In Dumas’ method 292 mg of an organic compound released 50 mL of nitrogen gas () at 300 K temperature and 715 mm Hg pressure. The percentage composition of ‘N’ in the organic compound is _____ % (Nearest integer). (Aqueous tension at 300 K = 15 mm Hg)

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Solution

Q.
In Dumas’ method 292 mg of an organic compound released 50 mL of nitrogen gas ( $N_{2}$ ) at 300 K temperature and 715 mm Hg pressure. The percentage composition of ‘N’ in the organic compound is _____ % (Nearest integer).

(Aqueous tension at 300 K = 15 mm Hg) [2025]

Ans.
(18)

Pressure of nitrogen $(P_{N_{2}})$ = Total pressure – aqueous tension

$= (715 - 15) mmHg = \frac{700}{760} atm$

Volume of nitrogen $(V_{N_{2}}) = 50 mL = 0.05 L$

Moles of nitrogen $(n_{N_{2}}) = \frac{P_{N_{2}} V_{N_{2}}}{R T} = \frac{\frac{700}{760} \times 0.05}{0.0821 \times 300} mol$

Mass of nitrogen $(W_{N_{2}}) = moles of nitrogen \times molar mass of nitrogen$

$= \frac{\frac{700}{760} \times 0.05}{0.0821 \times 300} \times 28 g$

Mass of organic compound $(W_{o . c .}) = 292 mg = 0.292 g$

Percentage of nitrogen in organic compound $= \frac{W_{N_{2}}}{W_{o . c .}} \times 100$

$= \frac{\frac{700}{760} \times 0.05 \times 28}{0.0821 \times 300 \times 0.292} \times 100 = 17.92 % \approx 18 %$

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