CHAPTER 3 STOICHIOMETRY 49 36. One method is to assume each quantity of reactant is limiting, then calculate the amount of product that could be produced from each 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 = 69.2 = 185 u (A = 184.95 u without rounding to proper significant figures.) 0.3740 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) 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.
50 CHAPTER 3 STOICHIOMETRY 151.96 = x(150.9196) + (100 − x)(152.9209) 100 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 = 51.82(106.905) + 48.18(A) 100 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. 500. atoms Fe × 1 mol Fe × 55.85 g Fe = 4.64 × 10−20 g Fe 6.022 × 1023 atoms Fe mol Fe 46. 500.0 g Fe × 1 mol Fe = 8.953 mol Fe 55.85 g Fe 8.953 mol Fe × 6.022 × 1023 atoms Fe = 5.391 × 1024 atoms Fe mol Fe
CHAPTER 3 STOICHIOMETRY 51 47. 1.00 carat × 0.200 g C × 1 mol C × 6.022 × 1023 atoms C = 1.00 × 1022 atoms C carat 12.01 g C mol C 48. 5.0 × 1021 atoms C × 1 mol C = 8.3 × 10−3 mol C × 1023 atoms C 6.022 8.3 × 10−3 mol C × 12.01 g C = 0.10 g C mol 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 × 1 mol NH3 = 0.0587 mol NH3 17.03 g NH3 b. 1.00 g N2H4 × 1 mol N 2H 4 = 0.0312 mol N2H4 32.05 g N 2H 4 c. 1.00 g (NH4)2Cr2O7 × 1 mol (NH4 )2 Cr2O7 = 3.97 × 10−3 mol (NH4)2Cr2O7 252.08 g (NH4 )2 Cr2O7 54. a. 1.00 g P4O6 × 1 mol P 4 O6 = 4.55 × 10−3 mol P4O6 219.88 g b. 1.00 g Ca3(PO4)2 × 1 mol Ca3 (PO4 )2 = 3.22 × 10−3 mol Ca3(PO4)2 310.18 g c. 1.00 g Na2HPO4 × 1 mol Na 2HPO 4 = 7.04 × 10−3 mol Na2HPO4 141.96 g
52 CHAPTER 3 STOICHIOMETRY 55. a. 5.00 mol NH3 × 17.03 g NH3 = 85.2 g NH3 mol NH3 b. 5.00 mol N2H4 × 32.05 g N 2H 4 = 160. g N2H4 mol N 2 H 4 c. 5.00 mol (NH4)2Cr2O7 × 252.08 g (NH 4 )2 Cr2O7 = 1260 g (NH4)2Cr2O7 1 mol (NH 4 )2 Cr2O7 56. a. 5.00 mol P4O6 × 219.88 g = 1.10 × 103 g P4O6 1 mol P 4 O6 b. 5.00 mol Ca3(PO4)2 × 310.18 g = 1.55 × 103 g Ca3(PO4)2 mol Ca3 (PO4 )2 c. 5.00 mol Na2HPO4 × 141.96 g = 7.10 × 102 g Na2HPO4 mol Na 2HPO 4 57. Chemical formulas give atom ratios as well as mole ratios. a. 5.00 mol NH3 × 1 mol N × 14.01 g N = 70.1 g N mol NH3 mol N b. 5.00 mol N2H4 × 2 mol N × 14.01 g N = 140. g N mol N2H4 mol N c. 5.00 mol (NH4)2Cr2O7 × 2 mol N × 14.01 g N = 140. g N mol (NH4 )2 Cr2O7 mol N 58. a. 5.00 mol P4O6 × 4 mol P × 30.97 g P = 619 g P mol P4O6 mol P b. 2 mol P × 30.97 g P = 310. g P 5.00 mol Ca3(PO4)2 × mol Ca3(PO4 )2 mol P c. 5.00 mol Na2HPO4 × 1 mol P × 30.97 g P = 155 g P mol Na2HPO4 mol P 59. a. 1.00 g NH3 × 1 mol NH3 × 6.022 × 1023 molecules NH3 17.03 g NH3 mol NH3 = 3.54 × 1022 molecules NH3 b. 1.00 g N2H4 × 1 mol N2H4 × 6.022 × 1023 molecules N2H4 32.05 g N2H4 mol N2H4 = 1.88 × 1022 molecules N2H4
CHAPTER 3 STOICHIOMETRY 53 c. 1.00 g (NH4)2Cr2O7 × 1 mol (NH4 )2 Cr2O7 252.08 g (NH4 )2 Cr2O7 × 6.022 × 1023 formula units (NH4 )2 Cr2O7 = 2.39 × 1021 formula units (NH4)2Cr2O7 mol (NH4 )2 Cr2O7 60. a. 1.00 g P4O6 × 1 mol P4O6 × 6.022 × 1023 molecules = 2.74 × 1021 molecules P4O6 219.88 g mol P4O6 b. 1.00 g Ca3(PO4)2 × 1 mol Ca3(PO4 )2 × 6.022 × 1023 formula units 310.18 g mol Ca3 (PO4 )2 = 1.94 × 1021 formula units Ca3(PO4)2 c. 1.00 g Na2HPO4 × 1 mol Na 2HPO4 × 6.022 × 1023 formula units 141.96 g mol Na 2HPO4 = 4.24 × 1021 formula units Na2HPO4 61. Using answers from Exercise 59: a. 3.54 × 1022 molecules NH3 × 1 atom N = 3.54 × 1022 atoms N molecule NH3 b. 1.88 × 1022 molecules N2H4 × 2 atoms N = 3.76 × 1022 atoms N molecule N 2H 4 c. 2.39 × 1021 formula units (NH4)2Cr2O7 × 2 atoms N formula unit (NH 4 )2 Cr2O7 = 4.78 × 1021 atoms N 62. Using answers from Exercise 60: a. 2.74 × 1021 molecules P4O6 × 4 atoms P = 1.10 × 1022 atoms P molecule P4O6 b. 1.94 × 1021 formula units Ca3(PO4)2 × 2 atoms P = 3.88 × 1021 atoms P formula unit Ca 3 (PO 4 )2 c. 4.24 × 1021 formula units Na2HPO4 × 1 atom P = 4.24 × 1021 atoms P formula unit Na 2HPO4 63. Molar mass of CCl2F2 = 12.01 + 2(35.45) + 2(19.00) = 120.91 g/mol 5.56 mg CCl2F2 × 1 g × 1 mol × 6.022 × 1023 molecules 1000 mg 120.91 g mol = 2.77 × 1019 molecules CCl2F2
