Question

$10$mL of a gaseous hydrocarbon was burnt completely in $80$mL of $O_2$ at NTP. The remaining gas occupied $70$mL at NTP. This volume became $50$mL on treatment with KOH solution. What is the formula of the hydrocarbon?

Answer 1

Let formula of hydrocarbon $=C_x{H}_y$

Volume of hydrocarbon taken $=10mL$

Volume of oxygen gas added $=80mL$

Combustion of hydrocarbon-

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

From the above reaction-

$1$ mole of $C_x{H}_y$ reacts with $(x+\frac{y}{4})$ mole of $O_2$ to produce $x$ mole of $C{O}_2$.

Therefore by avogadro's law-

$1$ mL of $C_x{H}_y$ reacts with $(x+\frac{y}{4})$ mL of $O_2$ to produce $x$ mL of $C{O}_2$.

$10$ mL of $C_x{H}_y$ reacts with $10(x+\frac{y}{4})$ mL of $O_2$ to produce $10x$ mL of $C{O}_2$.

Therefore,

Volume of residual gas $(C{O}_2+\text{unused}{O}_2)=70mL$

Since $KOH$ absorbs whole $C{O}_2$ and volume of gas left after $KOH$ treatment is $50$ mL

Volume of $C{O}_2$ formed $=70-50=20mL$

From the above reaction of combustion of $C_x{H}_y$-

$10x\text{mL of}C{O}_2=20\text{mL of}C{O}_2$

$x=\frac{20}{10}=2$

Now,

Volume of unused $O_2=50mL$

$\therefore$ Volume of $O_2$ used $4=80-50=30mL$

Again from the above combustion reaction-

$10(x+\frac{y}{4})=30$

$2+\frac{y}{4}=3$

$\Rightarrow y=4$

Hence the molecular formula of hydrocarbo will be $C_x{H}_y=C_2{H}_4$.