Chapter 12. Answers to Questions. 1. (a) nitrogen dioxide (b) hydrogen sulfide (c) hydrogen chloride

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1 Chapter 12 Answers to Questions 1. (a) nitrogen dioxide (b) hydrogen sulfide (c) hydrogen chloride 2. (a) hydrogen cyanide (b) dinitrogen oxide (c) ammonia 3. X = sulfur, Y = oxygen, Z = sulfur dioxide 4. X = carbon monoxide, Y = oxygen, Z = carbon dioxide 5. V = nrt P = 1.00 mol (8.31 kpa L mol 1 K 1 )(743 K) ( = 0.66 L kpa) 6. V = nrt P = 1.00 mol (8.31 kpa L mol 1 K 1 )(263 K) = L (0.60 kpa) 7. Press/vol/temp of O 2 mol O 2 Mol O 2 mass O 2 O g RT = 100 kpa (25.0 L) (8.31 kpa L mol 1 K 1 = 1.0 )(298 K)

2 Mass O 2 = 1.0 O g = 32.3 g O 2 8. Mass CO 2 mol CO 2 Mol/press/temp CO 2 vol CO 2 CO g Mol CO 2 = 30.0 g CO g = mol CO 2 V = nrt P = mol (8.31 kpa L mol 1 K 1 )(298 K) = 16.9 L (100 kpa) 9. Press/vol/temp of C 2 H 2 mol C 2 H 2 Mol C 2 H 2 mass C 2 H 2 C 2 H g RT = kpa (87.0 L) (8.31 kpa L mol 1 K 1 = 59.8 mol )(301 K) Mass C 2 H 2 = 59.8 mol C 2 H g = g C 2 H 2 = 1.56 kg C 2 H Press/vol/temp of He mol He Mol He mass He He 4.00 g RT = 102 kpa ( L) (8.31 kpa L mol 1 K 1 )(290 K) = mol Mass He = mol He = tonne He 4.00 g = g He

3 11. Press/vol/temp of X 2 H 6 mol X 2 H 6 Mol/mass X 2 H 6 molar mass X 2 H 6 mm X 2 H 6 = m/n RT = 196 kpa (1.26 L) (8.31 kpa L mol 1 K 1 )(356 K) = mol 5.20 g mol = 62.3 g mol 1 2 X = [62.3 6(1.01)] g mol 1 = 56.2 g mol 1 X = 28.1 g mol 1 According to the Periodic Table, this molar mass corresponds to the element silicon 12. Press/vol/temp of XH 3 mol XH 3 Mol/mass XH 3 molar mass XH 3 mm XH 3 = m/n RT = 115 kpa (0.775 L) (8.31 kpa L mol 1 K 1 )(312 K) = mol 2.68 g mol = 77.9 g mol 1 X = [77.9 3(1.01)] g mol 1 = 74.9 g mol 1 According to the Periodic Table, this molar mass corresponds to the element arsenic 13. Vol/mass of gas density of gas d = m/v

4 Molar V = nrt P = 1.00 mol (8.31 kpa L mol 1 K 1 )(298 K) = 24.8 L (100 kpa) Molar mass, H 2 = 2.02 g Density = 2.02 g 24.8 L = g L Vol/mass of gas density of gas d = m/v Molar V = nrt P = 1.00 mol (8.31 kpa L mol 1 K 1 )(298 K) = 24.8 L (100 kpa) Molar mass, UF 6 = g Density = g 24.8 L = 14.2 g L Part 1: Find Empirical Formula Assume g of compound. This will contain 92.3 g of carbon and 7.7 g of hydrogen. Mol of C = 92.3 g C 12.0 g = 7.69 mol C Mol of H = 7.7 g H 1.01 g = 7.7 mol H Ratio = 7.69 mol C 7.69 mol 7.7 mol H 7.69 mol = 1 C 1.0 H Rounding to the nearest whole number of 1:1, gives the empirical formula of CH. Part 2: Find molar mass of Compound

5 Press/vol/temp mol Mol/mass molar mass mm = m/n RT = 101 kpa (0.226 L) (8.31 kpa L mol 1 K 1 )(373 K) = mol g mol = 77.8 g mol 1 Part 3: Find Molar Mass and Molecular Formula The molecular formula will be some multiple of the empirical formula: (CH) n. To find n, the ratio of the molar mass (77.8 g) and the empirical formula mass ( g g = 13.0 g) is found. The value n is always an integer. Mass ratio, n = 77.8 g 13.0 g = 6 The molecular formula is (CH) 6 or more correctly, C 6 H Part 1: Find Empirical Formula Assume g of compound. This will contain g of carbon, g of hydrogen, and g of oxygen. Mol of C = g C g = mol C Mol of H = g H g = mol H Mol of O = g H g = mol O

6 Ratio = mol C mol mol H mol mol O mol = C H 1 O Rounding off, it is: 3.5 C : 7 H : 1 O Multiplying through by 2 gives: 7 C : 14 H : 2 O giving the empirical formula of C 7 H 14 O 2. Part 2: Find molar mass of Compound Press/vol/temp mol Mol/mass molar mass mm = m/n RT = kpa (1.880 L) (8.314 kpa L mol 1 K 1 )(448.1 K) = mol g mol = g mol 1 Part 3: Find Molar Mass and Molecular Formula The molecular formula will be some multiple of the empirical formula: (C 7 H 14 O 2 ) n. To find n, the ratio of the molar mass (130.3 g) and the empirical formula mass ( g g = g) is found. The value n is always an integer. Mass ratio, n = g g = 1 The molecular formula is (C 7 H 14 O 2 ) 1 or more correctly, C 7 H 14 O 2