A dibasic acid containing {C}, {H} and {O} was found to contain {C}=26.7\% and {H}=2.2\%. The vapour density of diethyl ester was found to be 73 . What is the molecular formula of the acid?

Percentage of elements present	Elements	Convert $\%$ to grams	Divide mass by molar masses to find moles	Divide all moles with smallest number of moles
$26.7\%$	C	$0.267\,\mathrm{g}$	$\frac{0.267}{16} = 0.022$	$\frac{0.022}{0.022} = 1$
$2.2\%$	H	$0.022\,\mathrm{g}$	$\frac{0.022}{1} = 0.022$	$\frac{0.022}{0.022} = 1$
$71.7\%$	O	$0.71\,\mathrm{g}$	$\frac{0.71}{16} = 0.044$	$\frac{0.044}{0.022} = 2$

∴

\mathrm{C}:\mathrm{H}:\mathrm{O}=1:1:2

Hence, empirical formula is

\mathrm{CHO}_2

Molecular mass of diethyl ether is

(\mathrm{C}_2\mathrm{H}_5 \mathrm{OC}_2\mathrm{H}_5)

2 \times 73 = 146\,\mathrm{u}

∴ Formula mass of ethyl group

=29\,\mathrm{u}

In diethyl ether two ethyl group must have replaced the

\mathrm{H}

-atom of dibasic acid.

∴ The molecular mass of acid

=

Molecular mass of diethyl ether

-(2 \times ,)+

mass of ethyl group

)+

mass of hydrogen

=149-(2 \times 29)+2=90\,\mathrm{u}

∴ Molecular formula of acid

=\frac{\text{Molecular mass of acid}}{\text{Empirical mass formula}} \times \mathrm{CHO}_2

= \frac{90}{(12+1+16 \times 2)} \times \mathrm{CHO}_2

= \frac{90}{45} \times \mathrm{CHO}_2 = 2 \times \mathrm{CHO}_2

∴ Molecular formula of acid

= \mathrm{C}_2\mathrm{H}_2\mathrm{O}_4

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