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Question
An organic compound, whose vapour density is 45, has the following percentage composition,
H = 2.22%, O = 71.19% and remaining carbon. Calculate its empirical formula.
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Solution
As we know, molecular weight is 2 x vapour density.
∴ Molecular weight of compound = 2 x 45 = 90
∵ H2 + O2 = 2.22 + 71.19 = 73.41 < 100
The third element of organic compound C = 100 − 73.41 = 26.59%
| Element | Percentage | Atomic Weight | Ratio | Simplest Ratio |
| C | 26.59 | 12 | `26.59/12` = 2.22 | `2.22/2.22` = 1 |
| H | 2.22 | 1 | `2.22/1` = 2.22 | `2.22/2.22` = 1 |
| O | 71.19 | 16 | `71.19/16` = 4.44 | `4.44/2.22` = 2 |
∴ The empirical formula of the given organic compound = CHO2
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