54 CHAPTER 3 STOICHIOMETRY 5.56 × 10−3 g CCl2F2 × 1 mol CCl2F2 × 2 mol Cl × 35.45 g Cl 120.91 g 1 mol CCl2F mol Cl = 3.26 × 10−3 g = 3.26 mg Cl 64. The •2H2O is part of the formula of bauxite (they are called waters of hydration). Combining elements together, the chemical formula for bauxite would be Al2O5H4. a. Molar mass = 2(26.98) + 5(16.00) + 4(1.008) = 137.99 g/mol b. 0.58 mol Al2O3•2H2O × 2 mol Al × 26.98 g Al = 31 g Al mol Al2O3 • 2H 2O mol Al c. 0.58 mol Al2O3•2H2O × 2 mol Al × 6.022 × 1023 atoms mol Al2O3 • 2H 2O mol Al = 7.0 × 1023 atoms Al d. 2.1 × 1024 formula units Al2O3•2H2O × 1 mol Al2O3 • 2H2O × 137.99 g 6.022 × 1023 formula units mol = 480 g Al2O3•2H2O 1 mol 65. a. 150.0 g Fe2O3 × 159.70 g = 0.9393 mol Fe2O3 b. 1g × 1 mol = 2.17 × 10−4 mol NO2 10.0 mg NO2 × 1000 mg 46.01 g c. 1.5 × 1016 molecules BF3 × 1 mol = 2.5 × 10−8 mol BF3 6.02 × 1023 molecules 66. a. 20.0 mg C8H10N4O2 × 1g × 1 mol = 1.03 × 10−4 mol C8H10N4O2 1000 mg 194.20 g b. 2.72 × 1021 molecules C2H5OH × 1 mol × 1023 molecules 6.022 = 4.52 × 10−3 mol C2H5OH c. 1.50 g CO2 × 1 mol = 3.41 × 10−2 mol CO2 44.01 g 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. Molar mass of C2H5O2N = 2(12.01) + 5(1.008) + 2(16.00) + 14.0l = 75.07 g/mol
CHAPTER 3 STOICHIOMETRY 55 5.00 g C2H5O2N × 1 mol C2H5O2N × 6.022 × 1023 molecules C2H5O2N 75.07 g C2H5O2N mol C2H5O2N × 1 atom N = 4.01 × 1022 atoms N molecule C2H5O2N b. Molar mass of Mg3N2 = 3(24.31) + 2(14.01) = 100.95 g/mol 5.00 g Mg3N2 × 1 mol Mg3N2 × 6.022 × 1023 formula units Mg3N2 100.95 g Mg3N2 mol Mg3N2 × 2 atoms N = 5.97 × 1022 atoms N mol Mg3N2 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 × 1 mol Ca(NO3)2 × 2 mol N × 6.022 × 1023 atoms N 164.10 g Ca(NO3)2 mol Ca(NO3)2 mol N = 3.67 × 1022 atoms N d. Molar mass of N2O4 = 2(14.01) + 4(16.00) = 92.02 g/mol 5.00 g N2O4 × 1 mol N2O4 × 2 mol N × 6.022 × 1023 atoms N 92.02 g N2O4 mol N2O4 mol N = 6.54 × 1022 atoms N 68. 4.24 g C6H6 × 1 mol = 5.43 × 10−2 mol C6H6 78.11 g 5.43 × 10−2 mol C6H6 × 6.022 × 1023 molecules = 3.27 × 1022 molecules C6H6 mol Each molecule of C6H6 contains 6 atoms C + 6 atoms H = 12 atoms total. 3.27 × 1022 molecules C6H6 × 12 atoms total = 3.92 × 1023 atoms total molecule 0.224 mol H2O × 18.02 g = 4.04 g H2O mol 0.224 mol H2O × 6.022 × 1023 molecules = 1.35 × 1023 molecules H2O mol 1.35 × 1023 molecules H2O × 3 atoms total = 4.05 × 1023 atoms total molecule
56 CHAPTER 3 STOICHIOMETRY 2.71 × 1022 molecules CO2 × 6.022 1 mol = 4.50 × 10−2 mol CO2 × 1023 molecules 4.50 × 10−2 mol CO2 × 44.01 g = 1.98 g CO2 mol 2.71 × 1022 molecules CO2 × 3 atoms total = 8.13 × 1022 atoms total molecule CO2 3.35 × 1022 atoms total × 1 molecule = 5.58 × 1021 molecules CH3OH 6 atoms total 5.58 × 1021 molecules CH3OH × 6.022 1 mol = 9.27 × 10−3 mol CH3OH × 1023 molecules 9.27 × 10−3 mol CH3OH × 32.04 g = 0.297 g CH3OH mol 69. Molar mass of C6H8O6 = 6(12.01) + 8(1.008) + 6(16.00) = 176.12 g/mol 500.0 mg × 1g × 1 mol = 2.839 × 10−3 mol C6H8O6 1000 mg 176.12 g 2.839 × 10−3 mol × 6.022 × 1023 molecules = 1.710 × 1021 molecules C6H8O6 mol 70. a. 9(12.01) + 8(1.008) + 4(16.00) = 180.15 g/mol b. 500. mg × 1g × 1 mol = 2.78 × 10−3 mol C9H8O4 1000 mg 180.15 g 2.78 × 10−3 mol × 6.022 × 1023 molecules = 1.67 × 1021 molecules C9H8O4 mol 71. a. 2(12.01) + 3(1.008) + 3(35.45) + 2(16.00) = 165.39 g/mol 1 mol b. 500.0 g × 165.39 g = 3.023 mol C2H3Cl3O2 c. 2.0 × 10-2 mol × 165.39 g = 3.3 g C2H3Cl3O2 mol d. 5.0 g C2H3Cl3O2 × 1 mol × 6.022 × 1023 molecules × 3 atoms Cl 165.39 g mol molecule = 5.5 × 1022 atoms of chlorine
