Question

$0.30g$ of an organic compound containing $C,H$ and $O$ on combustion yielded $0.44gC{O}_2$ and $0.18g{H}_{2}O$. If $1$ mole of compound weighs $60g$, then molecular formula of the compound is

• A. $C{H}_{2}O$
• B. $C_3{H}_{8}O$
• C. $C_4{H}_{6}O$
• D. $C_2{H}_4{O}_2$

Answer 1

Answer: (D)

$\%$ of $C$

$=\frac{12}{44}\times \frac{\text{ Weight of }C{O}_2\text{ formed }}{\text{ Weight of organic compound }}\times 100$

$=\frac{12}{44}\times \frac{0.44}{0.30}\times 100=40$

$\%$ of $H=\frac{2}{18}\times \frac{\text{ Weight of }{H}_{2}O\text{ formed }}{\text{ Weight of organic compound }}\times 100$

$=\frac{2}{18}\times \frac{0.18}{0.30}\times 100=6.66$

$\%$ of $O=100-(\%$ of $C+\%$ of $H)$

$=100-(40+6.66)=53.34$

For empirical formula

Element	Atomic mass	Relative number of moles	Simplest ratio
C	12	$\frac{40}{12}=3.33$	1
H	1	$\frac{6.66}{1}=6.66$	2
O	16	$\frac{53.34}{16}=3.33$	1

So, empirical formula (EF) is $C{H}_{2}O$.

Molecular formula (MF) $=n\times EF$

$n=\frac{\text{ MF mass }}{\text{ EF mass }}=\frac{60}{30}=2$

So, MF is ${\left(C{H}_{2}O\right)}_2=C_2{H}_4{O}_2$