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Chapter 9

Chapter 9. Problems 9.120 a, 9.18 b and 9.38 a. 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 g 12.01 g

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Chapter 9

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  1. Chapter 9 Problems 9.120 a, 9.18 b and 9.38 a

  2. 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 g 12.01 g • H: 100.0 g x 4.21 g x 1 mole = 4.1683 100.0 g 1.01 g

  3. 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 g 16.00 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

  4. Citric acid with a molecular mass 192 amu – cont’d • C6HxOy • x = 1.33 * 6 = 7.98 • y = 1.166 * 6 = 6.996 • C6H8O7 • This is the …..

  5. 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

  6. Citric acid C6H8O7Going Backward • Knowing the formula, can we calculate % mass composition? • Does it matter if we have empirical or molecular formula?

  7. Problem 9.18 b) % compositionof 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%

  8. Problem 9.38 a) • Mass in grams of 0.981 mole of SO2 • 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 SO2 in 0.981 mole?

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