CHAPTER 3 STOICHIOMETRY 57 e. 1.0 g Cl × 1 mol Cl × 1 mol C2H3Cl3O2 × 165.39 g C2H3Cl3O2 = 1.6 g chloral hydrate 35.45 g 3 mol Cl mol C2 H3Cl3O2 f. 500 molecules × 1 mol × 165.39 g = 1.373 × 10−19 g C2H3Cl3O2 6.022 × 1023 molecules mol 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. a. 2(12.01) + 6(1.008) + 2(14.01) + 1(16.00) = 74.09 g/mol b. 250 mg × 1 g × 1 mol = 3.4 × 10−3 mol c. 0.050 mol × 74.09 g = 3.7 g 1000 mg 74.09 g mol d. 1.0 mol C2H6N2O × 6.022 × 1023 molecules C2H6N2O × 6 atoms of H mol C2H6N2O molecule C2H6N2O = 3.6 × 1024 atoms of hydrogen e. 1.0 × 106 molecules × 1 mol × 74.09 g = 1.2 × 10−16 g 6.022 × 1023 molecules mol f. 1 molecule × 1 mol × 74.09 g = 1.230 × 10−22 g C2H6N2O 6.022 × 1023 molecules mol 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 = 36.03 g C × 100 = 50.00% C 72.06 g compound Mass % H = 4.032 g H × 100 = 5.595% H 72.06 g compound Mass % O = 100.00 − (50.00 + 5.595) = 44.41% O or: %O= 32.00 g × 100 = 44.41% O 72.06 g 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 = 48.04 g × 100 = 55.80% C; mass % H = 6.048 g × 100 = 7.025% H 86.09 g 86.09 g Mass % O = 100.00 − (55.80 + 7.025) = 37.18% O
58 CHAPTER 3 STOICHIOMETRY 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 = 36.03 g × 100 = 67.90% C; mass % H = 3.024 g × 100 = 5.699% H 53.06 g 53.06 g 14.01 g × 100 = 26.40% N or % N = 100.00 − (67.90 + 5.699) Mass % N = 53.06 g = 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 88.91 gY + 2 mol Ba 137.3 g Ba mol Y mol Ba + 3 mol Cu 63.55 g Cu + 7 mol O 16.00 gO mol Cu mol O Molar mass = 88.91 + 274.6 + 190.65 + 112.00 = 666.2 g/mol Mass % Y = 88.91 g × 100 = 13.35% Y; mass % Ba = 274.6 g × 100 = 41.22% Ba 666.2 g 666.2 g Mass % Cu = 190.65 g × 100 = 28.62% Cu; mass % O = 112.0 g × 100 = 16.81% O 666.2 g 666.2 g 75. NO: Mass % N = 14.01 g N × 100 = 46.68% N 30.01 g NO 14.01 g N NO2: Mass % N = 46.01 g NO2 × 100 = 30.45% N N2O: Mass % N = 2(14.01) g N × 100 = 63.65% N 44.02 g N2O 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. 76. a. C8H10N4O2: Molar mass = 8(12.01) + 10(1.008) + 4(14.0l) + 2(16.00) = 194.20 g/mol Mass % C = 8(12.01) g C × 100 = 96.08 g × 100 = 49.47% C 194.20 g C8H10N4O2 194.20 g b. C12 H22O11: Molar mass = 12(12.01) + 22(1.008) + 11(16.00) = 342.30 g/mol Mass % C = 12(12.01) g C × 100 = 42.10% C 342.30 g C12H22O11
CHAPTER 3 STOICHIOMETRY 59 c. C2H5OH: Molar mass = 2(12.01) + 6(1.008) + 1(16.00) = 46.07 g/mol Mass % C = 2(12.01) g C × 100 = 52.14% C 46.07 g C2H5OH The order from lowest to highest mass percentage of carbon is: sucrose (C12H22O11) < caffeine (C8H10N4O2) < ethanol (C2H5OH) 77. There are 0.390 g Cu for every 100.000 g of fungal laccase. Assuming 100.00 g fungal laccase: Mol fungal laccase = 0.390 g Cu × 1 mol Cu × 1 mol fungal laccase = 1.53 × 10−3 mol 63.55 g Cu 4 mol Cu x g fungallaccase = 100.000 g , x = molar mass = 6.54 × 104 g/mol mol fungal laccase 1.53 × 10−3 mol 78. There are 0.347 g Fe for every 100.000 g hemoglobin (Hb). Assuming 100.000 g hemoglobin: Mol Hb = 0.347 g Fe × 1 mol Fe × 1 mol Hb = 1.55 × 10−3 mol Hb 55.85 g Fe 4 mol Fe x g Hb = 100.000 g Hb , x = molar mass = 6.45 × 104 g/mol mol Hb 1.55 × 10−3 mol Hb Empirical and Molecular Formulas 79. a. Molar mass of CH2O = 1 mol C 12.01 g C + 2 mol H 1.008 gH mol C mol H + 1 mol O 16.00 gO = 30.03 g/mol mol O %C= 12.01 g C × 100 = 39.99% C; % H = 2.016 g H × 100 = 6.713% H 30.03 g CH 2O 30.03 g CH 2O % O = 16.00 g O × 100 = 53.28% O or % O = 100.00 − (39.99 + 6.713) = 53.30% 30.03 g CH 2O b. Molar mass of C6H12O6 = 6(12.01) + 12(1.008) + 6(16.00) = 180.16 g/mol %C= 76.06 g C × 100 = 40.00%; % H = 12.(1.008) g × 100 = 6.714% 180.16 g C6 H12O6 180.16 g % O = 100.00 − (40.00 + 6.714) = 53.29%
