.5 g O2 2NaHCO3 3 Na2CO3 H2O CO2 ; 1.31 g CO2 11. 12. 3O2 3 2ZnO 2SO2; ZnS is limiting a. Al is limiting b. 4.25 10 3 mol Al2O3 c. O2 is limiting a. CuS is limiting b. 15.6 g CuO Fe is limiting; 0.158 mol Cu 54 g Ba(NO3)2 a. 38 g Br2 b. 510 g I2 a. Ni is in excess b. 60.2 g Ni(NO3)2 CS2( g) 3O2( g) 3 2SO4( g) CO2( g) 0.80 mol O2 remain a. 0.84 g Hg(NH2)Cl b. 0.84 g a. 2Al(s) 2NaOH(aq) 2H2O(l) 3 2NaAlO2(aq) 3H2(g) b. NaOH is limiting; 0.56 mol H2 c. Al should be limiting because you would not want aluminum metal remaining in the drain. a. 0.0422 mol Cu; 0.169 mol HNO3 b. Cu is in excess c. 3.32 g H2O a. 2.90 mol NO; 4.35 mol H2O b. NH3 is limiting c. NH3 is limiting; 1.53 103 kg NO Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt Chemistry 326 Answer Key Back Print 13. 565 g CH3CHO; 16. 6CO2 29 g CH3CH2OH remains 14. 630 g HBr 15. 12.7 g SO2 16. a. 18.4 g Tb b. 2.4 g TbF3 6H2O 3 C6H12O6 6.32 103 g O2 17. 27.6 kg CaO 6O2 Causes of Change THERMOCHEMISTRY 1. 260.8 kJ/mol 2. 385.9 kJ/mol 3. 390. kJ/mol 4. 492.3 kJ/mol 5. 107.6 kJ/mol 6. 121.8 kJ/mol 7. 384.9 kJ/mol 8. 74.2 kJ/mol 9. 169.0 kJ/mol 10. a. CH4( g) 2O2( g) 3 PERCENTAGE YIELD 1. a. b. c. d. 2. a. b. c. d. 3. a. b. c. 4. a. b. c. 5. a. b. c. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 64.3% yield 58.0% yield 69.5% yield CH3CH2OH is limiting; 79% yield 69.5% yield 79.0% yield 48% yield 85% yield 59% yield 81.0% yield 2.3 105 mol P 91.8% yield 0.0148 mol W 16.1 g WO3 86.8% yield 92.2% yield 2.97 104 kg CS2; 4.39 104 kg S2Cl2 a. 81% yield b. 2.0 102 g N2O5 80.1% yield a. 95% yield b. 9.10 102 g Au c. 9 104 kg ore a. 87.5% yield b. 0.25 g CO a. 71% yield b. 26 metric tons c. 47.8 g NaCl d. 500 kg per hour NaOH a. 2Mg O2 3 2MgO b. 87.7% yield c. 3Mg N2 3 Mg3N2 d. 56% yield a. 80.% yield b. 66.2% yield c. 57.1% yield 2C3H6( g) 2NH3( g) 3O2( g) 3 2C3H3N( g) 6H2O( g) 91.0% yield a. CO 2H2 3 CH3OH 3.41 103 kg b. 91.5% yield 96.9% yield 11. 12. 13. 14. 15. 16. CO2( g) 2H2O( g) C3H8( g) 5O2( g) 3 3CO2( g) 4H2O( g) b. for methane: H 802.2 kJ/mol for propane: H 2043 kJ/mol c. outputmethane 4.998 104 kJ/kg 4.632 104 kJ/kg outputpropane Methane yields more energy per mass. 132.7 kJ/mol 7171.4 kJ/mol 141.1 kJ/mol 20.2 kJ/mol 285 kJ/mol K a. 786.8 kJ/mol b. 36 kJ/mol c. 2154 kJ/mol d. 496 kJ/mol e. 1346.4 kJ/mol Gases GAS LAWS 1. a. 105 kPa b. 5.0 mL c. 42.4 kPa d. 6.78 10 3 dm3 e. 1.24 atm f. 1.5 m3 2. 8.0 m3 3. 0.0258 atm 4. 8.01 10 2 dm3 5. a. 234 K b. 1.2 dm3 c. 269.17°C d. 8.10 10 2 L e. 487 cm3 f. 67.9 m3 Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt Chemistry 327 Answer Key Back Print 6. 1.45 cm3 7. 40.°C 8. a. 208.6°C b. 5.5 kPa c. 2.61 atm d. 297°C e. 35.6 atm f. 39 K 9. 0.899 atm 10. 2.23 atm 11. 