Question

A gaseous mixture containing propane, $C _{3} H _{8}$ and an olefinic hydrocarbon occupied $12 mL .57 mL$ of oxygen was required to burn the mixture completely. After combustion $36 mL$ of $CO _{2}$ were left. Calculate

(i) The formula of the olefin and

(ii) The composition of the mixture by volume.

All volume measurements were made at the same temperature and pressure.

Answer 1

The olefinic hydrocarbon corresponds to the general formula $C _{n} H _{2 n}$.

Vol. of mixture $\left( C _{3} H _{8}+ C _{n} H _{2 n}\right) \quad=12 mL$

Vol. of $O _{2}$ for complete combustion $\quad=57 mL$

Vol. of $CO _{2}$ formed $=36 mL$

Suppose that vol. of $C _{n} H _{2 n}$ gas $=x mL$

and that of $C _{3} H _{8}=(12-x) mL$

$C _{n} H _{2 n}( g )+\frac{3 n}{2} O _{2}( g )=n CO _{2}( g )+n H _{2} O (l)$

1 vol. $\quad \frac{3 n}{2}$ vol. $n$ vol.

$x mL \quad \frac{3 n x}{2} mL \quad n x mL \quad$ oml.

From these equations :

Total $CO _{2}: \quad 3(12-x)+n x=36$

or, $36-3 x+n x=36 \quad$ or, $n x=3 x \quad \therefore \quad n=3$

$\therefore$ Formula of olefinic hydrocarbon $= C _{3} H _{6}$

To find the value of $x$., now the

total $O_{2}$ reacted is: $5(12-x)+\frac{3 n x}{2}=57$

or, $\quad 60-5 x+\frac{3 \times 3 x}{2}=57 \quad(\because n=3)$

or, $\quad-\frac{x}{2}=57-60=-3 \quad \therefore \quad x=6$.

$\therefore$ Vol. of olefinic hydrocarbon $=6 mL$. and vol. of $C_{3} H_{8}=6 mL$.