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Question
On analysis, a substance was found to contain:
C = 54.54%, H = 9.09%, O = 36.36%
The vapour density of the substance is 44, calculate its empirical formula.
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Solution
| Element | Percentage | Atomic Weight | Ratio | Simplest Ratio |
| C | 54.54 | 12 | `54.54/12` = 4.54 | `4.54/2.27` = 2 |
| H | 9.09 | 1 | `9.09/1` = 9.09 | `9.09/2.27` = 4 |
| O | 36.36 | 16 | `36.36/16` = 2.27 | `2.27/2.27` = 1 |
So, its empirical formula = C2H4O
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