7.98 K 12. a. 2.02 L b. 75.8 kPa c. 110 K d. 4.69 103 mm3 e. 72°C f. 2.25 atm 13. 379 cm3 14. 98 kPa 15. 1.00 atm; use Boyle’s law to find the c. 6°C d. 2.24 atm or 227 kPa 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 33.7 atm 3.06 g/L 663 g/mol 204 L 0.0101 mol ethane 5.16 g NO 77.0% yield 10.5 g/mol 171 g/mol 6.55 atm 326 kPa 479 K or 206°C 1210 L at 75°C; 1620 L at 1.11 104 kPa 168 mL 3.85 103 L 4.05 103 L 29.0 g/mol 8°C 16. 17. 18. 19. 20. 21. 22. 23. 24. pressure of each gas in the whole space; add the partial pressures of both gases when they occupy the whole space. 285 mL 89 cm3 The pressure in the bottle on top of the mountain is the sum of PO2 dry at the temperature of the mountaintop and PH2O vapor at the temperature on top of the mountain. 59 cm3 Solve the equation: Vtotal/293 K ( Vtotal 0.20 cm3)/294 K 118 kPa 935 mL 26.4 kL 3.76 atm 115 mL 10.9 atm STOICHIOMETRY OF GASES 1. a. 19.0 mL N2 b. 4.26 104 L NH3 c. 896 L NH3 d. 899 L H2 2. a. 23 L H2O b. 1070 L O2 c. 326 mL CO2 d. 25.2 L total products 3. 1550 L O2 at STP 4. 0.894 L SiF4 5. a. 3.36 L H2 b. 488 g Fe c. 112 L H2 6. 0.013 L H2 or 13 mL H2 7. 7.50 L O2 at STP; 4.14 g diethyl ether 8. a. 3.36 L N2; 6.72 L CO2; 5.60 L H2O; 0.560 L O2 b. 15.0 L total volume all gases 9. 10. 11. 12. 13. 14. 15. IDEAL GAS LAW 1. a. 34.2 mol b. 6.68 103 kPa c. 148°C d. 1.1 105 L 2. a. 55.9 g/mol b. 0.111 g c. 4.46 10 2 L d. 0.846 atm e. 391 g/mol 3. 2.71 103 L 4. a. 48.4 g/mol b. 9.38 g/L 0.894 g NH4NO3 2.19 L PH3 1.2 103 kg Al; 7.4 105 L HCl 7.08 107 L NH3 3.77 g BaO2 5.85 L Cl2 28.0 kL NH3; 28.0 kL NO2 overall reaction is: 4NH3 7O2 3 4NO2 6H2O 16. 18.2 g KClO3 17. a. 38 000 L NH3 b. 1.30 105 g NaHCO3 c. 12.3 L NH3 d. 5.60 103 L Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt Chemistry 328 Answer Key Back Print 18. a. b. c. d. 9.93 L CO2 1.63 L O2 26.0 L CO2 3.29 103 g H2O 21. 22. 23. 24. Solutions CONCENTRATION OF SOLUTIONS 1. a. 60.0 g KMnO4; 440.0 g H2O b. 220 g BaCl2 c. 457 g glycerol d. 0.0642 M K2Cr2O7 e. 1.27 m CaCl2 f. 0.234 g NaCl g. 541 g glucose; 2040 g total 2. 10.6 mol H2SO4 3. 0.486 m linoleic acid 4. a. 13.0 g Na2S2O3 b. 0.0820 mol c. 0. 328 M 5. 338 g CoCl2 6. 0.442 L 7. 203 g urea 8. 18.8 g Ba(NO3)2 9. add 3.5 g (NH4)2SO4 to 96.5 g H2O 10. 54 g CaCl2 11. 1.25 mol; 1.25 M 12. 93.6 g/mol 13. 49.6 kg water; 0.5 kg NaCl 14. 8.06% 15. 1.4 L ethyl acetate 16. CdCl2(aq) Na2S(aq) 3 CdS(s) 2 contributes 315 g of H2O, so only 685 g H2O must be added. 383 g CaCl2 6H2O 0.446 g arginine 3254 g H2O; the hydrate contributes 987.9 g H2O. 9.646 g KAl(SO4)2 12H2O; 25.35 g H2O DILUTIONS 1. 0.0948 M 2. 0.44 mL 3. a. 3.0 M b. 0.83 L c. 1.5 103 g 4. 6.35 mL 5. 348 mL 6. 