Chapter 8 Answers to Questions 1. (a) formula mass Na2S = (2 × 23.0 u) + (1 × 32.1 u) = 78.1 u (b) formula mass CaCO3 = (1 × 40.1 u) + (1 × 12.0 u) + (3 × 16.0 u) = 100.1 u (c) formula mass Al(NO3)3 = (1 × 27.0 u) + (3 × 14.0 u) + (9 × 16.0 u) = 213.0 u 2. (a) molecular mass NO2 = (1 × 14.0 u) + (2 × 16.0 u) = 46.0 u (b) molecular mass P4O6 = (4 × 31.0 u) + (6 × 16.0 u) = 220.0 u (c) molecular mass IF7 = (1 × 126.9 u) + (7 × 19.0 u) = 259.9 u 3. (a) molar mass SnI4 = (1 × 118.7 g) + (4 × 126.9 g) = 626.3 g∙mol−1 (b) molar mass PF5 = (1 × 31.0 g) + (5 × 19.0 g) = 126.0 g∙mol−1 (c) molar mass (NH4)2SO4 = (2 × 14.0 g) + (8 × 1.01 g) + (1 × 32.1 g) + (4 × 16.0 g) = 132.1 g∙mol−1 4. (a) molar mass SF6 = (1 × 32.1 g) + (6 × 19.0 g) = 146.1 g∙mol−1 (b) molar mass SnCl2∙2H2O = (1 × 118.7 g) + (2 × 35.5 g) + (4 × 1.01 g) + (2 × 16.0 g) = 225.6 g∙mol−1 (c) molar mass Mg3(PO4)2 = (3 × 24.3 g) + (2 × 31.0 g) + (8 × 16.0 g) = 262.9 g∙mol−1 5. (a) First, calculate the molar mass of PbCl2 = (1 × 207.2 g) + (2 × 35.5 g) = 278.1 g∙mol−1 Strategy mass PbCl2 → mol PbCl2 (b) Relationship 1 mol ≡ 278.1 g PbCl2 First, calculate the molar mass of SiO2 = (1 × 28.1 g) + (2 × 16.0 g) = 60.1 g∙mol−1 Strategy mass SiO2 → mol CaI2 6. Relationship 1 mol ≡ 293.9 g SiO2 (a) First, calculate the molar mass of MgSO4 = (1 × 24.3 g) + (1 × 32.1 g) + (4 × 16.0 g) = 120.4 g∙mol−1 Strategy Relationship mass(kg) MgSO4 → mass(g) MgSO4 1 kg = 1000 g mass MgSO4 → mol MgSO4 1 mol ≡ 120.4 g MgSO4 (b) First, calculate the molar mass of Ag2S = (2 × 107.9 g) + (1 × 32.1 g) = 247.9 g∙mol−1 Strategy Relationship mass(mg) Ag2S → mass(g) Ag2S 1 kg = 1000 g mass Ag2S → mol Ag2S 1 mol ≡ 247.9 g Ag2S 7. (a) First, calculate the molar mass of BaBr2 = (1 × 137.3 g) + (2 × 79.9 g) = 297.1 g∙mol−1 Strategy Relationship Mol of BaBr2 → mass BaBr2 1 mol ≡ 297.1 g BaBr2 (b) First, calculate the molar mass of Cu2O = (2 × 63.5 g) + (1 × 16.0 g) = 143.0 g∙mol−1 Strategy Mol of Cu2O → mass Cu2O 8. Relationship 1 mol ≡ 132.1 g Cu2O (a) First, calculate the molar mass of PbCrO4 = (1 × 207.2 g) + (1 × 52.0 g) + (4 × 16.0 g) = 323.2 g∙mol−1 Strategy Relationship Mol of PbCrO4 → mass PbCrO4 1 mol ≡ 323.2 g PbCrO4 (b) First, calculate the molar mass of NCl3 = (1 × 14.0 g) + (3 × 35.5 g) = 120.4 g∙mol−1 Strategy Mol of NCl3 → mass NCl3 Relationship 1 mol ≡ 132.1 g NCl3 9. Strategy Mol of N2O3 → # of molecules N2O3 # of molecules N2O3 → # of atoms O Relationship 1 mol ≡ 6.02×1023 molecules 1 molecule N2O3 ≡ 3 atom O 10. Strategy Mol of Fe2O3 → # of formula units Fe2O3 # of formula units Fe2O3 → # of ions O2− Relationship 1 mol ≡ 6.02×1023 formula units 1 formula unit Fe2O3 ≡ 3 ions O2− 11. First, calculate the molar mass of SF4 = (1 × 32.1 g) + (4 × 19.0 g) = 108.1 g∙mol−1 Strategy Mass of SF4 → mol SF4 Mol of SF4 → # of molecules SF4 # of molecules SF4 → # of atoms F Relationship 1 mol ≡ 108.1 g SF4 1 mol ≡ 6.02×1023 molecules 1 molecule SF4 ≡ 4 atom F 12. First, calculate the molar mass of PbF4 = (1 × 207.2 g) + (4 × 19.0 g) = 283.2 g∙mol−1 