60 CHAPTER 3 STOICHIOMETRY c. Molar mass of HC2H3O2 = 2(12.01) + 4(1.008) + 2(16.00) = 60.05 g/mol % C = 24.02 g × 100 = 40.00%; % H = 4.032 g × 100 = 6.714% 60.05 g 60.05 g % 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. 81. 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 82. a. SNH: Empirical formula mass = 32.07 + 14.01 + 1.008 = 47.09 g/mol 188.35 g 47.09 g = 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 347.64 g 115.88 g = 3.0000; molecular formula is (NPCl2)3 or N3P3Cl6. c. CoC4O4: 58.93 + 4(12.01) + 4(16.00) = 170.97 g/mol 341.94 g 170.97 g = 2.0000; molecular formula: Co2C8O8 184.32 g d. SN: 32.07 + 14.01 = 46.08 g/mol; 46.08 g = 4.000; molecular formula: S4N4 83. Out of 100.00 g of compound, there are: 48.64 g C × 1 mol C = 4.050 mol C; 8.16 g H × 1 mol H = 8.10 mol H 12.01 g C 1.008 g H % O = 100.00 – 48.64 – 8.16 = 43.20%; 43.20 g O × 1 mol O = 2.700 mol O 16.00 g O Dividing each mole value by the smallest number:
CHAPTER 3 STOICHIOMETRY 61 4.050 = 1.500; 8.10 = 3.00; 2.700 = 1.000 2.700 2.700 2.700 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: 63.68 g C × 1 mol C = 5.302 mol C; 12.38 g N × 1 mol N = 0.8837 mol N 12.01 g C 14.01 g N 9.80 g H × 1 mol H = 9.72 mol H; 14.14 g O × 1 mol O = 0.8838 mol O 1.008 g H 16.00 g O Dividing each mole value by the smallest number: 5.302 = 6.000; 9.72 = 11.0; 0.8838 = 1.000 0.8837 0.8837 0.8837 The empirical formula for nylon-6 is C6H11NO 85. Compound I: Mass O = 0.6498 g HgxOy − 0.6018 g Hg = 0.0480 g O 0.6018 g Hg × 1 mol Hg = 3.000 × 10−3 mol Hg 200.6 g Hg 0.0480 g O × 1 mol O = 3.00 × 10−3 mol O 16.00 g 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 × 1 mol Hg = 2.00 × 10−3 mol Hg; 0.016 g O × 1 mol O = 1.0 × 10−3 mol O 200.6 g Hg 16.00 g O The mole ratio between Hg and O is 2 : 1, so the empirical formula is Hg2O. 86. 1 mol N = 8.001 × 10−2 mol N; 0.161 g H × 1 mol H = 1.60 × 10−1 mol H 1.121 g N × 14.01 g N 1.008 g H 1 mol C = 4.00 × 10−2 mol C; 0.640 g O × 1 mol O = 4.00 × 10−2 mol O 0.480 g C × 12.01 g C 16.00 g O Dividing all mole values by the smallest number:
62 CHAPTER 3 STOICHIOMETRY 8.001 × 10−2 1.60 × 10−1 4.00 × 10−2 4.00 × 10−2 = 2.00; 4.00 × 10−2 = 4.00; 4.00 × 10−2 = 1.00 The empirical formula is N2H4CO. 87. Out of 100.0 g, there are: 69.6 g S × 1 mol S = 2.17 mol S; 30.4 g N × 1 mol N = 2.17 mol N 32.07 g S 14.01 g N 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: 26.7 g P × 1 mol P = 0.862 mol P; 12.1 g N × 1 mol N = 0.864 mol N 30.97 g P 14.01 g N 61.2 g Cl × 1 mol Cl = 1.73 mol Cl 35.45 g Cl 1.73 = 2.01; the empirical formula is PNCl2. 0.862 The empirical formula mass is ≈ 31.0 + 14.0 + 2(35.5) = 116 g/mol. Molar mass = 580 = 5.0; the molecular formula is (PNCl2)5 = P5N5Cl10. Empirical formula mass 116 89. Assuming 100.00 g of compound: 47.08 g C × 1 mol C = 3.920 mol C; 6.59 g H × 1 mol H = 6.54 mol H 12.01 g C 1.008 g H 46.33 g Cl × 1 mol Cl = 1.307 mol Cl 35.45 g Cl Dividing all mole values by 1.307 gives: 3.920 = 2.999; 6.54 = 5.00; 1.307 = 1.000 1.307 1.307 1.307 The empirical formula is C3H5Cl. The empirical formula mass is 3(12.01) + 5(1.008) + 1(35.45) = 76.52 g/mol.
