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प्रश्न
A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapour density is 29, find its molecular formula.
[C = 12, H = 1]
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उत्तर
% of carbon = 82.76%
% of hydrogen = 100 - 82.76 = 17.24%
| Element | % weight | Atomic weight | Relative No. of moles | Simplest Ratio |
| C | 82.76 | 12 | 82.76/12 = 6.89 | 6.89/6.8 = 1 x 2 = 2 |
| H | 17.24 | 1 | 17.24/1 = 17.24 | 17.24/6.89 =2.5 x 2 = 5 |
Empirical formula = C2H5
Empirical formula weight = 2 × 12 + 1 × 5 = 24 + 5 = 29
Vapour density = 29
Relative molecular mass = 29 × 2 = 58
N = `"Relative molecular mass"/"Empirical weight"`
= `58/29` = 2
Molecular formula = n x empirical formula
= 2 x C2H5
= C4H10
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