Question

$40$ ml of a hydrocarbon undergoes combustion in $260$ ml of oxygen and gives $160$ ml of carbon dioxide. If all gases are measured under similar conditions of temperature and pressure, the formula of hydrocarbon is:

• A. $C_3{H}_8$
• B. $C_4{H}_8$
• C. $C_6{H}_{14}$
• D. $C_4{H}_{10}$

Answer 1

Answer: (D)

$C_4{H}_{10}$

The combustion reaction can be represented as:

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

$40ml$ $260ml$ $160ml$

At constant temperature and pressure (at STP), volume is proportionla to moles.

Hence, the combustion of $1$ mole of hydrocarbon will require $6.5$ moles $(\frac{260}{40}=6.5)$ of oxygen and will produce $4$ moles $(\frac{160}{40}=4)$ of carbon dioxide.

i,e $x=4$
Thus, the molecular formula contains $4$ carbon atoms.

$x+\frac{y}{4}=6.5$

$\therefore y=10$

Thus, the molecular formula contains $10$ hydrogen atoms.

Out of $6.5$ moles of oxygen, $4$ moles will combine with carbon to form $4$ moles of carbon dioxide.
The remaining $2.5$ moles of oxygen will combine with $10$ moles of hydrogen atoms (from the hydrocarbon) to form $5$ moles of water.
The hydrocarbon is $C_4{H}_{10}$ and the combustion reaction is $C_4{H}_{10}+6.5O_2\to 4C{O}_2+5H_{2}O$.

Hence, option $D$ is correct.