SGCh03 - C hapter Three MASS RELATIONSHIPS IN CHEMICAL...

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Unformatted text preview: C hapter Three MASS RELATIONSHIPS IN CHEMICAL REACTIONS • • • • • • Atomic and Molecular Masses The Molar Mass of Elements and Compounds Chemical Formulas and Percent Composition Chemical Reactions and Equations Stoichiometry: Amounts of Reactants and Products Limiting Reagents and Reaction Yields ATOMIC AND MOLECULAR MASSES STUDY OBJECTIVES 1. 2. 3. Describe the atomic mass unit scale and the carbon-12 standard of the scale. Determine the average atomic mass of an element given the masses and percent abundances of its isotopes. Determine the molecular mass of a compound given its molecular formula. The Atomic Mass Unit Scale. The atomic mass unit scale works like this. The masses of individual atoms cannot be measured with a balance; but relative masses of the atoms of different elements can be 4 measured. For instance, it is possible to determine that an atom of 2 He is very close to 1/3 the mass of an atom 12 of 6 C. Next we assign a certain value for the mass of a carbon-12 atom. By international agreement an atom of carbon-12 has a mass of exactly 12 atomic mass units (amu). Carbon-12 is the standard of the amu scale. On this scale a helium-4 atom has a mass of 4.00 amu (1/3 of 12). Atoms heavier than carbon-12 have masses greater than 12 amu. Fluorine atoms, for instance, are 1.583 times heavier than carbon-12 atoms. Thus the mass of a fluorine atom is 19.00 amu. Other experiments showed that the lightest element is hydrogen, H atoms having about 1/12 the mass of carbon-12 atoms. In this way, the relative masses of atoms of all the elements have been established. Modern Atomic Masses. The atomic masses that appear in the present-day periodic table are determined by taking into account the fact that atoms of the elements exist as isotopes. First the mass of each isotope of an element is measured with a mass spectrometer relative to the atomic mass of carbon-12. Then the percent abundance of each isotope is measured. This information permits calculation of the average atomic mass. 6 For example, the element lithium has two isotopes that occur in nature: 3 Li with 7.5 percent abundance, Li with 92.5 percent abundance. The atomic mass of lithium-6 is 6.01513 amu, and that of lithium-7 is 7.01601 amu. The average mass of such a mixture of Li atoms is given by: 7 and 3 average atomic mass = (fraction of isotope X)(mass of isotope X) + (fraction of isotope Y)(mass of isotope Y) = (0.075)(6.01513 amu) + (0.925)(7.0161 amu) = 0.45 amu + 6.49 amu = 6.94 amu Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 8 / Mass Relationships in Chemical Reactions 6 7 Note that neither 3 Li nor 3 Li has an atomic mass of 6.94 amu. This is the average value for the mixture of the two Li isotopes. Molecular Masses. Compounds result when atoms of different elements bond to each other. There are two kinds of compounds: molecular and ionic. M olecules are tiny particles that are formed when atoms combine and are held together by chemical forces. A molecule of a compound is formed by the combination of atoms of different elements. The m olecular mass is the sum of the atomic masses of the atoms in the molecular formula. For example, the molecular masses of two nitrogen oxides NO 2 and N2 O5 are as follows: molecular mass of NO2 = atomic mass of N + 2(atomic mass of O) = 14.01 amu + 2(16.00 amu) = 46.01 amu atomic mass of N2 O5 = 2( atomic mass of N) + 5(atomic mass of O) = 2(14.01 amu) + 5(16.00 amu) = 108.02 amu _______________________________________________________________________________ EXAMPLE 3.1 Average Atomic Mass The element boron (B) consists of two stable isotopes with atomic masses of 10.012938 amu and 11.009304 amu. The average atomic mass of B is 10.81 amu. Which isotope is more abundant, boron-10 or boron-11? •Method of Solution The atomic mass is an average of the masses of the naturally occurring isotopes of an element. This means that each isotopic mass is multiplied by its percent abundance. If both isotopes in this example had a natural abundance of 50 percent, then the atomic mass would be just the average of the two masses, or approximately 10.5 amu. Answer: Because the average mass of boron (10.81 amu) is between 10.5 and 11.0 amu, the abundance of the boron-11 isotope must be greater than that of the boron-10 isotope. The actual abundances of boron-10 and boron-11 are 19.6% and 80.4%, respectively. _______________________________________________________________________________ EXAMPLE 3.2 Molecular Mass Calculate the molecular mass of carbon tetrachloride (CCl4 ). •Method of Solution The molecular mass is the sum of the atomic masses of all the atoms in the molecule: molecular mass CCl 4 = (12.01 amu) + 4(35.45 amu) = 153.81 amu _______________________________________________________________________________ EXERCISES 1. 2. Back An atom of oxygen-18 is 1.49992 times heavier than and atom of carbon-12. What is the mass of an oxygen-18 atom in amu? The element silver consists of two isotopes: 47 Ag with an atomic mass of 106.905 amu and a natural 109 abundance of 51.83%, and 47 Ag with an atomic mass of 108.905 amu and natural abundance of 48.17%. Calculate the average atomic mass of silver. Forward 107 Main Menu TOC Study Guide TOC Textbook Website MHHE Website Mass Relationships in Chemical Reactions / 3 9 THE MOLAR MASSES OF ELEMENTS AND COMPOUNDS STUDY OBJECTIVES 1. 2. 3. Define the mole as a base unit of the SI. Calculate the number of moles in a given amount of an element or a compound. Calculate the number of atoms, molecules, or formula units in a given amount of a substance. The Mole. In the laboratory, amounts of elements and compounds are measured in units of grams. In order to measure out equal numbers of atoms of two elements, say carbon and silicon, we cannot simply weigh out equal masses of the two elements. Instead, we must measure a gram ratio of the two that is the same as the mass ratio of one Si atom to one C atom. The atomic masses of Si and C are 28.09 amu and 12.01 amu, respectively. Thus, one Si atom has a mass 2.3 times greater than that of a C atom. It follows that any amounts of Si and C that have a mass ratio of Si to C of 2.3 : 1.0, will have equal numbers of Si and C atoms. The m olar mass of an element is a mass equal to its atomic mass expressed in grams. A molar mass of any element will contain the same number of atoms as a molar mass of any other element. Therefore, 28.09 g Si contains the same number of atoms as 12.01 g C. The number of atoms in a molar mass, called Avogadro's number , is equal to 6.022 × 10 23 atoms. The quantity that contains