CHM 201 Problems Solution for Chap 3 Exercises, Fall 2013

17 CHAPTER 3 STOICHIOMETRY Atomic Masses and the Mass Spectrometer 23. 186.207 = 0.6260(186.956) + 0.3740(A), 186.207  117.0 = 0.3740(A) A = 3740 .0 69 2. = 185 amu (A = 184.95 amu without rounding to proper significant figures) 24. 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 amu; from the periodic table, the element is Pb. 25. Average atomic mass = A = 0.0800(45.95269) + 0.0730(46.951764) + 0.7380(47.947947) + 0.0550(48.947841) + 0.0540(49.944792) = 47.88 amu This is element Ti (titanium). 26. 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 amu. So the masses are about: 54Fe, 54.00 amu; 56Fe, 56.00 amu; 57Fe, 57.00 amu; 58Fe, 58.00 amu. Using these masses and the abundances given in the mass spectrum, 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 amu The average atomic mass listed in the periodic table is 55.85 amu. 27. If silver is 51.82% 107Ag, then the remainder is 109Ag (48.18%). Determining the atomic mass (A) of 109Ag: 107.868 = 100 48.18(A) 51.82(106.905)  10786.8 = 5540. + (48.18)A, A = 108.9 amu = atomic mass of 109Ag 28. Let x = % of 151Eu and y = % of 153Eu, then x + y = 100 and y = 100  x. 151.96 = 100 )(152.9209) (100 (150.9196) x x   15196 = (150.9196)x + 15292.09  (152.9209)x, 96 = (2.0013)x x = 48%; 48% 151Eu and 100  48 = 52% 153Eu 18 CHAPTER 3 STOICHIOMETRY 29. GaAs can be either 69GaAs or 71GaAs. The mass spectrum for GaAs will have two peaks at 144 (= 69 + 75) and 146 (= 71 + 75) with intensities in the ratio of 60 : 40 or 3 : 2. Ga2As2 can be 69Ga2As2, 69Ga71GaAs2, or 71Ga2As2. The mass spectrum will have three 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: 69Ga (0.60) 71Ga (0.40) 69Ga (0.60) 0.36 0.24 71Ga (0.40) 0.24 0.16 288 290 292 30. Scaled Intensity Compound Mass Intensity Largest Peak = 100 __________________________________________________________________ H2 120Te 121.92 0.09 0.3 H2 122Te 123.92 2.46 7.1 H2 123Te 124.92 0.87 2.5 H2 124Te 125.92 4.61 13.4 H2 125Te 126.92 6.99 20.3 H2 126Te 127.92 18.71 54.3 H2 128Te 129.92 31.79 92.2 H2 130Te 131.93 34.48 100.0 144 146 CHAPTER 3 STOICHIOMETRY 19 31. 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 at about equal abundance. The actual abundance is 50.68% 79Br and 49.32% 81Br. Moles and Molar Masses 32. 4.24 g C6H6 × .11g 78 1mol = 5.43 × 2 10 mol C6H6 5.43 × 2 10 mol C6H6 × mol molecules 10 .6 022 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 12atomstotal = 3.92 × 1023 atoms total 0.224 mol H2O × mol 18.02g = 4.04 g H2O 0.224 mol H2O × mol molecules 10 .6 022 23  = 1.35 × 1023 molecules H2O 1.35 × 1023 molecules H2O × molecule 3 atomstotal = 4.05 × 1023 atoms total 2.71 × 1022 molecules CO2 × molecules 10 022 .6 mol 1 23  = 4.50 × 2 10 mol CO2 100 50 0 122 124 126 128 130 132 134 20 CHAPTER 3 STOICHIOMETRY 4.50 × 2 10 mol CO2 × mol 44.01g = 1.98 g CO2 2.71 × 1022 molecules CO2 × CO2 molecule 3 atomstotal = 8.13 × 1022 atoms total 3.35 × 1022 atoms total × atomstotal 6 1molecule = 5.58 × 1021 molecules CH3OH 5.58 × 1021 molecules CH3OH × molecules 10 022 .6 mol 1 23  = 9.27 × 3 10 mol CH3OH 9.27 × 3 10 mol CH3OH × mol 32.04g = 0.297 g CH3OH 33. 