Chapter 8 Answers to Questions

Transcription

1 Chapter 8 Answers to Questions 1. (a) formula mass Na 2 S = ( u) + ( u) = 78.1 u (b) formula mass CaCO 3 = ( u) + ( u) + ( u) = u (c) formula mass Al(NO 3 ) 3 = ( u) + ( u) + ( u) = u 2. (a) molecular mass NO 2 = ( u) + ( u) = 46.0 u (b) molecular mass P 4 O 6 = ( u) + ( u) = u (c) molecular mass IF 7 = ( u) + ( u) = u 3. (a) molar mass SnI 4 = ( g) + ( g) = g mol 1 (b) molar mass PF 5 = ( g) + ( g) = g mol 1 (c) molar mass (NH 4 ) 2 SO 4 = ( g) + ( g) + ( g) + ( g) = g mol 1 4. (a) molar mass SF 6 = ( g) + ( g) = g mol 1 (b) molar mass SnCl 2 2H 2 O = ( g) + ( g) + ( g) + ( g) = g mol 1 (c) molar mass Mg 3 (PO 4 ) 2 = ( g) + ( g) + ( g) = g mol 1 5. (a) First, calculate the molar mass of PbCl 2 = ( g) + ( g) = g mol 1 mass PbCl 2 mol PbCl 2 1 mol g PbCl 2 (b)

2 First, calculate the molar mass of SiO 2 = ( g) + ( g) = 60.1 g mol 1 mass SiO 2 mol CaI 2 1 mol g SiO 2 6. (a) First, calculate the molar mass of MgSO 4 = ( g) + ( g) + ( g) = g mol 1 mass(kg) MgSO 4 mass(g) MgSO 4 1 kg = 1000 g mass MgSO 4 mol MgSO 4 1 mol g MgSO 4 (b) First, calculate the molar mass of Ag 2 S = ( g) + ( g) = 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 g Ag 2 S 7. (a) First, calculate the molar mass of BaBr 2 = ( g) + ( g) = g mol 1

3 Mol of BaBr 2 mass BaBr 2 1 mol g BaBr 2 (b) First, calculate the molar mass of Cu 2 O = ( g) + ( g) = g mol 1 Mol of Cu 2 O mass Cu 2 O 1 mol g Cu 2 O 8. (a) First, calculate the molar mass of PbCrO 4 = ( g) + ( g) + ( g) = g mol 1 Mol of PbCrO 4 mass PbCrO 4 1 mol g PbCrO 4 (b) First, calculate the molar mass of NCl 3 = ( g) + ( g) = g mol 1 Mol of NCl 3 mass NCl 3 1 mol g NCl 3 9. Mol of N 2 O 3 # of molecules N 2 O 3 1 mol molecules # of molecules N 2 O 3 # of atoms O 1 molecule N 2 O 3 3 atom O

4 10. Mol of Fe 2 O 3 # of formula units Fe 2 O 3 1 mol formula units # of formula units Fe 2 O 3 # of ions O 2 1 formula unit Fe 2 O 3 3 ions O First, calculate the molar mass of SF 4 = ( g) + ( g) = g mol 1 Mass of SF 4 mol SF 4 1 mol g SF 4 Mol of SF 4 # of molecules SF 4 1 mol molecules # of molecules SF 4 # of atoms F 1 molecule SF 4 4 atom F

5 12. First, calculate the molar mass of PbF 4 = ( g) + ( g) = g mol 1 Mass of PbF 4 mol PbF 4 1 mol g PbF 4 Mol of PbF 4 # of formula units PbF 4 1 mol 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 = ( g) + (3 35.5) = g mol First, calculate the molar mass of CaCO 3 = ( g) + ( g) + ( g) = g mol Assume g of compound. This will contain 48.0 g of zinc and 52.0 g of chlorine.

6 Rounding to the nearest whole number of 1:2, gives the empirical formula of ZnCl Assume 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 Percent O = 100.0% 62.6% 8.5% = 28.9% Assume 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.

7 18. Percent O = 100.0% 36.5% 25.4% = 38.1% Assume 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 Assume 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 ( g g g = 31.0 g) is found. The value n is always an integer.

8 The molecular formula is (CH 3 O) 2 or more correctly, C 2 H 6 O Assume 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 ( g g 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 First, the mass of oxygen can be found by subtraction = (4.626 g g) = g Then the empirical formula type solution is employed.

9 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 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.

10 Empirical formula is CaCl 2 6H 2 O