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Chapter 9 Problems 9.120 a, 9.18 b and 9.38 a
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Citric acid with a molecular mass 192 amu Mass %: C 37.50, H 4.21 and O 58.29 Determine molecular formula starting from 100.0 g citric acid How many moles of each in 100.0 g? C: 100.0 g x 37.50 g x 1 mole = 3.122 100.0 g12.01 g H: 100.0 g x 4.21 g x 1 mole = 4.1683 100.0 g1.01 g
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Citric acid with a molecular mass 192 amu – cont’d O: 100.0 g x 58.29 g x 1 mole = 3.643 100.0 g16.00 g Empirical formula C: 3.125/3.125 = 1 H: 4.17/3.125 = 1.33 O: 3.643/3.125 = 1.166 What is the multiplier? 6
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Citric acid with a molecular mass 192 amu – cont’d C6HxOyC6HxOyC6HxOyC6HxOy x = 1.33 * 6 = 7.98 y = 1.166 * 6 = 6.996 C6H8O7C6H8O7C6H8O7C6H8O7 This is the …..
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Citric acid with a molecular mass 192 amu – cont’d Empirical formula Empirical formula mass is … 6 * 12.01 + 8 * 1.01 + 7 * 16.00 = 192.14 amu What is the molecular formula? Same as empirical formula
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Citric acid C 6 H 8 O 7 Going Backward Knowing the formula, can we calculate % mass composition? Does it matter if we have empirical or molecular formula?
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Problem 9.18 b) % composition of Na in NaCN Na: 1 * 22.99 = 22.99 amu C: 1 * 12.01 = 12.01 amu N: 1 * 14.01 = 14.01 amu Total = 49.01 amu %Na by mass = (22.99/49.01) * 100 = 46.90879%
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Problem 9.38 a) Mass in grams of 0.981 mole of SO 2 S: 1 * 32.06 = 32.06 O: 2 * 16.00 = 32.00 Sum = 64.06 0.981 mole * (64.06 g/1mole) = 62.84286 g Significant figures? How many molecules of SO 2 in 0.981 mole?
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