4.0 g H2 × molH 1 atoms H 10 022 .6 molH 1 molH 2 016g H .2 1molH 23 2 2 2    = 2.4 × 1024 atoms 4.0 g He × molHe 1 atoms He 10 022 .6 003g He .4 1molHe 23   = 6.0 × 1023 atoms 1.0 mol F2 × molF 1 atomsF 10 022 .6 molF 1 2 molF 23 2   = 1.2 × 1024 atoms 44.0 g CO2 × molatoms 1 atoms 10 022 .6 molCO 1 2 O) molatoms 1( C 3 .01g CO 44 1molCO 23 2 2 2     = 1.81 × 1024 atoms 146 g SF6 × molatoms 1 atoms 022 10 .6 molSF 1 6 F) molatoms 1( S 7 .07 g SF 146 1molSF 23 6 6 6     = 4.21 × 1024 atoms The order is: 4.0 g He < 1.0 mol F2 < 44.0 g CO2 < 4.0 g H2 < 146 g SF6 34. Molar mass of C6H8O6 = 6(12.011) + 8(1.0079) + 6(15.999) = 176.123 g/mol 500.0 mg × .12g 176 mol 1 mg 1000 1g  = 2.839 × 3 10 mol 2.839 × 10−3 mol mol molecules 10 .6 022 23   = 1.710 × 1021 molecules 35. a. 9(12.011) + 8(1.0079) + 4(15.999) = 180.158 g/mol b. 500. mg × g 180.16 mol 1 mg 1000 1g  = 2.78 × 3 10 mol CHAPTER 3 STOICHIOMETRY 21 2.78 × 3 10 mol × mol molecules 10 .6 022 23  = 1.67 × 1021 molecules 36. a. 14 mol C × C mol 12.011g + 18 mol H × H mol .1 0079g + 2 mol N × N mol 14.007g + 5 mol O × O mol 15.999g = 294.305 g/mol b. 10.0 g C14H18N2O5 × 5 2 18 14 5 2 18 14 3. g C H N O 294 1molC H N O = 3.40 × 102 mol C14H18N2O5 c. 1.56 mol × mol 294 3. g = 459 g C14H18N2O5 d. 5.0 mg × mol molecles 02 10 .6 3. g 294 mol 1 mg 1000 1g 23    = 1.0 × 1019 molecules C14H18N2O5 e. 1.2 g C14H18N2O5 × N mol atomsN 10 02 .6 C H N O mol molN 2 3. g C H N O 294 1molC H N O 23 5 2 18 14 5 2 18 14 5 2 14 18    = 4.9 × 1021 atoms N f. 1.0 × 109 molecules × mol 3. g 294 atoms 10 02 .6 mol 1 23   = 4.9 × 1013 g g. 1 molecule mol .305g 294 atoms 10 022 .6 mol 1 23    = 4.887 × 1022 g C14H18N2O5 37. a. 2(12.01) + 3(1.008) + 3(35.45) + 2(16.00) = 165.39 g/mol b. 500.0 g × .39 g 165 1 mol = 3.023 mol C2H3Cl3O2 c. 2.0 × 10-2 mol × mol 165.39g = 3.3 g C2H3Cl3O2 d. 5.0 g C2H3Cl3O2 × molecule atomsCl 3 mol molecules 10 02 .6 .39g 165 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 C H Cl O mol .39g C H Cl O 165 molCl 3 molC H Cl O 1 .45g 35 1molCl   = 1.6 g chloral hydrate 22 CHAPTER 3 STOICHIOMETRY f. 500 molecules × mol .39g 165 molecules 10 022 .6 mol 1 23   = 1.373 × 1019 g 38. 1.0 lb flour × 2 4 2 2 4 2 2 4 2 9 9. g C H Br 187 molC H Br 1 flour g g C H Br 10 0. 30 flour lb 454g flour     2 4 2 23 C H Br mol molecules 10 .6 02   = 4.4 × 1016 molecules C2H4Br2 39. a. 20.0 mg C8H10N4O2 × .20g 194 mol 1 mg 1000 1g  = 1.03 × 4 10 mol C8H10N4O2 b. 2.72 × 1021 molecules C2H5OH × molecules 10 022 .6 mol 1 23  = 4.52 × 3 10 mol C2H5OH c. 1.50 g CO2 × .01g 44 1mol = 3.41 × 2 10 mol CO2 40. a. A chemical formula gives atom ratios as well as mole ratios. We will use both ratios to illustrate how these conversion factors can be used. Molar mass of C2H5O2N = 2(12.01) + 5(1.008) + 2(16.00) + 14.0l = 75.07 g/mol 5.00 g C2H5O2N × C H O N mol 10 molecules C H O N 022 .6 .07 g C H O N 75 molC H O N 1 2 5 2 2 5 2 23 2 5 2 2 5 2   × C H O N molecule atomN 1 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 Mg N mol formulaunitsMg N 10 022 .6 .95g Mg N 100 1molMg N   Mg3N2 mol 2 