Chemistry 9th edition zumdahl solutions manual 1 by james.reed414

Solution Manual for Chemistry 9th Edition by Zumdahl

ISBN 1133611095 9781133611097

Full download link at: Solution manual: https://testbankpack.com/p/solution-manual-for-chemistry9th-edition-by-zumdahl-isbn-1133611095-9781133611097/

CHAPTER 3 STOICHIOMETRY

Questions

23.

Average mass = 0.9889 0000) (12 + 0.0111(13.034) = 12.01 u

Note: u is an abbreviation for amu (atomic mass units).

From the relative abundances, there would be 9889 atoms of 12C and 111 atoms of 13C in the 10,000 atom sample. The average mass of carbon is independent of the sample size; it will always be 12.01 u.

By using the carbon-12 standard to define the relative masses of all of the isotopes, as well as to define the number of things in a mole, then each element’s average atomic mass in units of grams is the mass of a mole of that element as it is found in nature.

24 Consider a sample of glucose, C6H12O6. The molar mass of glucose is 180.16 g/mol. The chemical formula allows one to convert from molecules of glucose to atoms of carbon,

Isotope Mass Abundance

C 12.0000u 98.89% 13C 13.034 u 1.11%

12.01u

1.201

105 u

Total mass = 10,000 atoms × atom

0.9889(6.0221 ×

23) = 5.955 ×

0.0111(6.0221 × 1023) = 6.68 × 1021 atoms

mass = 6.0221 × 1023 atoms × atom 12.01u = 7.233 × 1024 u Total mass in g = 6.0221 × 1023 atoms × atom 12.01u × u 10 6.0221 1g 23  = 12.01 g/mol

For 1 mole of carbon (6.0221 × 1023 atoms C), the average mass would still be 12.01 u. The number of 12C atoms would be

atoms 12C, and the number of 13C atoms would be

Total

hydrogen, or oxygen present and vice versa. The chemical formula also gives the mole relationship in the formula. One mole of glucose contains 6 mol C, 12 mol H, and 6 mol O. Thusmoleconversionsbetweenmoleculesandatoms arepossibleusingthe chemicalfor-mula. The molar mass allows one to convert between mass and moles of compound, and Avogadro’s number (6.022 × 1023) allows one to convert between moles of compound and number of molecules.

25 Avogadro’s number of dollars = 6.022 × 1023 dollars/mol dollars

1 trillion = 1,000,000,000,000 = 1 × 1012; each person would have 90 trillion dollars.

26. Molar mass of CO2 = 12.01 + 2(16.00) = 44.01 g/mol

One mol of CO2 contains 6.022  1023 molecules of CO2, 6.022  1023 atoms of C, and 1.204  1024 atoms of O. We could also break down 1 mol of CO2 into the number of protons and the number of electrons present (1.325  1025 protons and 1.325  1025 electrons). In order to determine the number of neutrons present, we would need to know the isotope abundances for carbon and oxygen.

The mass of 1 mol of CO2 would be 44.01 g. From the molar mass, one mol of CO2 would contain 12.01 g C and 32.00 g O. We could also break down 1 mol of CO2 into the mass of protons and mass of electrons present (22.16 g protons and 1.207  10 2 g electrons). This assumes no mass loss when the individual particles come together to form the atom. This is not a great assumption as will be discussed in Chapter 19 on Nuclear Chemistry.

27. Only in b are the empirical formulas the same for both compounds illustrated. In b, general formulas of X2Y4 and XY2 are illustrated, and both have XY2 for an empirical formula.

For a, general formulas of X2Y and X2Y2 are illustrated. The empirical formulas for these two compounds are the same as the molecular formulas. For c, general formulas of XY and XY2 are illustrated; these general formulas are also the empirical formulas. For d, general formulas of XY4 and X2Y6 are illustrated. XY4 is also the molecular formula, but X2Y6 has the empirical formula of XY3.

28. The molar mass is the mass of 1 mole of the compound. The empirical mass is the mass of 1 mole of the empirical formula. The molar mass is a whole-number multiple of the empirical mass. The masses are the same when the molecular formula = empirical formula, and the masses are different when the two formulas are different. When different, the empirical mass must be multiplied by the same whole number used to convert the empirical formula to the molecular formula. For example, C6H12O6 is the molecular formula for glucose, and CH2O is the empirical formula. The whole-number multiplier is 6. This same factor of 6 is the multiplier used to equate the empirical mass (30 g/mol) of glucose to the molar mass (180 g/mol).

29 The mass percent of a compound is a constant no matter what amount of substance is present. Compounds always have constant composition.

CHAPTER 3 STOICHIOMETRY 47

people 10 7 dollars mol dollars 10 6.022 dollars 1mol 9 23    = 8.6 × 1013 = 9 × 10

13 dollars/person

30 A balanced equation starts with the correct formulas of the reactants and products. The coefficients necessary to balance the equation give molecule relationships as well as mole relationships between reactants and products. The state (phase) of the reactants and products is also given. Finally, special reaction conditions are sometimes listed above or below the arrow. These can include special catalysts used and/or special temperatures required for a reaction to occur.

31. Only one product is formed in this representation. This product has two Y atoms bonded to an X. The other substance present in the product mixture is just the excess of one of the reactants (Y). The best equation has smallest whole numbers. Here, answer c would be this smallest whole number equation (X + 2 Y  XY2). Answers a and b have incorrect products listed, and for answer d, an equation only includes the reactants that go to produce the product; excess reactants are not shown in an equation.

32. A balanced equation must have the same number and types of atoms on both sides of the equation, but it also needs to have correct formulas. The illustration has the equation as:

H + O  H2O

Under normal conditions, hydrogen gas and oxygen gas exist as diatomic molecules. So the first change to make is to change H + O to H2 + O2. To balance this equation, we need one more oxygen atom on the product side. Trial and error eventually gives the correct balanced equation of:

2 H2 + O2  2 H2O

This equation uses the smallest whole numbers and has the same number of oxygen atoms and hydrogen atoms on both sides of the equation (4 H + 2 O atoms). So in your drawing, there should be two H2 molecules, 1 O2 molecule, and 2 H2O molecules.

33 The theoretical yield is the stoichiometric amount of product that should form if the limiting reactant is completely consumed and the reaction has 100% yield.

34 A reactant ispresentinexcessifthereis more ofthat reactant present thanisneededto combine with the limiting reactant for the process. By definition, the limiting reactant cannot be present in excess. An excess of any reactant does not affect the theoretical yield for a process; the theoretical yield is determined by the limiting reactant.

35 The specific information needed is mostly the coefficients in the balanced equation and the molar masses of the reactants and products. For percent yield, we would need the actual yield of the reaction and the amounts of reactants used.

CHAPTER 3 STOICHIOMETRY

Mass of CB produced = 1.00 × 104 molecules A2B2 2 2 2 2 23 2 2 B A 1mol CB 2mol B A molecules 10 6.022 B A 1mol    CB mol CB of mass molar 

Atoms of A produced = 1.00 × 104 molecules A2B2 × 2 2B A 1molecule A 2atoms c. Moles of C reacted = 1.00 × 104 molecules A2B2 × 2 2 23 2 2 B A molecules 10 022 6 B A 1mol 

d. Percent yield = mass theoretical actualmass × 100; the theoretical mass of CB produced was calculated in part a. If the actual mass of CB produced is given, then the percent yield can be determined for the reaction using the percent yield equation.

36. One method is to assume each quantity of reactant is limiting, then calculate the amount of product that could be produced fromeach reactant. This gives two possible answers (assuming two reactants). The correct answer (the amount of product that could be produced) is always the smaller number. Even though there is enough of the other reactant to form more product, once the smaller quantity is reached, the limiting reactant runs out, and the reaction cannot continue.

A second method would be to pick one of the reactants and then calculate how much of the other reactant would be required to react with all of it. How the answer compares to the actual amount of that reactant present allows one to deduce the identity of the limiting reactant. Once the identity is known, one would take the limiting reactant and convert it to mass of product formed.

Exercises

Atomic Masses and the Mass Spectrometer

37. Let A = average atomic mass

A = 0.0140(203.973) + 0.2410(205.9745) + 0.2210(206.9759) + 0.5240(207.9766)

A = 2.86 + 49.64 + 45.74 + 109.0 = 207.2 u; from the periodic table, the element is Pb.

Note: u is an abbreviation for amu (atomic mass units).

38. Average atomic mass = A = 0.0800(45.952632) + 0.0730(46.951764) + 0.7380(47.947947) + 0.0550(48.947841) + 0.0540(49.944792) = 47.88 amu

This is element Ti (titanium).

39 Let A = mass of 185Re: 186.207 = 0.6260(186.956) + 0.3740(A), 186.207 117.0 = 0.3740(A)

A = 0.3740 69.2 = 185 u (A = 184.95 u without rounding to proper significant figures.)

40 Abundance 28Si = 100.00 (4.70 + 3.09) = 92.21%; from the periodic table, the average atomic mass of Si is 28.09 u.

28.09 = 0.9221(27.98) + 0.0470(atomic mass 29Si) + 0.0309(29.97)

CHAPTER 3 STOICHIOMETRY 49 × 2 2B A 1mol C mol 2

Atomic mass 29Si = 29.01 u

The mass of 29Si is actually a little less than 29 u. There are other isotopes of silicon that are considered when determining the 28.09 u average atomic mass of Si listed in the atomic table.

41 Let x = % of 151Eu and y = % of 153Eu, then x + y = 100 and y = 100 x

151.96 = 100 9209) )(152 (100 9196) (150 x x 

15196 = (150.9196)x + 15292.09 (152.9209)x, 96 = (2.0013)x

x = 48%; 48% 151Eu and 100 48 = 52% 153Eu

42 If silver is 51.82% 107Ag, then the remainder is 109Ag (48.18%). Determining the atomic mass (A) of 109Ag:

107.868 = 100 18(A) 48 905) 82(106 51 

10786.8 = 5540. + (48.18)A, A = 108.9 u = atomic mass of 109Ag

43 There are three peaks in the mass spectrum, each 2 mass units apart. This is consistent with two isotopes differing in mass by two mass units. The peak at 157.84 corresponds to a Br2 molecule composed of two atoms of the lighter isotope. This isotope has mass equal to 157.84/2 or 78.92. This corresponds to 79Br. The second isotope is 81Br with mass equal to 161.84/2 = 80.92. The peaks in the mass spectrum correspond to 79Br2, 79Br81Br, and 81Br2 in order of increasing mass. The intensities of the highest and lowest masses tell us the two isotopes are present in about equal abundance. The actual abundance is 50.68% 79Br and 49.32% 81Br.

44. Because we are not given the relative masses of the isotopes, we need to estimate the masses of the isotopes. A good estimate is to assume that only the protons and neutrons contribute to the overall mass of the atom and that the atomic mass of a proton and neutron are each 1.00 u. So the masses are about: 54Fe, 54.00 u; 56Fe, 56.00 u; 57Fe, 57.00 u; 58Fe, 58.00 u. Using these masses, the calculated average atomic mass would be:

0.0585(54.00) + 0.9175(56.00) + 0.0212(57.00) + 0.0028(58.00) = 55.91 u

The average atomic mass listed in the periodic table is 55.85 u.

Moles and Molar Masses

45 When more than one conversion factor is necessary to determine the answer, we will usually put all the conversion factors into one calculation instead of determining intermediate answers. This method reduces round-off error and is a time saver.

