Section 6.7: Empirical Formulas Tutorial 1 Practice, page 292 1. (a) Given: Al 20.2 ; Cl 79.8 Solution: A 100.0 g sample of this compound contains 20.2 g of aluminum and 79.8 g of chlorine. (20.2 g) # " 26.98 g 0.748 70 mol [2 extra digits carried] l (79.8 g) # " 35.45 g l 2.2511 mol [2 extra digits carried] 0.748 70 mol 0.748 70 mol 1.00 l 2.2511 mol 0.748 70 mol 3.01 The ratio of aluminum to chlorine is 1:3. This ratio suggests an empirical formula of AlCl 3. Statement: The empirical formula of the compound is AlCl 3. (b) Given: C 18.4 ; N 21.5; remainder is K Solution: A 100.0 g sample of this compound contains 18.4 g of carbon, 21.5 g of nitrogen, and 60.01 g of potassium. Step 1. Calculate the percentage of potassium in the compound. K 100! ( C + N) 100! (18.4 + 21.5 ) 100! 39.9 K 60.1 A 100.0 g sample of this compound contains 18.4 g of carbon, 21.5 g of nitrogen, and 60.01 g of potassium. Step 2. Calculate the amount of each element in the 100.0 g sample. (18.4 g) # 1.5321 mol [2 extra digits carried] Copyright 2011 Nelson Education Ltd. Chapter 6: Quantities ihemical Formulas 6.7-1
(21.5 g) # " 14.01 g 1.5346 mol [2 extra digits carried] n K (60.1 g) # " 39.10 g n K 1.5371 mol [2 extra digits carried] Since the amount of each element, to two decimal places, is the same, the three elements are in the ratio 1:1:1. This ratio suggests an empirical formula of KCN. Statement: The empirical formula of the compound is KCN. (c) Given: Al 52.9 ; remainder is O Solution: A 100.0 g sample of this compound contains 52.9 g of aluminum and 47.1 g oxygen. (52.9 g) # " 26.98 g 1.9607 mol [2 extra digits carried] (47.1 g) # 2.9438 mol [2 extra digits carried] 1.9607 mol 1.9607 mol 1.00 2.9438 mol 1.9607 mol 1.50 The ratio of aluminum to oxygen is 1:1.50. This ratio suggests an empirical formula of Al 1 O 1.5. Multiplying the subscripts of this formula by 2 gives Al 2 O 3. Statement: The empirical formula of the compound is Al 2 O 3. (d) Given: C 50.85 ; H 8.47 ; O 40.68 Solution: A 100.0 g sample of this compound contains 50.85 g of carbon, 8.47 g of hydrogen, and 40.68 g of oxygen. (50.85 g) # 4.233 97 mol [2 extra digits carried] Copyright 2011 Nelson Education Ltd. Chapter 6: Quantities ihemical Formulas 6.7-2
(8.47 g) # " 1.01 g 8.3861 mol [2 extra digits carried] (40.68 g) # 2.542 50 mol [2 extra digits carried] 4.233 97 mol 2.542 50 mol 1.67 8.3861 mol 2.542 50 mol 3.30 The ratio of carbon to hydrogen to oxygen is 1.67 : 3.30 : 1. This ratio suggests an empirical formula of C 1.67 H 3.3 O 1. Multiplying the subscripts of this formula by 3 gives C 5 H 10 O 3. Statement: The empirical formula of the compound is C 5 H 10 O 3. Section 6.7 Questions, page 293 1. (a) I would need the percentage composition of the compound. (b) I would need the molar masses of the elements in the compound. 2. The compounds H 2 CO 3, K 2 Cr 2 O 7, C 3 H 6 O 3 N are empirical formulas because the elements in each compound are in the simplest whole-number ratio. C 2 H 4 O 2 is not an empirical formula because its elements are not in the simplest whole-number ratio. The empirical formula of this compound is CH 2 O. 3. (a) Both compounds have the same empirical formula NO 2. (b) The two compounds do not have the same empirical formula. The empirical formula of C 3 H 6 is CH 2 and C 4 H 7 is already an empirical formula. (c) Both compounds have the same empirical formula CH. (d) Both compounds have the same empirical formula C 6 H 5 O. 4. (a) The multiplier is 2. The empirical formula is C 2 H 7. (b) The multiplier is 3. The empirical formula is C 2 H 2 O 3. (c) The multiplier is 2. The empirical formula is Na 2 S 2 O 3. (d) The multiplier is 3. The empirical formula is C 4 H 11 O 3. Copyright 2011 Nelson Education Ltd. Chapter 6: Quantities ihemical Formulas 6.7-3
5. Given: C 40.9 ; H 4.55 ; remainder is O Required: empirical formula of ascorbic acid Solution: A 100.0 g sample of this compound contains 40.9 g of carbon, 4.55 g of hydrogen, and 54.55 g of oxygen. (40.9 g) # 3.4055 mol [2 extra digits carried] (4.55 g) # " 1.01 g 4.5050 mol [2 extra digits carried] (54.55 g) # 3.409 38 mol [2 extra digits carried] 3.409 38 mol 1.00 3.4055 mol 4.5050 mol 3.4055 mol 1.32 The ratio of carbon to hydrogen to oxygen is 1:1.32:1. This ratio suggests an empirical formula of C 1 H 1.32 O 1. Multiplying the subscripts of this formula by 3 gives C 3 H 4 O 3. Statement: The empirical formula of ascorbic acid is C 3 H 4 O 3. 6. Given: Compound 1: N 46.7 ; O 53.3 Compound 2: N 30.4 ; O 69.6 Required: empirical formulas of Compound 1 and Compound 2 Solution: A 100.0 g sample of Compound 1 contains 46.7 g of nitrogen and 53.3 g oxygen. A 100.0 g sample of Compound 2 contains 30.4 g of nitrogen and 69.6 g oxygen. Compound 1: (46.7 g) # " 14.01 g 3.3333 mol [2 extra digits carried] Copyright 2011 Nelson Education Ltd. Chapter 6: Quantities ihemical Formulas 6.7-4