atoms  = 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 × 2 3 2 3 .10g Ca(NO ) 164 molCa(NO ) 1 Ca(NO3)2 mol 2 molN  N mol atomsN 10 .6 022 23   = 3.67 × 1022 atoms N CHAPTER 3 STOICHIOMETRY 23 d. Molar mass of N2O4 = 2(14.01) + 4(16.00) = 92.02 g/mol 5.00 g N2O4 × 4 2 4 2 4 2 N O mol molN 2 .02g N O 92 1molN O  N mol atomsN 10 .6 022 23   = 6.54 × 1022 atoms N Percent Composition 41. NO: Mass % N = .01g NO 30 14.01g N × 100 = 46.68% N NO2: Mass % N = .01g NO2 46 14.01g N × 100 = 30.45% N N2O: Mass % N = .02g N O 44 (14.01) g N 2 2 × 100 = 63.65% N 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. 42. a. C8H10N4O2: Molar mass = 8(12.01) + 10(1.008) + 4(14.0l) + 2(16.00) = 194.20 g/mol Mass % C = 2 4 8 10 .20g C H N O 194 8(12.01) g C × 100 = .20 194 96.08 × 100 = 49.47% C b. C12 H22O11: Molar mass = 12(12.01) + 22(1.008) + 11(16.00) = 342.30 g/mol Mass % C = 11 22 12 .30g C H O 342 12(12.01) g C × 100 = 42.10% C c. C2H5OH: Molar mass = 2(12.01) + 6(1.008) + 1(16.00) = 46.07 g/mol Mass % C = .07 g C H OH 46 (12.01) g C 2 5 2 × 100 = 52.14% C The order from lowest to highest mass percentage of carbon is: sucrose (C12H22O11) < caffeine (C8H10N4O2) < ethanol (C2H5OH) 43. Molar mass = 20(12.01) + 29(1.008) + 19.00 + 3(16.00) = 336.43 g/mol 24 CHAPTER 3 STOICHIOMETRY Mass % C = .43g compound 336 20(12.01) g C × 100 = 71.40% C Mass % H = .43g compound 336 29 .1( 008) g H × 100 = 8.689% H Mass % F = .43g compound 336 19.00g F × 100 = 5.648% F Mass % O = 100.00  (71.40 + 8.689 + 5.648) = 14.26% O or: Mass % O = .43g compound 336 (3 16.00) g O × 100 = 14.27% O 44. a. C3H4O2: Molar mass = 3(12.011) + 4(1.0079) + 2(15.999) = 36.033 + 4.0316 + 31.998 = 72.063 g/mol Mass % C = g compound 72.063 36.033g C × 100 = 50.002% C Mass % H = g compound 72.063 4.0316g H × 100 = 5.5945% H Mass % O = 100.000  (50.002 + 5.5945) = 44.404% O or: % O = g 72.063 31.998g × 100 = 44.403% O b. C4H6O2: Molar mass = 4(12.011) + 6(1.0079) + 2(15.999) = 48.044 + 6.0474 + 31.998 = 86.089 g/mol Mass % C = g 86.089 48.044g × 100 = 55.807% C; mass % H = g 86.089 6.0474g × 100 = 7.0246% H Mass % O = 100.000  (55.807 + 7.0246) = 37.168% O c. C3H3N: Molar mass = 3(12.011) + 3(1.0079) + 1(14.007) = 36.033 + 3.0237 + 14.007 = 53.064 g/mol Mass % C = g 53.064 36.033g × 100 = 67.905% C; mass % H = g 53.064 3.0237g × 100 = 5.6982% H Mass % N = g 53.064 14.007g × 100 = 26.396% N or % N = 100.000  (67.905 + 5.6982) = 26.397% N 45. In 1 mole of YBa2Cu3O7, there are 1 mole of Y, 2 moles of Ba, 3 moles of Cu, and 7 moles of O. CHAPTER 3 STOICHIOMETRY 25 Molar mass = 1 mol Y              Ba mol 2 molBa 1373. g Ba Y mol 88.91g Y + 3 mol Cu       Cu mol 63.55g Cu + 7 mol O       O mol 16.00g O Molar mass = 88.91 + 274.6 + 190.65 + 112.00 = 666.2 g/mol Mass % Y = 2. g 666 88.91g × 100 = 13.35% Y; mass % Ba = 2. g 666 274 6. g × 100 = 41.22% Ba Mass % Cu = 2. g 666 190.65g × 100 = 28.62% Cu; mass % O = 2. g 666 112 0. g × 100 = 16.81% O 46. If we have 100.0 g of Portland cement, we have 50. g Ca3SiO5, 25 g Ca2SiO4, 12 g Ca3Al2O6, 8.0 g Ca2AlFeO5, and 3.5 g CaSO42H2O. Mass percent Ca: 50. g Ca3SiO5 × molCa 1 .08g Ca 40 