CHAPTER 3 STOICHIOMETRY

CHAPTER 3 STOICHIOMETRY 51 Fe mol Fe 55.85g Fe atoms 10 6.022 Fe 1mol Fe atoms 500 23    = 4.64 × 20 10 g Fe 46. 500.0 g Fe × Fe 85g 55 Fe 1mol = 8.953 mol Fe 8.953 mol Fe × Fe mol Fe atoms 10 022 6 23  = 5.391 × 1024 atoms Fe 47. 1.00 carat × C mol atomsC 10 6.022 C 01g 12 C 1mol carat C 0.200g 23    = 1.00 × 1022 atoms C 48. 5.0 × 1021 atoms C × atomsC 10 022 6 C 1mol 23  = 8.3 × 3 10 mol C 8.3 × 3 10 mol C × C mol C 12.01g = 0.10 g C 49. Al2O3: 2(26.98) + 3(16.00) = 101.96 g/mol Na3AlF6: 3(22.99) + 1(26.98) + 6(19.00) = 209.95 g/mol 50. HFC 134a, CH2FCF3: 2(12.01) + 2(1.008) + 4(19.00) = 102.04 g/mol HCFC 124, CHClFCF3: 2(12.01) + 1(1.008) + 1(35.45) + 4(19.00) = 136.48 g/mol 51 a. The formula is NH3. 14.01 g/mol + 3(1.008 g/mol) = 17.03 g/mol b. The formula is N2H4. 2(14.01) + 4(1.008) = 32.05 g/mol c. (NH4)2Cr2O7: 2(14.01) + 8(1.008) + 2(52.00) + 7(16.00) = 252.08 g/mol 52 a. The formula is P4O6. 4(30.97 g/mol) + 6(16.00 g/mol) = 219.88 g/mol b. Ca3(PO4)2: 3(40.08) + 2(30.97) + 8(16.00) = 310.18 g/mol c. Na2HPO4: 2(22.99) + 1(1.008) + 1(30.97) + 4(16.00) = 141.96 g/mol 53 a. 1.00 g NH3 × 3 3 NH 03g 17 NH 1mol = 0.0587 mol NH3 b. 1.00 g N2H4 × 4 2 4 2 H N 32.05g H N 1mol = 0.0312 mol N2H4 c. 1.00 g (NH4)2Cr2O7 × 7 2 2 4 7 2 2 4 O Cr ) (NH 08g 252 O Cr ) (NH 1mol = 3.97 × 3 10 mol (NH4)2Cr2O7

CHAPTER 3 STOICHIOMETRY

54. a. 1.00 g P4O6 × 88g 219 O P 1mol 6 4 = 4.55 × 3 10 mol P4O6 b. 1.00 g Ca3(PO4)2 × 18g 310 ) (PO Ca 1mol 2 4 3 = 3.22 × 3 10 mol Ca3(PO4)2 c. 1.00 g Na2HPO4 × 141.96g HPO Na 1mol 4 2 = 7.04 × 3 10 mol Na2HPO4 55. a. 5.00 mol NH3 × 3 3 NH mol NH 17.03g = 85.2 g NH3 b. 5.00 mol N2H4 × 4 2 4 2 H N mol H N 05g 32 = 160. g N2H4 c. 5.00 mol (NH4)2Cr2O7 × 7 2 2 4 7 2 2 4 O Cr ) (NH 1mol O Cr ) (NH 252.08g = 1260 g (NH4)2Cr2O7 56. a. 5.00 mol P4O6 × 6 4 O P 1mol 219.88g = 1.10 × 103 g P4O6 b. 5.00 mol Ca3(PO4)2 × 2 4 3 ) (PO Ca mol 18g 310 = 1.55 × 103 g Ca3(PO4)2 c. 5.00 mol Na2HPO4 × 4 2HPO Na mol 141.96g = 7.10 × 102 g Na2HPO4 57. Chemical formulas give atom ratios as well as mole ratios. a. 5.00 mol NH3 × N mol N 14.01g NH mol N 1mol 3  = 70.1 g N b. 5.00 mol N2H4 × N mol N 01g 14 H N mol N mol 2 4 2  = 140. g N c. 5.00 mol (NH4)2Cr2O7 × N mol N 01g 14 O Cr ) (NH mol N mol 2 7 2 2 4  = 140. g N 58. a. 5.00 mol P4O6 × P mol P g 30.97 O P mol P mol 4 6 4  = 619 g P b. 5.00 mol Ca3(PO4)2 × P mol P g 30.97 ) (PO Ca mol P mol 2 2 4 3  = 310. g P c. 5.00 mol Na2HPO4 × P mol P g 97 30 HPO Na mol P 1mol 4 2  = 155 g P

CHAPTER 3 STOICHIOMETRY 53 59 a. 1.00 g NH3 × 3 3 23 3 3 NH mol NH molecules 10 022 6 NH 03g 17 NH 1mol   = 3.54 × 1022 molecules NH3 b. 1.00 g N2H4 × 4 2 4 2 23 4 2 4 2 H N mol H N molecules 10 6.022 H N 32.05g H N 1mol   = 1.88 × 1022 molecules N2H4 c. 1.00 g (NH4)2Cr2O7 × 7 2 2 4 7 2 2 4 O Cr ) (NH 08g 252 O Cr ) (NH 1mol 7 2 2 4 7 2 2 4 23 O Cr ) (NH mol O Cr ) (NH units formula 10 6.022   = 2.39 × 1021 formula units (NH4)2Cr2O7 60. a. 1.00 g P4O6 × 6 4 23 6 4 O P mol molecules 10 022 6 219.88g O P 1mol   = 2.74 × 1021 molecules P4O6 b. 1.00 g Ca3(PO4)2 × 2 4 3 23 2 4 3 ) (PO Ca mol units formula 10 6.022 18g 310 ) (PO Ca 1mol   = 1.94 × 1021 formula units Ca3(PO4)2 c. 1.00 g Na2HPO4 × 4 2 23 4 2 HPO Na mol units formula 10 6.022 141.96g HPO Na 1mol   = 4.24 × 1021 formula units Na2HPO4 61. Using answers from Exercise 59: a. 3.54 × 1022 molecules NH3 × 3NH molecule N 1atom = 3.54 × 1022 atoms N b. 1.88 × 1022 molecules N2H4 × 4 2H N molecule N 2atoms = 3.76 × 1022 atoms N c. 2.39 × 1021 formula units (NH4)2Cr2O7 × 7 2 2 4 O Cr ) (NH unit formula N 2atoms = 4.78 × 1021 atoms N 62. Using answers from Exercise 60: a. 2.74 × 1021 molecules P4O6 × 6 4O P molecule P 4atoms = 1.10 × 1022 atoms P b. 1.94 × 1021 formula units Ca3(PO4)2 × 2 4 3 ) (PO Ca unit formula P 2atoms = 3.88 × 1021 atoms P

CHAPTER 3 STOICHIOMETRY

The •2H2O is part of the formula of bauxite (they are called waters of hydration). Combining elements together, the chemical formula for bauxite would

c. 4.24 × 1021 formula units Na2HPO4 × 4 2HPO Na unit formula P 1atom = 4.24 × 1021 atoms P 63 Molar mass of CCl2F2 = 12.01 + 2(35.45) + 2(19.00) = 120.91 g/mol 5.56 mg CCl2F2 mol molecules 10 6.022 91g 120 1mol mg 1000 1g 23     = 2.77 × 1019 molecules CCl2F2 5.56 × 10−3 g CCl2F2 Cl mol Cl 35.45g F CCl 1mol Cl mol 2 91g 120 F CCl 1mol 2 2 2    = 3.26 × 10−3 g = 3.26 mg Cl 64.

be Al2O5H4. a. Molar mass = 2(26.98) + 5(16.00) + 4(1.008) = 137.99 g/mol b. 0.58 mol Al2O3•2H2O O 2H O Al mol Al mol 2 2 3 2   Al mol Al 26.98g  = 31 g Al c. 0.58 mol Al2O3•2H2O Al mol atoms 10 6.022 O 2H O Al mol Al 2mol 23 2 3 2     = 7.0 × 1023 atoms Al d. 2.1 × 1024 formula units Al2O3•2H2O mol 99g 137 units formula 10 022 6 O 2H O Al 1mol 23 2 3 2     = 480 g Al2O3•2H2O 65 a. 150.0 g Fe2O3 × 159.70g 1mol = 0.9393 mol Fe2O3 b. 10.0 mg NO2 × 46.01g 1mol mg 1000 1g  = 2.17 × 4 10 mol NO2 c. 1.5 × 1016 molecules BF3 × molecules 10 02 6 1mol 23  = 2.5 × 8 10 mol BF3 66 a. 20.0 mg C8H10N4O2 × 194.20g 1mol mg 1000 1g  = 1.03 × 4 10 mol C8H10N4O2 b. 2.72 × 1021 molecules C2H5OH × molecules 10 022 6 1mol 23  = 4.52 × 3 10 mol C2H5OH

67. a. A chemical formula gives atom ratios as well as mole ratios. We will use both ideas to show how these conversion factors can be used.

CHAPTER 3 STOICHIOMETRY 55 c. 1.50 g CO2 × 01g 44 1mol = 3.41 × 2 10 mol CO2

Molar mass of C2H5O2N = 2(12.01) + 5(1.008) + 2(16.00) + 14.0l = 75.07 g/mol 5.00 g C2H5O2N N O H C mol N O H moleculesC 10 6.022 N O H C g 75.07 N O H C 1mol 2 5 2 2 5 2 23 2 5 2 2 5 2    N O H C molecule N 1atom 2 5 2  = 4.01 × 1022 atoms N b. Molar mass of Mg3N2 = 3(24.31) + 2(14.01) = 100.95 g/mol 5.00 g Mg3N2 2 3 2 3 23 2 3 2 3 N Mg mol N Mg units formula 10 6.022 N Mg 100.95g N Mg 1mol    2 3N Mg mol N 2atoms  = 5.97 × 1022 atoms N c. Molar mass of Ca(NO3)2 = 40.08 + 2(14.01) + 6(16.00) = 164.10 g/mol 5.00 g Ca(NO3)2 N mol N atoms 10 6.022 ) Ca(NO mol N 2mol ) Ca(NO 164.10g ) Ca(NO 1mol 23 2 3 2 3 2 3     = 3.67 × 1022 atoms N d. Molar mass of N2O4 = 2(14.01) + 4(16.00) = 92.02 g/mol 5.00 g N2O4 N mol N atoms 10 6.022 O N mol N 2mol O N 02g 92 O N 1mol 23 4 2 4 2 4 2     = 6.54 × 1022 atoms N 68 4.24 g C6H6 × 78.11g 1mol = 5.43 × 2 10 mol C6H6 5.43 × 2 10 mol C6H6 × mol molecules 10 022 6 23  = 3.27 × 1022 molecules C6H6 Each molecule of C6H6 contains 6 atoms C + 6 atoms H = 12 atoms total 3.27 × 1022 molecules C6H6 × molecule total 12atoms = 3.92 × 1023 atoms total 0.224 mol H2O × mol 18.02g = 4.04 g H2O

0.224 mol H2O × mol molecules 10 022 6 23  = 1.35 × 1023 molecules H2O 1.35 × 1023 molecules H2O × molecule total 3atoms = 4.05 × 1023 atoms total 2.71 × 1022 molecules CO2 × molecules 10 022 6 1mol 23  = 4.50 × 2 10 mol CO2 4.50 × 2 10 mol CO2 × mol 01g 44 = 1.98 g CO2 2.71 × 1022 molecules CO2 × 2CO molecule total 3atoms = 8.13 × 1022 atoms total 3.35 × 1022 atoms total × total 6atoms 1molecule = 5.58 × 1021 molecules CH3OH 5.58 × 1021 molecules CH3OH × molecules 10 6.022 1mol 23  = 9.27 × 3 10 mol CH3OH 9.27 × 3 10 mol CH3OH × mol 32.04g = 0.297 g CH3OH 69. Molar mass of C6H8O6 = 6(12.01) + 8(1.008) + 6(16.00) = 176.12 g/mol 500.0 mg × 12g 176 1mol mg 1000 1g  = 2.839 × 3 10 mol C6H8O6 2.839 × 10−3 mol mol molecules 10 6.022 23   = 1.710 × 1021 molecules C6H8O6 70. a. 9(12.01) + 8(1.008) + 4(16.00) = 180.15 g/mol b. 500. mg × 15g 180 1mol mg 1000 1g  = 2.78 × 3 10 mol C9H8O4 2.78 × 3 10 mol × mol molecules 10 6.022 23  = 1.67 × 1021 molecules C9H8O4 71. a. 2(12.01) + 3(1.008) + 3(35.45) + 2(16.00) = 165.39 g/mol b. 500.0 g × g 165.39 1mol = 3.023 mol C2H3Cl3O2

CHAPTER 3 STOICHIOMETRY

72 As we shall see in later chapters, the formula written as (CH3)2N2O tries to tell us something about how the atoms are attached to each other. For our purposes in this problem, we can write the formula as C2H6N2O.

