Question

$500mL$ of a hydrocarbon gas , burnt in excess og oxygen , yields $2500mLC{O}_2$ and 3 liters of water vapour, all volumes being mesured at the same temperature and pressure, what is the formula of the hydrocarbon?

• A. $C_4{H}_8$
• B. $C_5{H}_{12}$
• C. $C_5{H}_{1}0$
• D. $C_2{H}_6$

Answer 1

The correct option is B $C_5{H}_{12}$

Volume of hydrocarbons= $500ml$

Volume of $C{O}_2=2500ml$

Volume of water vapour= $3L=3000ml$

Let the formula of hydrocarbons= $C_x{H}_y$

$C_x{H}_y+(x+\frac{y}{4})O_2⟶xC{O}_2+\frac{y}{2}{H}_{2}O$

$500ml$ $500(x+\frac{y}{4})ml$ $500xml$ $500\times \frac{y}{2}=250yml$

$2500ml$ $3000ml$

now $500x=2500ml$

$\Rightarrow x=5$

Again $250y=3000=1y=12$

$\therefore$ Formula of hydrocarbon= $C_5{H}_{12}$