molCa SiO 1 molCa 3 .33g Ca SiO 228 molCa SiO 1 5 3 5 3 5 3   = 26 g Ca 25 g Ca2SiO4 × .25g Ca2SiO 4 172 80.16g Ca = 12 g Ca 12 g Ca3Al2O6 × 6 2 3 .20g Ca Al O 270 120.24g Ca = 5.3 g Ca 8.0 g Ca2AlFeO5 × .99g Ca2AlFeO5 242 80.16g Ca = 2.6 g Ca 3.5 g CaSO42H2O × 2 H O .18g CaSO 172 .08g Ca 40 2 4  = 0.81 g Ca Mass of Ca = 26 + 12 + 5.3 + 2.6 + 0.81 = 47 g Ca % Ca = 0. g cement 100 47 g Ca × 100 = 47% Ca Mass percent Al: 12 g Ca3 Al2O6 × 6 2 3 .20 g Ca Al O 270 53.96 g Al = 2.4 g Al 26 CHAPTER 3 STOICHIOMETRY 8.0 g Ca2AlFeO5 × .99g Ca2AlFeO5 242 26.98g Al = 0.89 g Al % Al = 0. g 100 .0 89g g 4.2  × 100 = 3.3% Al Mass percent Fe: 8.0 g Ca2AlFeO5 × .99g Ca2AlFeO5 242 55.85g Fe = 1.8 g Fe; % Fe = 0. g 100 g 8.1 × 100 = 1.8% Fe 47. There are 0.390 g Cu for every 100.000 g of fungal laccase. Let’s assume 100.000 g fungal laccase. Mol fungal laccase = 0.390 g Cu molCu 4 molfungal laccase 1 g Cu 63.55 1molCu   = 1.53 × 103 mol molfungal laccase 1 x g fungal laccase = mol 10 53 .1 .000g 100 3  , x = molar mass = 6.54 × 104 g/mol 48. Assuming 100.00 g cyanocobalamin: mol cyanocobalamin = 4.34 g Co Co mol amin molcyanocobal 1 g Co 58.93 1molCo   = 7.36 × 102 mol cyanocobalamin mol 36 10 .7 .00g 100 amin molcyanocobal 1 amin cyanocobal g 2   x , x = molar mass = 1360 g/mol Empirical and Molecular Formulas 49. a. SNH: Empirical formula mass = 32.07 + 14.01 + 1.008 = 47.09 g/mol .09g 47 188.35g = 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 .88g 115 347.64g = 3.0000; molecular formula is (NPCl2)3 or N3P3Cl6. c. CoC4O4: 58.93 + 4(12.01) + 4(16.00) = 170.97 g/mol .97 g 170 341.94g = 2.0000; molecular formula: Co2C8O8 CHAPTER 3 STOICHIOMETRY 27 d. SN: 32.07 + 14.01 = 46.08 g/mol; .08g 46 184.32g = 4.000; molecular formula: S4N4 50. 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 51. a. Molar mass of CH2O = 1 mol C       C mol 12.011g + 2 mol H       H mol .1 0079g + 1 mol O       O mol 15.999g = 30.026 g/mol % C = .026g CH O 30 .011g C 12 2 × 100 = 40.002% C; % H = .026g CH O 30 0158g H .2 2 × 100 = 6.7135% H % O = .026g CH O 30 .999g O 15 2 × 100 = 53.284% O or % O = 100.000  (40.002 + 6.7135) = 53.285% b. Molar mass of C6H12O6 = 6(12.011) + 12(1.0079) + 6(15.999) = 180.155 g/mol % C = 6 6 12 .155g C H O 180 72.066g C × 100 = 40.002%; % H = .155g 180 12 .1( 0079) g × 100 = 6.7136% % O = 100.00  (40.002 + 6.7136) = 53.284% c. Molar Mass of HC2H3O2 = 2(12.011) + 4(1.0079) + 2(15.999) = 60.052 g/mol % C = .052g 60 24.022g × 100 = 40.002%; % H = .052g 60 .4 0316g × 100 = 6.7135% % O = 100.000  (40.002 + 6.7135) = 53.285% 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. 28 CHAPTER 3 STOICHIOMETRY 52. Assuming 100.00 g of compound (mass oxygen = 100.00 g  41.39 g C  3.47 g H = 55.14 g O): 41.39 g C × .011g C 12 1molC = 3.446 mol C; 3.47 g H × 008g H .1 1molH = 3.44 mol H 55.14 g O × .999g O 15 1molO = 3.446 mol O All are the same mole values, so the empirical formula is CHO. The empirical formula mass is 12.01 + 1.008 + 16.00 = 29.02 g/mol. Molar mass = 129mol .0 15 0. g = 116 g/mol mass Empirical Molar mass = 29.02 116 = 4.00; molecular formula = (CHO)4 = C4H4O4 53. Assuming 100.0 g of compound: 26.7 g P × .97 g P 30 1molP = 0.862 mol P; 12.1 g N × .01g N 14 1molN = 0.864 mol N 61.2 g Cl × .45g Cl 35 1molCl = 1.73 mol Cl .0 862 .1 73 = 2.01; the empirical formula is PNCl2. The empirical formula mass is  31.0 + 14.0 + 2(35.5) = 116 g/mol. 116 580 formulamass Empirical Molar mass  = 5.0; the molecular formula is (PNCl2)5 = P5N5Cl10. 