Avogadro's number of atoms or other entities is called a m ole . The following are examples of a mole of an element: One mole of carbon contains 6.022 × 10 23 C atoms and has a mass of 12.01 g (the molar mass). One mole of silicon contains 6.022 × 10 23 Si atoms and has a mass of 28.09 g (the molar mass). The term mole can be used in relation to any kind of particle, such as atoms, ions, or molecules. For clarity the particle must always be specified. We say 1 mole of O3 (ozone), or 1 mole of O 2 (diatomic oxygen), or 1 mole of Na+ (sodium ions). Other examples of molar amounts are: 1 mole Na + ions = 6.022 × 10 23 Na+ ions = 23.00 g Na+ 1 mole O 2 molecules = 6.022 × 10 23 O2 molecules = 32.00 g O2 1 mole O 3 molecules = 6.022 × 10 23 O3 molecules = 48.00 g O3 The mole is an SI unit that is defined in relation to the mass of the carbon-12 isotope. A m ole is the amount of substance of a system that contains as many elementary entities as there are atoms in 0.012 kg of carbon-12. In 0.012 kg of carbon-12 there are 6.022 × 10 23 carbon-12 atoms. The symbol for mole is m ol. Compounds. The m olar mass of a compound is its molecular mass expressed in grams. The molar masses of NO2 and N2 O5 are 46.01 g and 108.0 g, respectively. Since 46.01 g of NO2 contains 6.022 × 10 23 molecules of NO2 , this amount is 1 mole of NO2 . On the other hand, 1 mole of N2 O5 has a mass of 108.0 g. 1 mol NO 2 = 6.022 × 10 23 NO2 molecules = 46.01 g NO2 1 mol N2 O5 = 6.022 × 10 23 N2 O5 molecules = 108.0 g N2 O5 Also note that 1 mole of NO2 consists of 1 mole of N atoms, and 2 moles of O atoms. 1 mol NO 2 contains 1 mol N atoms = 14.01 g N 1 mol NO 2 contains 2 mol O atoms = 32.00 g O 1 mol NO 2 = 46.01 g NO2 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 0 / Mass Relationships in Chemical Reactions Just as the molecular mass is the sum of the atomic masses, the molar mass of a compound is the sum of the molar masses of the atoms in the molecule. _______________________________________________________________________________ EXAMPLE 3.3 The Molar Mass Naturally occurring lithium consists of 7.5 percent Li-6 atoms, and 92.5 percent Li-7 atoms. Complete the following sentences: a. b. The number of Li atoms in one mole of Li is ________________. One mole of Li atoms consists of _______ Li-6 atoms and ________ Li-7 atoms and will have a mass of ____ g. •Method of Solution a. b. One mole of Li contains 6.022 × 10 23 Li atoms (a mixture of Li-6 atoms and Li-7 atoms). One mole of Li atoms contains (0.074)(6.022 × 10 23 ) Li-6 atoms, which is equal to 4.45 × 10 22 6 Li atoms; and (0.925)(6.022 × 10 23 ) Li-7 atoms, which is 5.57 × 10 23 Li-7 atoms. Note that the total number of Li atoms is 6.022 × 10 23 , and thus will have a mass of 6.94 g. _______________________________________________________________________________ EXAMPLE 3.4 Mass of a Given Number of Moles What is the mass of 2.25 moles of iron (Fe)? •Method of Solution To convert moles of Fe to grams of Fe, first write the relationship between moles, atoms, and mass. 1 mol Fe = 6.022 × 10 23 Fe atoms = 55.85 g Fe The problem asks: ? g Fe = 2.25 mol Fe Since the atomic mass of iron is 55.85 amu, then 1 mol of iron has a mass of 55.85 g. The unit conversion factor is 55.85 g Fe =1 1 mol Fe •Calculation Since 1 mol Fe has a mass of 55.85 g, then 2.25 mol Fe must have a mass of: ? g Fe = 2.25 mol Fe × 5 5.85 g = 126 g Fe 1 mol Fe _______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Mass Relationships in Chemical Reactions / 4 1 _______________________________________________________________________________ EXAMPLE 3.5 Number of Atoms in a Given Mass How many zinc atoms are present in 20.0 g Zn? •Method of Solution We know that 1 mole of Zn contains 6.022 × 10 23 atoms. First find how many moles of Zn atoms are in 20.0 g Zn, and then multiply by Avogadro's number. Finding the number of moles of zinc is the key to finding the number of atoms. The relationship between moles, atoms, and grams is: 1 mol Zn = 6.022 × 10 23 Zn atoms = 65.39 g Zn The problem asks: ? Zn atoms = 20.0 g Zn Our road map is: g Zn → mol Zn → number of Zn atoms •Calculation One mol of Zn has a mass of 65.39 g. Therefore, 20.0 g Zn corresponds to: ? Zn mol = 20.0 g Zn × 1 m ol Zn = 0.3058 mol Zn 65.39 g Zn Here we will carry one more digit than the correct number of significant figures. To convert moles of zinc to atoms of zinc, we use Avogadro's number, the number of atoms per mole, as the unit factor: 6.022 × 1 0 23 Z n atoms =1 1 mol Zn ? Zn atoms = 0.3058 mol Zn × 6 .022 × 1 0 23 Z n atoms = 1.84 × 10 23 Zn atoms 1 mol Zn •Comment Instead of calculating the number of moles separately, we could have strung the conversions factors together into one calculation: ? Zn atoms = 20.0 g Zn × 1 m ol Zn 6 .022 × 1 0 23 Z n atoms × 65.39 g Zn 1 mol Zn = 1.84 × 10 23 Zn atoms _______________________________________________________________________________ EXAMPLE 3.6 Moles of a Molecular Compound a. b. How many molecules of ethane (C2 H6 ) are present in 50.3 g of ethane? How many atoms each of H and C are in this sample? •Method of Solution (a) The problem asks: ? C 2 H6 molecules = 50.3 g C2 H6 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 2 / Mass Relationships in Chemical Reactions To convert grams to molecules, you need the molar mass. The molecular mass of C 2 H6 is: 2(12.01 amu) + 6(1.008 amu) = 30.07 amu Therefore, we can write: 1 mol C2 H6 = 6.022 × 10 23 C 2 H6 molecules = 30.07 g C2 H6 Finding the number of moles of C2 H6 is the key to finding the number of molecules. g C 2 H6 → mol C2 H6 → number of C 2 H6 molecules •Calculation for (a) ? mol C2 H6 = 50.3 g C2 H6 Since 1 mol C2 H6 is 30.1 g, then 50.3 g C2 H6 must be: 1 m ol C 2 H6 ? mol C2 H6 = 50.3 g C2 H6 × = 1.67 mol C2 H6 30.07 g C 2 H6 Since 1 mol C2 H6 contains 6.022 × 10 23 molecules, then 1.67 mol must contain: ? C 2 H6 molecules = 1.67 mol C2 H6 × 6 .022 × 1 0 23 molecules 1 mol C 2 H6 = 1.01 × 10 24 C 2 H6 molecules •Method of Solution (b) The molecular formula shows the number and kinds of atoms in a molecule. 6 H atoms C 2 H6 molecule a nd 2 C atoms C 2 H6 molecule • Calculation for (b) Multiplying the number of C2 H6 molecules by the number of atoms of each kind per molecule gives the number of each kind of atom present. 