CHAPTER 3 STOICHIOMETRY 63 Molar mass 153 Empirical formula mass = 76.52 = 2.00 ; the molecular formula is (C3H5Cl)2 = C6H10Cl2. 90. 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 × 1 mol C = 3.446 mol C; 3.47 g H × 1 mol H = 3.44 mol H 12.01 g C 1.008 g H 55.14 g O × 1 mol O = 3.446 mol O 16.00 g 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 = 15.0 g = 116 g/mol 0.129 mol Molar mass = 116 = 4.00; molecular formula = (CHO)4 = C4H4O4 Empirical mass 29.02 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: mass C = 2.20 g CO2 × 1 mol CO 2 × 1 mol C × 12.01g C = 0.600 g C 44.01 g CO 2 mol CO2 mol C mass H = 0.900 g H2O × 1 mol H 2O × 2 mol H × 1.008 g H = 0.101 g H 18.02 g H 2O mol H 2O mol H Mass O = 1.50 g fructose − 0.600 g C − 0.101 g H = 0.799 g O So, in 1.50 g of the fructose, we have: 0.600 g C × 1 mol C = 0.0500 mol C; 0.101 g H × 1 mol H = 0.100 mol H 12.01 g C 1.008 g H 0.799 g O × 1 mol O = 0.0499 mol O 16.00 g O Dividing by the smallest number: 0.100 = 2.00; the empirical formula is CH2O. 0.0499 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 H2O came from the 35.0 mg of compound.
64 CHAPTER 3 STOICHIOMETRY 3.35 × 10−2 g CO2 × 1 mol CO2 × 1 mol C × 12.01 g C = 9.14 × 10−3 g C 44.01 g CO2 mol CO2 mol C Mass % C = 9.14 × 10−3 g C × 100 = 26.1% C 3.50 × 10−2 g compound 4.11 × 10−2 g H2O × 1 mol H2O × 2 mol H × 1.008 g H = 4.60 × 10−3 g H 18.02 g H2O mol H2O mol H Mass % H = 4.60 × 10−3 g H × 100 = 13.1% H 3.50 × 10−2 g compound The mass percent of nitrogen is obtained by difference: Mass % N = 100.0 − (26.1 + 13.1) = 60.8% N Now perform the empirical formula determination by first assuming 100.0 g of compound. Out of 100.0 g of compound, there are: 26.1 g C × 1 mol C = 2.17 mol C; 13.1 g H × 1 mol H = 13.0 mol H 12.01 g C 1.008 g H 60.8 g N × 1 mol N = 4.34 mol N 14.01 g N Dividing all mole values by 2.17 gives: 2.17 = 1.00; 13.0 = 5.99; 4.34 = 2.00 2.17 2.17 2.17 The empirical formula is CH6N2. 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 (100.0 − mass % H = mass % C). 1g × 2.016 g H × 1000 mg = 4.79 mg H 42.8 mg H2O × 1000 mg 18.02 g H2O g Mass % H = 4.79 mg H × 100 = 10.1% H; mass % C = 100.0 − 10.1 = 89.9% C 47.6 mg cumene Now solve the empirical formula problem. Out of 100.0 g cumene, we have: 89.9 g C × 1 mol C = 7.49 mol C; 10.1 g H × 1 mol H = 10.0 mol H 12.01 g C 1.008 g H
CHAPTER 3 STOICHIOMETRY 65 10.0 = 1.34 ≈ 4 ; the mole H to mole C ratio is 4 : 3. The empirical formula is C3H4. 7.49 3 Empirical formula mass ≈ 3(12) + 4(1) = 40 g/mol. The molecular formula must be (C3H4)3 or C9H12 because the molar mass of this formula will be between 115 and 125 g/mol (molar mass ≈ 3 × 40 g/mol = 120 g/mol). 94. There are several ways to do this problem. We will determine composition by mass percent: 1g × 12.01 g C × 1000 mg = 4.369 mg C 16.01 mg CO2 × 1000 mg 44.01 g CO2 g %C= 4.369 mg C × 100 = 40.91% C 10.68 mg compound 1g × 2.016 g H × 1000 mg = 0.489 mg H 4.37 mg H2O × 1000 mg 18.02 g H2O g % H = 0.489 mg × 100 = 4.58% H; % O = 100.00 − (40.91 + 4.58) = 54.51% O 10.68 mg So, in 100.00 g of the compound, we have: 40.91 g C × 1 mol C = 3.406 mol C; 4.58 g H × 1 mol H = 4.54 mol H 12.01 g C 1.008 g H 54.51 g O × 1 mol O = 3.407 mol O 16.00 g O Dividing by the smallest number: 4.54 = 1.33 ≈ 4 3.406 3 ; the empirical formula is C3H4O3. The empirical formula mass of C3H4O3 is ≈ 3(12) + 4(1) + 3(16) = 88 g/mol. Because 176.1 = 2.0, the molecular formula is C6H8O6. 88 Balancing Chemical Equations 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. 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)