54. Out of 100.00 g of adrenaline, there are: 56.79 g C × .011g C 12 1molC = 4.728 mol C; 6.56 g H × 008g H .1 1molH = 6.51 mol H 28.37 g O × .999g O 15 1molO = 1.773 mol O; 8.28 g N × .01g N 14 1molN = 0.591 mol N Dividing each mole value by the smallest number: .0 591 .4 728 = 8.00; .0 591 .6 51 = 11.0; .0 591 .1 773 = 3.00; .0 591 .0 591 = 1.00 This gives adrenaline an empirical formula of C8H11O3N. CHAPTER 3 STOICHIOMETRY 29 55. Compound I: mass O = 0.6498 g HgxOy  0.6018 g Hg = 0.0480 g O 0.6018 g Hg × 6. g Hg 200 1 mol Hg = 3.000 × 103 mol Hg 0.0480 g O × .00g O 16 1molO = 3.00 × 103 mol O The mole ratio between Hg and O is 1 : 1, so the empirical formula of compound I is HgO. Compound II: mass Hg = 0.4172 g HgxOy  0.016 g O = 0.401 g Hg 0.401 g Hg × 6. g Hg 200 1 mol Hg = 2.00 × 103 mol Hg; 0.016 g O × .00g O 16 1molO = 1.0 × 103 mol O The mole ratio between Hg and O is 2 : 1, so the empirical formula is Hg2O. 56. 1.121 g N × g N 14.007 1molN = 8.003 × 2 10 mol N; 0.161 g H × 008g H .1 1molH = 1.60 × 1 10 mol H 0.480 g C × .01g C 12 1molC = 4.00 × 2 10 mol C; 0.640 g O × .00g O 16 1molO = 4.00 × 2 10 mol O Dividing all mole values by the smallest number: 2 2 10 00 .4 10 003 .8     = 2.00; 2 1 10 00 .4 10 60 .1     = 4.00; 2 2 10 00 .4 10 00 .4     = 1.00 The empirical formula is N2H4CO. 57. First, we will determine composition in mass percent. We assume that all the carbon in the 0.213 g CO2 came from the 0.157 g of the compound and that all the hydrogen in the 0.0310 g H2O came from the 0.157 g of the compound. 0.213 g CO2 × .01g CO2 44 12.01g C = 0.0581 g C; % C = 157g compound .0 .0 0581g C × 100 = 37.0% C 0.0310 g H2O × .02 g H O 18 016g H .2 2 = 3.47 × 103 g H; % H = 157g .0 g 10 .3 47 3   100 = 2.21% H We get the mass percent of N from the second experiment: 0.0230 g NH3 × .03g NH3 17 14.01g N = 1.89 × 102 g N 30 CHAPTER 3 STOICHIOMETRY % N = 103 g .0 g .1 89 10 2  × 100 = 18.3% N The mass percent of oxygen is obtained by difference: % O = 100.00  (37.0 + 2.21 + 18.3) = 42.5% O So, out of 100.00 g of compound, there are: 37.0 g C × .01g C 12 1 mol C = 3.08 mol C; 2.21 g H × 008g H .1 1molH = 2.19 mol H 18.3 g N × .01g N 14 1molN = 1.31 mol N; 42.5 g O × .00g O 16 1molO = 2.66 mol O Lastly, and often the hardest part, we need to find simple whole number ratios. Divide all mole values by the smallest number: .1 31 .3 08 = 2.35; .1 31 .2 19 = 1.67; .1 31 .1 31 = 1.00; .1 31 .2 66 = 2.03 Multiplying all these ratios by 3 gives an empirical formula of C7H5N3O6. 