1.01 × 10 24 molecules × 6 H a toms = 6.06 × 10 24 H atoms 1 molecule 2 C a toms = 2.02 × 10 24 C atoms 1 molecule _______________________________________________________________________________ 1.01 × 10 24 molecules × EXERCISES 3. Back Calculate the molecular mass of the following: a. carbon tetrachloride, CCl4 b. formaldehyde, H2 CO c. xenon difluoride, XeF2 Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Mass Relationships in Chemical Reactions / 4 3 4. Calculate the molar mass of the following: a. Na2 SO4 , sodium sulfate b. FeCl3 , iron(III) chloride c. Ba(OH)2 , barium hydroxide d. SO3 , sulfur trioxide 5. What is the mass of 1 mole of each of the following? a. Cu b. CuO c. CuSO4 d. CuSO 4 ·5H2 O 6. a. How many moles of silver are in 5.00 g Ag? b. How many moles of sodium are in 5.00 g Na? 7. a. How many silver atoms are in 5.0 g Ag? b. How many H2 molecules are there in 4.0 g H2 ? c. How many molecules are in 25.0 g of methane (CH4 )? 8. a. What is the mass of 2.0 moles of H2 ? b. What is the mass of 2.79 × 10 22 atoms of Ag? c. What is the average mass in grams of one atom of aluminum? 9. a. How many moles of NaNO 3 are in 8.72 g NaNO 3 ? b. How many moles of O atoms are in 8.72 g NaNO3 ? CHEMICAL FORMULAS AND PERCENT COMPOSITION STUDY OBJECTIVES 1. 2. 3. Determine the percent composition of a compound with a known formula. Determine the empirical formula of a compound given the percent composition. Determine a molecular formula when the molar mass is given. Percent Composition. The p ercent composition of a compound is the percentage by mass of each element in the compound. The percent composition can be determined by chemical analysis or it can be calculated directly if the formula of the compound is known. The percent is: percent of element = mass of element per mole of compound × 100% molar mass of compound Let us take sodium chloride (NaCl) as an example. The molar mass, 58.44 g, is the sum of the mass of 1 mole of Na, 22.99 g, and the mass of 1 mole of Cl, 35.45 g. The percentage of Na by mass is %Na = 22.99 g Na × 100% = 39.33% 58.44 g NaCl The percentage of Cl by mass is %Cl = Back Forward 35.45 g Cl × 100% = 60.66% 58.44 g NaCl Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 4 / Mass Relationships in Chemical Reactions Empirical and Molecular Formulas. There are two types of formulas that give information about compounds. The e mpirical formula gives the simplest whole-number ratios between the numbers of atoms of the elements making up the compound. In sulfur dioxide (SO2 ) there are twice as many O atoms in the compound as S atoms. The m olecular formula indicates the numbers of atoms of each element in a molecule of a substance. The molecular formula and the empirical formula may be the same for some substances, as in the case of SO2 , where one molecule contains two atoms of oxygen and one atom of sulfur. On the other hand, the molecular formula of benzene is C6 H6 , but the empirical formula is CH because the smallest or simplest ratio of C atoms to H atoms is 1 : 1. Chemical formulas of compounds are determined from a knowledge of the percent composition of the elements. For example, aluminum oxide is a solid with a percent composition of 52.92% Al and 47.08% O. Its empirical formula will give the ratio of Al atoms to O atoms. A convenient starting point for determining its empirical formula is to assume you have 100 g of aluminum oxide. Then, 100 g of compound will contain 52.92 g Al, and 47.08 g O. The number of moles of each element present is: nAl = 52.92 g Al × nO = 47.08 g O × 1 m ol Al = 1.96 mol Al 26.98 g Al 1 m ol O = 2.94 mol O 16.00 g O Dividing each number of moles by the smaller of the two molar amounts yields the following mole ratios: 2.94 mol O = 1.50 mol O to 1.00 mol Al 1.96 mol Al Which is the same as: 1.5 O atoms to 1 Al atom This is not the simplest whole-number ratio and suggests a formula of Al1 O1.5 . Multiplying both kinds of atoms by 2 gives the simplest whole-number ratio of 3.0 to 2.0. The empirical formula of aluminum oxide is Al2 O3 . The road map of the solution is: percent composition # moles each element m ole ratio atom ratio e mpirical formula Molecular Formula. When the molar mass is known, the molecular formula can be deduced from the empirical formula. For example, in the case of benzene, which has an empirical formula of CH, the molar mass is known to be 78.11 g. Since the empirical mass is 13.01 g, then there are 78.11/13.01 or 6.00 CH units per molecule, and so the molecular formula is C 6 H6 . _______________________________________________________________________________ EXAMPLE 3.7 Percent Composition Calculate the percent composition of glucose (C6 H12 O6 ). •Method of Solution First the molar mass of glucose must be found from the atomic masses of each element. molar mass of C6 H12 O6 = 6 mol C × 1 2.01 g 1 .008 g 1 6.00 g + 12 mol H × + 6 m ol O × 1 mol C 1 mol H 1 mol O = 72.06 g + 12.10 g + 96.00 g = 180.16 g Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Mass Relationships in Chemical Reactions / 4 5 Then the percent of each element present is found by dividing the mass of each element in 1 mole of glucose by the total mass of 1 mole of glucose. •Calculation mass percent of C = %C = mass of C in 1 mol C 6 H12 O6 × 100% mass of 1 mole C 6 H12 O6 6 m o l C × ( 12.01 g/mol C) 72.06 g × 100% = × 100% = 40.0% C 180.16 g C 6 H12 O6 180.16 g After practicing the calculation several times, you can simplify the notation: %H = 12 × 1 .008 g 12.1 g × 100% = × 100% = 6.72% H 180.16 g 180.16 g %O = 6 × 1 6.00 g 96.0 g × 100% = × 100% = 53.3% O 180.16 g 180.16 g Thus glucose (C 6 H12 O6 ) contains 40.0% C, 6.72% H, and 53.3% O by mass. •Comment A good way to check the results is to sum the percentages of the elements, because their total must add to 100%. Here we get 100.02%. _______________________________________________________________________________ EXAMPLE 3.8 Using Percent Composition Calculate the mass of carbon in 10.00 g of glucose (C6 H12 O6 ). •Method of Solution In Example 3.7 we calculated that glucose contains 40.00% carbon. Therefore: 40 g C mass C = 40.00% of 10.00 g C6 H12 O6 = × 10.00 g C6 H12 O6 = 4.000 g C 100 g C 6 H12 O6 •Calculation This calculation would have to be done from scratch if we did not already know the percentage of carbon. Our road map is: mass of glucose → mol glucose → mol carbon → g carbon 1 m ol C 6 H12 O6 6 m ol C 1 2.01 g C ? mass C = 10.00 g C6 H12 O6 × × × = 4.000 g C 180.16 g C 6 H12 O6 1 mol C 6 H12 O6 1 mol C _______________________________________________________________________________ EXAMPLE 3.9 Empirical Formula An elemental analysis of a new compound reveals its percent composition to be: 50.7% C, 4.23% H, and 45.1% O. Determine the empirical formula. •Method of Solution The empirical formula simply gives the relative numbers of atoms of each element in a formula unit. To start, assume you have exactly 100 g of compound, and then determine the number of moles of each element present. Then find the ratios of the number of moles. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 6 / Mass Relationships in Chemical Reactions •Calculation 1 m ol C = 4.22 mol C 12.01 g C 1 m ol H 4.23 g H × = 4.20 mol H 1.008 g H 1 m ol O 45.1 g O × = 2.82 mol O 16.00 g O 50.7 g C × This gives the mole ratio for C : H : O of 4.22 : 4.20 : 2.82. Dividing by the smallest of the molar amounts gives: 4.22 mol C = 1.50 C to 1.00 O 2.82 mol O and 4.20 mol H = 1.50 H to 1.00 O 2.82 mol O Therefore, C1.5 H1.5 O1.0 is a possible formula. This is not an acceptable formula because it does not contain all whole numbers. To convert these fractions to whole numbers, multiply each subscript by the same number, in this case a 2. The empirical formula therefore is C 3 H3 O2 _______________________________________________________________________________ EXAMPLE 3.10 Molecular Formula Mass spectrometer experiments on the compound in Example 3.9 show its molecular mass to be about 140 amu. What is the molecular formula? •Method of Solution First find how many empirical formula units there are in 1 mole of compound which is 140 g. The molar mass of the empirical formula C 3 H3 O2 is 71 g. •Calculation Dividing 140 g by 71 g shows that there are two empirical units per molecule. Therefore, the molecular formula must show twice as many atoms as the empirical formula. molecular formula = C6 H6 O4 _______________________________________________________________________________ EXAMPLE 3.11 Carbon–Hydrogen Analysis When a 0.761-g sample of a compound of carbon and hydrogen is burned in a C–H combustion apparatus (see Figure 3.5 in the text), 2.23 g CO2 and 1.37 g H 2 O are produced. Determine the percent composition of the compound. •Method of Solution Here we want the masses of carbon and hydrogen in the original compound. All the carbon is now present in 2.23 g of CO2 . All the hydrogen is now present in 1.37 g of H2 O. How many grams of C are there in 2.23 g of CO2 , and how many grams of H are there in 1.37 g of H2 O? We know there is 1 mole of C atoms (12.01 g) in every mole of CO 2 , or 44.01 g. Our road map is: g CO 2 → mol CO2 → g C Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Mass Relationships in Chemical Reactions / 4 7 •Calculation 1 m ol CO 2 1 2.01 g C × = 0.608 g C 44.01 g CO 2 1 mol CO 2 1 m ol H 2 O 2 .01 g H ? g H = 1.37 g H2 O × × = 0.153 g H 18.0 g H 2 O 1 mol H 2 O ? g C = 2.23 g CO2 × The percentage composition is %H = 0.153 g H × 100% = 20.1% H 0.761 g compound 0.608 g C × 100% = 79.9% C 0.761 g compound _______________________________________________________________________________ %C = EXERCISES 10. Calculate the percent composition by mass of the following compounds: a. CO 2 b. H 3 AsO4 c. CHCl3 d. NaNO2 e. H 2 SO4 11. What is the empirical formula of each of the following compounds? a. H 2 O2 b . CaF 2 c . C 2 H4 O2 d. B2 H6 12. Determine the empirical formula of each compound from the given percent compositions. a. 46.7% N, 53.3% O c. 55.3% K, 14.6% P, 30.1% O b. 63.6% N, 36.4% O d. 26.6% K, 35.4% Cr, 38.1% O 13. Cyclohexane has the empirical formula CH2 . Its molecular mass is 84.16 amu. What is its molecular formula? 14. When 2.65 mg of the substance responsible for the green color on the yolk of a boiled egg is analyzed, it is found to contain 1.42 mg Fe and 1.23 mg S. What is the empirical formula of the compound? 15. Ascorbic acid is a compound consisting of three elements: C, H, and O. When a 0.214 g sample is burned in oxygen, 0.320 g CO2 and 0.0874 g H 2 O are formed. What is the empirical formula of ascorbic acid? CHEMICAL REACTIONS AND EQUATIONS STUDY OBJECTIVES 1. 2. Back Write a chemical equation from a written description of the reaction that includes the names of reactants and products. Balance a chemical equation when given a "skeletal" equation. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 8 / Mass Relationships in Chemical Reactions Chemical Equations. A chemical equation is a symbolic way of representing a chemical reaction. The formulas of the reactants are written on the left side of the equation, and the formulas of the products on the right side. Equations are written according to the following format: 1. 2. Reactants and products are separated by an arrow. The arrow means "produces" or "yields." Plus signs are placed between reactants and between products. A plus sign between reactants implies that a mixture of two (and sometimes more) reactants is required for reaction. A plus between products implies that a mixture of two or more products is formed by the reaction. The plus sign can also be interpreted to mean "and" or just "plus." Table 3.1 Symbols Used in Writing Chemical Equations _______________________________________________ Symbol Meaning _______________________________________________ → yields, produces reversible reaction, equilibrium (s) solid phase (l) liquid phase (g) gas phase (aq) aqueous solution + added to, and _______________________________________________ 3. 4. Abbreviations are used to indicate states of matter. These symbols, given in Table 3.1, are written in parentheses and follow the elements or compounds to which they refer. Symbols such as ∆ indicating the conditions of a reaction are placed above the arrow. Equations are balanced. That is, the number of atoms of a certain element appearing in the reactants must be equal to the number of atoms of that element appearing in the products. A balanced equation shows the same number of each type of atom on both sides of the equation. This means that mass is conserved in chemical reactions. Balancing of equations is accomplished by making use of coefficients. A c oefficient is a number placed before a chemical formula in an equation which changes the amount of the substance. The coefficient is a multiplier for the formula. For example, 2H2 O means two molecules of water. Two molecules of water consists of 4 hydrogen atoms and 2 oxygen atoms. Similarly, 3H2 O means three molecules of water, which stands for 6 H atoms and 3 O atoms. Absence of a coefficient is understood to mean one. Balancing Equations. When sodium metal is heated in the presence of chlorine gas, a bright yellow flame is produced along with a white solid, sodium chloride. The unbalanced chemical equation representing the reaction is: Na + Cl2 → NaCl Writing a "balance sheet" of the numbers of atoms in the reactants and products is helpful. We see that the sodium atoms are balanced, but the chlorine atoms are not balanced. Reactants Na (1) Cl (2) Products Na (1) Cl (1) Two Cl atoms appear in the reactants, but only one appears in the products. To balance the equation, we must adjust the coefficients of the formulas. Since there are already two Cl atoms in the reactants, we can make two Cl atoms appear in the product by placing a 2 in front of NaCl. Two formula units of NaCl contain two Na atoms and two Cl atoms. The equation is now balanced for chlorine, but sodium is no longer balanced. