Unformatted text preview: Molecularity CH1010
Lecture 1 James P. Dittami August 28, 2008 General Information MyWPI Course Site Syllabus General Text Grading Laboratory Chemistry Overview Chemist's language: Special Terms we use Chemistry is the study of Matter Matter the "Stuff" of the Universe We look at the Properties of Matter ... the Changes that matter undergoes ... the Energy associated with those changes Anything with Mass and Volume Chemistry Overview We study the Composition of matter The type and amount of simpler substances that make up matter. We are concerned with Properties Substances Matter with a defined fixed composition Special characteristics that give a substance a unique identity Chemistry Overview We are concerned with Physical Properties Characteristics absent of other substances We examine Chemical Properties Color, density, electrical conductivity, melting point Characteristics as substance changes to or interacts with other substances Flammability, corrosiveness Chemistry Overview We talk about Physical Changes Alteration of Physical form not composition
Melting, Freezing, hardness, viscosity We examine Chemical Changes Reactions Conversion of one substance to another water > hydrogen + oxygen Sample Problem 1.1 Distinguishing Between Physical and Chemical Change PROBLEM: Decide whether each of the following process is primarily a physical or a chemical change, and explain briefly: (a) Frost forms as the temperature drops on a humid winter night. (b) A cornstalk grows from a seed that is watered and fertilized. (c) Dynamite explodes to form a mixture of gases. (d) Perspiration evaporates when you relax after jogging. (e) A silver fork tarnishes slowly in air. PLAN: "Does the substance change composition or just change form?" SOLUTION: (a) physical change (b) chemical change (c) chemical change (d) physical change (e) chemical change Macrosopic vs Microscopic Matter is composed of Atoms Macroscopic World Cars Tables Rocks Oceans Microscopic World Atoms Molecules Macroscopic View The Beach Continuous solid surface Closer Inspection Microscopic View Macroscopic vs. Microscopic We Study Observable Changes on the Macroscopic Scale We attempt to understand these changes based on their Unobservable causes which are occuring at the molecular or atomic scale Atoms and Elements Fundamental Building Blocks Elements combine to make molecules also called compounds Everything in the Universe is composed of these basic elements E.g. Sugar is composed of Carbon, Hydrogen and Oxygen in fixed proportions About 116 different elements The physical states of matter. Energy is the capacity to do work.
energy due to the position of the object or energy from a chemical reaction Potential Energy Kinetic Energy energy due to the motion of the object Potential and kinetic energy can be interconverted. Energy is the capacity to do work. Figure 1.3A less stable
change in potential energy EQUALS kinetic energy more stable
A gravitational system. The potential energy gained when a lifted weight is converted to kinetic energy as the weight falls. Scientific Approach: Developing a Model Observations : Natural phenomena and measured events; universally consistent ones can be stated as a natural law. Tentative proposal that explains observations.
revised if experiments do not support it Hypothesis: Experiment: Procedure to test hypothesis; measures one variable at a time. Set of conceptual assumptions that explains data from accumulated experiments; predicts related phenomena. Model (Theory): altered if predictions do not support it Further Experiment: Tests predictions based on model. Units of Measurement Measurement a quantitative observation Consists of two parts A Number A scale or a Unit SI System of Units International System of Units Based on Metric System Includes several categories of Units Table 1. 2 SI Base Units
Unit Name kilogram meter second kelvin ampere mole candela Unit Abbreviation kg m s K A mol cd Physical Quantity (Dimension) mass length time temperature electric current amount of substance luminous intensity Table 1.3 Common Decimal Prefixes Used with SI Units