58. a. Only acrylonitrile contains nitrogen. If we have 100.00 g of polymer: 8.80 g N × molC H N 1 .06g C H N 53 .01g N 14 molC H N 1 3 3 3 3 3 3  = 33.3 g C3H3N % C3H3N = .00g polymer 100 33 3. g C H N 3 3 = 33.3% C3H3N Only butadiene in the polymer reacts with Br2: 0.605 g Br2 × 6 4 6 4 2 6 4 2 2 C H mol .09g C H 54 Br mol molC H 1 8. g Br 159 1molBr   = 0.205 g C4H6 % C4H6 = 20g .1 .0 205g × 100 = 17.1% C4H6 b. If we have 100.0 g of polymer: 33.3 g C3H3N × .06g 53 1molC H N 3 3 = 0.628 mol C3H3N 17.1 g C4H6 × 6 4 6 4 .09g C H 54 1molC H = 0.316 mol C4H6 CHAPTER 3 STOICHIOMETRY 31 49.6 g C8H8 × 8 8 8 8 .14g C H 104 1molC H = 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 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 (A4B2S3)n. 59. First, we will determine composition by mass percent: 16.01 mg CO2 × g mg 1000 .009g CO 44 .011g C 12 mg 1000 g 1 2   = 4.369 mg C % C = .68mg compound 10 .4 369mg C × 100 = 40.91% C 4.37 mg H2O × g mg 1000 .02g H O 18 016g H .2 mg 1000 g 1 2   = 0.489 mg H % H = .68mg 10 .0 489mg × 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 × .011g C 12 1molC = 3.406 mol C; 4.58 g H × 008g H .1 1molH = 4.54 mol H 54.51 g O × .999g O 15 1molO = 3.407 mol O Dividing by the smallest number: 406 .3 .4 54 = 1.33  3 4 ; the empirical formula is C3H4O3. The empirical formula mass of C3H4O3 is ≈ 3(12) + 4(1) + 3(16) = 88 g. Because 88 1. 176 = 2.0, the molecular formula is C6H8O6. 60. 41.98 mg CO2 × .009 mg CO2 44 12.011mg C = 11.46 mg C; % C = .81mg 19 11.46 mg × 100 = 57.85% C 6.45 mg H2O × .02 mg H O 18 016mg H .2 2 = 0.722 mg H; % H = .81mg 19 .0 722mg × 100 = 3.64% H 32 CHAPTER 3 STOICHIOMETRY % O = 100.00  (57.85 + 3.64) = 38.51% O Out of 100.00 g terephthalic acid, there are: 57.85 g C × .011g C 12 1molC = 4.816 mol C; 3.64 g H × 008g H .1 1molH = 3.61 mol H 38.51 g O × .999g O 15 1molO = 2.407 mol O .2 407 .4 816 = 2.001; 407 .2 .3 61 = 1.50; .2 407 .2 407 = 1.000 The 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) = 83. Molar mass = 250mol .0 41 5. g = 166 g/mol; 83 166 = 2.0; the molecular formula is C8H6O4. Balancing Chemical Equations 61. 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) d. 2 Eu(s) + 6 HF(g)  2 EuF3(s) + 3 H2(g) e. 2 C6H6(l) + 15 O2(g)  12 CO2(g) + 6 H2O(g) 62. An important part to this problem is writing out correct formulas. If the formulas are incorrect, then the balanced reaction is incorrect. 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) 63. Only one product is formed in this representation. This product has two Ys 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. CHAPTER 3 STOICHIOMETRY 33 64. 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) 65. 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. 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) C and S are balanced; balance N: CS2 + 2 NH3  H2S + NH4SCN H is also balanced. CS2(l) + 2 NH3(g)  H2S(g) + NH4SCN(s). 66. a. SiO2(s) + C(s)  Si(s) + CO(g); Si is balanced. Balance oxygen atoms: SiO2 + C  Si + 2 CO Balance carbon atoms: SiO2(s) + 2 C(s)  Si(s) + 2 CO(g) b. SiCl4(l) + Mg(s)  Si(s) + MgCl2(s); Si is balanced. Balance Cl atoms: SiCl4 + Mg  Si + 2 MgCl2 Balance Mg atoms: SiCl4(l) + 2 Mg(s)  Si(s) + 2 MgCl2(s) 34 CHAPTER 3 STOICHIOMETRY c. Na2SiF6(s) + Na(s)  Si(s) + NaF(s); Si is balanced. Balance F atoms: Na2SiF6 + Na  Si + 6 NaF Balance Na atoms: Na2SiF6(s) + 4 Na(s)  Si(s) + 6 NaF(s) Reaction Stoichiometry 67. 