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Mass Relationships in Chemical Reactions / 4 9 Na + Cl2 → 2NaCl Reactants Products Na (1) Na (2) Cl (2) Cl (2) Next sodium can be balanced by placing the coefficient 2 in front of Na. 2Na + Cl2 → 2NaCl Reactants Products Na (2) Na (2) Cl (2) Cl (2) The equation is now balanced, and tells us that 2 Na atoms react with one chlorine molecule to yield 2 formula units of NaCl. When balancing an equation, you should never alter the subscripts in the formulas. This would change the substances involved. For instance, removing the 2 for Cl 2 in order to make the equation balance is incorrect. Na + Cl → NaCl The reactant is diatomic chlorine (Cl2 ), a yellow-green gas, not atomic chlorine (Cl). These are two very different chemical species. Never change the formulas of the reactants or products. Adjust only their coefficients. In the examples that follow a few hints will be given that will make balancing easier. _______________________________________________________________________________ EXAMPLE 3.12 Balancing Chemical Equations Balance the following displacement reaction. In this reaction zinc displaces H2 from chemical combination with chlorine. Zn(s) + HCl(aq) → ZnCl 2 (aq) + H2 (g) •Method of Solution Step 1: As a general way to begin, identify elements that occur in only two compounds in the equation. In this case, each of the three elements Zn, H, and Cl occurs in only two compounds. Step 2: Of the elements that occur in only two compounds, identify and balance first that element with the largest subscript. Both Cl and H have the subscript 2, so balance either one of them first. Let's balance H by placing the coefficient 2 before HCl: Zn + 2HCl → ZnCl 2 + H2 Taking stock: Reactants H (2) Cl (2) Zn (1) Products H (2) Cl (2) Zn (1) At this point both H and Cl are balanced. Further inspection reveals that Zn is also balanced. Thus the equation is balanced! Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 5 0 / Mass Relationships in Chemical Reactions _______________________________________________________________________________ EXAMPLE 3.13 Balancing Chemical Equations The reaction of a hydrocarbon with oxygen is an example of a combustion reaction. Balance the equation for the reaction of pentane with oxygen gas: C 5 H12 + O2 → CO 2 + H2 O •Method of Solution Following step 1 given in the previous example, we note that both H and C appear in only two compounds. According to step 2, we note that of these, H has the largest subscript. Balance H first by placing the coefficient 6 in front of H2 O. Next balance the carbon atoms by placing a 5 in front of CO2 . C 5 H12 + O2 → 5CO 2 + 6H2 O Taking stock: Reactants C (5) H (12) O (2) Products C (5) H (12) O (16) Finally the oxygen atoms can be balanced. The products already have their coefficients determined, and so the reactants must be adjusted to supply 16 O atoms. Eight molecules of O2 contain 16 O atoms, and so write the coefficient 8 in front of O2 . C 5 H12 + 8O2 → 5CO 2 + 6H2 O Reactants Products C (5) C (5) H (12) H (12) O (16) O (16) _______________________________________________________________________________ EXAMPLE 3.14 Balancing Chemical Equations Balance the following decomposition reaction: NaClO3 (s) → NaCl(s) + O 2 (g) •Method of Solution According to steps 1 and 2, we should balance O first. There are 3 O atoms on the reactant side and 2 on the product side. A third step is helpful in this situation. Step 3: For an element that occurs in only two substances in the equation, note whether the number of atoms of that element is even on one side and odd on the other side of the equation. If this is the case, use two coefficients which increase the number of atoms on each side of the equation to the least common multiple. NaClO3 → NaCl + O2 Reactants O (3) Back Forward Products O (2) Main Menu TOC Study Guide TOC Textbook Website MHHE Website Mass Relationships in Chemical Reactions / 5 1 The least common multiple for balancing O is 3 × 2 = 6. The two coefficients are 2 and 3. 2NaClO3 → NaCl + 3O2 Reactants O (6) Products O (6) Balancing the Na and Cl atoms we get: 2NaClO3 → 2NaCl + 3O2 _______________________________________________________________________________ EXAMPLE 3.15 Balancing Chemical Equations Balance the following precipitation reaction in which sulfuric acid reacts with barium chloride in aqueous solution to yield hydrochloric acid and a precipitate of barium sulfate. H2 SO4 (aq) + BaCl2 (aq) → HCl(aq) + BaSO4 (s) •Method of Solution Here it helps to recognize certain groups of atoms that maintain their identity during the reaction. In this 2– equation SO4 , the sulfate ion, appears in the reactants and products as a unit. H2 SO4 + BaCl 2 → HCl + BaSO4 Reactants Products H (2) H (1) 2– 2– SO4 (1) SO4 (1) Ba (1) Cl (2) Ba (1) Cl (1) Either H or Cl could be balanced first. Balance the chlorine by placing the coefficient 2 in front of HCl: H2 SO4 + BaCl 2 → BaSO4 + 2HCl A check of the H atoms shows they are now balanced along with the other elements. _______________________________________________________________________________ EXERCISES 16. Rust (Fe 2 O3 ) forms readily when iron is exposed to air. Write a balanced chemical equation for the formation of rust. 17. Write a balanced chemical equation for the reaction between hydrogen gas and carbon monoxide to yield methanol (CH3 OH). 18. Balance the following equations: . a. __ CH4 + __ H2 O → __ CO2 + __ H2 b. __ H2 SO4 + __ NaOH → __ Na 2 SO4 + __ H2 O c. __ NH 3 + __ O2 → __ NO + __ H2 O Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 5 2 / Mass Relationships in Chemical Reactions STOICHIOMETRY: AMOUNTS OF REACTANTS AND PRODUCTS STUDY OBJECTIVES 1. 