Strategy Mass of PbF4 → mol PbF4 Mol of PbF4 → # of formula units PbF4 # of formula units PbF4 → # of ions F− Relationship 1 mol ≡ 283.2 g PbF4 1 mol ≡ 6.02×1023 formula units 1 formula unit PbF4 ≡ 4 ions F− 13. First, calculate the molar mass of PCl3 = (1 × 31.0 g) + (3 × 35.5) = 137.5 g∙mol−1 14. First, calculate the molar mass of CaCO3 = (1 × 40.1 g) + (1 × 12.0 g) + (3 × 16.0 g) = 100.1 g∙mol−1 15. Assume 100.0 g of compound. This will contain 48.0 g of zinc and 52.0 g of chlorine. Rounding to the nearest whole number of 1:2, gives the empirical formula of ZnCl2. 16. Assume 100.0 g of compound. This will contain 19.0 g of tin and 81.0 g of iodine. Rounding to the nearest whole number of 1:4, gives the empirical formula of SnI4. 17. Percent O = 100.0% – 62.6% − 8.5% = 28.9% Assume 100.0 g of compound. This will contain 62.6 g of lead; 8.5 g of nitrogen; and 28.9 g of oxygen. Rounding to the nearest whole number of 1:2:6, gives the empirical formula of PbN2O6. 18. Percent O = 100.0% – 36.5% − 25.4% = 38.1% Assume 100.0 g of compound. This will contain 36.5 g of sodium; 25.4 g of sulfur; and 38.1 g of oxygen. Rounding to the nearest whole number of 2:1:3, gives the empirical formula of Na2SO3. 19. Assume 100.0 g of compound. This will contain 38.7 g of carbon; 9.8 g of hydrogen; and 51.5 g of oxygen. Rounding to the nearest whole number of 1:3:1, gives the empirical formula of CH3O. The molecular formula will be some multiple of this: (CH3O)n. To find n, the ratio of the molar mass (62.1 g) and the empirical formula mass (1 × 12.0 g + 3 × 1.01 g + 1 × 16.0 g = 31.0 g) is found. The value n is always an integer. The molecular formula is (CH3O)2 or more correctly, C2H6O2 20. Assume 100.0 g of compound. This will contain 48.6 g of carbon; 8.2 g of hydrogen; and 43.2 g of oxygen. To obtain integers, each number must be multiplied by two: 3.00 C : 6.0 H : 2 O Rounding to the nearest whole number of 3:6:2, gives the empirical formula of C3H6O2. The molecular formula will be some multiple of this: (C3H6O2)n. To find n, the ratio of the molar mass (74.1 g) and the empirical formula mass (3 × 12.0 g + 6 × 1.01 g + 2 × 16.0 g = 74.1 g) is found. The value n is always an integer. The molecular formula is (C3H6O2)1 or more correctly, C3H6O2 21. First, the mass of oxygen can be found by subtraction = (4.626 g – 4.306 g) = 0.320 g Then the empirical formula type solution is employed. Empirical formula is Ag2O. 22. First, the mass of oxygen can be found by subtraction = (7.59 g – 3.76 g) = 3.83 g Then the empirical formula type solution is employed. To obtain integers, each number must be multiplied by two: 2 Mn : 6.98 O Rounding to the nearest whole number of 2:7, gives the empirical formula of Mn2O7. 23. First, the mass of water is needed = (7.52 g – 5.54 g) = 1.98 g Then the empirical formula type solution is employed. Empirical formula is FePO4∙3H2O 24. The empirical formula type solution is employed. Empirical formula is CaCl2∙6H2O

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