CHAPTER 3 STOICHIOMETRY 57 c. 2.0 × 10-2 mol × mol 165.39g = 3.3 g C2H3Cl3O2 d. 5.0 g C2H3Cl3O2 × molecule 3atomsCl mol molecules 10 022 6 165.39g 1mol 23    = 5.5 × 1022 atoms of chlorine e. 1.0 g Cl × 2 3 3 2 2 3 3 2 2 3 3 2 O Cl H C mol O Cl H C 165.39g Cl 3mol O Cl H C 1mol 35.45g Cl 1mol   = 1.6 g chloral hydrate f. 500 molecules × mol 165.39g molecules 10 022 6 1mol 23   = 1.373 × 10 19 g C2H3Cl3O2

a. 2(12.01) + 6(1.008) + 2(14.01) + 1(16.00) = 74.09 g/mol b. 250 mg 74.09g 1mol mg 1000 1g   = 3.4 × 10−3 mol c. 0.050 mol × mol 09g 74 = 3.7 g d. 1.0 mol C2H6N2O O N H C molecule H 6atomsof O N H C mol O N H C molecules 10 6.022 2 6 2 2 6 2 2 6 2 23    = 3.6 × 1024 atoms of hydrogen e. 1.0 × 106 molecules mol 74.09g molecules 10 022 6 1mol 23    = 1.2 × 10−16 g f. 1 molecule mol 74.09g molecules 10 6.022 1mol 23    = 1.230 × 10−22 g C2H6N2O Percent Composition 73. a. C3H4O2: Molar mass = 3(12.01) + 4(1.008) + 2(16.00) = 36.03 + 4.032 + 32.00 = 72.06 g/mol Mass % C = compound 06g 72 C 03g 36 × 100 = 50.00% C Mass % H = compound 06g 72 H 032g 4 × 100 = 5.595% H Mass % O = 100.00 (50.00 + 5.595) = 44.41% O or:

CHAPTER 3 STOICHIOMETRY

From the calculated mass percents, only NO is 46.7% N by mass, so NO could be this species. Any other compound having NO as an empirical formula could also be the compound.

% O = 06g 72 00g 32 × 100 = 44.41% O b. C4H6O2: Molar mass = 4(12.01) + 6(1.008) + 2(16.00) = 48.04 + 6.048 + 32.00 = 86.09 g/mol Mass % C = 09g 86 48.04g × 100 = 55.80% C; mass % H = g 09 86 6.048g × 100 = 7.025% H Mass % O = 100.00 (55.80 + 7.025) = 37.18% O c. C3H3N: Molar mass = 3(12.01) + 3(1.008) + 1(14.01) = 36.03 + 3.024 + 14.01 = 53.06 g/mol Mass % C = g 53.06 36.03g × 100 = 67.90% C; mass % H = g 53.06 g 3.024 × 100 = 5.699% H Mass % N = g 53.06 14.01g × 100 = 26.40% N or % N = 100.00 (67.90 + 5.699) = 26.40% N 74 In 1 mole of YBa2Cu3O7, there are 1 mole of Y, 2 moles of Ba, 3 moles of Cu, and 7 moles of O. Molar mass = 1 mol Y                  Ba mol Ba 3g 137 Ba mol 2 Y mol Y 91g 88 + 3 mol Cu         Cu mol Cu 63.55g + 7 mol O         O mol O 16.00g Molar mass = 88.91 + 274.6 + 190.65 + 112.00 = 666.2 g/mol Mass % Y = 2g 666 88.91g × 100 = 13.35% Y; mass % Ba = 2g 666 274.6g × 100 = 41.22% Ba Mass % Cu = 2g 666 65g 190 × 100 = 28.62% Cu; mass % O = 2g 666 0g 112 × 100 = 16.81% O 75. NO: Mass % N = NO 30.01g N 14.01g × 100 = 46.68% N NO2: Mass % N = 2NO 01g 46 N 14.01g × 100 = 30.45% N N2O: Mass % N = O N 02g 44 N g 01) 2(14 2 × 100 = 63.65% N

The order from lowest to highest mass percentage of carbon is:

77 There are 0.390 g Cu for every 100.000 g of fungal laccase. Assuming 100.00 g fungal laccase:

78. There are 0.347 g Fe for every 100.000 g hemoglobin (Hb). Assuming 100.000 g hemoglobin:

Empirical and Molecular Formulas

CHAPTER 3 STOICHIOMETRY 59 76 a. C8H10N4O2: Molar mass = 8(12.01) + 10(1.008) + 4(14.0l) + 2(16.00) = 194.20 g/mol Mass % C = 2 4 10 8 O N H C 20g 194 C g 8(12.01) × 100 = 20g 194 08g 96 × 100 = 49.47% C

C12 H22O11: Molar mass = 12(12.01) + 22(1.008) + 11(16.00) = 342.30 g/mol Mass % C = 11 22 12 O H C 30g 342 C 01)g 12(12 × 100 = 42.10% C

C2H5OH: Molar mass = 2(12.01) + 6(1.008) + 1(16.00) = 46.07 g/mol Mass % C = OH H C g 46.07 C g 2(12.01) 5 2 × 100 = 52.14% C

sucrose (C12H22O11) < caffeine (C8H10N4O2) < ethanol (C2H5OH)

Mol fungal laccase = 0.390 g Cu × Cu mol 4 laccase fungal 1mol Cu 55g 63 Cu 1mol  = 1.53 × 3 10 mol laccase fungal mol fungallaccase g x = mol 10 1.53 000g 100 3  , x = molar mass = 6.54 × 104 g/mol

Mol Hb = 0.347 g Fe Fe mol 4 Hb 1mol Fe 85g 55 Fe 1mol   = 1.55 × 10−3 mol Hb Hb mol Hb g x = Hb mol 10 55 1 Hb 000g 100 3  , x = molar mass = 6.45 × 104 g/mol

79 a. Molar mass of CH2O = 1 mol C         C mol C 01g 12 + 2 mol H         H mol H 008g 1 + 1 mol O         O mol O 16.00g = 30.03 g/mol % C = O CH 03g 30 C 01g 12 2 × 100 = 39.99% C; % H = O CH 03g 30 H 016g 2 2 × 100 = 6.713% H

CHAPTER 3 STOICHIOMETRY

% O = O CH 03g 30 O 00g 16 2 × 100 = 53.28% O or % O = 100.00 (39.99 + 6.713) = 53.30%

b. Molar mass of C6H12O6 = 6(12.01) + 12(1.008) + 6(16.00) = 180.16 g/mol

% O = 100.00 (40.00 + 6.714) = 53.29%

c. Molar mass of HC2H3O2 = 2(12.01) + 4(1.008) + 2(16.00) = 60.05 g/mol

% C = 60.05g 24.02g × 100 = 40.00%; % H = 60.05g 4.032g × 100 = 6.714%

% O = 100.00 (40.00 + 6.714) = 53.29%

80 All three compounds have the same empirical formula, CH2O, and different molecular formulas. The composition of all three in mass percent is also the same (within rounding differences). Therefore, elemental analysis will give us only the empirical formula.

a. The molecular formula is N2O4. The smallest whole number ratio of the atoms (the empirical formula) is NO2

b. Molecular formula: C3H6; empirical formula: CH2

c. Molecular formula: P4O10; empirical formula: P2O5

d. Molecular formula: C6H12O6; empirical formula: CH2O

a. SNH: Empirical formula mass = 32.07 + 14.01 + 1.008 = 47.09 g/mol 47.09g 188.35g = 4.000; so the molecular formula is (SNH)4 or S4N4H4

b. NPCl2: Empirical formula mass = 14.01 + 30.97 + 2(35.45) = 115.88 g/mol 115.88g 347.64g = 3.0000; molecular formula is (NPCl2)3 or N3P3Cl6.

c. CoC4O4: 58.93 + 4(12.01) + 4(16.00) = 170.97 g/mol g 97 170 94g 341 = 2.0000; molecular formula: Co2C8O8

d. SN: 32.07 + 14.01 = 46.08 g/mol; 46.08g 184.32g = 4.000; molecular formula: S4N4

6 12 6 O H C 16g

008)g

% C =

180

06g 76 × 100 = 40.00%; % H = 16g 180

12.(1 × 100 = 6.714%

83 Out of 100.00 g of compound, there are:

Dividing each mole value by the smallest number:

Because a whole number ratio is required, the C : H : O ratio is 1.5 : 3 : 1 or 3 : 6 : 2. So the empirical formula is C3H6O2.

84 Assuming 100.00 g of nylon-6:

Dividing each mole value by the smallest number:

The empirical formula for nylon-6 is C6H11NO

The mole ratio between Hg and O is 1 : 1, so the empirical formula of compound I is HgO.

CHAPTER 3 STOICHIOMETRY 61

48.64 g C × C 01g 12 C 1mol = 4.050 mol C; 8.16 g H × H 008g 1 H 1mol = 8.10 mol H % O = 100.00 – 48.64 – 8.16 = 43.20%; 43.20 g O × O 16.00g O 1mol = 2.700 mol O

2.700 4.050 = 1.500; 2.700 8.10 = 3.00; 2.700 2.700 = 1.000

63.68 g C × C 01g 12 C 1mol = 5.302 mol C; 12.38 g N × N 01g 14 N 1mol = 0.8837 mol N 9.80 g H × H 1.008g H 1mol = 9.72 mol H; 14.14 g O × O 16.00g O 1mol = 0.8838 mol O

0.8837 5.302 = 6.000; 0.8837 9.72 = 11.0; 0.8837 0.8838 = 1.000

85. Compound I: Mass O = 0.6498 g HgxOy 0.6018 g Hg = 0.0480 g O 0.6018 g Hg × Hg g 6 200 Hg 1mol = 3.000 × 10 3 mol Hg 0.0480 g O × O 00g 16 O 1mol = 3.00 × 10 3 mol O

Compound II: Mass Hg = 0.4172 g HgxOy 0.016 g O = 0.401 g Hg 0.401 g Hg × Hg g 200.6 Hg 1mol = 2.00 × 10 3 mol Hg; 0.016 g O × O 16.00g O 1mol = 1.0 × 10 3 mol O

CHAPTER 3 STOICHIOMETRY

The mole ratio between Hg and O is 2 : 1, so the empirical formula is Hg2O.

Dividing all mole values by the smallest number:

The empirical formula is N2H4CO 87 Out of

The empirical formula is SN because the mole values are in a 1 : 1 mole ratio.

The empirical formula mass of SN is ~ 46 g/mol. Because 184/46 = 4.0, the molecular formula is S4N4.