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 g CO 0.4 g air 100 CO mol .01g CO 44 molLiOH 2 molCO 1 .95g LiOH 23 1molLiOH min 60 h 1 L air 140 min 1 mL 1000 L 1 0010g air .0 1mLair    = 68 h = 2.8 days 68. 1.0 × 104 kg waste ×        4 2 7 5 4 4 4 molNH 55 molC H O N 1 .04g NH 18 molNH 1 kg g 1000 kg waste 100 kg NH 0.3  C H O N mol 1. g C H O N 113 2 7 5 2 7 5 = 3.4 × 104 g tissue if all NH4 + converted Because only 95% of the NH4 + ions react: mass of tissue = (0.95)(3.4 × 104 g) = 3.2 × 104 g or 32 kg bacterial tissue 69. 1.0 × 103 g phosphorite ×   4 2 3 4 2 3 4 2 3 .18g Ca (PO ) 310 molCa (PO ) 1 e g phosphorit 100 75g Ca (PO ) 4 2 3 4 molCa (PO ) 2 molP 1 × 4 4 P mol 123.88g P = 150 g P4 70. Total mass of copper used: 10,000 boards × 3 cm 96g .8 board .0 060cm) 16 0. cm cm ( 0.8    = 6.9 × 105 g Cu Amount of Cu to be recovered = 0.80 × (6.9 × 105 g) = 5.5 × 105 g Cu 5.5 × 105 g Cu × 2 4 3 2 3 4 2 4 3 Cu(NH ) Cl mol 6. g Cu(NH ) Cl 202 Cu mol molCu(NH ) Cl 1 .55g Cu 63 1molCu   = 1.8 × 106 g Cu(NH3)4Cl2 CHAPTER 3 STOICHIOMETRY 35 5.5 × 105 g Cu × 3 3 3 NH mol .03g NH 17 Cu mol molNH 4 .55g Cu 63 1molCu   = 5.9 × 105 g NH3 71. 1.000 kg Al × 4 4 4 4 4 4 NH ClO mol .49g NH ClO 117 molAl 3 molNH ClO 3 .98g Al 26 molAl 1 Al kg 1000g Al    = 4355 g NH4ClO4 72. 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 P O mol .88g P O 283 molKClO 10 molP O 3 .55g KClO 122 1molKClO   = 36.8 g P4O10 73. Fe2O3(s) + 2 Al(s)  2 Fe(l) + Al2O3(s) 15.0 g Fe × .85g Fe 55 1molFe = 0.269 mol Fe; 0.269 mol Fe × Al mol .98g Al 26 molFe 2 2 molAl  = 7.26 g Al 0.269 mol Fe × 3 2 3 2 3 2 Fe O mol .70g Fe O 159 molFe 2 1molFe O  = 21.5 g Fe2O3 0.269 mol Fe × 3 2 3 2 3 2 Al O mol .96g Al O 101 molFe 2 1molAl O  = 13.7 g Al2O3 74. 1.0 × 106 kg HNO3 3 3 3 3 0. g HNO 63 molHNO 1 HNO kg 1000g HNO   = 1.6 × 107 mol HNO3 We need to get the relationship between moles of HNO3 and moles of NH3. We have to use all three equations: 3 2 2 3 molNH 4 molNO 4 molNO 2 molNO 2 molNO 3 2 molHNO   = 3 3 molNH 24 16 molHNO Thus we can produce 16 mol HNO3 for every 24 mol NH3 that we begin with: 1.6 × 107 mol HNO3 × 3 3 3 3 NH mol 0. g NH 17 molHNO 16 24 molNH  = 4.1 × 108 g or 4.1 × 105 kg NH3 This is an oversimplified answer. In practice, the NO produced in the third step is recycled back continuously into the process in the second step. If this is taken into consideration, then the conversion factor between mol NH3 and mol HNO3 turns out to be 1 : 1; that is, 1 mol of NH3 produces 1 mol of HNO3. Taking into consideration that NO is recycled back gives an answer of 2.7 × 105 kg NH3 reacted. 36 CHAPTER 3 STOICHIOMETRY Limiting Reactants and Percent Yield 75. One method to solve limiting-reagent problems is to assume that each reactant is limiting and then 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. 