In Dumas’ method for estimation of nitrogen, 0.5 gram of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound (Aqueous tension at 300 K = 15 mm Hg) is

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Solution

Q.
In Dumas’ method for estimation of nitrogen, 0.5 gram of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound (Aqueous tension at 300 K = 15 mm Hg) is [2025]

1 1.257

2 20.87

3 18.67

4 12.57

Ans.
(4)

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

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

Volume of nitrogen $(V_{N_{2}})$ = 60 mL = 0.06 L

$n_{N_{2}} = \frac{P_{N_{2}} V_{N_{2}}}{R T} = \frac{700 \times 0.06}{760 \times 0.0821 \times 300} = 2.24 \times 10^{- 3} mol$

Mass of nitrogen = $n_{N_{2}} \times {molar mass of N}_{2}$

$= 2.24 \times 10^{- 3} \times 28 g = 0.06282 g$

Percentage of nitrogen in organic compound:

$= \frac{Mass of nitrogen}{Mass of organic compound} \times 100$

$= \frac{0.06282}{0.5} \times 100 \approx 12.57 %$

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