1 / 8

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

tuyet
Download Presentation

Chapter 9

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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?

More Related