88. Assuming 100.0 g of compound:

86. 1.121 g N × N 01g 14 N 1mol = 8.001 × 2 10 mol N; 0.161 g H × H 008g 1 H 1mol = 1.60 × 1 10 mol H 0.480 g C × C 01g 12 C 1mol = 4.00 × 2 10 mol C; 0.640 g O × O 00g 16 O 1mol = 4.00 × 2 10 mol O

2 2 10 4.00 10 8.001   = 2.00; 2 1 10 4.00 10 1.60   = 4.00; 2 2 10 4.00 10 4.00   = 1.00

100.0 g,

69.6 g S × gS 07 32 1molS = 2.17 mol S; 30.4 g N × N 01g 14 N 1mol = 2.17 mol N

there are:

26.7 g P × P g 30.97 P 1mol = 0.862 mol P; 12.1 g N × N 14.01g N 1mol = 0.864 mol N 61.2 g Cl × Cl 45g 35 Cl 1mol = 1.73 mol Cl 862 0 73 1 = 2.01; the empirical formula is PNCl2

empirical

 31.0 + 14.0 + 2(35.5) = 116 g/mol. 116 580 mass formula Empirical mass Molar  = 5.0; the molecular formula is (PNCl2)5 = P5N5Cl10

47.08 g C × C 01g 12 C 1mol = 3.920 mol C; 6.59 g H × H 008g 1 H 1mol = 6.54 mol H 46.33 g Cl × Cl 35.45g Cl 1mol = 1.307 mol Cl

The

formula mass is

89 Assuming 100.00 g of compound:

Dividing all mole values by 1.307 gives:

The

90. Assuming 100.00 g of compound (mass oxygen = 100.00 g 41.39 g C − 3.47 g H

All are the same mole values, so the empirical formula is CHO. The empirical formula mass is

91 When combustion data are given, it is assumed that all the carbon in the compound ends up as carbon in CO2 and all the hydrogen in the compound ends up as hydrogen in H2O. In the sample of fructose combusted, the masses of C and H are:

we have:

CHAPTER 3 STOICHIOMETRY 63

1.307 3.920 = 2.999; 1.307 6.54 = 5.00; 1.307 1.307 = 1.000

empirical

g/mol mass Empiricalformula mass Molar = 52 76 153 = 2.00 ; the molecular formula is (C3H5Cl)2 = C6H10Cl2

empirical formula is C3H5Cl. The

formula mass is 3(12.01) + 5(1.008) + 1(35.45) = 76.52

41.39 g C × C 12.01g C 1mol = 3.446 mol C; 3.47 g H × H 1.008g H 1mol = 3.44 mol H 55.14 g O × O 00g 16 O 1mol = 3.446 mol O

55.14 g O):

Molar mass = mol 129 0 15.0g = 116 g/mol 29.02 116 mass Empirical mass Molar  = 4.00; molecular formula = (CHO)4 = C4H4O4

12.01 + 1.008 + 16.00 = 29.02 g/mol.

mass C = 2.20 g CO2 × molC C 12.01g CO mol C 1mol CO 44.01g CO 1mol 2 2 2   = 0.600 g C mass H = 0.900 g H2O × H mol H 1.008g O H mol H mol 2 O H 18.02g O H 1mol 2 2 2   = 0.101 g H Mass O = 1.50 g fructose 0.600 g C 0.101 g H = 0.799 g O

fructose,

0.600 g C × C 12.01g C 1mol = 0.0500 mol C; 0.101 g H × H 1.008g H 1mol = 0.100 mol H

So, in 1.50

of the

Dividing by the smallest number: 0.0499

= 2.00; the empirical formula is CH2O.

92 This compound contains nitrogen, and one way to determine the amount of nitrogen in the compound is to calculate composition by mass percent. We assume that all the carbon in 33.5 mg CO2 came from the 35.0 mg of compound and all the hydrogen in 41.1 mg H2

came from the 35.0 mg of compound.

The mass percent of nitrogen is obtained by difference:

Now perform the empirical formula determination by first assuming 100.0 g of compound. Out of 100.0 g of compound, there are:

93 The combustion data allow determination of the amount of hydrogen in cumene. One way to determine the amount of carbon in cumene is to determine the mass percent of hydrogen in the compound from the data in the problem; then determine the mass percent of carbon by difference

mass % H = mass % C).

g O × O 00g 16 O 1mol = 0.0499 mol O

CHAPTER 3 STOICHIOMETRY 0.799

0.100

3.35 × 2 10 g CO2 × C mol C 01g 12 CO mol C 1mol CO 01g 44 CO 1mol 2 2 2   = 9.14 × 3 10 g C Mass % C = compound g 10 3.50 C g 10 9.14 2 3   × 100 = 26.1% C 4.11 × 2 10 g H2O × H mol H 1.008g O H mol H mol 2 O H 18.02g O H 1mol 2 2 2   = 4.60 × 3 10 g H Mass % H = compound g 10 50 3 H g 10 60 4 2 3   × 100 = 13.1% H

Mass % N

100.0

(26.1 + 13.1) = 60.8% N

26.1 g C × C 01g 12 C 1mol = 2.17 mol C; 13.1 g H × H 008g 1 H 1mol = 13.0 mol H 60.8 g N × N 01g 14 N 1mol = 4.34 mol N Dividing all mole values by 2.17 gives: 17 2 17 2 = 1.00; 17 2 0 13 = 5.99; 17 2 34 4 = 2.00

The empirical formula is CH6N2

(100.0

42.8 mg H2O × g mg 1000 O H 02g 18 H 016g 2 mg 1000 1g 2   = 4.79 mg H

Now solve the empirical formula problem. Out of 100.0

are several ways to do this problem. We will determine composition by mass percent:

Balancing Chemical Equations

CHAPTER 3 STOICHIOMETRY 65 Mass % H = cumene mg 6 47 H mg 79 4 × 100 = 10.1% H; mass % C = 100.0 10.1 = 89.9% C

89.9 g C × C 12.01g C 1mol = 7.49 mol C; 10.1 g H × H 1.008g H 1mol = 10.0 mol H 7.49 0 10 = 1.34  3 4 ; the mole H to mole C ratio is 4 : 3. The empirical formula is C3H4.

formula mass  3(12) + 4(1) = 40 g/mol.

molecular

(C3H4)3 or C9H12

molar

115 and 125 g/mol (molar mass  3 × 40 g/mol = 120 g/mol). 94

16.01 mg CO2 × g mg 1000 CO 44.01g C 01g 12 mg 1000 1g 2   = 4.369 mg C % C = compound 10.68mg C mg 4.369 × 100 = 40.91% C 4.37 mg H2O × g mg 1000 O H 02g 18 H 2.016g mg 1000 1g 2   = 0.489 mg H % H = 68mg 10 mg 489 0 × 100 = 4.58% H; % O = 100.00 (40.91 + 4.58) = 54.51% O So, in 100.00 g of the compound, we have: 40.91 g C × C 01g 12 C 1mol = 3.406 mol C; 4.58 g H × H 008g 1 H 1mol = 4.54 mol H 54.51 g O × O 00g 16 O 1mol = 3.407 mol O Dividing by the smallest number: 406 3 54 4 = 1.33  3 4 ; the empirical formula is C3H4O3 The empirical formula mass of C3H4O3 is  3(12) + 4(1) + 3(16) = 88 g/mol

88 176.1 = 2.0, the molecular formula is C6H8O6.

g cumene, we have:

Empirical

The

formula must be

because the

mass of this formula will be between

There

Because

CHAPTER 3 STOICHIOMETRY

95 When balancing reactions, start with elements that appear in only one of the reactants and one of the products, and then go on to balance the remaining elements.

96. An important part to this problem is writing out correct formulas. If the formulas are incorrect, then the balanced reaction is incorrect.

a. C6H12O6(s) + O2(g)  CO2(g) + H2O(g) Balance C atoms: C6H12O6 + O2  6 CO2 + H2O Balance H atoms: C6H12O6 + O2  6 CO2 + 6 H2O Lastly, balance O atoms: C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(g) b. Fe2S3(s) + HCl(g)  FeCl3(s) + H2S(g) Balance Fe atoms: Fe2S3 + HCl  2 FeCl3 + H2S Balance S atoms: Fe2S3 + HCl  2 FeCl3 + 3 H2S There are 6 H and 6 Cl on right, so balance with 6 HCl on left: Fe2S3(s) + 6 HCl(g)  2 FeCl3(s) + 3 H2S(g). c. CS2(l) + NH3(g)  H2S(g) + NH4SCN(s)

and S balanced; balance N: CS2 + 2 NH3  H2S + NH4SCN H is also balanced. CS2(l) + 2 NH3(g)  H2S(g) + NH4SCN(s)

a. C2H5OH(l) + 3 O2(g)  2 CO2(g) + 3 H2O(g) b. 3 Pb(NO3)2(aq) + 2 Na3PO4(aq)  Pb3(PO4)2(s) + 6 NaNO3(aq) c. Zn(s) + 2 HCl(aq)  ZnCl2(aq) + H2(g) d. Sr(OH)2(aq) + 2 HBr(aq)  2H2O(l) + SrBr2(aq) 97. 2 H2O2(aq) MnO2 catalyst 2 H2O(l) + O2(g) 98. Fe3O4(s) + 4 H2(g)  3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4 CO(g)  3 Fe(s) + 4 CO2(g) 99. a. 3 Ca(OH)2(aq) + 2 H3PO4(aq)  6 H2O(l) + Ca3(PO4)2(s) b. Al(OH)3(s) + 3 HCl(aq)  AlCl3(aq) + 3 H2O(l)

a. The formulas of the reactants and products are C

To balance this combustion reaction, notice that all of the carbon in C6H6 has to end up as carbon in CO2 and all of the hydrogen in C6H6 has to end up as hydrogen in H

To balance C and H, we need 6 CO

molecules and 3 H2O molecules for every 1 molecule of C

H6. We do oxygen last. Because we have 15 oxygen atoms in 6 CO2 molecules and 3 H2O molecules, we need 15/2 O2 molecules in order to have 15 oxygen atoms on the reactant side.

CHAPTER 3 STOICHIOMETRY 67 c. 2 AgNO3(aq) + H2SO4(aq)  Ag2SO4(s) + 2 HNO3(aq) 100 a. 2 KO2(s) + 2 H2O(l)  2 KOH(aq) + O2(g) + H2O2(aq) or 4 KO2(s) + 6 H2O(l)  4 KOH(aq) + O2(g) + 4 H2O2(aq) b. Fe2O3(s) + 6 HNO3(aq)  2 Fe(NO3)3(aq) + 3 H2O(l) c. 4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g) d. PCl5(l) + 4 H2O(l)  H3PO4(aq) + 5 HCl(g) e. 2 CaO(s) + 5 C(s)  2 CaC2(s) + CO2(g) f. 2 MoS2(s) + 7 O2(g)  2 MoO3(s) + 4 SO2(g) g. FeCO3(s) + H2CO3(aq)  Fe(HCO3)2(aq)

101.

2(g)

H2O(g).

C6H6(l) + 2 15 O2(g)  6 CO2(g) + 3 H2O(g); multiply by two to give whole numbers. 2 C6H6(l) + 15 O2(g)  12 CO2(g) + 6 H2O(g)

6H6(l) + O

(g) → CO

2(g) → CO2(g) + H2O(g). C4H10(g) + 2 13 O2(g)  4 CO2(g) + 5 H2O(g); multiply by two to give whole numbers. 2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(g) c. C12H22O11(s) + 12 O2(g)  12 CO2(g) + 11 H2O(g) d. 2 Fe(s) + 2 3 O2(g)  Fe2O3(s); for whole numbers: 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) e. 2 FeO(s) + 2 1 O2(g)  Fe2O3(s); for whole numbers, multiply by two. 4 FeO(s) + O2(g)  2 Fe2O3(s) 102. a. 16 Cr(s) + 3 S8(s)  8 Cr2S3(s) b. 2 NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(g) c. 2 KClO3(s)  2 KCl(s) + 3 O2(g)

b. The formulas of the reactants and products are C4H10(g) + O

103

Balance

Reaction Stoichiometry

105. The stepwise method to solve stoichiometry problems is outlined in the text. Instead of calculating intermediate answers for each step, we will combine conversion factors into one calculation. This practice reduces round-off error and saves time.

CHAPTER 3 STOICHIOMETRY

HF(g)  2 EuF3(s) + 3 H2(g)

2 Eu(s) + 6

SiO2(s) + C(s)  Si(s) + CO(g); Si is balanced.

oxygen atoms: SiO2

C  Si

carbon atoms: SiO2(s) + 2 C(s)  Si(s) + 2 CO(g)

+ 2

Balance

SiCl4(l) + Mg(s)  Si(s) + MgCl2(s); Si

Cl atoms: SiCl4 + Mg  Si + 2 MgCl2

Mg atoms: SiCl4(l) + 2 Mg(s)  Si(s) + 2 MgCl2(s)

is balanced. Balance

Balance

6(s)

Na(s)  Si(s) + NaF(s); Si

Balance F atoms: Na2SiF6 + Na  Si + 6 NaF Balance Na atoms: Na2SiF6(s) + 4 Na(s)  Si(s) + 6 NaF(s) 104 CaSiO3(s) + 6 HF(aq)  CaF2(aq) + SiF4(g) + 3 H2O(l)

Na2SiF

is balanced.

