Question

$40ml$ of hydrocarbon on combustion gave $120mlC{O}_2$ and $80ml$ water vapour. The molecular formula of hydrocarbon is:

• A. $C_2{H}_6$
• B. $C_3{H}_6$
• C. $C_3{H}_8$
• D. $C_3{H}_4$

Answer 1

Answer: (D)

$C_3{H}_4$

The balanced equation for the combustion is:

$C_x{H}_y+[x+\frac{y}{4}]O_2\to xC{O}_2+\left[\frac{y}{2}\right]H_{2}O$

40 ml of $C_x{H}_y$ gives = 120 ml of $C{O}_2$

1 vol. of $C_x{H}_y$ = 3 vol. of $C{O}_2$

$x=3$

40 ml of $C_x{H}_y$ gives = 80 ml of $H_{2}O$

1 vol. of $C_x{H}_y$ = 2 vol. of $H_{2}O$

$\therefore \frac{y}{2}$ = 2 ; $y=4$

$\therefore$ formula of $C_x{H}_y$ =$C_3{H}_4$