Prefix Symbol T G M k h da d c m n p f Prefix
tera giga mega kilo hecto deka deci centi milli micro nano pico femto Word
trillion billion million thousand hundred ten one tenth hundredth thousandth millionth billionth trillionth quadrillionth Conventional Notation
1,000,000,000,000 1,000,000,000 1,000,000 1,000 100 10 1 0.1 0.01 0.001 0.000001 0.000000001 0.000000000001 0.000000000000001 Exponential Notation 1x1012 1x109 1x106 1x103 1x102 1x101 1x100 1x101 1x102 1x103 1x106 1x109 1x1012 1x1015 SI Units Volume not included in the SI System Volume is derived from length cm3(Cubic centimeter) = cc = mL (milliliter) 1000 mL = 1 L (liter) Common SIEnglish Equivalent Quantities Quantity SI Unit SI Equivalent English Equivalent Length 1 kilometer(km) English to 1000(103)m SI Equivalent 0.62miles(mi)
Table 1.4 1 mi = 1.61km 1 meter(m) 100(102)cm 1000(103)mm 1.094yards(yd) 39.37inches(in) 1 yd = 0.9144m 1 foot (ft) = 0.3048m 1 centimeter(cm) 0.01(102)m 0.3937in 1 in = 2.54cm (exactly!) 1 kilometer(km) 1000(103)m 0.62mi Common SIEnglish Equivalent Quantities Quantity SI Unit SI Equivalent English Equivalent Volume English to SI Equivalent 1 cubic meter(m3)
Table 1.4 1,000,000(106) cubic centimeters 35.2cubic feet (ft3) 1 cubic decimeter(dm3) 1000cm3 0.2642 gallon (gal) 1.057 quarts (qt) 1 gal = 3.785 dm3 1 qt = 0.9464 dm3 1 ft3 = 0.0283m3 1 cubic centimeter (cm3) 0.001 dm3 0.0338 fluid ounce 1 qt = 946.4 cm3 1 fluid ounce = 29.6 cm3 Common SIEnglish Equivalent Quantities Quantity SI Unit SI Equivalent English Equivalent Mass English to 1 kilogram (kg) SI Equivalent 1000 grams 2,205 pounds (lb)
Table 1.4 1 (lb) = 0.4536 kg 1 gram (g) 1000 milligrams 0.03527 ounce(oz)
1 lb = 453.6 g 1 ounce = 28.35 g Mass vs. Weight Mass refers to quantity of matter Weight refers response of Mass to gravity Mass is the same on Earth or the Moon Weight differs on Earth vs. Moon due to gravitational field differences The SI Unit of Mass or kilogram is the only unit whose standard is a physical object a platinumiridium cylinder in France Uncertainty in Measurement Every measurement has a degree of uncertainty. Uncertainty depends upon the "precision" of measuring device. Precision defined as the degree of agreement among several measurements of the same quantity Five people read volume in syringe
1 2 3 4 5 2.15 mL 2.14 mL 2.16 mL 2.17 mL 2.16 mL Five people read volume in syringe
1 2 3 4 5 2.15 mL 2.14 mL 2.16 mL 2.17 mL 2.16 mL
Here 2.1 are "Certain Digits" The last number is "Uncertain" We have two certain digits Weigh and Object Five Times
1 2 3 4 5 2.486 g 2.487 g 2.485 g 2.484 g 2.488 g Five people read volume in syringe
1 2 3 4 5 2.486 g 2.487 g 2.485 g 2.484 g 2.488 g
Here 2.48 are "Certain Digits" The last number is "Uncertain" We have three certain digits Significant Figures Always indicate the degree of uncertainty Record all "certain digits" and the first "uncertain digit" Assume the "uncertain digit" is 1 E.g. 2.17 mL is read 2.17 0.01mL Precision vs. Accuracy Measurement is Accurate if it agrees with true value Errors in Accuracy can be Measurement is Precise if there is good agreement among several measurements of same quantity Random or indeterminate equal chance of being high or low Systematic or determinate consistently high or low Precision vs. Accuracy a) Neither precise nor accurate b) Precise but not accurate c) Accurate and Precise Counting Significant Figures 1. All non zero integers count as sig. figs. 2458 has 4 significant figures 27 has 2 significant figures 2.98 has 3 significant figures Counting Significant Figures 2. Zeros 3 types. Some count, some don't Leading zeros that precede non zero digits Do Not Count e.g. 0.0025 has only 2 sig. figs. Captive zeros occur between nonzero digits Always Count!! e.g. 2.09 has 3 sig. figs. Trailing Zeros occur to the right of the number Count only if there is a decimal point 200 has 1 sig. fig. but 200.0 has 4 sig. figs. 200. and 2.00 x 103 each have 3 