In Dumas method for estimation of nitrogen, 0.50 g of an organic compound gave 70 mL of nitrogen collected at 300 K and 715 mm pressure. The percentage of nitrogen in the organic compound is __ %.
(Aqueous tension at 300 K is 15 mm).

We need to find the percentage of nitrogen in an organic compound using Dumas method.

Given mass of compound = 0.50 g, volume of $$N_2$$ collected = 70 mL, temperature = 300 K, pressure = 715 mm Hg, aqueous tension at 300 K = 15 mm Hg.

The actual pressure of dry nitrogen is obtained by subtracting the aqueous tension from the total pressure: $$P_{N_2} = P_{total} - P_{water} = 715 - 15 = 700 \text{ mm Hg}$$.

To convert the collected volume to STP conditions, we use the relation $$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$ with $$P_2 = 760$$ mm Hg and $$T_2 = 273$$ K. Thus, $$V_{STP} = \frac{700 \times 70 \times 273}{760 \times 300} = \frac{13377000}{228000} = 58.67 \text{ mL}$$.

At STP, 22400 mL of $$N_2$$ has a mass of 28 g, so the mass of the collected nitrogen is $$\text{Mass of } N_2 = \frac{28 \times 58.67}{22400} = \frac{1642.76}{22400} = 0.07333 \text{ g}$$.

The percentage of nitrogen in the compound is then $$\% N = \frac{0.07333}{0.50} \times 100 = 14.67\%$$.

Rounding to the nearest integer: $$\% N \approx$$ 15%.