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Q.

$3.9 g$ of an organic compound on combustion gives $13.2 g$ of ${CO}_{2}$ and $2.7 g$ of $H_{2} O$ . The empirical formula of the compound is

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a

CH

b

$C_{2} H_{4}$

c

${CH}_{2}$

d

$C_{6} H_{6}$

answer is A.

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Detailed Solution

$C : H :: \frac{13.2}{44} : \frac{2 \times 7.2}{15} :: 0.3 : 0.3 :: 1 : 1$ Empirical Formula : CH

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