Fe2O3(s) + 2 Al(s)  2 Fe(l) + Al2O3(s) 15.0 g Fe × Fe 55.85g Fe 1mol = 0.269 mol Fe; 0.269 mol Fe × Al mol Al 98g 26 Fe mol 2 Al mol 2  = 7.26 g Al 0.269 mol Fe × 3 2 3 2 3 2 O Fe mol O Fe 159.70g Fe mol 2 O Fe 1mol  = 21.5 g Fe2O3 0.269 mol Fe × 3 2 3 2 3 2 O Al mol O Al 96g 101 Fe mol 2 O Al 1mol  = 13.7 g Al2O3 106 10 KClO3(s) + 3 P4(s)  3 P4O10(s) + 10 KCl(s) 52.9 g KClO3 × 10 4 10 4 3 10 4 3 3 O P mol O P 88g 283 KClO mol 10 O P 3mol KClO 55g 122 KClO 1mol   = 36.8 g P4O10 107 1.000 kg Al × 4 4 4 4 4 4 ClO NH mol ClO NH 117.49g Al 3mol ClO NH 3mol Al 26.98g Al 1mol Al kg Al g 1000   

Because only 95% of the NH4 + ions react:

CHAPTER 3 STOICHIOMETRY 69 = 4355 g = 4.355 kg NH4ClO4 108 a. Ba(OH)2•8H2O(s) + 2 NH4SCN(s)  Ba(SCN)2(s) + 10 H2O(l) + 2 NH3(g) b. 6.5 g Ba(OH)2•8H2O × 315.4g O 8H Ba(OH) 1mol 2 2  = 0.0206 mol = 0.021 mol 0.021 mol Ba(OH)2•8H2O × SCN NH mol SCN NH 76.13g O 8H Ba(OH) 1mol SCN NH mol 2 4 4 2 2 4   = 3.2 g NH4SCN 109. a. 1.0 × 102 mg NaHCO3 3 7 8 6 3 3 NaHCO 3mol O H C 1mol NaHCO 01g 84 NaHCO 1mol mg 1000 1g    7 8 6 7 8 6 O H C mol O H C 12g 192  = 0.076 g or 76 mg C6H8O7 b. 0.10 g NaHCO3 2 2 3 2 3 3 CO mol CO 44.01g NaHCO 3mol CO 3mol NaHCO 84.01g NaHCO 1mol    = 0.052 g or 52 mg CO2 110. a. 1.00 × 102 g C7H6O3 × 3 6 4 3 6 4 3 6 7 3 6 4 3 6 7 3 6 7 O H C 1mol O H C 09g 102 O H C 1mol O H C 1mol O H C 12g 138 O H C 1mol   = 73.9 g C4H6O3 b. 1.00 × 102 g C7H6O3 × 4 8 9 4 8 9 3 6 7 4 8 9 3 6 7 3 6 7 O H C mol O H C 180.15g O H C 1mol O H C 1mol O H C 138.12g O H C 1mol   = 1.30 × 102 g aspirin 111. 1.0 × 104 kg waste ×        4 2 7 5 4 4 4 NH mol 55 N O H C 1mol NH 04g 18 NH 1mol kg 1000g waste kg 100 NH kg 0 3 × N O H C mol N O H C 12g 113 2 7 5 2 7 5 = 3.4 × 104 g tissue if all NH4 + converted

mass of tissue

(0.95)(3.4 × 104 g) = 3.2 × 104 g or 32 kg bacterial tissue 112. 1.0 × 103 g phosphorite × 2 4 3 2 4 3 2 4 3 ) (PO Ca 310.18g ) (PO Ca 1mol phosphorite 100g ) (PO Ca 75g  × 2 4 3 4 ) (PO Ca 2mol P 1mol × 4 4 P mol P 88g 123 = 150 g P4

Limiting Reactants and Percent Yield

115. The product formed in the reaction is NO2; the other species present in the product representtationisexcessO2. Therefore, NO isthe limitingreactant. In the pictures, 6 NO moleculesreact with 3 O2 molecules to form 6 NO2 molecules.

6 NO(g) + 3 O2(g)  6 NO2(g)

For smallest whole numbers, the balanced reaction is:

2 NO(g) + O2(g)  2 NO2(g)

116. In the following table we have listed three rows of information. The “Initial” row is the number of molecules present initially, the “Change” row is the number of molecules that react to reach completion, and the “Final” row is the number of molecules present at completion. To determine the limiting reactant, let’s calculate how much of one reactant is necessary to react with the other.

Because we have 10 molecules of NH3 and only 8 molecules of NH3 are necessaryto react with all of the O2, O2 is limiting. Now use the 10 molecules of O2 and the molecule relationships given in the balanced equation to determine the number of molecules of each product formed, then complete the table.

The total number of molecules present after completion

molecules

2 + 8 molecules NO + 12 molecules H2O = 22 molecules.

117. a. The strategy we will generally use to solve limiting reactant problems is to assume each reactant is limiting, and then calculate the quantity of product each reactant could produce if it were limiting. The reactant that produces the smallest quantity of product is the

STOICHIOMETRY 113. 1.0 ton CuO C 95g coke 100.g C mol C 12.01g CuO mol 2 C 1mol CuO 55g 79 CuO 1mol kg 1000g ton kg 907       = 7.2 × 104 g or 72 kg coke 114 2 LiOH(s) + CO2(g)  Li2CO3(aq) + H2O(l) The total volume of air exhaled each minute for the 7 astronauts is 7 × 20. = 140 L/min. 25,000 g LiOH × 2 2 2 2 CO 4.0g air 100g CO mol CO 44.01g LiOH mol 2 CO 1mol LiOH 23.95g LiOH 1mol    min 60 1h air L 140 1min mL 1000 1L air g 0010 0 air 1mL     = 68 h = 2.8 days

CHAPTER 3

10 molecules O2 × 2 3 O 5molecules NH molecules 4 = 8 molecules NH3 to react with all of the O2

4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g) Initial 10 molecules 10 molecules 0 0 Change 8 molecules 10 molecules +8 molecules +12 molecules Final 2 molecules 0 8 molecules 12 molecules

2 molecules

+ 0

limiting reactant (runs out first) and therefore determines the mass of product that can be produced.

Assuming N2 is limiting:

Assuming H2 is limiting:

Because N2 produces the smaller mass of product (1220 g vs. 2820 g NH3), N2 is limiting and 1220 g NH3 can be produced. As soon as 1220 g of NH3 is produced, all of the N2 has run out. Even though we have enough H2 to produce more product, there is no more N2 present as soon as 1220 g of NH3 have been produced.

Because Ca3(PO4)2 produces the smaller quantity of product, Ca3(PO4)2 is limiting and 1300 g CaSO4 can be produced.

119. Assuming BaO2 is limiting:

CHAPTER 3 STOICHIOMETRY 71

1.00 × 103 g N2 3 3 2 3 2 2 NH mol NH 03g 17 N mol NH mol 2 N 02g 28 N 1mol    = 1.22 × 103 g NH3

5.00 × 102 g H2 3 3 2 3 2 2 NH mol NH 17.03g H 3mol NH mol 2 H 2.016g H 1mol    = 2.82 × 103 g NH3

b. 1.00  103 g N2 2 2 2 2 2 2 H mol H 016g 2 N mol H 3mol N 02g 28 N 1mol    = 216 g H2 reacted Excess H2 = 500. g H2 initially – 216 g H2 reacted = 284 g H2 in excess (unreacted) 118. Ca3(PO4)2 + 3 H2SO4  3 CaSO4 + 2 H3PO4 Assuming Ca3(PO4)2 is limiting: 1.0 × 103 g Ca3(PO4)2 × 4 4 2 4 3 4 2 4 3 2 4 3 CaSO mol CaSO 136.15g ) (PO Ca mol CaSO 3mol ) (PO Ca 18g 310 ) (PO Ca 1mol   = 1300 g CaSO4 Assuming concentrated H2SO4 reagent is limiting: 1.0 × 103 g conc. H2SO4 × 4 2 4 2 4 2 4 2 SO H 98.09g SO H 1mol SO conc.H 100g SO H 98g  4 4 4 2 4 CaSO mol CaSO 136.15g SO H 3mol CaSO 3mol   = 1400 g CaSO4

1.0 × 103 g Ca3(PO4)2  2 4 3 2 4 3 ) (PO Ca 310.18g ) (PO Ca 1mol  4 3 4 3 2 4 3 4 3 PO H mol PO H 97.99g ) (PO Ca mol PO H mol 2  = 630 g H3PO4 produced

CHAPTER 3 STOICHIOMETRY

Because C10H10N4SO2 produces the smaller amount of product, it is limiting and 71.4 g of silver sulfadiazine can be produced.

121. Tosolve limiting-reagent problems, we will generallyassume eachreactantislimitingandthen calculate how much product could be produced from each reactant. The reactant that produces the smallest amount of product will run out first and is the limiting reagent.

1.50 g BaO2 × 2 2 2 2 2 2 2 2 2 O H mol O H 34.02g BaO mol O H 1mol BaO 3g 169 BaO 1mol   = 0.301 g H2O2 Assuming HCl is limiting: 25.0 mL × 2 2 2 2 2 2 O H mol O H g 34.02 HCl mol 2 O H 1mol HCl g 36.46 HCl 1mol mL HCl g 0.0272    = 0.317 g H2O2 BaO2 produces the smaller amount of H2O2, so it is limiting and a mass of 0.301 g of H2O2 can be produced. Initial mol HCl present: 25.0 mL HCl 36.46g HCl 1mol mL HCl 0.0272g   = 1.87 × 10 2 mol HCl The amount of HCl reacted: 1.50 g BaO2 × 2 2 2 BaO mol HCl mol 2 BaO 169.3g BaO 1mol  = 1.77 × 2 10 mol HCl Excess mol HCl = 1.87 × 2 10 mol 1.77 × 2 10 mol = 1.0 × 3 10 mol HCl Mass of excess HCl = 1.0 × 3 10 mol HCl × HCl mol HCl 46g 36 = 3.6 × 2 10 g HCl unreacted 120. Assuming

25.0 g Ag2O × O Ag 231.8g O Ag 1mol 2 2  2 4 9 10 2 4 9 10 2 2 4 9 10 SO N H AgC mol SO N H AgC 357.18g O Ag mol SO N H AgC mol 2  = 77.0 g AgC10H9N4SO2 Assuming C10H10N4SO2

50.0 g C10H10N4SO2 × 2 4 10 10 2 4 10 10 SO N H C 250.29g SO N H C 1mol  2 4 10 10 2 4 9 10 SO N H C mol 2 SO N H AgC mol 2 2 4 9 10 2 4 9 10 SO N H AgC mol SO N H AgC 357.18g  = 71.4 g AgC10H9N4SO2

Ag2O is limiting:

is limiting:

5.00 × 106 g NH3 × 3 3 3 NH mol 2 HCN mol 2 NH 03g 17 NH 1mol  = 2.94 × 105 mol HCN

O2 is limiting because it produces the smallest amount of HCN. Although more product could be produced from NH3 and CH4, only enough O2 is present to produce 1.04 × 10

mol HCN. The mass of HCN produced is:

O2 produces the smallest amount of product; thus O2 is limiting, and 11.1 g C3H3N can be produced.

limiting:

If Cl2 is limiting:

Cl2 is limiting because it produces the smaller quantity of product. Hence, the theoretical yield for this reaction is 591 g

The percent yield is:

CHAPTER 3 STOICHIOMETRY 73 5.00 × 106 g O2 × 2 2 2 O 3mol HCN mol 2 O 00g 32 O 1mol  = 1.04 × 105 mol HCN 5.00 ×

CH4 × 4 4 4 CH mol 2 HCN mol 2 CH 04g 16 CH 1mol  = 3.12 × 105 mol HCN

1.04 × 105 mol HCN × HCN mol HCN 03g 27 = 2.81 × 106 g HCN 5.00 × 106 g O2 × O H 1mol O H 02g 18 O 3mol O H mol 6 O 32.00g O 1mol 2 2 2 2 2 2   = 5.63 × 106 g H2O 122. If C3H6 is limiting: 15.0 g C3H6 × N H C mol N H C 53.06g H C mol 2 N H C mol 2 H C 42.08g H C 1mol 3 3 3 3 6 3 3 3 6 3 6 3   = 18.9 g C3H3N

NH3

5.00 g NH3 × N H C mol N H C 06g 53 NH mol 2 N H C mol 2 NH 03g 17 NH 1mol 3 3 3 3 3 3 3 3 3   = 15.6 g C3H3N

10.0 g O2 × N H C mol N H C 06g 53 O 3mol N H C mol 2 O 32.00g O 1mol 3 3 3 3 2 3 3 2 2   = 11.1 g C3H3N

is limiting:

123. C2H6(g) + Cl2(g)  C2H5Cl(g) + HCl(g)

300. g C2H6 × Cl H C mol Cl H C 51g 64 H C mol Cl H C 1mol H C g 30.07 H C 1mol 5 2 5 2 6 2 5 2 6 2 6 2   = 644 g C2H5Cl

If C

H6 is

650. g Cl2 × Cl H C mol Cl H C 51g 64 Cl mol Cl H C 1mol Cl 90g 70 Cl 1mol 5 2 5 2 2 5 2 2 2   = 591 g C2H5Cl

C2H5Cl.