2. Write mole ratios relating the amounts of reactants and products in a chemical reaction. Calculate the mass of a substance that is produced or consumed in a chemical reaction, given the mass of any other species involved in the reaction. Mass Relationships in Reactions. Chemical equations contain information about the relative masses of reactants involved in the reaction. The balanced equation can be used to decide how many grams of one substance will be needed to react with a given mass of another, or how many grams of product can be produced by the reaction of a specific mass of a reactant. For instance, the balanced equation 2Al + Fe2 O3 → A l2 O3 + 2Fe 2 mol 1 mol 1 mol 2 mol 53.96 g 159.70 g 101.96 g 111.70 g states that 2 moles of Al will react completely with 1 mole of Fe2 O3 , and will yield 1 mole of Al2 O3 and 2 moles of Fe. Converting these molar amounts to grams, we find that 53.96 g Al reacts with 159.70 g Fe2 O3 , producing 101.96 g Al 2 O3 and 111.70 g Fe. The mass relationships as discussed in the above reaction can be determined only from the balanced equation. The equation, however, is written in terms of moles, not grams. The quantitative relationships between elements and compounds in chemical reactions is a part of chemistry called s toichiometry . In all stoichiometry problems the balanced chemical equation provides the "bridge" that allows us to relate the amount of a reactant to a given amount of another reactant, or to find the amounts required to yield a given amount of a product. Mole Ratios. To calculate the amount of a specific product that is formed when a given amount of one reactant is consumed, we need quantitative relationships. In the preceding equation, note that 2 moles of Al is consumed whenever 1 mole of Fe2 O3 reacts; and that whenever 1 mole of Al2 O3 is formed, 2 moles of Al must react. We can write the following equivalences: 2 mol Al 1 mol Fe2 O3 1 mol Al2 O3 2 mol Al The symbol means "is stoichiometrically equivalent to." This symbol is used because the equivalency cannot be universally applied to every reaction. These amounts are equivalent only in the context of this specific reaction. We can write conversion factors called mole ratios relating any two quantities that are equivalent. 1 m ol Al 2 O3 2 mol Al 2 mol Fe , , and 1 mol Fe 2 O3 2 mol Al 1 mol Fe 2 O3 One of these mole ratios will be used in the following example. How many grams of iron are produced when 25.0 g Fe2 O3 reacts with aluminum? 2Al + Fe2 O3 → Al 2 O3 + 2Fe Three steps are required. Since the balanced equation relates the number of moles of Fe2 O3 to moles of Fe, (1) convert the grams of Fe2 O3 to moles of Fe2 O3 using its molar mass. (2) Use a conversion factor (2 mol Fe formed per 1 mol Fe2 O3 consumed) to find the number of moles of Fe produced. (3) Convert the moles of Fe to grams using its molar mass. 2Al + Fe2 O3 → Al 2 O3 + 2Fe Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Mass Relationships in Chemical Reactions / 5 3 The road map outlined above is: step 1 step 2 step 3 g Fe 2 O3 → mol Fe2 O3 → mol Fe → g Fe produced The three steps are: 1 m ol Fe 2 O3 = 0.156 mol Fe2 O3 159.7 g Fe 2 O3 1. 25.0 g Fe2 O3 × 2. 0.156 mol Fe2 O3 × 3. 2 m ol Fe = 0.312 mol Fe 1 mol Fe 2 O3 5 5.58 g Fe 2 O3 0.312 mol Fe × = 17.3 g Fe 1 mol Fe This calculation can be set up and carried out in one equation. 1 m ol Fe 2 O3 5 5.58 g Fe 2 O3 2 m ol Fe 25.0 g Fe2 O3 × × × = 17.3 g Fe 159.7 g Fe 2 O3 1 mol Fe 2 O3 1 mol Fe The amount of product predicted by these calculations is called the theoretical yield. S ummary. Once you have the balanced chemical equation, you can work all stoichiometry problems using the three steps of the mole method. S t e p 1 . Convert the amounts of given substances into moles if necessary. S t e p 2 . Use the coefficients in the balanced equation to calculate the number of moles of the sought or unknown substance in the problem from the appropriate mole ratio. S t e p 3 . Convert the number of moles of the needed substance into mass units if required. Always check your answer for reasonableness. _______________________________________________________________________________ EXAMPLE 3.16 The Mole Method Sulfur dioxide can be removed from stack gases by reaction with quicklime, CaO. SO2 (g) + CaO(s) → CaSO 3 (s) If 975 g of SO2 is to be removed from stack gases by the above reaction, what mass of CaO is required to completely react with it? •Method of Solution First be certain that the equation is correctly balanced. Then follow the three steps as outlined in the above section. SO2 + C aO → CaSO3 step 1 step 2 step 3 g SO 2 → mol SO2 → mol CaO → g CaO 1. 2. 3. Back Convert the 975 g of SO2 to moles using its molar mass. Use the chemical equation to find the number of moles of CaO that reacts per mole of SO 2 . From the balanced equation we can write: 1 mol CaO 1 mol SO2 Convert moles of CaO to grams of CaO using the molar mass of CaO. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 5 4 / Mass Relationships in Chemical Reactions •Calculation Write out an equation which restates the problem: ? g CaO = 9.75 × 10 2 g SO2 String the conversion factors from the three steps one after the other. 1 m ol SO 2 1 m ol CaO 5 6.08 g CaO ? g CaO = 9.75 × 10 2 g SO2 × × × 64.07 g SO 2 1 mol SO 2 1 mol CaO = 8.53 × 10 2 g CaO _______________________________________________________________________________ EXERCISES 19. Silicon tetrachloride (SiCl4 ) can be prepared by heating silicon in the presence of chlorine gas: Si(s) + 2Cl2 (g) → S iCl 4 (l) a. How many moles of SiCl4 are produced when 4.24 moles of Cl2 gas reacts? b. How many grams of SiCl4 are produced when 4.24 moles of Cl2 gas reacts? 20. The following reaction can be used to prepare hydrogen gas: CaH2 + 2H2 O → Ca(OH)2 + 2H2 How many grams of H 2 will result from the reaction of 100 g CaH2 with an excess of H 2 O? 21. Oxygen gas can be produced by the decomposition of mercury(II) oxide, HgO. How many grams of O2 will be produced by the reaction of 24.2 g of the oxide? 2HgO(s) → 2Hg(l) + O 2 22. Carbon dioxide in the air of a spacecraft can be removed by its reaction with lithium hydroxide. CO2 (g) + 2LiOH(aq) → Li 2 CO3 (aq) + H2 O(l) On average, a person will exhale about one kg of CO 2 per day. How many kg of LiOH are required to react with 1.0 kg of CO2 ? LIMITING REAGENTS AND REACTION YIELDS STUDY OBJECTIVES 1. 2. Determine which reactant is the limiting reagent in a chemical reaction. Calculate the percent yield of a product. Limiting Reagents. Often the two reactants are not present in the exact ratio prescribed by the balanced chemical equation. In this situation, one reactant will be completely consumed before the other runs out. When this occurs, the reaction will stop and no more product will be made. The amount of this reactant determines the amount of product formed. The reactant that is consumed first is called the l imiting reagent . The reactant that is not completely consumed is called the e xcess reagent . Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Mass Relationships in Chemical Reactions / 5 5 Before reaction has started After reaction is completed Limiting reactant Excess reactant Figure 3.1 The limiting reagent reacts completely. Unreacted portions of the excess reagent remain. Take for example, the reaction: 2Al + Fe2 O3 → Al 2 O3 + 2Fe If 1.2 mol Al and 1 mol Fe2 O3 are mixed together, the reactant that is used up first is Al. We can tell this because the 1.0 mol Fe2 O3 given requires 2.0 mol Al for complete reaction. The 1.2 mol Al given will react with only 0.6 mol Fe2 O3 . When this reaction ceases, 0.4 mol Fe2 O3 will be left unreacted. Fe2 O3 is present in excess. In this situation aluminum is consumed first. Therefore, aluminum is the limiting reagent. In any stoichiometry problem, it is important to determine which reactant is the limiting one, in order to correctly predict the yield of products. Note, in this case, that the limiting reactant (1.2 mol Al) was not the reactant in lesser molar amount (1.0 mol Fe 2 O3 ). It is necessary to make use of the mole ratio of the reactants given by the balanced equation (the stoichiometric ratio). The amount of Fe2 O3 required to react with 1.2 mol Al is: 1 m ol Fe 2 O3 1.2 mol Al × = 0.6 mol Fe2 O3 2 mol Al Since 1.0 mol Fe2 O3 is given, there is more Fe2 O3 than needed to react completely with the given amount of Al. Fe 2 O3 is the excess reagent. Therefore, Al is the limiting reagent. Percent Yield. In practice it is not unusual for the actual yield to be less than the calculated, or "theoretical" yield. The theoretical yield is calculated by assuming that all of the limiting reagent reacts according to the balanced equation. The percent yield is defined as % yield = actual yield × 100% theoretical yield Limiting reagents and theoretical yields are illustrated in Examples 3.17 and 3.18. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 5 6 / Mass Relationships in Chemical Reactions _______________________________________________________________________________ EXAMPLE 3.17 Limiting Reagent Phosphine (PH 3 ) burns in oxygen to produce phosphorus pentoxide (P2 O5 ) and water. 2PH3 (g) + 4O 2 (g) → P 2 O5 (s) + 3H 2 O(l) How many grams of P2 O5 will be produced when 17.0 g of phosphine is mixed with 16.0 g of O2 and reaction occurs? •Method of Solution When the amounts of both reactants are given, it is possible that one will be completely used up before the other. The limiting reagent will be completely consumed and will determine the amount of products. To find the limiting reagent, first find the number of moles of each reagent available. •Calculation 1 m ol PH 3 = 0.500 mol PH3 33.99 g PH 3 1 m ol O 2 16.0 g O2 × = 0.500 mol O2 32.00 g O 2 17.0 g PH3 × The balanced equation gives the stoichiometric ratio of moles of O2 required to react per mole of PH3 . 2 mol O 2 stoichiometric ratio = 1 mol PH 3 The amount of O 2 required for complete reaction of the given amount of PH3 is: 2 m ol O 2 0.500 mol PH3 × = 1.00 mol O2 1 mol PH 3 Since only 0.50 mol O2 is given, then O2 is the limiting reagent. When all of the O2 is consumed, some PH3 will be left unreacted (the excess reagent). The theoretical yield of P2 O5 is calculated by assuming complete reaction of the limiting reagent. The road map is: g O2 → mol O2 → m ol P 2 O5 → g P 2 O5 1 m ol P 2 O5 1 41.9 g P 2 O5 ? g P 2 O5 = 0.500 mol O2 × × 4 mol O 2 1 mol P 2 O5 = 17.7 g P2 O5 _______________________________________________________________________________ EXAMPLE 3.18 Percent Yield In the reaction of 4.0 moles of N2 with 6.0 moles of H2 , a chemist obtained 1.6 moles of NH3 . What is the percent yield of NH3 ? 3H2 + N2 → 2NH3 •Method of Solution To calculate the percent yield, you must first calculate the theoretical yield. The theoretical yield of NH3 is determined by the limiting reagent. From the balanced equation we see that the stoichiometric ratio is 3 mol H 2 to 1 mol N2 . The 4.0 mol N2 given require 12.0 mol H 2 for complete reaction. Since only 6.0 mol H2 are Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Mass Relationships in Chemical Reactions / 5 7 given, then H 2 is the limiting reactant. In this case, 6.0 mol H2 will react with 2.0 mol N2 yielding 4.0 mol NH3 . 2 m ol NH 3 theoretical yield = 6.0 mol H 2 × = 4.0 mol NH3 3 mol H 2 The percent yield is found by dividing the actual yield by the theoretical yield. actual yield NH3 % yield NH3 = × 100% theoretical yield NH3 1.6 mol NH 3 = × 100% 4.0 mol NH 3 = 40% _______________________________________________________________________________ EXERCISES 23. Given 5.00 mol KOH and 2.00 mol H3 PO4 , how many moles of K3 PO4 can be prepared? H3 PO4 + 3KOH → K3 PO4 + 3H2 O 24. The reaction of iron ore with carbon follows the equation: 2Fe2 O3 + 3 C → 4Fe + 3CO2 a. How many grams of Fe can be produced from a mixture of 200 g Fe2 O3 and 300 g C? b. If the actual yield of Fe is 110 g, what is the percent yield of iron? 25. Hydrofluoric acid (HF) can be prepared according to the reaction: CaF 2 + H2 SO4 → 2HF + CaSO4 In one experiment 42.0 g CaF2 was treated with excess H2 SO4 and a yield of 14.2 g of HF was obtained. a. What is the theoretical yield of HF? b. Calculate the percent yield of HF. _______________________________________________________________________________ CONCEPTUAL QUESTIONS 1. 2. A mole is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12 g of the carbon-12 isotope. What is an elementary entity of a substance? 3. Sulfur atoms, on average, have twice the mass of O oxygen atoms. Why is it important to stress the wording "on average"? 4. Back Describe why the atomic mass unit scale is referred to as a relative scale. Describe how a compound differs from a mixture of elements at the molecular level? Draw two boxes about 5 cm on a side. In one box sketch a representation of hydrogen gas and oxygen gas at the level of atoms and molecules. In the other box show molecules of water in the liquid state. The sketches should show at least 12 particles in each box. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 5 8 / Mass Relationships in Chemical Reactions 5. An unknown element combines with oxygen to form a compound with a formula MO2 . If 25.0 g of the element combines with 4.50 g of oxygen, what is the atomic mass of M? PRACTICE TEST 1. If the standard of the atomic mass unit scale had been defined such that a fluorine atom was 10.0 amu, what would the atomic masses of oxygen and argon be? 2. How many times heavier are bromine atoms on the average than neon atoms? 3. A sample of magnesium could one day be brought back from a planet circling a distant star. What is the atomic mass of this magnesium if it has two isotopes with the following masses and abundances? magnesium-24 23.985 amu; 18.05% magnesium-26 25.982 amu; 81.95% 4. a. What is the mass in grams of 1.5 × 10 22 molecules of PCl5 ? b. What is the mass in grams of 1012 carbon atoms? 5. What is the mass of a single atom of fluorine in grams? 6. How many O atoms are in each of the following amounts? a. 1 mol ozone, O3 b. 1 mol CuO c. 20 g CuSO4 ·5H2 O 7. What mass of Ca 3 (PO4 )2 would contain 0.500 mol Ca? 8. a. How many moles of CaSO4 are there in 600 g CaSO4 ? b. How many moles of oxygen are there in 600 g CaSO4 ? c. How many oxygen atoms are in 600 g CaSO4 ? 