Calculation of the percentage composition :

i)  Percentage of carbon

                        $=\frac{12}{44}\times \frac{{\text{Mass of CO}}_2\text{Produced}}{\text{Mass of substance taken}}\times 100$

                        $=\frac{12}{44}\times \frac{0\text{.}792}{0\text{.}45}\times 100=48\text{.}0$

ii)  Percentage of hydrogen

                           $=\frac{2}{18}\times \frac{{\text{Mass of H}}_2\text{O produced}}{\text{Mass of substance taken}}\times 100$

                          $=\frac{2}{18}\times \frac{0\text{.}324}{0\text{.}45}\times 100=8\text{.}0$

iii)  Percentage of nitrogen

mEq. of H₂SO₄ $=50\times \frac{\text{M}}{8}\times 2=12\text{.}5$

mEq. of NaOH $=72\times \frac{1}{10}\times 1=7\text{.}7$

Excess of H₂SO₄ used to neutralise ammonia

                       = 12.5 - 7.7 = 4.8

Percentage of N₂ $=\frac{1\text{.}4\times 4\text{.}8}{0\text{.}24}=28\text{.}0\text{\%}$

iv)  Percentage of oxygen = 100 - (percentage of C + percentage of H + percentage of N) = 100 - (48.0 + 8.0 + 28.0) = 16.0.

b.  Calculation of empirical formula :

 

Element	Percentage	Atomic mass number of atoms	Relative number of atoms	Simlpest atomic ratio	Simplest whole number atomic ratio
Carbon	48.8	12	$\frac{48\text{.}0}{12}=4$	$\frac{4}{1}=4$	4
Hydrogen	8.0	1	$\frac{8\text{.}0}{1}=8$	$\frac{8\text{.}0}{1}=8$	8
Nitrogen	28.0	14	$\frac{28\text{.}0}{14}=2$	$\frac{2}{1}=2$	2
Oxygen	16.0	16	$\frac{16\text{.}0}{16}=1$	$\frac{1}{1}=1$	1

Hence, the empirical formula of the compound is C₄H₈N₂O.