CHAPTER 3 STOICHIOMETRY

yield = theoretical actual × 100 = 591g 490.g × 100 = 82.9% 124. a. 1142 g C6H5Cl × 5 9 14 5 9 14 5 6 5 9 14 5 6 5 6 Cl H C mol Cl H C 46g 354 Cl H C mol 2 Cl H C 1mol Cl H C 55g 112 Cl H C 1mol   = 1798 C14H9Cl5 485 g C2HOCl3 × 5 9 14 5 9 14 3 2 5 9 14 3 2 3 2 Cl H C mol Cl H C g 354.46 HOCl C mol Cl H C mol 1 HOCl C 147.38g HOCl C 1mol   = 1170 g C14H9Cl5

the masses of product calculated, C2HOCl3 is limiting and 1170 g C14H9Cl5 can be produced. b. C2HOCl3 is limiting, and C6H5Cl is in excess. c. 485 g C2HOCl3 × Cl H C mol Cl H C 112.55g HOCl C mol Cl H C mol 2 HOCl 147.38gC HOCl C 1mol 5 6 5 6 3 2 5 6 3 2 3 2   = 741 g C6H5Cl reacted 1142 g 741 g = 401 g C6H5Cl in excess d. Percent yield = DDT 1170g DDT 0g 200 × 100 = 17.1% 125. 2.50 metric tons Cu3FeS3 3 3 3 3 FeS Cu 1mol Cu 3mol 342.71g FeS Cu 1mol kg 1000g ton metric kg 1000     Cu mol 55g 63  = 1.39 × 106 g Cu (theoretical) 1.39 × 106 g Cu (theoretical) × (theoretical) Cu g 100 (actual) Cu 86.3g = 1.20 × 106 g Cu = 1.20 × 103 kg Cu = 1.20 metric tons Cu (actual) 126. P4(s) + 6 F2(g)  4 PF3(g); the theoretical yield of PF3 is: 120. g PF3 (actual) (actual) PF 78.1g (theoretical) PF 100.0g 3 3  = 154 g PF3 (theoretical) 154 g PF3 2 2 3 2 3 3 F mol F 00g 38 PF mol 4 F mol 6 PF g 97 87 PF 1mol    = 99.8 g F2 99.8 g F2 is needed to actually produce 120. g of PF3 if the percent yield is 78.1%.

percent

From

Additional Exercises

127 12C2 1H6: 2(12.000000) + 6(1.007825) = 30.046950 u 12C1H2 16O: 1(12.000000) + 2(1.007825) + 1(15.994915) = 30.010565 u

14N16O: 1(14.003074) + 1(15.994915) = 29.997989 u

CHAPTER 3 STOICHIOMETRY 75

12C1H2 16O. 128 We would see the peaks corresponding

10B35Cl3 [mass  10 + 3(35) = 115 u], 10B35Cl2 37Cl (117), 10B35Cl37Cl2 (119), 10B37Cl3 (121), 11B35Cl3 (116), 11B35Cl2 37Cl (118), 11B35Cl37Cl2 (120), 11B37Cl3 (122)

The peak results from

to:

115, 116, 117, 118, 119, 120,

129 Molar mass XeFn = molecules 10 6.022 XeF 1mol XeF molecules 10 03 9 XeF 368g 0 23 20    n n n = 245 g/mol 245 g = 131.3 g + n(19.00 g), n = 5.98; formula = XeF6 130. a. 14 mol C × C mol 12.01g + 18 mol H × H mol 1.008g + 2 mol N × N mol 14.01g + 5 mol O × O mol 16.00g = 294.30 g b. 10.0 g C14H18N2O5 × 5 2 18 14 5 2 18 14 O N H C 30g 294 O N H C 1mol = 3.40 × 10 2 mol C14H18N2O5 c. 1.56 mol × mol 294.3g = 459 g C14H18N2O5 d. 5.0 mg × mol molecules 10 6.022 294.30g 1mol mg 1000 1g 23    = 1.0 × 1019 molecules C14H18N2O5

1.2 g C14H18N2O5 × 5 2 18 14 5 2 18 14 5 2 18 14 O N H C mol N mol 2 O N H C 294.30g O N H C 1mol  N mol N atoms 10 022 6 23   = 4.9 × 1021 atoms N f. 1.0 × 109 molecules × mol 30g 294 atoms 10 022 6 1mol 23   = 4.9 × 10 13 g

We would see a total of eight peaks at approximate masses of

121, and 122.

e. The chemical formula tells us that 1 molecule of C14H18N2O5 contains 2 atoms of N. If we have 1 mole of C14H18N2O5 molecules, then 2 moles of N atoms are present.

CHAPTER 3 STOICHIOMETRY

g. 1

131. Molar mass = 20(12.01) + 29(1.008) + 19.00 + 3(16.00) = 336.43 g/mol

g 01) 20(12 × 100 = 71.40% C

Mass % C = compound 336.43g

Mass % H = compound 336.43g H g 008) 29(1 × 100 = 8.689% H

19.00g × 100 = 5.648% F

Mass % F = compound 43g 336

Mass % O = 100.00 (71.40 + 8.689 + 5.648) = 14.26% O or: % O = compound 336.43g

g 3(16.00) × 100 = 14.27% O

132. In 1 hour, the 1000. kg of wet cereal produced contains 580 kg H2O and 420 kg of cereal. We want the final product to contain 20.% H2O. Let x = mass of H2O in final product.

The amount of water to be removed is 580 110 = 470 kg/h.

133. Out of 100.00 g of adrenaline, there are:

Dividing each mole value by the smallest number:

This gives adrenaline an empirical formula of

134. Assuming 100.00 g of compound (mass hydrogen = 100.00 g

1mol 23    =

H18N2

molecule mol 294.30g atoms 10 022 6

4.887 × 10 22 g C14

x x  420 = 0.20, x = 84 + (0.20)x, x = 105  110 kg H2O

56.79 g C × C 01g 12 C 1mol = 4.729 mol C; 6.56 g H × H 008g 1 H 1mol = 6.51 mol H 28.37 g O × O 00g 16 O 1mol = 1.773 mol O; 8.28 g N × N 01g 14 N 1mol = 0.591 mol N

591 0 729 4 = 8.00; 591 0 51 6 = 11.0; 591 0 773 1 = 3.00; 591 0 591 0 = 1.00

C8H11O3N.

49.31 g C 43.79

= 6.90 g H): 49.31 g C × C 01g 12 C 1mol = 4.106 mol C; 6.90 g H × H 008g 1 H 1mol = 6.85 mol H

g O

Dividing

135. There are many valid methods to solve this problem. We will assume 100.00 g of compound, and then determine from the information in the problem how many moles of compound equals 100.00 g of compound. From this information, we can determine the mass of one mole of compound (the molar mass) by setting up a ratio. Assuming 100.00 g

CHAPTER 3 STOICHIOMETRY 77 43.79 g O × O 00g 16 O 1mol = 2.737 mol O

all mole values

2.737

737 2 106 4 = 1.500; 737 2 85 6 = 2.50; 737 2 737 2 = 1.000

a whole number ratio is required, the empirical formula is C3H5O2. Empirical formula mass: 3(12.01) + 5(1.008) +2(16.00) = 73.07 g/mol mass formula Empirical mass Molar = 73.07 146.1 = 1.999; molecular formula = (C3H5O2)2 = C6H10O4

gives:

Because

cyanocobala-min: mol cyanocobalamin = 4.34 g Co × Co mol cyanocobalamin 1mol Co 93g 58 Co 1mol  = 7.36 × 2 10 mol cyanocobalamin cyanocobalamin 1mol cyanocobalamin g x = mol 10 7.36 100.00g 2  , x = molar mass = 1.36 × 103 g/mol 136. 2 tablets × Bi mol Bi 209.0g BiO H C 1mol Bi 1mol BiO H C 11g 362 BiO H C 1mol tablet BiO H C 262g 0 4 5 7 4 5 7 4 5 7 4 5 7    = 0.302 g Bi consumed 137. Empirical formula mass = 12.01 + 1.008 = 13.02 g/mol; because 104.14/13.02 = 7.998  8, the molecular formula for styrene is (CH)8 = C8H8. 2.00 g C8H8 × H mol H atoms 10 6.022 H C mol H 8mol H C 104.14g H C 1mol 23 8 8 8 8 8 8    = 9.25 × 1022 atoms H 138. 41.98 mg CO2 × 2CO 01mg 44 C 01mg 12 = 11.46 mg C; % C = 81mg 19 mg 46 11 × 100 = 57.85% C 6.45 mg H2O × O H mg 02 18 H mg 2.016 2 = 0.722 mg H; % H = 81mg 19 mg 0.772 × 100 = 3.64% H % O = 100.00 (57.85 + 3.64) = 38.51% O Out of 100.00 g terephthalic acid, there are:

CHAPTER 3 STOICHIOMETRY

The empirical formula mass is ~29 g/mol, so two times the empirical formula would put the compound in the correct range of the molar mass.

Note: From the periodic table, element E is silicon, Si.

57.85 g C × C 01g 12 C 1mol = 4.817 mol C; 3.64 g H × H 008g 1 H 1mol = 3.61 mol H 38.51 g O × O 00g 16 O 1mol = 2.407 mol O 2.407 4.817 = 2.001; 2.407 3.61 = 1.50; 2.407 2.407 = 1.000

C : H : O mole ratio is 2 : 1.5 : 1 or 4 : 3 : 2. The empirical formula is C4H3O2. Mass of C4H3O2  4(12) + 3(1) + 2(16) = 83g/mol

mass = mol 0.250 41.5g = 166 g/mol; 83 166 = 2.0; the molecular formula is C8H6O4 139. 17.3 g H × H 1.008g H 1mol = 17.2 mol H; 82.7 g C × C 12.01g C 1mol = 6.89 mol C 6.89 2 17 = 2.50; the empirical formula is C2H5.

The

Molar

Molecular formula = (C2H5)2 = C4H10 2.59 × 1023 atoms H × molecules 10 6.022 H C 1mol H 10atoms H C 1molecule 23 10 4 10 4   = 4.30 × 2 10 mol C4H10 4.30 × 2 10 mol C4H10 × 10 4H C mol 58.12g = 2.50 g C4H10 140. Assuming 100.00 g E3H8: mol E = 8.73 g H × H 8mol E 3mol H 008g 1 H 1mol  = 3.25 mol E E 25mol 3 E g 27 91 E 1mol E g  x , x = molar mass of E = 28.1 g/mol; atomic mass of E = 28.1 u

141. Mass of H2O = 0.755 g CuSO4•xH2O 0.483 g CuSO4 = 0.272 g H2O 0.483 g CuSO4 × 4 4 CuSO 159.62g CuSO 1mol = 0.00303 mol CuSO4 0.272 g H2O × O H 02g 18 O H 1mol 2 2 = 0.0151 mol H2O

a. Only acrylonitrile contains nitrogen. If we have 100.00 g of polymer:

Only butadiene in the polymer reacts with

have

This is close to a mole ratio of 4 : 2 : 3. Thus there are 4 acrylonitrile to 2 butadiene to 3 styrene molecules in the polymer, or (A

)

CHAPTER 3 STOICHIOMETRY 79 4 2 CuSO 00303g 0 O H 0.0151mol = 4 2 CuSO 1mol O H mol 98 4 ; compound formula = CuSO4•5H2O, x = 5 142.