(53.3 g) # 3.3313 mol [2 extra digits carried] Since the amount of each element, to two decimal places, is the same, nitrogen and oxygen are in the ratio 1:1. This ratio suggests an empirical formula of NO. Compound 2: (30.4 g) # " 14.01 g 2.1699 mol [2 extra digits carried] (69.6 g) # 4.3500 mol [2 extra digits carried] Step 2. Divide the amount of each element iompound 2 by the smallest amount. 4.3500 mol 2.1699 mol 2.00 The ratio of nitrogen to oxygen is 1:2. This ratio suggests an empirical formula of NO 2. Statement: The empirical formula of Compound 1 is NO and the empirical formula of Compound 2 is NO 2. 7. Given: C 63.68 ; H 9.80 ; N 12.38 ; remainder is O Required: empirical formula of nylon-6 Solution: A 100.0 g sample of this compound contains 63.68 g of carbon, 9.80 g of hydrogen, 12.38 g of nitrogen, and 14.14 g of oxygen. Compound 1: (63.68 g) # 5.302 25 mol [2 extra digits carried] (9.80 g) # " 1.01 g 9.7030 mol [2 extra digits carried] (12.38 g) # " 14.01 g 0.883 655 mol [2 extra digits carried] Copyright 2011 Nelson Education Ltd. Chapter 6: Quantities ihemical Formulas 6.7-5
(14.14 g) # 0.883 750 mol [2 extra digits carried] 5.302 25 mol 0.883 655 mol 6.00 9.7030 mol 0.883 655 mol 11.0 0.883 750 mol 0.883 655 mol 1.00 The ratio of carbon to hydrogen to oxygen to nitrogen is 6 : 11 : 1 : 1. This ratio suggests an empirical formula of C 6 H 11 ON. Statement: The empirical formula of Nylon-6 is C 6 H 11 ON. 8. Given: Al 75 ; remainder is C Required: empirical formula of aluminum carbide Solution: A 100.0 g sample of this compound contains 75 g of aluminum and 25 g of carbon. (75 g) # " 26.98 g 2.780 mol [2 extra digits carried] (25 g) # 2.082 mol [2 extra digits carried] 2.780 mol 2.082 mol 1.34 The ratio of aluminum to carbon is 1:1.34. This ratio suggests an empirical formula of Al 1.34 C. Multiplying the subscripts of this formula by 3 gives Al 4 C 3. Statement: The empirical formula of aluminum carbide is Al 4 C 3. 9. Given: m compound 21.7 g ; m Hg 20.1 g Solution: Step 1. Calculate the mass of oxygen in the compound. m O 21.7 g! 20.1 g 1.6 g Copyright 2011 Nelson Education Ltd. Chapter 6: Quantities ihemical Formulas 6.7-6
Step 2. Calculate the amount of each element in the 21.7 g sample. g (20.1 g) # " 200.59 g g 0.100 20 mol [2 extra digits carried] (1.6 g) # 0.1000 mol [2 extra digits carried] Since the amount of each element is the same to three decimal places, mercury and oxygen are in the ratio 1 : 1. This ratio suggests an empirical formula of HgO. Statement: The empirical formula of the compound is HgO. 10. (a) Given: O 25.8 ; remainder is X Required: percentage of element X Solution: X 100 25.8 74.2 Statement: The percentage of element X in the compound is 74.2. (b) Given: X 74.2 ; O 25.8 ; formula of compound X 2 O Required: identity of element X Solution: A 100.0 g sample of X 2 O contains 74.2 g X and 25.8 g of oxygen. Step 1. Calculate the amount of oxygen in the 100.0 g sample. (25.8 g) # 1.6125 mol [2 extra digits carried] Since the formula of the compound is X 2 O, the amount of X in 100.0 g sample must be twice the amount of oxygen, which is 3.2250 mol. Step 2. Use the formula n m M to calculate the molar mass of element X. n X m X M X M X m X n X 74.2 g 3.2250 mol M X 23.0 g/mol The element with molar mass closest to this value is sodium. Statement: The unknown element is sodium. Copyright 2011 Nelson Education Ltd. Chapter 6: Quantities ihemical Formulas 6.7-7
11. Answers may vary. Sample answers: (a) Glucose Fructose empirical formula CH 2 O CH 2 O molecular formula C 6 H 12 O 6 C 6 H 12 O 6 structural formula (b) There is considerable evidence to suggest a link or correlation between the consumption of fructose and obesity. In one study, laboratory rats were given water sweetened with high-fructose corn syrup. The concentration of the fructose was half the concentration of fructose in soft drinks. Another group of rats was given water sweetened with sucrose (table sugar) at twice the concentration used for the fructose group. Researchers found that the fructose group showed greater signs of obesity than the sucrose group. Sucrose consists of one molecule of glucose and one molecule of fructose bonded together. Glucose in the bloodstream stimulates the production of the hormone leptin, which helps to suppress the appetite and insulin, which helps the body process glucose. Fructose, however, has no effect on leptin or insulin. Some human studies have shown that drinking beverages sweetened with fructose led to greater buildup of fat around the belly than the consumption of glucose-sweetened beverages. Copyright 2011 Nelson Education Ltd. Chapter 6: Quantities ihemical Formulas 6.7-8