Question

$0.078\, g$ of hydrocarbon occupies $22.4\, ml$ of volume at $1 $ atm and $0^{\circ} C$. The empirical formula of the hydrocarbon is $CH$. The molecular formula is:

• A. $C _{2} H _{2}$
• B. $C _{4} H _{4}$
• C. $C _{6} H _{6}$
• D. $C _{8} H _{8}$

Answer 1

Answer: (C)

$22.4 \,mL$ of hydrocarbon weighs $0.078 \,g$
Then, $22400\, mL$ of hydrocarbon weighs $=\frac{0.078}{22.4} \times 22400 \,g$
$=78\, g =$ molar mass
Emprical mass $=$ at. mass of $C +$ at. mass of $H$
$=12+1=13 g$
$n=\frac{78}{13}=6$
$\therefore M.F =( E \cdot F )_{ n }=( CH )_{6}$
or $C _{6} H _{6}$