sig. figs. Counting Significant Figures 3. Exact Numbers determined by counting or definition have infinite # sig.figs. So for example 12 apples, 8 molecules, 3 experiments Circle Circumferance = 2r (2 is exact) Sphere Volume=4/3 r2 (4 & 3 are exact) 1 inch = 2.54 cm (1 & 2.54 are exact) Calculations with Significant Figures 1. For addition and subtraction. The answer has the same number of decimal places as there are in the measurement with the fewest decimal places. Example: adding two volumes 83.5 mL + 23.28 mL 106.78 mL = 106.8 mL Example: subtracting two volumes 865.9 mL  2.8121 mL 863.0879 mL = 863.1 mL Rules for Significant Figures in Answers
2. For multiplication and division. The number with the least certainty limits the certainty of the result. Therefore, the answer contains the same number of significant figures as there are in the measurement with the fewest significant figures. Multiply the following numbers: 9.2 cm x 6.8 cm x 0.3744 cm = 23.4225 cm3 = 23 cm3 Rules for Rounding Off Numbers
1. If the digit removed is more than 5, the preceding number increases by 1. 5.379 rounds to 5.38 if three significant figures are retained and to 5.4 if two significant figures are retained. 2. If the digit removed is less than 5, the preceding number is unchanged. 0.2413 rounds to 0.241 if three significant figures are retained and to 0.24 if two significant figures are retained. 3.If the digit removed is 5, the preceding number increases by 1 if it is odd and remains unchanged if it is even. 17.75 rounds to 17.8, but 17.65 rounds to 17.6. If the 5 is followed only by zeros, rule 3 is followed; if the 5 is followed by nonzeros, rule 1 is followed: 17.6500 rounds to 17.6, but 17.6513 rounds to 17.7 4. Be sure to carry two or more additional significant figures through a multistep calculation and round off only the final answer. Convert 2.85 cm to inches Dimensional Analysis Converting from one system of units to another Look up equivalence statement So 1 in 2.54 cm = 1 in or 2.54 cm NOTE: Exact Numbers Now set up so units cancel 2.85 cm x 1 in = 1.12 in 2.54 cm cm cancels NOTE answer has 3 sig. figs. 2.54 cm 1 in Convert 7.00 inches to cm 7.00 in x 2.54 cm = 17.8 cm 7.00 1 in Note 3 sig. figs. in 7.00 and answer 17.8 2.54 cm/1 in has infinite sig.figs. 2.54 cm/1 in and 1 in/ 2.54 cm are "Unit Factors" Dimensional Analysis Converting from one system of units to another Dimensional Analysis For Practice doing conversions see sample problems 1.31.7 in text For Significant Figures see sample problems 1.81.9 Temperature Scales Three Scales Celsius Symbol = C Kelvin Symbol = K Fahrenheit Symbol = F Can relate all to b.p. or f.p. of water The freezing and boiling points of water. Temperature Scales and Interconversions Kelvin ( K )  The "Absolute temperature scale" begins at
absolute zero and only has positive values.
o o T (in K) = T (in oC) + 273.15 T (in oC) = T (in K)  273.15 T (in oF) = 9/5 T (in oC) + 32 T (in oC) = [ T (in oF)  32 ] 5/9 Celsius vs. Kelvin Temperature Unit is same size Zero Point Differs So in general for conversions: 0 oC = 273.15 K 0 K = 273.15 oC Tk = Tc + 273.15 Tc = Tk 273.15 Celsius vs. Fahrenheit Degree size is different must adjust Range of freezing to boiling is 180 oF Units = 100 oC Units So 180 oF /100 oC = 9 oF /5 oC 180 /100 = 9 /5 Zero Point is also different 32 oF = 0 oC 32 = 0 Celsius vs. Fahrenheit TC = 5/9 (TF32) TF= 9/5 TC + 32 oF Problems from Text Sample Problem 1.3 Converting Units of Length PROBLEM: To wire your stereo equipment, you need 325 centimeters (cm) of speaker wire that sells for $0.15/ft. What is the price of the wire? PLAN: Known length (in cm) of wire and cost per length ($/ft) We have to convert cm to inches and inches to feet followed by finding the cost for the length in ft. length (cm) of wire 2.54 cm = 1 in length (in) of wire 12 in = 1 ft length (ft) of wire 1 ft = $0.15 Price ($) of wire SOLUTION: Length (in) = length (cm) x conversion factor = 325 cm x in = 128 in 2.54 cm Length (ft) = length (in) x conversion factor ft = 10.7 