Question

Ten millilitres of a gaseous hydrocarbon was exploded with oxygen. After the explosion, there was a contraction of 20 ml in volume. On shaking the residual gaseous mixture with KOH, there was a further concentration of 20 ml in volume. Molecular formula is $C_x{H}_y$. All the volumes were recorded at the same temperature and pressure. Here $\text{x+y}$ is:

Answer 1

Let the formula of the hydrocarbon be $C_x{H}_y$. The combustion of the hydrocarbon can be shown as:

$C_x{H}_y+(x+\frac{y}{4})O_2⟶xC{O}_2+\frac{y}{2}{H}_{2}O$

The first reduction in volume after the explosion

$=10+10(x+\frac{y}{4})-10x=20$
$=10+\frac{10y}{4}=20$

Thus, $y=\frac{10\times 4}{10}=4$

The volume of carbon dioxide produced = 20 mI

Thus, $10x=20$

$⟹x=\frac{20}{10}=2$

Hence, the molecular formula of the hydrocarbon is $C_2{H}_4$

Therefore, $x+y=2+4=6$.