0.558 M 7. a. 850 mL; 2.4 mL; 86 mL b. 1.3 L concentrated HNO3 c. 1.16 L concentrated HCl 8. 0.48 M 9. 2.72 M 10. Dilute 4.73 mL of the 6.45 M acetic 11. 12. 13. 14. 15. 17. 18. 19. 20. NaCl(aq) a. 0.196 mol CdCl2 b. 0.196 mol CdS c. 28.3 g CdS 34.4 g H2SO4 1.54 105 mol HCl 85.7 mL BaCl2 solution a. Measure out 9.39 g CuSO4 5H2O and add 90.61 g H2O to make 100. g of solution. The 9.39 g of CuSO4 5H2O contributes the 6.00 g of CuSO4 needed. b. Measure out 200. g CuSO4 5H2O, dissolve in water, then add water to make 1.00 L. Water of hydration does not have to be considered here as long as the molar mass of the hydrate is used in determining the mass to weigh out. c. Measure out 870 g CuSO4 5H2O and add 685 g H2O. The hydrate 16. 17. 18. 19. 20. acid to 25 mL. This means adding 20.27 mL of water. 39.7 g/mol 0.27 M 1.9 102 g 0.667 L For 1.00 L of 0.495 M urea solution, take 161 mL of 3.07 M stock solution of urea and dilute with water (839 mL of water) to make 1.00 L. a. 17.8 m b. 1080 g, 6.01 M 47.17 g Na2CO3 per 50.00 g sample 94.3% Na2CO3 151 g CuCl2 Add 2.3 volumes of H2O per volume of stock solution. a. 14.9 g b. 7.02 10 2 mol c. 0.167 M COLLIGATIVE PROPERTIES 1. 2. 3. 4. 5. 6. 7. 8. 9. 9.88°C ; 102.7°C 103.3°C 201.6°C 50. g ethanol 82 g/mol 18.2°C 15.5 g/mol 66.8 g/mol 183.3°C Answer Key Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt Chemistry 329 Back Print 10. a. 15.0°C b. 104.1°C 11. 292 g/mol 12. 47.0°C 13. 190 g/mol 14. 107.2°C 15. 27.9°C 16. 204.4 g/mol; C16H10 17. 9.9 g CaCl2; 49 g glucose 18. C3H6O3 19. 29.2°C 20. 2.71 kg; 104.9°C 21. a. 2.2 m b. 1.7 m 22. 92.0 g H2O 23. 150 g/mol; C5H10O5 24. 124 g/mol; C6H6O3 EQUILIBRIUM OF SALTS, Ksp 1. 1.0 10 4 2. 1.51 10 7 3. a. 1.6 10 11 b. 0.49 g 4. 2.000 10 7 5. a. 8.9 10 4 M b. 0.097 g 6. 0.036 M 7. a. 1.1 10 4 M b. 4.6 L 8. a. 5.7 10 4 M b. 2.1 g 9. 0.83 g remains 10. 2.8 10 3 M 11. a. 1.2 10 5 b. Yes 12. a. 1.1 10 8 b. No 13. a. Mg(OH)2 3 Mg2 2OH b. 11 L c. A suspension contains undissolved Chemical Equilibrium EQUILIBRIUM 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 1.98 10 7 2.446 10 12 3.97 10 5 a. 8.13 10 4 M b. 0.0126 M a. The concentrations are equal. b. K will increase. 0.09198 M a. 0.7304 M b. 6.479 10 4 M a. [A] [B] [C] 1/2[A]initial b. [A], [B], and [C] will increase equally. K remains the same. a. Keq [HBr]2/[H2][Br2] b. 2.11 10 10 M c. Br2 and H2 will still have the same concentration. HBr will have a much higher concentration than the two reactants; at equilibrium, essentially only HBr will be present. 1.281 10 6 4.61 10 3 a. Keq [HCN]/[HCl] b. 3.725 10 7 M a. The reaction yields essentially no products at 25°C; as a result, the equilibrium constant is very small. At 110 K, the reaction proceeds to some extent. b. 2.51 M 0.0424 0.0390 14. 