Question

10mL of a gaseous hydrocarbon was exploded with 33mL of oxygen. After cooling the volume of the residual gas was 28mL and on treatment with KOH, the volume decreased to 8mL. Find the formula of hydrocarbon.

Answer 1

SOLUTION : Given that :

Vol. of hydrocarbon $=10 mL$

Vol. of $O_{2}$ added $=33 mL \quad \therefore \quad$ Total volume $=(10+33) mL =43 mL$

Vol. after explosion $=28 mL \quad \therefore \quad$ Contraction $=(43-28) mL =15 mL$

Vol. after $KCH$ treatment $=8 mL \therefore \quad$ Vol. of $CO _{2}$ formed $=(28-8) mL =20 mL$

This vol. $8 mL$ is unused $O_{2} . \quad \therefore \quad$ Vol. of $O_{2}$ reacted $=(33-8) mL =25 mL$

Let us suppose that the formula of hydrocarbon is $C_{x} H_{y}$.

COMBUSTION REACTION :

$C_{x} H_{y}(g)+\left(x+\frac{y}{4}\right) O_{2}(g)=x CO _{2}(g)+\frac{y}{2} H _{2} O (l)$

1 vol. $\quad\left(x+\frac{y}{4}\right)$ vol. $\quad x$ vol. $\quad$ ovol.

$10 mL \quad 10\left(x+\frac{y}{4}\right) mL \quad 10 x mL \quad$ oml.

$\therefore \quad 10 x=20 \quad\left(\right.$ vol. of $\left.CO _{2}\right) \quad x=2 .$.

Again, $\quad 10\left(x+\frac{y}{4}\right)=25$ (vol. of $O_{2}$ reacted) or, $2+\frac{y}{4}=\frac{25}{10}=2 \cdot 5$

or, $\quad \frac{y}{4}=0 \cdot 5 ; \quad \therefore \quad y=0.5 \times 4=2$

Hence the formula of hydrocarbon $= C _{2} H _{2}$.

[Note : We may also make the use of contraction formula and still we get the same result.]

Contraction in volume $=10+10\left(x+\frac{y}{4}\right)-10 x=10+10 x+\frac{5}{2} y-10 x$

$=10+\frac{5}{2} y \quad \therefore 10+\frac{5}{2} y=15 ; \quad \text { or, } \frac{5}{2} y=5, \quad \therefore \quad y=2$