Question

$15\text{mL}$ of a gaseous hydrocarbon was required for complete combustion in $357.15\text{mL}$ of air ($21\%$ of oxygen by volume) and the gaseous products occupied $327\text{mL}$ (all volumes being measured at NTP). What is the formula of the hydrocarbon?

• A. $C_3{H}_8$
• B. $C_4{H}_8$
• C. $C_5{H}_{10}$
• D. $C_4{H}_{10}$

Answer 1

The correct option is A $C_3{H}_8$

Let the formula of hydrocarbon be $C_x{H}_y$
$C_x{H}_y+(x+\frac{y}{4})O_2\to xC{O}_2+\frac{y}{2}{H}_{2}O\text{According to the question}\text{Volume of}{O}_2=21\%\text{of volume of air}\Rightarrow \frac{21}{100}\times 357.15=75\text{mL}(x+\frac{y}{4})\times 15=75\Rightarrow x+\frac{y}{4}=\frac{75}{15}\Rightarrow x+\frac{y}{4}=\mathrm{5...}\left(i\right)$
Now,
$\text{Volume of product}=327\text{mL}(\text{at NTP)}$
$\text{Volume of air (remained ) + Volume of}C{O}_2=327\text{mL}$
$15x+282=327\Rightarrow x=3$
Put this in equation (i),
$\Rightarrow y=8$
Formula $=C_3{H}_8$