0.45 gm of an organic compound gave on combustion 0.792 gm of CO 2 and 0.324 gm of water. 0.24 gm of the same substance was Kjeldahlised and the ammonia liberated was absorbed in 50.0 ml of M/8 H 2 SO

Question

0.45 gm of an organic compound gave on combustion 0.792 gm of CO 2 and 0.324 gm of water. 0.24 gm of the same substance was Kjeldahlised and the ammonia liberated was absorbed in 50.0 ml of M/8 H 2 SO 4 . The excess acid required 77.0 ml of N/10 NaOH for complete neutralisation. Calculate the empirical formula of the compound.

TestHub · Accepted Answer

Calculation of the percentage composition :

i) Percentage of carbon

$= \frac{12}{44} \times \frac{{Mass of CO}_{2} Produced}{Mass of substance taken} \times 100$

$= \frac{12}{44} \times \frac{0.792}{0.45} \times 100 = 48.0$

ii) Percentage of hydrogen

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

$= \frac{2}{18} \times \frac{0.324}{0.45} \times 100 = 8.0$

iii) Percentage of nitrogen

mEq. of H₂SO₄ $= 50 \times \frac{M}{8} \times 2 = 12.5$

mEq. of NaOH $= 72 \times \frac{1}{10} \times 1 = 7.7$

Excess of H₂SO₄used to neutralise ammonia

= 12.5 - 7.7 = 4.8

Percentage of N₂ $= \frac{1.4 \times 4.8}{0.24} = 28.0 %$

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	$\frac{48.0}{12} = 4$	$\frac{4}{1} = 4$	4
Hydrogen	8.0	1	$\frac{8.0}{1} = 8$	$\frac{8.0}{1} = 8$	8
Nitrogen	28.0	14	$\frac{28.0}{14} = 2$	$\frac{2}{1} = 2$	2
Oxygen	16.0	16	$\frac{16.0}{16} = 1$	$\frac{1}{1} = 1$	1

Hence, the empirical formula of the compound is C₄H₈N₂O.

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