Ten millilitres of a gaseous hydrocarbon was exploded with oxygen. After the explosion, there was a contraction of $20 m l$ in volume. On shaking the residual gaseous mixture with $K O H$ , there was a further concentration of $20 m l$ in volume. Calculate the molecular formula. All the volumes were recorded at same temperature and pressure.

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Solution

Let the formula of the hydrocarbon be $C_{x} H_{y}$ . The combustion of the hydrocarbon can be shown as:
$C_{x} H_{y} 10 m l + (x + \frac{y}{4}) O_{2} 10 (x \frac{y}{4}) m l \to x C O_{2} 10 x m l + \frac{y}{2} H_{2} O$
The first reduction in volume after explosion
$= 10 + 10 (s + \frac{y}{4}) - 10 x = 20$
$= 10 + \frac{10 y}{4} = 20$
Thus, $y = \frac{10 \times 4}{10} = 4$
$V o l u m e o f c a r b o n d i o x i d e p r o d u c e d = 20 m l$
Thus, $10 x = 20$
$⟹ x = \frac{20}{10} = 2$
Hence, the molecular formula of the hydrocarbon is $C_{2} H_{4}$ .

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