9. How many antimony atoms are there in 5.00 mol Sb2 O5 ? 10. Answer the following questions concerning sulfur trioxide. a. What is the mass of 17.5 mol of SO3 ? b. What is the mass of 7.5 × 10 20 SO 3 molecules? 11. What is the empirical formula of each of the following compounds? a. C 6 H8 O6 b. P 4 O10 c. MgCl2 12. Calculate the percent composition by mass of the following compounds: a. Al 2 O3 b. HNO 3 c. CCl2 F 2 13. Calculate the percentage of fluorine by mass in the refrigerant HFC-134a. Its formula is C 2 H2 F 4 ? 14. The empirical formula of styrene is CH, and its molar mass is 104 g/mol. What is its molecular formula? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Mass Relationships in Chemical Reactions / 5 9 15. Chromic oxide is a compound of chromium and oxygen. Determine the empirical formula of chromic oxide given that 2.00 g of the compound contains 1.04 g Cr. 16. The compound glycerol contains three elements: C, H, and O. When a 0.740 g sample is burned in oxygen, 1.06 g CO2 and 0.580 g H 2 O are formed. What is the empirical formula of glycerol? 17. Balance the following equations: a. P 4 O10 + H2 O → H 3 PO4 b. Ga + H2 SO4 → Ga 2 (SO4 )3 + H2 c. C 4 H10 + O2 → CO2 + H2 O 18. Complete lines a, b, and c of the following table for the reaction: 2SO2 + O2 → 2SO 3 _______________________________________________ moles SO 2 grams O2 moles SO 3 grams SO3 _______________________________________________ a. 1.50 _____ _____ ______ b.______ 20.0 _____ ______ c.______ _____ 5.21 ______ _______________________________________________ 19. Sodium thiosulfate, the "hypo" used in photographic developing, can be made by the reaction: Na2 CO3 + 2Na 2 S + 4SO2 → 3Na2 S 2 O3 + CO2 a. How many grams of Na 2 CO3 (sodium carbonate) are required to produce 321 g of sodium thiosulfate, Na2 S 2 O3 ? b. How many grams of Na 2 S are required to react with 25.0 g Na2 CO3 if SO2 is present in excess? 20. Hydrofluoric acid (HF) can be prepared according to the following equation: CaF 2 + H2 SO4 → 2HF + CaSO4 a. How many grams of HF can be prepared from 75.0 g H 2 SO4 and 63.0 g CaF2 ? b. How many grams of the excess reagent will remain after the reaction ceases? c. If the actual yield of HF is 26.2 g, what is the percent yield? General Questions 21. The density of silver is 10.5 g/cm3 . How many Ag atoms are present in a silver bar that measures 0.10 m × 0.05 m × 0.01 m? 22. The pesticide malathion has the chemical formula C10 H19 O6 PS 2 . a. What is the molecular mass? b. The dose that is lethal to 50 percent of a human population is about 1.25 g per kilogram of body mass. How many molecules are in a dose lethal to an adult male weighing 70 kg? To an adult female of 58 kg? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 6 0 / Mass Relationships in Chemical Reactions ANSWERS Exercises 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 17.999 amu 107.87 amu a. 153.81 amu b. 30.03 amu c. 169.3 amu a. 142.04 g b. 162.21 g c. 171.3 g d. 80.08 g a. 63.546 g b. 79.545 g c. 159.6 g d. 249.7 g a. 0.0463 mol Ag b. 0.217 mol Na a. 2.79 × 10 22 atoms b. 1.2 × 10 24 atoms c. 9.41 × 10 23 molecules a. 4.0 g b. 5.0 g c. 4.48 × 10 –23 g a. 0.103 mol NaNO3 b. 0.308 mol O a. 27.29% C, 72.71% O b. 2.40% H, 59.5% As, 38.1% O c. 10.06% C, 0.844% H, 89.09% Cl d. 33.32% Na, 20.30% N, 46.38% O e. 2.06% H, 32.70% S, 65.24% O a. HO b. CaF2 c . CH 2 O d. BH3 a. NO b. N2 O c. K3 PO4 d. K2 Cr2 O7 C 6 H12 Fe2 S 3 C 3 H4 O3 4Fe + 3O2 → 2Fe 2 O3 2H2 + C O → CH3 OH a. CH4 + 2H2 O → CO 2 + 4H2 b. H2 SO4 + 2NaOH → Na2 SO4 + 2H2 O c. 4NH3 + 5O2 → 4NO + 6H2 O a. 2.12 mol b. 360 g 9.58 g H2 1.79 g 1.09 kg 1.67 mol a. Fe 2 O3 is the limiting reactant; 1.40 g Fe b. 78.6% yield a. 21.5 g b. 66% Conceptual Questions 1. 2. Back Atoms cannot be weighed directly. However, how many times heavier an atom of one element is compared to an atom of another element can be measured. These relative masses allow one to order the elements by their increasing atomic mass. Assigning a mass to one of the atoms makes it the standard for the scale. The individual masses of atoms of all the elements are assigned by comparison to the standard. The present standard of the atomic mass unit scale is the carbon-12 atom. See Section 3.1 of the text. The elementary entity in a mole depends on the nature of the substance we are referring to. For metallic iron the elementary entity is an iron atom, for water the elementary entity is an H2 O molecule. For sodium ions it would be a Na+ ion. The elementary entity is the species that is repeated many times to make up the given mass of the substance. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Mass Relationships in Chemical Reactions / 6 1 3. 4. 5. The average atomic mass of sulfur is 32.07 amu and that of oxygen is 16.00 amu. All sulfur atoms do not weigh 32.07 amu, and all oxygen atoms do not weigh 16.00 amu. So only on average do S atoms weigh 2.0 times as much as the average O atom. Models of H2 and O2 molecules are depicted in Figure 2.9of the text. A sketch should show at least 6 H2 molecules and 6 O2 molecules with large separations between them because they are both gases. H2 O molecules are also shown in the same figure. Sketch in 12 H2 O molecules placing them in the bottom half of the square because water is a liquid. Clearly water is not just a mixture of H2 and O2 . 178 amu Practice Test 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. Oxygen would be 8.4, and argon would be 21.0 3.96 times 25.62 amu a. 5.2 g b. 2.0 × 10 –11 g 3.15 × 10 –23 g a. 1.81 × 10 24 atoms b. 6.022 × 10 23 atoms c. 4.3 × 10 23 atoms 51.69 g a. 4.41 mol CaSO4 b. 17.6 mol O c. 1.06 × 10 25 O atoms 6.022 × 10 24 Sb atoms a. 1.40 × 10 3 g b. 0.0997 g a. C 3 H4 O3 b. P 2 O5 c. MgCl2 a. 52.91% Al, 47.08% O b. 1.60% H, 22.23% N, 76.17% O c. 9.93% C, 58.63% Cl, 31.43% F 74.5% F C 8 H8 CrO3 C 3 H8 O3 a. P 4 O10 + 6H2 O → 4H 3 PO4 b. 2Ga + 3H2 SO4 → Ga 2 (SO4 )3 + 3H2 c. 2C 4 H10 + 13O2 → 8CO2 + 10H2 O _______________________________________________ mol SO 2 g O 2 m ol SO 3 g SO3 _______________________________________________ a. 1.50 24.0 1.50 120 b. 1.25 2 0.0 1.25 100 c. 5.21 83.4 5 .21 417 _______________________________________________ a. 71.8 g Na2 CO3 b. 36.8 g Na2 S a. H 2 SO4 is the limiting reactant; 30.6 g HF b. 3.27 g CaF2 c. 85.6% yield 2.93 × 10 24 Ag atoms a. 330 amu b. male, 1.6 × 10 23 molecules; female, 1.3 × 10 23 molecules. _______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
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This note was uploaded on 09/15/2009 for the course CHEM 102 taught by Professor Bastos during the Spring '08 term at Adelphi.

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