Question

Combustion of $0.277gm$ of an organic compound gave $0.66g$ Carbon dioxide and $0.337gm$ Water vapour density of the compound is equal to $37$ Calculate its molecular formula .

Answer 1

Let the molecular formula be $C_x{H}_y$

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

Moles of $C{O}_2=\frac{0.66}{44}=0.015$

Moles of $H_{2}O=\frac{0.337}{18}=0.0187$

Molecular weight of the compound $=2\times$ vapor density$=74g$

Moles of $C_x{H}_y=\frac{0.277}{74}=0.00374$

$\therefore$ $0.00374$ mol $C_x{H}_y$ gives $0.015$ mol $C{O}_2$.

$\therefore 1$ mol $C_x{H}_y$ gives $x$ mol of $C{O}_2$.

$\therefore x=\frac{0.015}{0.00374}=4$

$0.00374$ mol $C_x{H}_y$ gives $0.0187$ mol $H_{2}O$

$\therefore$ $1$ mol of $C_x{H}_y$ gives $\frac{y}{2}$ mol $H_{2}O$.

$\therefore \frac{y}{2}=\frac{0.0187}{0.000374}=5⟹y=10$

$\therefore$ Molecular formula is $C_4{H}_{10}$