8.80 g N × N H C 1mol N H C 06g 53 N 01g 14 N H C 1mol 3 3 3 3 3 3  = 33.3 g C3H3N % C3H3N = polymer 100.00g N H C 33.3g 3 3 = 33.3% C3H3N

Br2: 0.605 g Br2 × 6 4 6 4 2 6 4 2 2 H C mol H C 09g 54 Br mol H C 1mol Br 8g 159 Br 1mol   = 0.205 g C4H6 % C4H6 = 20g 1 205g 0 × 100 = 17.1% C4H6 b.

33.3 g C3H3N × 53.06g N H C 1mol 3 3 = 0.628 mol C3H3N 17.1 g C4H6 × 6 4 6 4 H C 09g 54 H C 1mol = 0.316 mol C4H6 49.6 g C8H8 × 8 8 8 8 H C 104.14g H C 1mol = 0.476 mol C8H8 Dividing by 0.316: 0.316 0.628 = 1.99; 0.316 0.316 = 1.00; 0.316 0.476 = 1.51

If we

100.0 g of polymer:

4B2S3

n. 143. 1.20 g CO2 × O N H C mol 376.51g C mol 24 O N H C 1mol CO mol C 1mol 44.01g CO 1mol 3 30 24 3 30 24 2 2    = 0.428 g C24H30N3O sample 00g 1 O N H C 0.428g 3 30 24 × 100 = 42.8% C24H30N3O (LSD) 144. a. CH4(g) + 4 S(s) → CS2(l) + 2 H2S(g) or 2 CH4(g) + S8(s) → 2 CS2(l) + 4 H2S(g)

CHAPTER 3 STOICHIOMETRY

Because S produces the smaller quantity of CS2, sulfur is the limiting reactant and 71.2 g CS2 can be produced. The same amount of CS2 would be produced using the balanced equation with S8.

Because O2 produces the smallest quantity of product, O2 is limiting and 81.1g H2O can be produced.

b. The Cu remains unreacted. After filtering, washing, and drying, the mass of the unreacted copper could be measured.

b. 120. g CH4 × 2 2 4 2 4 4 molCS 76.15gCS molCH 1molCS CH 16.04g CH 1mol   = 570. g CS2 120. g S × 2 2 2 molCS 76.15gCS 4molS 1molCS gS 32.07 1molS   = 71.2 g CS2

145. 126 g B5H9 × O H mol O H 18.02g H B mol 2 O H mol 9 H B 63.12g H B 1mol 2 2 9 5 2 9 5 9 5   = 162 g H2O 192 g O2 × O H mol O H 18.02g O mol 12 O H mol 9 O 00g 32 O 1mol 2 2 2 2 2 2   = 81.1 g H2O

146. 2 NaNO3(s)  2 NaNO2(s) + O2(g); the amount of NaNO3 in the impure sample is: 0.2864 g NaNO2 3 3 2 3 2 2 NaNO mol NaNO 00g 85 NaNO mol 2 NaNO mol 2 NaNO 00g 69 NaNO 1mol    = 0.3528 g NaNO3 Mass percent NaNO3 = sample 0.4230g NaNO 0.3528g 3 × 100 = 83.40% 147. 453 g Fe × 3 2 3 2 3 2 O Fe mol O Fe g 159.70 Fe mol 2 O Fe 1mol Fe 85g 55 Fe 1mol   = 648 g Fe2O3 Mass percent Fe2O3 = ore 752g O Fe 648g 3 2 × 100 = 86.2% 148 a. Mass of Zn in alloy = 0.0985 g ZnCl2 × 2ZnCl 136.28g Zn 65.38g = 0.0473 g Zn % Zn = brass 0.5065g Zn 0473g 0 × 100 = 9.34% Zn; % Cu = 100.00 9.34 = 90.66% Cu

149

mol C = 286.4 g vitamin A × C 12.01g C 1mol A vitamin g C 0.8386g  = 20.00 mol C

Assuming 1 mole of vitamin A (286.4 g vitamin A):

mol H = 286.4 g vitamin A ×

0.1056g

vitamin g

1.008g

= 30.00 mol H

Because 1 mole of vitamin A contains 20 mol C and 30 mol H, the molecular formula of vitamin A is C20H30E. To determine E, let’s calculate the molar mass of E:

286.4 g = 20(12.01) + 30(1.008) + molar mass E, molar mass E = 16.0 g/mol

From the periodic table, E = oxygen, and the molecular formula of vitamin A is C20H30O.

150. a. At 40.0 g of Na added, Cl2 and Na both run out at the same time (both are limiting reactants). Past 40.0 g of Na added, Cl2 is limiting, and because the amount of Cl2 present in each experiment was the same quantity, no more NaCl can be produced. Before 40.0 g of Na added, Na was limiting. As more Na was added (up to 40.0 g Na), more NaCl was produced.

c. At 40.0 g Na added, both Cl2 and Na are present in stoichiometric amounts.

g Cl2 was present at 40.0 g Na added, and from the problem, the same 61.7 g Cl2 was present in each experiment.

d. At 50.0 g Na added, Cl2 is limiting:

Excess Cl2 = 61.7 g Cl2 initially

Note: We know that 40.0 g Na is the point where Na and the 61.7 g of Cl2 run out at the same time. So if 20.0 g of Na are reacted, one-half of the Cl2 that was present at 40.0 g Na reacted will be in excess. The previous calculation confirms this.

For 50.0 g Na reacted, Cl2 is limiting and 40.0 g Na will react as determined previously.

Excess Na = 50.0 g Na initially – 40.0 g Na reacted = 10.0 g Na in excess. 151 X2Z: 40.0% X and 60.0% Z by mass;

where A = molar mass.

CHAPTER 3 STOICHIOMETRY 81



1mol

 NaCl mol NaCl 58.44g Na mol 2 NaCl mol 2 Na 22.99g Na 1mol   = 50.8 g NaCl

20.0 g Na

40.0 g Na  2 2 2 Cl mol Cl 70.90g Na mol 2 Cl 1mol Na 22.99g Na 1mol   = 61.7 g Cl2

61.7

61.7 g Cl2  NaCl mol NaCl 44g 58 Cl 1mol NaCl mol 2 Cl 70.90g Cl 1mol 2 2 2   = 101.7 g = 102 g NaCl

g Na  2 2 2 Cl mol Cl 90g 70 Na mol 2 Cl 1mol Na 99g 22 Na 1mol   = 30.8 g Cl2 reacted

e. 20.0

– 30.8 g Cl2 reacted = 30.9 g Cl2 in excess

x z z x 0)A (60 0)A (40 60.0/A 40.0/A 2 Z mol X mol    or Az = 3Ax,

For XZ2, molar mass = Ax + 2Az = Ax + 2(3Ax) = 7Ax

Mass percent X = x

x 7A A × 100 = 14.3% X; % Z = 100.0 14.3 = 85.7% Z

ChemWork Problems

The answers to the problems 152-159 (or a variation to these problems) are found in OWL. These problems are also assignable in OWL.

Challenge Problems

160. GaAs can be either 69GaAs or 71GaAs. The mass spectrum for GaAs will have 2 peaks at 144 (= 69 + 75) and 146 (= 71 + 75) with intensities in the ratio of 60 : 40 or 3 : 2.

144146

Ga2As2 can be 69Ga2As2, 69Ga71GaAs2, or 71Ga2As2. The mass spectrum will have 3 peaks at 288, 290, and 292 with intensities in the ratio of 36 : 48 : 16 or 9 : 12 : 4. We get this ratio from the following probability table:

161 The volume of a gas is proportional to the number of molecules of gas. Thus the formulas are:

I: NH3 ; II: N2H4; III: HN3

The mass ratios are:

CHAPTER 3 STOICHIOMETRY

69Ga (0.60) 71Ga (0.40) 69Ga (0.60) 0.36 0.24 71Ga (0.40) 0.24 0.16 288

290 292

If we set the atomic mass of H equal to 1.008, then the atomic mass, A, for nitrogen is:

162

if we had exactly 100 atoms, x = number of 85

163 First, we will determine composition in mass percent. We assume that all the carbon in the 0.213

CO2 came from the 0.157 g of the compound and that all the hydrogen in the 0.0310

came from the 0.157 g of the compound.

We get the mass percent of N from the second experiment:

The mass percent of oxygen is obtained by difference:

So, out of 100.00 g of compound, there are:

CHAPTER 3 STOICHIOMETRY 83 I: H 17.75g N 82.25g = H g N 4.634g ; II: H g N 6.949g ; III: H g N g 41.7

I: 14.01;

14.0

example,

I: 008) 3(1 A = 1 4.634 , A = 14.01

II: 14.01; III.

For

for compound

atoms Rb atoms Rb 87 85

x x 100 = 2.591, x = 259.1 (2.591)x, x = 3.591 259.1 = 72.15; 72.15% 85Rb 0.7215(84.9117) + 0.2785(A) = 85.4678, A = 0.2785 26 61 4678 85 = 86.92 u

= 2.591;

Rb atoms, and 100 x = number of 87Rb atoms.

g H

0.213 g CO2 × 2CO 44.01g C 01g 12 = 0.0581 g C; % C = compound 0.157g C 0581g 0 × 100 = 37.0% C 0.0310 g H2O × O H g 18.02 H 2.016g 2 = 3.47 × 10 3 g H; % H = 0.157g g 10 47 3 3   100 = 2.21% H

0.0230 g NH3  3NH 03g 17 N 01g 14 = 1.89 × 10 2 g N % N = g 0.103 g 10 89 1 2  × 100 = 18.3% N

% O = 100.00 (37.0 + 2.21 + 18.3) = 42.5% O

37.0 g C × C 12.01g C 1mol = 3.08 mol C; 2.21 g H × H 1.008g H 1mol = 2.19 mol H

Lastly, and often the hardest part, we need to find simple whole-number ratios. Divide all mole values by the smallest number:

Multiplying all these ratios by 3 gives

We need to get the relationship between moles of HNO3 and moles of NH3. We have to use all three equations.

Thus we can produce 16 mol HNO

for every 24 mol NH3, we begin with:

Thisisan oversimplified answer. In practice,the NO producedinthe third stepis recycled back continuously into the process in the second step. If this is taken into consideration, then the conversion factor betweenmol NH3 and mol HNO3 turns out to be 1 : 1; that is, 1 mole of NH3 produces 1 mole of HNO3 Taking into consideration that NO is recycled back gives an answer of 2.7 × 105 kg NH3 reacted.