ft 12 in Price ($) = length (ft) x conversion factor = 10.7 ft x $0.15 ft = $1.60 = 128 in x Sample Problem 1.4 Converting Units of Volume PROBLEM: When a small piece of galena, an ore of lead, is submerged in the water of a graduated cylinder that originally reads 19.9 mL, the volume increases to 24.5 mL. What is the volume of the piece of galena in cm3 and in L? PLAN: The volume of galena is equal to the change in the water volume before and after submerging the solid. volume (mL) before and after addition subtract volume (mL) of galena 1 mL = 1 cm3 volume (cm3) of galena SOLUTION: 1 mL = 103 L volume (L) of galena (24.5 19.9) mL = volume of galena = 4.6 mL
3 4.6 mL x 1 cm mL = 4.6 cm 3 4.6 mL x 10 L mL
3 = 4.6x103 L Sample Problem 1.5 Converting Units of Mass PROBLEM: What is the total mass (in kg) of a cable made of six strands of optical fiber, each long enough to link New York and Paris (8.84 x 103 km)? One strand of optical fiber used to traverse the ocean floor weighs 1.19 x 103 lbs/m. length (km) of fiber PLAN: The sequence of steps may vary but 1 km = 103 m essentially you have to find the length of the entire cable and convert it to mass. length (m) of fiber SOLUTION: 103 m 8.84 x 10 km x km
3 6 = 8.84 x 106 m 1 m = 1.19x103 lb mass (lb) of fiber 6 fibers = 1 cable mass (lb) of cable 2.205 lb = 1 kg mass (kg) of cable 1.19 x 10 3 lbs 8.84 x 10 m x = 1.05 x 104 lb m 6 fibers = 1.05 x 104 lb x cable cable 6.30 x 104 lb cable 1kg 6.30 x 104 lb 2.86 x 104 kg x = cable 2.205 lb Sample Problem 1.6 Calculating Density from Mass and Length PROBLEM: If a rectangular slab of Lithium (Li) weighs 1.49 x 103 mg and has sides that measure 20.9 mm by 11.1 mm by 11.9 mm, what is the density of Li in g/cm3 ? PLAN: Density is expressed in g/cm3 so we need the mass in grams and the volume in cm3. mass (mg) of Li 103 mg = 1 g mass (g) of Li lengths (mm) of sides 10 mm = 1 cm 1 g = 1.49 g 1000 mg lengths (cm) of sides multiply lengths volume (cm3) divide mass by volume density (g/cm3) of Li 1.49 g 2.76 cm3 SOLUTION: 1.49x103 mg x 20.9 mm x 1 cm = 2.09 cm 10 mm Similarly the other sides will be 1.11 cm and 1.19 cm, respectively. 2.09 x 1.11 x 1.19 = 2.76 cm3 density of Li = = 0.540 g/cm3 Sample Problem 1.7 Converting Units of Temperature PROBLEM: A child has a body temperature of 38.7C. (a) If normal body temperature is 98.6F, does the child have a fever? (b) What is the child's temperature in kelvins? PLAN: We have to convert C to F to find out if the child has a fever and we use the C to Kelvin relationship to find the temperature in Kelvin. 9 5 SOLUTION: (a) Converting from C to F (b) Converting from C to K (38.7C) + 32 = 101.7F 38.7C + 273.15 = 311.8K Sample Problem 1.8 Determining the Number of Significant Figures PROBLEM: For each of the following quantities, underline the zeros that are significant figures (sf), and determine the number of significant figures in each quantity. For (d) to (f), express each in exponential notation first. (a) 0.0030 L (d) 0.00004715 m PLAN: (b) 0.1044 g (e) 57,600. s (c) 53,069 mL (f) 0.0000007160 cm3 Determine the number of sf by counting digits and paying attention to the placement of zeros. (b) 0.1044 g 4sf (e) 57,600. s 5.7600x104 s 5sf (c) 53.069 mL 5sf (f) 0.0000007160 cm3 4sf 7.160x107 cm3 SOLUTION: (a) 0.0030 L 2sf (d) 0.00004715 m 4sf 4.715x105 m Sample Problem 1.9 Significant Figures and Rounding PROBLEM: Perform the following calculations and round the answer to the correct number of significant figures: 1 g 4.80x104 mg 1000 mg 16.3521 cm2 1.448 cm2 (a) (b) 7.085 cm 11.55 cm3 PLAN: In (a) we subtract before we divide; for (b) we are using an exact number. (a) 16.3521 cm2 1.448 cm2 7.085 cm 4.80x104 mg 1 g 1000 mg = = 14.904 cm2 7.085 cm 48.0 g 11.55 cm3 = 2.104 cm SOLUTION: (b) 11.55 cm3 = 4.16 g/ cm3 ...
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This note was uploaded on 09/08/2008 for the course CH 1010 taught by Professor Kumar during the Spring '06 term at WPI.
 Spring '06
 Kumar
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