Question

One mole of hydrocarbon X is subjected to combustion. The product obtained is condensed and the resulting gaseous product occupied a volume of 89.6 $l$ at STP. Oxygen required for this combustion is 145.6 $l$. at STP. What should be the molecular formula of X?

• A. $C_2{H}_6$
• B. $C_4{H}_{10}$
• C. $C_3{H}_8$
• D. $C{H}_4$

Answer 1

Answer: (B)

$C_4{H}_{10}$

1 mole of any gas occupies 22.4 l

Volume of the product = $\frac{89.6}{22.4}=4mol$

Vol of oxygen is = $\frac{145.6}{22.4}$ = 6.5 mol $O_2$

From this value frame the hydrocarbon equation.

$C_x{H}_y+O_2\to xC{O}_2+y{H}_{2}O$

$C_4{H}_{10}+\frac{13}{2}{O}_2\to 4C{O}_2+5H_{2}O$

The hydrocarbon is $C_4{H}_{10}$ .

Hence, option B is correct.