Q. 500 ml of a hydrocarbon gas burnt in excess of oxygen yields 2500 ml of $CO_{2}$ and 3 litres of water vapour, all the volumes being measured at the same temperature and pressure. The formula of hydrocarbon is
NTA AbhyasNTA Abhyas 2022
Solution:
Suppose the formula of hydrocarbon $=C_{x}H_{y}$
$x=\frac{v o l u m e o f C O_{2}}{V o l u m e o f h y d r o c a r b o n}=\frac{2500}{500}=5$
$\underset{\underset{500 m l}{\underset{\underset{}{\underset{\underset{}{}}{}}}{1 v o l .}}}{C_{x} H_{y}} + \left(x + \frac{y}{4}\right) O_{2} = \underset{\underset{500 x m l}{\underset{\underset{}{\underset{}{\underset{}{}}}}{x v o l .}}}{x C O_{2}} + \underset{\underset{\frac{500 y}{2} m l}{\underset{}{\frac{y}{2} v o l .}}}{\frac{y}{2} H_{2} O}$
$\frac{500 y}{2}=3000$
So y = 12
Molecular formula of hydrocarbon $=C_{5}H_{12}$ .
$x=\frac{v o l u m e o f C O_{2}}{V o l u m e o f h y d r o c a r b o n}=\frac{2500}{500}=5$
$\underset{\underset{500 m l}{\underset{\underset{}{\underset{\underset{}{}}{}}}{1 v o l .}}}{C_{x} H_{y}} + \left(x + \frac{y}{4}\right) O_{2} = \underset{\underset{500 x m l}{\underset{\underset{}{\underset{}{\underset{}{}}}}{x v o l .}}}{x C O_{2}} + \underset{\underset{\frac{500 y}{2} m l}{\underset{}{\frac{y}{2} v o l .}}}{\frac{y}{2} H_{2} O}$
$\frac{500 y}{2}=3000$
So y = 12
Molecular formula of hydrocarbon $=C_{5}H_{12}$ .