66 CHAPTER 3 STOICHIOMETRY 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 balanced; balance N: CS2 + 2 NH3 → H2S + NH4SCN H is also balanced. CS2(l) + 2 NH3(g) → H2S(g) + NH4SCN(s) 96. 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) c. Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) d. Sr(OH)2(aq) + 2 HBr(aq) → 2H2O(l) + SrBr2(aq) MnO2 97. 2 H2O2(aq) 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) 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)
CHAPTER 3 STOICHIOMETRY 67 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. a. The formulas of the reactants and products are C6H6(l) + O2(g) → CO2(g) + H2O(g). 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 H2O. To balance C and H, we need 6 CO2 molecules and 3 H2O molecules for every 1 molecule of C6H6. 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. C6H6(l) + 15 O2(g) → 6 CO2(g) + 3 H2O(g); multiply by two to give whole numbers. 2 2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) b. The formulas of the reactants and products are C4H10(g) + O2(g) → CO2(g) + H2O(g). C4H10(g) + 13 O2(g) → 4 CO2(g) + 5 H2O(g); multiply by two to give whole numbers. 2 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) + 3 O2(g) → Fe2O3(s); for whole numbers: 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) 2 e. 2 FeO(s) + 1 O2(g) → Fe2O3(s); for whole numbers, multiply by two. 2 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) d. 2 Eu(s) + 6 HF(g) → 2 EuF3(s) + 3 H2(g) 103. 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)
68 CHAPTER 3 STOICHIOMETRY 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) 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) 104. CaSiO3(s) + 6 HF(aq) → CaF2(aq) + SiF4(g) + 3 H2O(l) 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. Fe2O3(s) + 2 Al(s) → 2 Fe(l) + Al2O3(s) 15.0 g Fe × 1 mol Fe = 0.269 mol Fe; 0.269 mol Fe × 2 mol Al × 26.98 g Al = 7.26 g Al 55.85 g Fe 2 mol Fe mol Al 0.269 mol Fe × 1 mol Fe2O3 × 159.70 g Fe2O3 = 21.5 g Fe2O3 2 mol Fe mol Fe2O3 0.269 mol Fe × 1 mol Al2O3 × 101.96 g Al2O3 = 13.7 g Al2O3 2 mol Fe mol Al2O3 106. 10 KClO3(s) + 3 P4(s) → 3 P4O10(s) + 10 KCl(s) 52.9 g KClO3 × 1 mol KClO3 × 3 mol P4O10 × 283.88 g P4O10 = 36.8 g P4O10 122.55 g KClO3 10 mol KClO3 mol P4O10 107. 1.000 kg Al × 1000 g Al × 1 mol Al × 3 mol NH4ClO4 × 117.49 g NH4ClO4 kg Al 26.98 g Al 3 mol Al mol NH4ClO4 = 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 × 1 mol Ba(OH)2 • 8H 2O = 0.0206 mol = 0.021 mol 315.4 g
CHAPTER 3 STOICHIOMETRY 69 0.021 mol Ba(OH)2•8H2O × 2 mol NH 4SCN × 76.13 g NH 4SCN 1 mol Ba(OH)2 • 8H 2O mol NH 4SCN = 3.2 g NH4SCN 109. a. 1.0 × 102 mg NaHCO3 × 1g × 1 mol NaHCO3 × 1 mol C6H8O7 1000 mg 84.01 g NaHCO3 3 mol NaHCO3 × 192.12 g C6H8O7 = 0.076 g or 76 mg C6H8O7 mol C6H8O7 b. 0.10 g NaHCO3 × 1 mol NaHCO3 × 3 mol CO2 × 44.01 g CO2 84.01 g NaHCO3 3 mol NaHCO3 mol CO2 = 0.052 g or 52 mg CO2 110. a. 1.00 × 102 g C7H6O3 × 1 mol C7H6O3 × 1 mol C4H6O3 × 102.09 g C4H6O3 138.12 g C7H6O3 1 mol C7H6O3 1 mol C4H6O3 = 73.9 g C4H6O3 b. 1.00 × 102 g C7H6O3 × 1 mol C7H6O3 × 1 mol C9H8O4 × 180.15 g C9H8O4 138.12 g C7H6O3 1 mol C7H6O3 mol C9H8O4 = 1.30 × 102 g aspirin 111. 1.0 × 104 kg waste × 3.0 kg NH4+ × 1000 g × 1 mol NH4+ × 1 mol C5H7O2N 100 kg waste kg 18.04 g NH4+ 55 mol NH + 4 × 113.12 g C5H7O2N = 3.4 × 104 g tissue if all NH4+ converted mol C5H7O2N 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 112. 1.0 × 103 g phosphorite × 75 g Ca3(PO4 )2 × 1 mol Ca3(PO4 )2 100 g phosphorite 310.18 g Ca3(PO4 )2 × 1 mol P4 × 123.88 g P4 = 150 g P4 2 mol Ca3 (PO4 )2 mol P4 113. 1.0 ton CuO × 907 kg × 1000 g × 1 mol CuO × 1 mol C × 12.01 g C × 100. g coke ton kg 79.55 g CuO 2 mol CuO mol C 95 g C = 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.