15. 16. 17. 18. 19. 20. Mg(OH)2 suspended in a saturated solution of Mg(OH)2. As hydroxide ions are depleted by titration, the dissociation equilibrium continues to replenish them until all of the Mg(OH)2 is used up. a. 0.184 M b. 46.8 g 5.040 10 3 a. 1.0 b. 0.25 c. 0.01 d. 1 10 6 4 10 4; 3 10 5; 3 10 5; 1 10 6; 3 10 7 7.1 10 7 M; 1.3 10 3 g a. 0.011 M b. 0.022 M c. 12.35 0.063 g Acids and Bases pH 1. a. [OH ] 0.05 M, [H3O ] 2 10 13 M b. [OH ] 2.0 10 12 M, [H3O ] 5.0 10 3 M c. [OH ] 0.013 M, [H3O ] 7.7 10 13 M Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt Chemistry 330 Answer Key Back Print 6.67 10 14 M, [H3O ] 0.150 M e. [OH ] 0.0400 M, [H3O ] 2.50 10 13 M f. [OH ] 2.56 10 14 M, [H3O ] 0.390 M g. 10, 2.3, 12.11, 0.824, 12.602, 0.409 2. [OH ] 0.160 M [H3O ] 6.25 10 14 M 3. 0.08 M 4. 2.903; 4.137 5. 10.0; 10.7 6. a. [H3O ] 0.020 M, [OH ] 5.0 10 13 M b. 1.7 7. a. 2.804 b. 2.755 8. 2.9 10 11 M 9. 10.54 10. [H3O ] 1 10 4 M [OH ] 1 10 10 M 11. 10.96 12. 0.2 g 13. [H3O ] 1.0 10 3 M [OH ] 1.0 10 4 M 14. a. 0.40% b. 1.3% c. 4.0% 15. 1.402 16. a. 4.50 b. 2.40 c. 3.83 d. 5.63 17. [H3O ] 1 10 9 M [OH ] 1 10 5 M 18. 11.70 19. [H3O ] 0.020 M [HCl] 0.020 M 20. 3.2 10 4 M 21.pH 0.000 10. kL 1.0 102 L 10. kL 22. 0.052 mol NaOH 2.1 g NaOH 23. 7.1 10–3 24. [H3O ] 7.1 10–4 M [OH ] 1.4 10 11 M d. [OH ] 4. a. 20.00 mL base b. 10.00 mL acid c. 80.00 mL acid 5. 5.066 M HF 6. 0.2216 M oxalic acid 7. 1.022 M H2SO4 8. 0.1705 M KOH 9. 0.5748 M citric acid 10. 0.3437 M KOH 11. 43.2 mL NaOH 12. 23.56 mL H2SO4 13. 0.06996 mol KOH; 3.926 g KOH; 14. 15. 16. 17. 18. 19. 20. 98.02% KOH 51.7 g Mg(OH)2 0.514 M NH3 20.85 mL oxalic acid a. 0.5056 M HCl b. 0.9118 M RbOH 570 kg Ca(OH)2 16.9 M HNO3 33.58 mL EQUILIBRIUM OF ACIDS AND BASES, Ka AND Kb 1. pH 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. TITRATIONS 1. 0.4563 M KOH 2. 2.262 M CH3COOH 3. 2.433 M NH3 2.857 Ka 1.92 10 5 pH 1.717 Ka 2.04 10 3 a. [H3O ] 3.70 10 11 M pH 10.431 Kb 1.82 10 7 b. [B] 4.66 10 3 M Kb 1.53 10 4 pH 10.93 c. [OH ] 1.9 10 3 M [H3O ] 5.3 10 12 M [B] 0.0331 M Kb 1.1 10 4 d. [B]initial 7.20 10 3 M Kb 1.35 10 4 pH 10.96 6.4 10 5 Kb 3.1 10 5 [H2NCH2CH2OH] 6.06 10 4 1.3 10 [OH ] 1.58 10 5 M pH 9.20 Ka 2.2 10 3 [HB]initial 0.0276 M 4.63 10 3 2.62 10 4 [H3O ] 0.0124 M pH 1.907 3 M Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt Chemistry 331 Answer Key Back Print 12. [H3O ] 13. 14. 15. 16. 17. 18. 19. 20. 0.0888 M pH 1.05 Ka 3.50 10 4 pH 2.241 a. 3.16 10 6 M b. 5.500 c. 0.025 M d. 4.0 10 10 e. HCN is a fairly weak acid. f. 5.8 10 6 M a. 0.740 b. 0.0325 c. 1.02 M d. It is moderately weak. 2.80 10 11 a. 0.200 M b. 0.233 pH 11.37 Kb 7.41 10 5 7.36 10 4 dimethylamine is the stronger base a. 8.3 10 7 b. 1.3 10 4 c. 10.11 15. KClO3 6HCl 3 3Cl2 16. 