Question

A sample of gaseous hydrocarbon occupies $0.56$L at STP. The complete combustion of the hydrocarbon produces $2.2$g of $C{O}_2$ and $0.9$g of $H_{2}O$. The molecular formula of the compound is?

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

Answer 1

The correct option is A $C_2{H}_4$

$22.4$ lit of hydrocarbon= $1$ mole

$0.5$ lit of hydrocarbon= $\frac{0.56}{22.4}=0.025$ mol

Number of moles of hydrocarbon= $0.025$ mol

Number of moles of $C{O}_2=\frac{mass}{\text{Molecular weight}}=\frac{2.2}{44}=0.05$ mol

Number of moles of $H_{2}O=\frac{0.9}{18}=0.05$ mol

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

$1$ mole of $C_x{H}_y=x$ moles of $C{O}_2$

$0.025$ moles of $C_x{H}_y=0.05x$ moles of $C{O}_2$

$\Rightarrow 0.025x=0.05$

$\Rightarrow x=2$

$0.025$ mol of $C_x{H}_x=0.05$ mol of $H_{2}O$

$1$ mol of $C_x{H}_x=2$ mol of $H_{2}O$

$=\frac{y}{2}$ mol of $H_{2}O$

$\Rightarrow \frac{y}{2}=2\Rightarrow y=4$

Thus compound is $C_2{H}_4$ .