Question

The vapour density of a hydrocarbon is $30$. if on complete combustion of hydrocarbon, $8.8gC{O}_2$ and ${5.4}_g{H}_{2}O$ are obtained then molecular formula of hydrocarbon is ?

• A. $C_3{H}_8$
• B. $C_4{H}_{12}$
• C. $C_2{H}_6$
• D. $C_2{H}_4$

Answer 1

The correct option is B $C_4{H}_{12}$

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

Moles of $C{O}_2=x=\frac{8.8}{44}=0.2$ moles.

Moles of $H_{2}O=\frac{y}{2}=\frac{5.4}{18}=0.3$ moles.

$\therefore x=0.2$ and $y=0.6⟹\frac{x}{y}=\frac{1}{3}$

Empirical formula of hydrocarbon is $C{H}_3$.

Now, vapor density $=\frac{M}{2}$

$\therefore$ $M=2\times 30=60g/mol$

$\therefore$ $(x\times 12)+(y\times 1)=6012x+y=6012x+3x=60$

$\therefore 15x=60$

$\therefore x=4$ and $y=12$

Therefore the molecular formula of hydrocarbon is $C_4{H}_{12}$.