70 CHAPTER 3 STOICHIOMETRY 25,000 g LiOH × 1 mol LiOH × 1 mol CO2 × 44.01 g CO2 × 100 g air 23.95 g LiOH 2 mol LiOH mol CO2 4.0 g CO2 × 1 mL air × 1 L × 1 min × 1 h = 68 h = 2.8 days 0.0010 g air 1000 mL 140 L air 60 min Limiting Reactants and Percent Yield 115. The product formed in the reaction is NO2; the other species present in the product represent- tation 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) 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. 10 molecules O2 × 4 molecules NH3 = 8 molecules NH3 to react with all of the O2 5 molecules O2 Because we have 10 molecules of NH3 and only 8 molecules of NH3 are necessary to 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. 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 8 molecules 12 molecules 0 The total number of molecules present after completion = 2 molecules NH3 + 0 molecules O2 + 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 limiting reactant (runs out first) and therefore determines the mass of product that can be produced. Assuming N2 is limiting: 1.00 × 103 g N2 × 1 mol N2 × 2 mol NH3 × 17.03 g NH3 = 1.22 × 103 g NH3 28.02 g N2 mol N2 mol NH3
CHAPTER 3 STOICHIOMETRY 71 Assuming H2 is limiting: 5.00 × 102 g H2 × 1 mol H2 × 2 mol NH3 × 17.03 g NH3 = 2.82 × 103 g NH3 2.016 g H2 3 mol H2 mol NH3 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. b. 1.00 × 103 g N2 × 1 mol N2 × 3 mol H2 × 2.016 g H2 = 216 g H2 reacted 28.02 g N2 mol N2 mol H2 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 × 1 mol Ca3(PO4 )2 × 3 mol CaSO4 × 136.15 g CaSO4 310.18 g Ca3(PO4 )2 mol Ca3 (PO4 )2 mol CaSO4 = 1300 g CaSO4 Assuming concentrated H2SO4 reagent is limiting: 1.0 × 103 g conc. H2SO4 × 98 g H2SO4 × 1 mol H2SO4 100 g conc. H2SO4 98.09 g H2SO4 × 3 mol CaSO4 × 136.15 g CaSO4 = 1400 g CaSO4 3 mol H2SO4 mol CaSO4 Because Ca3(PO4)2 produces the smaller quantity of product, Ca3(PO4)2 is limiting and 1300 g CaSO4 can be produced. 1.0 × 103 g Ca3(PO4)2 × 1 mol Ca3(PO4 )2 × 2 mol H3PO4 × 97.99 g H3PO4 310.18 g Ca3(PO4 )2 mol Ca3(PO4 )2 mol H3PO4 = 630 g H3PO4 produced 119. Assuming BaO2 is limiting: 1.50 g BaO2 × 1 mol BaO2 × 1 mol H2O2 × 34.02 g H2O2 = 0.301 g H2O2 169.3 g BaO2 mol BaO2 mol H2O2 Assuming HCl is limiting: 25.0 mL × 0.0272 g HCl × 1 mol HCl × 1 mol H 2O 2 × 34.02 g H 2O 2 = 0.317 g H2O2 mL 36.46 g HCl 2 mol HCl mol H 2O 2
72 CHAPTER 3 STOICHIOMETRY 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 × 0.0272 g HCl × 1 mol HCl = 1.87 × 10−2 mol HCl mL 36.46 g HCl The amount of HCl reacted: 1.50 g BaO2 × 1 mol BaO2 × 2 mol HCl = 1.77 × 10−2 mol HCl 169.3 g BaO2 mol BaO2 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 × 36.46 g HCl = 3.6 × 10−2 g HCl unreacted mol HCl 120. Assuming Ag2O is limiting: 25.0 g Ag2O × 1 mol Ag2O × 2 mol AgC10H9N4SO2 × 357.18 g AgC10H9N4SO2 231.8 g Ag2O mol Ag2O mol AgC10H9N4SO2 = 77.0 g AgC10H9N4SO2 Assuming C10H10N4SO2 is limiting: 50.0 g C10H10N4SO2 × 1 mol C10H10N4SO2 × 2 mol AgC10H9N4SO2 250.29 g C10H10N4SO2 2 mol C10H10N4SO2 × 357.18 g AgC10H9N4SO2 = 71.4 g AgC10H9N4SO2 mol AgC10H9N4SO2 Because C10H10N4SO2 produces the smaller amount of product, it is limiting and 71.4 g of silver sulfadiazine can be produced. 121. To solve limiting-reagent problems, we will generally assume 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 × 1 mol NH3 × 2 mol HCN = 2.94 × 105 mol HCN 17.03 g NH3 2 mol NH3 5.00 × 106 g O2 × 1 mol O2 × 2 mol HCN = 1.04 × 105 mol HCN 32.00 g O2 3 mol O2 5.00 × 106 g CH4 × 1 mol CH4 × 2 mol HCN = 3.12 × 105 mol HCN 16.04 g CH4 2 mol CH4
