Question

An organic compound containing $C,H$ and $O$ has $49.3\%$ carbon, $6.84\%$ hydrogen and its vapour density is $73$ . Molecular formula of the compound is

• A. $C_3{H}_5{O}_2$
• B. $C_4{H}_{10}{O}_2$
• C. $C_6{H}_{10}{O}_4$
• D. $C_3{H}_{10}{O}_2$

Answer 1

Answer: (C)

Element	$\%$	Relative number of atom	Simplest ratio
$C$	$49.3\%$	$\frac{49.3}{12}=4.1$	$\frac{4.1}{2.74}=1.5\times 2=3$
$H$	$6.84$	$\frac{6.84}{1}=6.84$	$\frac{6.84}{2.74}=2.5\times 2=5$
$O$	$43.86$	$\frac{43.86}{16}=2.74$	$\frac{2.74}{2.74}=1\times 2=2$

Thus, the empirical formula is $C_3{H}_5{O}_2$.
Empirical formula weight
$=3\times 12+5\times 1+2\times 16$
$=36+5+32=73$
Molecular weight of the compound $=2\times VD$
$=2\times 73=146$
$n=\frac{\text{ mol. }wt.}{\text{ empirical formula }wt.}=\frac{146}{73}=2$
Molecular formula $=$ Empirical formula $\times 2$
$=\left(C_3{H}_5{O}_2\right)\times 2=C_6{H}_{10}{O}_4$