Question

$16$ mL of a gaseous aliphatic compound $C_n{H}_{3n}{O}_m$ was mixed with $60$ mL $O_2$ and sparked .The gas mixture on cooling occupied $44$ mL. After treatment with $KOH$ solution, the volume of gas remaining was $12$ mL. Deduce the formula of compound.

• A. $C_2{H}_{6}O$
• B. $C_3{H}_{8}O$
• C. $C{H}_{4}O$
• D. None of these

Answer 1

The correct option is A $C_2{H}_{6}O$

$C_n{H}_{3n}{O}_m+[n+\frac{3n}{4}-\frac{m}{2}]O_2\to nC{O}_2+\frac{3n}{2}{H}_{2}O$
Volume before reaction $16$ $60$ $0$ $0$
Volume after reaction $0$ $60-16[n+\frac{3n}{4}-\frac{m}{2}]$ $16n$ $24n$
Volume of gases left = $60-16[\frac{7n}{4}-d\frac{m}{2}]+16n$
$44=60-12n+8m$
$12n-8m=16$ ...(i)
Volume of $O_2$ left = $60-16[\frac{7n}{4}-\frac{m}{2}]12=60-28n+8m$
$28n-8m=48$ ...(ii)
By equations (i) and (ii) , $n=2,m=1$
Therefore, the gas is $C_2{H}_{6}O$