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Question
Calculate the empirical and molecular formula of a compound containing 76.6% carbon, 6.38% hydrogen and rest oxygen its vapour density is 47.
Solution
| Element | Percentage | Atomic mass | Relative number of atoms | Simple ratio | Whole no. |
| C | 76.6 | 12 | `76.6/12` = 6.38 | `6.38/1.06` = 6 | 6 |
| H | 6.38 | 1 | `6.38/1` = 6.38 | `6.38/1.06` = 6 | 6 |
| O | 17.02 | 16 | `17.02/16` = 1.06 | `1.06/1.06` = 1 | 1 |
Empirical formula = C6H6O
η = `"Molar mass"/"Calculated empirical formula mass"`
= `(2 xx "vapour density")/94`
= `(2 xx 47)/94`
= 1
∴ molecular formula (C6H6O) × 1 = C6H6O
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