In Dumas' method for estimation of nitrogen 1g of an organic compound gave 150 mL of nitrogen collected at 300K temperature and 900 mm Hg pressure. The percentage composition of nitrogen in the compound is ________ % (nearest integer). (Aqueous tension at 300 K = 15 mm Hg)

In Dumas’ method the nitrogen gas is collected over water, so the measured pressure is the total (nitrogen + water-vapour) pressure. We must first obtain the partial pressure of dry nitrogen.

Total pressure of the gas mixture, $$P_{\text{total}} = 900 \text{ mm Hg}$$
Aqueous tension at $$300 \text{ K},\; P_{\text{H}_2O} = 15 \text{ mm Hg}$$
Therefore, pressure of dry $$N_2$$ is
$$P_{N_2} = P_{\text{total}} - P_{\text{H}_2O} = 900 - 15 = 885 \text{ mm Hg}$$

Convert this pressure into atmospheres for use in the ideal-gas equation:
$$P_{N_2} = \frac{885}{760} \text{ atm} = 1.164 \text{ atm (approx)}$$

Volume of nitrogen collected,
$$V = 150 \text{ mL} = 0.150 \text{ L}$$

Temperature,
$$T = 300 \text{ K}$$

Ideal-gas constant (in compatible units),
$$R = 0.0821 \text{ L atm mol}^{-1}\text{ K}^{-1}$$

Using the ideal-gas equation $$PV = nRT$$ to find the moles of $$N_2$$:
$$n = \frac{PV}{RT} = \frac{(1.164)\,(0.150)}{(0.0821)\,(300)}$$

Calculate $$n$$:
Numerator: $$1.164 \times 0.150 = 0.1746$$
Denominator: $$0.0821 \times 300 = 24.63$$
$$n = \frac{0.1746}{24.63} = 0.00709 \text{ mol}$$

Mass of nitrogen present (each mole of $$N_2$$ weighs $$28 \text{ g}$$):
$$m_{N} = n \times 28 = 0.00709 \times 28 = 0.1985 \text{ g}$$

The organic sample weighed $$1.00 \text{ g}$$, so the percentage of nitrogen is
$$\%N = \frac{0.1985}{1.00} \times 100 = 19.85\%$$

To the nearest integer, the percentage composition of nitrogen in the compound is **20 %**.