Chapter 8 Answers to Questions 1. (a) formula mass Na 2 S = (2 23.0 u) + (1 32.1 u) = 78.1 u (b) formula mass CaCO 3 = (1 40.1 u) + (1 12.0 u) + (3 16.0 u) = 100.1 u (c) formula mass Al(NO 3 ) 3 = (1 27.0 u) + (3 14.0 u) + (9 16.0 u) = 213.0 u 2. (a) molecular mass NO 2 = (1 14.0 u) + (2 16.0 u) = 46.0 u (b) molecular mass P 4 O 6 = (4 31.0 u) + (6 16.0 u) = 220.0 u (c) molecular mass IF 7 = (1 126.9 u) + (7 19.0 u) = 259.9 u 3. (a) molar mass SnI 4 = (1 118.7 g) + (4 126.9 g) = 626.3 g mol 1 (b) molar mass PF 5 = (1 31.0 g) + (5 19.0 g) = 126.0 g mol 1 (c) molar mass (NH 4 ) 2 SO 4 = (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 SF 6 = (1 32.1 g) + (6 19.0 g) = 146.1 g mol 1 (b) molar mass SnCl 2 2H 2 O = (1 118.7 g) + (2 35.5 g) + (4 1.01 g) + (2 16.0 g) = 225.6 g mol 1 (c) molar mass Mg 3 (PO 4 ) 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 PbCl 2 = (1 207.2 g) + (2 35.5 g) = 278.1 g mol 1 mass PbCl 2 mol PbCl 2 1 mol 278.1 g PbCl 2 (b)
First, calculate the molar mass of SiO 2 = (1 28.1 g) + (2 16.0 g) = 60.1 g mol 1 mass SiO 2 mol CaI 2 1 mol 293.9 g SiO 2 6. (a) First, calculate the molar mass of MgSO 4 = (1 24.3 g) + (1 32.1 g) + (4 16.0 g) = 120.4 g mol 1 mass(kg) MgSO 4 mass(g) MgSO 4 1 kg = 1000 g mass MgSO 4 mol MgSO 4 1 mol 120.4 g MgSO 4 (b) First, calculate the molar mass of Ag 2 S = (2 107.9 g) + (1 32.1 g) = 247.9 g mol 1 mass(mg) Ag 2 S mass(g) Ag 2 S 1 kg = 1000 g mass Ag 2 S mol Ag 2 S 1 mol 247.9 g Ag 2 S 7. (a) First, calculate the molar mass of BaBr 2 = (1 137.3 g) + (2 79.9 g) = 297.1 g mol 1
Mol of BaBr 2 mass BaBr 2 1 mol 297.1 g BaBr 2 (b) First, calculate the molar mass of Cu 2 O = (2 63.5 g) + (1 16.0 g) = 143.0 g mol 1 Mol of Cu 2 O mass Cu 2 O 1 mol 132.1 g Cu 2 O 8. (a) First, calculate the molar mass of PbCrO 4 = (1 207.2 g) + (1 52.0 g) + (4 16.0 g) = 323.2 g mol 1 Mol of PbCrO 4 mass PbCrO 4 1 mol 323.2 g PbCrO 4 (b) First, calculate the molar mass of NCl 3 = (1 14.0 g) + (3 35.5 g) = 120.4 g mol 1 Mol of NCl 3 mass NCl 3 1 mol 132.1 g NCl 3 9. Mol of N 2 O 3 # of molecules N 2 O 3 1 mol 6.02 10 23 molecules # of molecules N 2 O 3 # of atoms O 1 molecule N 2 O 3 3 atom O
10. Mol of Fe 2 O 3 # of formula units Fe 2 O 3 1 mol 6.02 10 23 formula units # of formula units Fe 2 O 3 # of ions O 2 1 formula unit Fe 2 O 3 3 ions O 2 11. First, calculate the molar mass of SF 4 = (1 32.1 g) + (4 19.0 g) = 108.1 g mol 1 Mass of SF 4 mol SF 4 1 mol 108.1 g SF 4 Mol of SF 4 # of molecules SF 4 1 mol 6.02 10 23 molecules # of molecules SF 4 # of atoms F 1 molecule SF 4 4 atom F
12. First, calculate the molar mass of PbF 4 = (1 207.2 g) + (4 19.0 g) = 283.2 g mol 1 Mass of PbF 4 mol PbF 4 1 mol 283.2 g PbF 4 Mol of PbF 4 # of formula units PbF 4 1 mol 6.02 10 23 formula units # of formula units PbF 4 # of ions F 1 formula unit PbF 4 4 ions F 13. First, calculate the molar mass of PCl 3 = (1 31.0 g) + (3 35.5) = 137.5 g mol 1 14. First, calculate the molar mass of CaCO 3 = (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 ZnCl 2. 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 SnI 4. 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 PbN 2 O 6.
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 Na 2 SO 3. 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 CH 3 O. The molecular formula will be some multiple of this: (CH 3 O) 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 (CH 3 O) 2 or more correctly, C 2 H 6 O 2 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 C 3 H 6 O 2. The molecular formula will be some multiple of this: (C 3 H 6 O 2 ) 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 (C 3 H 6 O 2 ) 1 or more correctly, C 3 H 6 O 2 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 Ag 2 O. 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 Mn 2 O 7. 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 FePO 4 3H 2 O 24. The empirical formula type solution is employed.
Empirical formula is CaCl 2 6H 2 O