An organic substance containing carbon, hydrogen and oxygen gave the percentage composition as $C = 40.687$ % $, H = 5.085$ % and the remaining oxygen. The vapour density of the compound is 59. Calculate the molecular formula of the compound.

C_{4} H_{6} O

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C_{4} H_{6} O_{4}

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C_{2} H_{6} O_{4}

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C_{4} H_{2} 6 O_{4}

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Solution

The correct option is A $C_{4} H_{6} O_{4}$
Consider $\to 100 g$ of compound
Masses Molar mass Moles Ratios Ratio
$C$ $40.3687$ $12$ $3.3916$ $3.4 / 3.4 = 1$ $2$
$H$ $5.085$ $1$ $5.085$ $5.1 / 3.4 = 1.5$ $3$
$O$ $54.228$ $16$ $3.392$ $3.4 / 3.4 = 1$ $2$
Emperical formula $= C_{2} H_{2} O_{2}$
Emperical mass $= 2 \times 12 + 3 \times 1 + 2 \times 16$
$= 24 + 3 + 32$
$= 59$

Molecular mass $= 2 \times$ vapor density $= 2 \times 59 = 118$
$n = \frac{M . M}{E . M} = \frac{118}{59} = 2$

Molecular formula $= n \times E . F = 2 \times C_{2} H_{3} O_{2}$
$= C_{4} H_{6} O_{4}$

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