5.00 × 106 g NH3 3 3 3 molNH 2 molHCN 2 .03g NH 17 1molNH   = 2.94 × 105 mol HCN 5.00 × 106 g O2 2 2 2 molO 3 molHCN 2 .00g O 32 1molO   = 1.04 × 105 mol HCN 5.00 × 106 g CH4 4 4 4 molCH 2 molHCN 2 .04g CH 16 1molCH   = 3.12 × 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 × 105 mol HCN. The mass of HCN that can be produced is: 1.04 × 105 mol HCN × HCN mol 27.03g HCN = 2.81 × 106 g HCN 5.00 × 106 g O2 molH O 1 .02g H O 18 molO 3 molH O 6 .00g O 32 molO 1 2 2 2 2 2 2    = 5.63 × 106 g H2O 76. 2 C3H6(g) + 2 NH3(g) + 3 O2(g)  2 C3H3N(g) + 6 H2O(g) a. We will solve this limiting reagent problem using the same method as described in Exercise 3.75. 1.00 × 103 g C3H6 6 3 3 3 6 3 6 3 molC H 2 molC H N 2 .08g C H 42 1molC H   = 23.8 mol C3H3N 1.50 × 103 g NH3 3 3 3 3 3 molNH 2 molC H N 2 .03g NH 17 1molNH   = 88.1 mol C3H3N 2.00 × 103 g O2 2 3 3 2 2 molO 3 molC H N 2 .00g O 32 1molO   = 41.7 mol C3H3N C3H6 is limiting because it produces the smallest amount of product, and the mass of acrylonitrile that can be produced is: 23.8 mol × mol 53.06g C H N 3 3 = 1.26 × 103 g C3H3N CHAPTER 3 STOICHIOMETRY 37 b. 23.8 mol C3H3N H O mol .02g H O 18 molC H N 2 molH O 6 2 2 3 3 2   = 1.29 × 103 g H2O Amount NH3 needed: 23.8 mol C3H3N 3 3 3 3 3 NH mol .03g NH 17 molC H N 2 2 molNH   = 405 g NH3 Amount NH3 in excess = 1.50 × 103 g  405 g = 1.10 × 103 g NH3 Amount O2 needed: 23.8 mol C3H3N 2 2 3 3 2 O mol .00g O 32 molC H N 2 3 molO   = 1.14 × 103 g O2 Amount O2 in excess = 2.00 × 103 g  1.14 × 103 g = 860 g O2 1.10 × 103 g NH3 and 860 g O2 are in excess. 77. The product formed in the reaction is NO2; the other species present in the product picture is excess O2. Therefore, NO is the limiting reactant. In the pictures, 6 NO molecules react 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) 78. 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. 10 molecules O2 × 2 3 moleculesO 5 4 moleculesNH = 8 molecules NH3 to react with all the O2 Because we have 10 molecules of NH3 and only 8 molecules of NH3 are necessary to react with all the O2, O2 is limiting. 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 38 CHAPTER 3 STOICHIOMETRY The total number of molecules present after completion = 2 molecules NH3 + 0 molecules O2 + 8 molecules NO + 12 molecules H2O = 22 molecules. 79. 1.50 g BaO2 × 2 2 3. g BaO 169 1molBaO = 8.86 × 103 mol BaO2 25.0 mL .46g HCl 36 molHCl 1 mL .0 0272g HCl   = 1.87 × 102 mol HCl The required mole ratio from the balanced reaction is 2 mol HCl to 1 mol BaO2. The actual mole ratio is: 2 3 2 molBaO 10 86 .8 molHCl 10 87 .1     = 2.11 Because the actual mole ratio is larger than the required mole ratio, the denominator (BaO2) is the limiting reagent. 8.86 × 103 mol BaO2 × 2 2 2 2 2 2 2 H O mol .02g H O 34 BaO mol 1molH O  = 0.301 g H2O2 The amount of HCl reacted is: 8.86 × 103 mol BaO2 × BaO2 mol 2 molHCl = 1.77 × 102 mol HCl Excess mol HCl = 1.87 × 102 mol  1.77 × 102 mol = 1.0 × 103 mol HCl Mass of excess HCl = 1.0 × 103 mol HCl × HCl mol 36.46g HCl = 3.6 × 102 g HCl 80. 25.0 g Ag2O × 8. g 231 1mol = 0.108 mol Ag2O 50.0 g C10H10N4SO2 × .29g 250 1mol = 0.200 mol C10H10N4SO2 Ag O mol C H N SO mol 2 2 4 10 10 (actual) = .0 108 .0 200 = 1.85 The actual mole ratio is less than the required mole ratio (2), so C10H10N4SO2 is limiting. 0.200 mol C10H10N4SO2 × 2 4 9 10 2 4 10 10 2 4 9 10 AgC H N SO mo