CHAPTER 3 STOICHIOMETRY 73 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 produced is: 1.04 × 105 mol HCN × 27.03 g HCN = 2.81 × 106 g HCN mol HCN 5.00 × 106 g O2 × 1 mol O2 × 6 mol H2O × 18.02 g H2O = 5.63 × 106 g H2O 32.00 g O2 3 mol O2 1 mol H2O 122. If C3H6 is limiting: 15.0 g C3H6 × 1 mol C3H6 × 2 mol C3H3N × 53.06 g C3H3N = 18.9 g C3H3N 42.08 g C3H6 2 mol C3H6 mol C3H3N If NH3 is limiting: 5.00 g NH3 × 1 mol NH3 × 2 mol C3H3N × 53.06 g C3H3N = 15.6 g C3H3N 17.03 g NH3 2 mol NH3 mol C3H3N If O2 is limiting: 10.0 g O2 × 1 mol O2 × 2 mol C3H3N × 53.06 g C3H3N = 11.1 g C3H3N 32.00 g O2 3 mol O2 mol C3H3N O2 produces the smallest amount of product; thus O2 is limiting, and 11.1 g C3H3N can be produced. 123. C2H6(g) + Cl2(g) → C2H5Cl(g) + HCl(g) If C2H6 is limiting: 300. g C2H6 × 1 mol C2H6 × 1 mol C2H5Cl × 64.51 g C2H5Cl = 644 g C2H5Cl 30.07 g C2H6 mol C2H6 mol C2H5Cl If Cl2 is limiting: 650. g Cl2 × 1 mol Cl2 × 1 mol C2H5Cl × 64.51 g C2H5Cl = 591 g C2H5Cl 70.90 g Cl2 mol Cl2 mol C2H5Cl Cl2 is limiting because it produces the smaller quantity of product. Hence, the theoretical yield for this reaction is 591 g C2H5Cl. The percent yield is: percent yield = actual × 100 = 490. g × 100 = 82.9% theoretical 591 g 124. a. 1142 g C6H5Cl × 1 mol C6H5Cl × 1 mol C14H9Cl5 × 354.46 g C14H9Cl5 112.55 g C6H5Cl 2 mol C6H5Cl mol C14H9Cl5 = 1798 C14H9Cl5
74 CHAPTER 3 STOICHIOMETRY 485 g C2HOCl3 × 1 mol C2HOCl3 × 1 mol C14H9Cl5 × 354.46 g C14H9Cl5 147.38 g C2HOCl3 mol C2HOCl3 mol C14H9Cl5 = 1170 g C14H9Cl5 From 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 × 1 mol C2HOCl3 × 2 mol C6H5Cl × 112.55 g C6H5Cl 147.38 g C2HOCl3 mol C2HOCl3 mol C6H5Cl = 741 g C6H5Cl reacted 1142 g − 741 g = 401 g C6H5Cl in excess d. Percent yield = 200.0 g DDT × 100 = 17.1% 1170 g DDT 125. 2.50 metric tons Cu3FeS3 × 1000 kg × 1000 g × 1 mol Cu3FeS3 × 3 mol Cu metric ton kg 342.71 g 1 mol Cu3FeS3 × 63.55 g = 1.39 × 106 g Cu (theoretical) mol Cu 1.39 × 106 g Cu (theoretical) × 86.3 g Cu (actual) = 1.20 × 106 g Cu = 1.20 × 103 kg Cu 100. g Cu (theoretical) = 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) × 100.0 g PF3 (theoretical) = 154 g PF3 (theoretical) 78.1 g PF3 (actual) 154 g PF3 × 1 mol PF3 × 6 mol F2 × 38.00 g F2 = 99.8 g F2 87.97 g PF3 4 mol PF3 mol F2 99.8 g F2 is needed to actually produce 120. g of PF3 if the percent yield is 78.1%. Additional Exercises 127. 12C21H6: 2(12.000000) + 6(1.007825) = 30.046950 u 12C1H216O: 1(12.000000) + 2(1.007825) + 1(15.994915) = 30.010565 u 14N16O: 1(14.003074) + 1(15.994915) = 29.997989 u The peak results from 12C1H216O.
CHAPTER 3 STOICHIOMETRY 75 128. We would see the peaks corresponding to: 10B35Cl3 [mass ≈ 10 + 3(35) = 115 u], 10B35Cl237Cl (117), 10B35Cl37Cl2 (119), 10B37Cl3 (121), 11B35Cl3 (116), 11B35Cl237Cl (118), 11B35Cl37Cl2 (120), 11B37Cl3 (122) We would see a total of eight peaks at approximate masses of 115, 116, 117, 118, 119, 120, 121, and 122. 129. Molar mass XeFn = 0.368 g XeFn = 245 g/mol 9.03 × 1020 molecules XeFn × 1 mol XeFn 6.022 × 1023 molecules 245 g = 131.3 g + n(19.00 g), n = 5.98; formula = XeF6 130. a. 14 mol C × 12.01 g + 18 mol H × 1.008 g + 2 mol N × 14.01 g mol C mol H mol N + 5 mol O × 16.00 g = 294.30 g mol O b. 10.0 g C14H18N2O5 × 1 mol C14H18N2O5 = 3.40 × 10−2 mol C14H18N2O5 294.30 g C14H18N2O5 294.3 g c. 1.56 mol × mol = 459 g C14H18N2O5 d. 5.0 mg × 1 g × 1 mol × 6.022 ×1023 molecules 1000 mg 294.30 g mol = 1.0 × 1019 molecules C14H18N2O5 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. 1.2 g C14H18N2O5 × 1 mol C14H18N2O5 × 2 mol N 294.30 g C14H18N2O5 mol C14H18N2O5 × 6.022 × 1023 atoms N = 4.9 × 1021 atoms N mol N f. 1.0 × 109 molecules × 1 mol × 294.30 g = 4.9 × 10−13 g 6.022 × 1023 atoms mol g. 1 molecule × 1 mol × 294.30 g = 4.887 × 10−22 g C14H18N2O5 6.022 × 1023 atoms mol 131. Molar mass = 20(12.01) + 29(1.008) + 19.00 + 3(16.00) = 336.43 g/mol Mass % C = 20(12.01) g C × 100 = 71.40% C 336.43 g compound
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