Let’s assume x moles of Fe reacts to form x moles of FeO. Then 0.3581 – x, the remaining moles of Fe, reacts to form Fe2O3. Balancing the two equations in terms of x:

18.3 g N × N 01g 14 N 1mol = 1.31 mol N; 42.5 g O × O 00g 16 O 1mol = 2.66 mol O

CHAPTER 3 STOICHIOMETRY

31 1 08 3 = 2.35; 31 1 19 2 = 1.67; 31 1 31 1 = 1.00; 31 1 66 2 = 2.03

an empirical formula of C7H5N3O6. 164 1.0 × 106 kg HNO3 × 3 3 3 3 HNO 63.02g HNO 1mol HNO kg HNO 1000g  = 1.6 × 107 mol HNO3

3 3 3 2 2 3 NH mol 24 HNO mol 16 NH mol 4 NO mol 4 NO mol 2 NO mol 2 NO 3mol HNO mol 2   

1.6 × 107 mol HNO3 × 3 3 3 3 NH mol NH g 17.03 HNO mol 16 NH mol 24  = 4.1 × 108 g or 4.1 × 105 kg NH3

165 Fe(s) + FeO(s) (g) O2 2 1  ; 2 Fe(s) + (s) O Fe (g) O 3 2 2 2 3  20.00 g Fe  55.85g Fe 1mol = 0.3581 mol 00g 32 O 1mol O g 24) 3 20 (11 2 2  = 0.2488 mol O2 consumed (1 extra sig. fig.)

x Fe + FeO O 2 1 2 x x  3 2 2 O Fe mol 2 3581 0 O mol 2 3581 0 2 3 Fe mol ) 3581 (0               x x x

Setting up an equation for total moles of

167 The two relevant equations are:

From the balanced equations,

Setting up four equations to solve for the four unknowns:

CHAPTER 3 STOICHIOMETRY 85

FeO 079mol 0 0791 0 , O 2488mol 0 ) 3581 (0 2 4 3 2 1     x x x 0.079 mol FeO  mol FeO 85g 71 = 5.7 g FeO produced Mol Fe2O3 produced = 2 079 0 3581 0 = 0.140 mol Fe2O3 0.140 mol Fe2O3  mol O Fe 159.70g 3 2 = 22.4 g Fe2O3 produced 166 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l); C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) 30.07 g/mol 44.09 g/mol Let x = mass C2H6, so 9.780 x = mass C3H8. Use the balanced equation to set up a mathematical expression for the moles of O2 required. 1 5 44.09 9.780 2 7 30.07    x x = 1.120 mol O2 Solving: x = 3.7 g C2H6; 780g 9 g 3.7 × 100 = 38% C2H6 by mass

Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) and Mg(s) + 2 HCl(aq)  MgCl2(aq) + H2(g)

x = mass Mg,

10.00

mass Zn.

moles Mg reacted Mol H2 = 0.5171 g H2 × 2 2 H 2.016g H 1mol = 0.2565 mol H2 0.2565 = 24.31 x + 65.38 10.00 x ; solving: x = 4.008 g Mg 00g 10 008g 4 × 100 = 40.08% Mg 168. a N2H4 + b NH3 + (10.00 4.062) O2  c NO2 + d H2O

consumed:

Let

x =

moles

produced

moles

reacted +

2a + b = c (N mol balance) 2c + d = 2(10.00 4.062) (O mol balance) 4a + 3b = 2d (H mol balance)

CHAPTER 3 STOICHIOMETRY

a(32.05) + b(17.03) = 61.00 (mass balance)

Solving the simultaneous equations gives a = 1.12 = 1.1 mol N2H4

169 We know that water is a product, so one of the elements in the compound is hydrogen. XaHb + O2  H2O + ?

To balance the H atoms, the mole ratio between XaHb and H2O = b 2

Mol compound = 09g/mol 62 1.39g = 0.0224 mol; mol H2O = 02g/mol 18 1.21g = 0.0671 mol 0671 0 0224 0 2  b , b = 6; XaH6 has a molar mass of 62.09 g/mol.

62.09 = a(molar mass of X) + 6(1.008), a(molar mass of X) = 56.04

Some possible identities for X could be Fe (a = 1), Si (a = 2), N (a = 4), and Li (a = 8). N fits the data best, so N4H6 is the most likely formula.

170. The balanced equation is 2 Sc(s) + 2x HCl(aq) → 2 ScClx(aq)+ x H2(g)

The mole ratio of Sc : H2 = x 2 .

171. Total mass of copper used:

g 00 61 H N 05g/mol 32 H N mol 1 1 4 2 4 2   100 = 58% N2H4

Mol

2.25

1molSc

0.0500

Mol H2 = 0.1502 g H2 × 2 2 H 016g 2 H 1mol = 0.07450 mol H2 x 2 =

0 0500 0

ScCl3

Sc =

g Sc × 44.96gSc

mol Sc

07450

, x = 3; the formula is

10,000 boards × 3 cm 96g 8 board 060cm) 0 0cm 16 cm 0 (8    = 6.9 × 105 g Cu Amount of Cu to be

0.80 × (6.9 × 105 g) = 5.5 × 105 g Cu. 5.5 × 105 g Cu × 2 4 3 2 4 3 2 4 3 Cl ) Cu(NH mol Cl ) Cu(NH 202.59g Cu mol Cl ) Cu(NH 1mol Cu g 63.55 Cu 1mol   = 1.8 × 106 g Cu(NH3)4Cl2

recovered =

172. a. From the reaction stoichiometry we would expect to produce 4 mol of acetaminophen for every 4 mol of C6H5O3N reacted. The actual yield is 3 mol of acetaminophen compared to a theoretical yield of 4 mol of acetaminophen. Solving for percent yield by mass (where M = molar mass acetaminophen):

b. The product of the percent yields of the individual steps must equal the overall yield, 75%.

(0.87)(0.98)(x) = 0.75, x = 0.88; step III has a percent yield of 88%.

173.

So 3.60 × 2 10 mol X has a mass equal to 7.45 g X. The molar mass of X is:

There must be twice as many moles of MO2 as moles of M2S3 in order to balance M in the reaction. Setting up an equation for 2(mol M2S3) = mol MO2 where A = molar mass M:

Note: From the periodic table, M is tungsten, W.

175. Consider the case of aluminum plus oxygen. Aluminum forms Al3+ ions; oxygen forms O2 anions.Thesimplestcompound betweenthetwoelementsisAl2O3 Similarly,wewouldexpect the formula of any Group 6A element with Al to be Al2X3 Assuming this, out of 100.00 g of compound, there are 18.56 g Al and 81.44 g of the unknown element, X. Let’s use this information to determine the molar mass of X, which will allow us to identify X from the periodic table.

CHAPTER 3 STOICHIOMETRY 87 5.5 × 105 g Cu × 3 3 3 NH mol NH 03g 17 Cu mol NH 4mol Cu g 55 63 Cu 1mol   = 5.9 × 105 g NH3

M mol 4 M mol 3   ×

percent yield =

100 = 75%

10.00

2 → 12.55 g XCl4; 2.55 g Cl

XCl2

XCl4. XCl4

g XCl2. From

g XCl2

2.55

Cl;

XCl2 = 10.00 2.55 = 7.45 g X. 2.55 g Cl × 2 2 XCl mol X mol 1 Cl mol 2 XCl mol 1 Cl g 35.45 Cl mol 1   = 3.60 × 2 10 mol X

g XCl

+ excess Cl

reacted with

to form

contains 2.55 g Cl and 10.00

the mole ratios, 10.00

must also contain

mass X in

X mol 10 60 3 X g 7.45 2  = 207

X; atomic

174 4.000 g M2S3  3.723 g MO2

g/mol

mass = 207 u, so X is Pb.

32.00 A 3.723 96.21 2A 8.000 2(16.00), A 3.723g 3(32.07) 2A 4.000g 2             (8.000)A + 256.0 = (7.446)A + 358.2, (0.554)A

= 102.2, A = 184 g/mol; atomic mass

184 u

18.56

81.44 g of X must contain 1.032 mol of X.

Molar mass of X = X mol 032 1 X 44g 81 = 78.91 g/mol X.

From the periodic table, the unknown element is selenium, and the formula is Al2Se3.

176. Let x = mass KCl and y = mass KNO3. Assuming 100.0 g of mixture, x + y = 100.0 g.

Molar mass KCl = 74.55 g/mol; molar mass KNO3 = 101.11 g/mol

Mol KCl = 74.55 x ; mol KNO3 = 101.11 y

Knowing that the mixture is 43.2% K, then in the 100.0 g mixture, an expression for the mass of K is:

We have two equations and two unknowns:

177. The balanced equations are:

Let 4x = number of moles of NO formed, and let 4y = number of moles of NO2 formed.

Solving by the method of simultaneous equations:

STOICHIOMETRY

CHAPTER 3

Al mol 2 X mol 3 Al g 98 26 Al mol 1  = 1.032 mol X

g Al ×

39.10        101.11 74.55 y x = 43.2

(0.5245)

43.2 x

x + (0.3867)y =

+ y = 100.0

KCl; 100.0g 9g 32  100 = 32.9% KCl

Solving, x = 32.9 g

4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g) and 4 NH3(g) + 7 O2(g)  4 NO2(g) + 6 H2O(g)

Then: 4x NH3 + 5x O2  4x NO + 6x H2O and 4y NH3 + 7y O2  4y NO2 + 6y H2O

NH3

4x + 4y = 2.00. 10.00 6.75 = 3.25 mol O2 reacted, so 5x + 7y = 3.25.

All the

reacted, so

20x + 28y = 13.0 20x 20y = 10.0 8y = 3.0, y = 0.38; 4x + 4 × 0.38 = 2.00, x = 0.12

Mass % O = 100.00 (60.0 + 4.48) = 35.5% O

Assuming 100.00 g aspirin:

Balance the aspirin synthesis reaction to determine the formula for salicylic acid.

Integrative Problems

b. The total number of platinum atoms is 14 × 20 = 280 atoms (exact number). The mass of these atoms is:

CHAPTER 3 STOICHIOMETRY 89

NO = 4x = 4 × 0.12

0.48 mol NO formed 178 CxHyOz + oxygen  x CO2 + y/2 H2O

% C in aspirin = aspirin 1.00g C mol C 01g 12 CO mol C 1mol CO 01g 44 CO 1mol CO 2.20g 2 2 2 2    = 60.0% C Mass % H in aspirin = aspirin 1.00g H mol H 1.008g O H mol H 2mol O H 18.02g O H 1mol O H 400g 0 2 2 2 2    = 4.48% H

Mol

Mass

60.0 g C × C 01g 12 C 1mol = 5.00 mol C; 4.48 g H × H 008g 1 H 1mol = 4.44 mol H 35.5 g O × O 00g 16 O 1mol = 2.22 mol O

22 2 00 5 = 2.25; 22 2 44 4 = 2.00

2.25 H2.00O)4 = C9H8O4. Empirical mass  9(12)

8(1)

170–190 g/mol

C9H8O4.

Dividing by the smallest number:

Empirical formula: (C

+ 4(16) = 180 g/mol; this is in the

range, so the molecular formula is also

CaHbOc + C4H6O3  C9H8O4 + C2H4O2, CaHbOc = salicylic acid = C7H6O3

179. a. 1.05 × 20 10 g Fe × Fe mol Fe atoms 10 022 6 Fe 85g 55 Fe 1mol 23   = 113 atoms Fe

280 atoms Pt × Pt mol Pt 1g 195 Pt atoms 10 6.022 Pt 1mol 23   = 9.071 × 20 10 g Pt c. 9.071 × 20 10 g Ru × Ru mol Ru atoms 10 022 6 Ru 101.1g Ru 1mol 23   = 540.3 = 540 atoms Ru

180 Assuming 100.00 g of tetrodotoxin:

CHAPTER 3 STOICHIOMETRY

To get whole numbers for each element, multiply through by 3.

Because the empirical mass and molar mass are the same, the molecular formula is the same as

The mass of X = 1/2(70.9 g/mol) = 35.5 g/mol. This is the element chlorine.

100.00 g of MX3 (= MCl3) compound:

M is the element yttrium (Y), and the name of YCl3 is yttrium(III) chloride.

41.38 g C × C 01g 12 C 1mol = 3.445 mol C; 13.16 g N × N 01g 14 N 1mol = 0.9393 mol N 5.37 g H × H 008g 1 H 1mol = 5.33 mol H; 40.09 g O × O 00g 16 O 1mol = 2.506 mol O

0.9393 445 3 = 3.668; 0.9393 33 5 = 5.67; 0.9393 506 2 = 2.668

Divide by the smallest number:

Empirical formula: (C3.668H5.67NO2 668)3 = C11H17N3O8; the mass of the empirical formula is 319.3 g/mol.

mass tetrodotoxin = molecules 10 022 6 mol 1 3molecules g 10 59 1 23 21    = 319 g/mol

Molar

the

C11H17N3O8 165 lb × 1mol molecules 10 6.022 3g 319 1mol gμ g 10 1 kg gμ10. 2046lb 2 1kg 23 6       = 1.4 × 1018 molecules tetrodotoxin is the LD50 dosage 181 Molar mass X2 = molecules 10 6.022 mol 1 molecules 10 8.92 0.105g 23 20    = 70.9 g/mol

empirical formula,

Assuming

54.47 g Cl × 35.45g 1mol = 1.537 mol Cl 1.537 mol Cl × Cl 3mol M 1mol = 0.5123 mol M Molar mass of M = M 5123mol 0 M 53g 45 = 88.87 g/mol M