0.45 gm of an organic compound gave on combustion 0.792 gm of CO_ 2...

Calculation of the percentage composition :

i)& Percentage of carbon

$= \dfrac{1 2}{4 4} × \dfrac{{Mass of CO}_{2} \text{Produced}}{\text{Mass of substance taken}} × 1 0 0$

$= \dfrac{1 2}{4 4} × \dfrac{0 \text{.} 7 9 2}{0 \text{.} 4 5} × 1 0 0 = 4 8 \text{.} 0$

ii) Percentage of hydrogen

$= \dfrac{2}{1 8} × \dfrac{{Mass of H}_{2} \text{O produced}}{\text{Mass of substance taken}} × 1 0 0$

$= \dfrac{2}{1 8} × \dfrac{0 \text{.} 3 2 4}{0 \text{.} 4 5} × 1 0 0 = 8 \text{.} 0$

iii) Percentage of nitrogen

mEq. of H₂SO₄ $= 5 0 × \dfrac{\text{M}}{8} × 2 = 1 2 \text{.} 5$

mEq. of NaOH $= 7 2 × \dfrac{1}{1 0} × 1 = 7 \text{.} 7$

Excess of H₂SO₄ used to neutralise ammonia

= 12.5 - 7.7 = 4.8

Percentage of N₂ = \dfrac{1 \text{.} 4 × 4 \text{.} 8}{0 \text{.} 2 4} = 2 8 \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	$\dfrac{4 8 . 0}{1 2} = 4$	$\dfrac{4}{1} = 4$	4
Hydrogen	8.0	1	$\dfrac{8 . 0}{1} = 8$	$\dfrac{8 . 0}{1} = 8$	8
Nitrogen	28.0	14	$\dfrac{2 8 . 0}{1 4} = 2$	$\dfrac{2}{1} = 2$	2
Oxygen	16.0	16	$\dfrac{1 6 . 0}{1 6} = 1$	$\dfrac{1}{1} = 1$	1

Hence, the empirical formula of the compound is C₄H₈N₂O.