6 Answer Key Chapter 3

Solution Videos

3.1 Formula Mass and the Mole Concept
1. (a) 12.01 amu
(b) 12.01 amu
(c) 144.12 amu
(d) 60.05 amu
3. (a) 123.896 amu
(b) 18.015 amu
(c) 164.086 amu
(d) 60.052 amu
(e) 342.297 amu
5. (a) 56.107 amu
(b) 54.091 amu
(c) 199.9976 amu
(d) 97.9950 amu
7. Use the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of compound by the molar mass of the compound expressed in grams.
9. Formic acid. Its formula has twice as many oxygen atoms as the other two compounds (one each). Therefore, 0.60 mol of formic acid would be equivalent to 1.20 mol of a compound containing a single oxygen atom.
11. The two masses have the same numerical value, but the units are different: The molecular mass is the mass of 1 molecule while the molar mass is the mass of 6.022 × 1023 molecules.
13. (a) 256.48 g/mol
(b) 72.150 g/mol
(c) 378.103 g/mol
(d) 58.080 g/mol
(e) 180.158 g/mol
15. (a) 197.382 g/mol
(b) 257.163 g/mol
(c) 194.193 g/mol
(d) 60.056 g/mol
(e) 306.464 g/mol
17. (a) 0.819 g
(b) 307 g
(c) 0.23 g
(d) 1.235 x 106 g (1235 kg)
(e) 765 g
19. (a) 99.41 g
(b) 2.27 g
(c) 3.5 g
(d) 222 kg
(e) 160.1 g
21. (a) 9.60 g
(b) 19.2 g
(c) 28.8 g
23. zirconium: 2.0381023 atoms, 30.87 g; silicon: 2.038 x 1023 atoms, 9.504 g; oxygen: 8.151 × 1023 atoms, 21.66 g
25. AlPO4: 1.000 mol, or 26.98 g Al; Al2Cl6: 1.994 mol, or 53.74 g Al; Al2S3: 3.00 mol, or 80.94 g Al; The Al2S3 sample thus contains the greatest mass of Al.
27. 3.113 x 1025 C atoms
29. 0.865 servings, or about 1 serving.
31. 20.0 g H2O represents the least number of molecules since it has the least number of moles.
3.2 Determining Empirical and Molecular Formulas
33. (a) % N = 82.24%, % H = 17.76%
(b) % Na = 29.08%, % S = 40.56%, % O = 30.36%
(c) % Ca2+ = 38.76%
35. % NH3 = 38.2%
37. (a) CS2
(b) CH2O
39. C6H6
41. Mg3Si2H3O8 (empirical formula), Mg6Si4H6O16 (molecular formula)
43. C15H15N3
3.3 Molarity
45. We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution.
47. (a) 0.679 M
(b) 1.00 M
(c) 0.06998 M
(d) 1.75 M
(e) 0.070 M
(f) 6.6 M
49. (a) determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass
(b) 27 g
51. (a) 37.0 mol H2SO4, 3.63 × 103 g H2SO4
(b) 3.8 × 10−7 mol NaCN, 1.9 × 10−5 g NaCN
(c) 73.2 mol H2CO, 2.20 kg H2CO
(d) 5.9 × 10−7 mol FeSO4, 8.9 × 10−5 g FeSO4
53. (a) Determine the molar mass of KMnO4; determine the number of moles of KMnO4 in the solution; from the number of moles and the volume of solution, determine the molarity
(b) 1.15 x 10−3 M
55. (a) 5.04 x 10−3 M
(b) 0.499 M
(c) 9.92 M
(d) 1.1 × 10−3 M
57. 0.025 M
59. 0.5000 L
61. 1.9 mL
63. (a) 0.125 M
(b) 0.04888 M
(c) 0.206 M
(d) 0.0056 M
65. 11.9 M
67. 1.6 L

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