2KClO3 4HCl 3 17. 18. 19. 20. 21. 22. 3H2O KCl Cl2 2ClO2 2H2O 2KCl 2MnCl3 2H2O 3 MnCl2 MnO2 4HCl 2NaOH 6H2O 2Al 3 2NaAl(OH)4 3H2 6Br2 6Ca(OH)2 3 5CaBr2 Ca(BrO3)2 6H2O N2O 2NaClO 2NaOH 3 2NaCl 2NaNO2 H2O 4HBr MnO2 3 MnBr2 2H2O Br2 Au 4HCl HNO3 3 HAuCl4 NO 2H2O E0 2.77 V; spontaneous 0 E 1.11 V; spontaneous E0 0.46 V; not spontaneous E0 1.50 V; spontaneous E0 2.46 V; spontaneous 0 E 1.28 V; spontaneous E0 3.71 V; spontaneous E0 3.41 V; not spontaneous E0 1.32 V; spontaneous 0 E 3.60 V; not spontaneous Overall reaction: 2Cl Cl2 Ni 3 Ni2 Cathode reaction: Cl2 2e 3 2Cl Anode reaction: Ni 3 Ni2 2e Cell voltage: 1.62 V Overall reaction: 2Fe 3 3Hg 2Fe3 3Hg2 Cathode reaction: Hg2 2e 3 Hg 3e Anode reaction: Fe 3 Fe3 Cell voltage: 0.89 V Overall reaction: Al 3 3MnO2 Al3 3MnO4 4 Cathode reaction: e 3 MnO2 MnO4 4 Anode reaction: Al 3 Al3 3e Cell voltage: 2.22 V Overall reaction: 6H 5H2S 3 2MnO4 8H2O 5S 2Mn2 Cathode reaction: 8H 5e 3 Mn2 4H2O MnO4 Anode reaction: H2S 3 S 2H 2e Cell voltage: 1.36 V Overall reaction: Ca2 2Li 3 Ca 2Li Cathode reaction: Ca2 2e 3 Ca ELECTROCHEMISTRY 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. Oxidation, Reduction, and Electronegativity REDOX EQUATIONS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 3Mg N2 3 Mg3N2 SO2 Br2 2H2O 3 2HBr H2SO4 H2S Cl2 3 S 2HCl PbO2 4HBr 3 PbBr2 Br2 2H2O S 6HNO3 3 6NO2 H2SO4 2H2O NaIO3 N2H4 2HCl 3 N2 NaICl2 3H2O MnO2 H2O2 2HCl 3 MnCl2 O2 2H2O 3AsH3 4NaClO3 3 3H3AsO4 4NaCl K2Cr2O7 3H2C2O4 8HCl 3 2CrCl3 6CO2 2KCl 7H2O 2Hg(NO3)2 3 2HgO 4NO2 O2 4HAuCl4 3N2H4 3 4Au 3N2 16HCl 5Sb2(SO4)3 4KMnO4 24H2O 3 10H3SbO4 2K2SO4 4MnSO4 9H2SO4 3Mn(NO3)2 5NaBiO3 9HNO3 3 5Bi(NO3)2 3HMnO4 5NaNO3 3H2O H3AsO4 4Zn 8HCl 3 AsH3 4ZnCl2 4H2O 12. 13. 14. 15. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt Chemistry 332 Answer Key Back Print Anode reaction: Li 3 Li e Cell voltage: 0.17 V Overall reaction: 16H 10Br 3 2MnO4 8H2O 5Br2 2Mn2 Cathode reaction: 8H 5e 3 MnO4 2 4H2O Mn Anode reaction: 2Br 3 Br2 2e Cell voltage: 0.43 V Overall reaction: Sn 3 2Fe2 Sn2 2Fe3 Cathode reaction: Fe3 e 3 Fe2 2 Anode reaction: Sn 3 Sn 2e Cell voltage: 0.91 V Overall reaction: 14H 3Zn 3 Cr2O2 7 7H2O 3Zn2 2Cr3 Cathode reaction: 14H 6e 3 Cr2O2 7 7H2O 2Cr3 2e Anode reaction: Zn 3 Zn2 Cell voltage: 1.99 V Overall reaction: Ca Ba Ca2 3 Ba2 2e 3 Ca Cathode reaction: Ca2 2e Anode reaction: Ba 3 Ba2 Cell voltage: 0.04 V Overall reaction: 2Hg Cd Hg2 3 Cd2 2 2e 3 2Hg Cathode reaction: Hg2 2 2 2e Anode reaction: Cd 3 Cd Cell voltage: 1.20 V 16. 17. 18. 19. 20. Copyright © by Holt, Rinehart and Winston. All rights reserved. Holt Chemistry 333 Answer Key ...
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