Given:

Volume of $N_2$, $V = 50$ mL

Pressure, $P = 715$ mm Hg

Temperature, $T = 300$ K

Aqueous Tension at 300 K = 15 mm Hg

Calculation:

Correct the Pressure for $N_2$:

$ P_{N_2} = 715 \text{ mmHg} - 15 \text{ mmHg} = 700 \text{ mmHg} $

Convert the pressure from mm Hg to atm:

$ P_{N_2} = \frac{700}{760} \text{ atm} $

Calculate the Moles of $N_2$ using the Ideal Gas Law:

$ n_{N_2} = \frac{P_{N_2} \cdot V}{R \cdot T} $

$ n_{N_2} = \frac{\frac{700}{760} \times \frac{50}{1000}}{0.0821 \times 300} $

Calculate the Moles of $N$:

$ n_{N} = 2 \times n_{N_2} $

Calculate the Mass of $N$:

$ \text{Mass of } N = 2 \times n_{N} \times 14 $

Determine the Percentage of Nitrogen in the Organic Compound:

$ \% N = \frac{\text{Mass of } N}{\text{Mass of organic compound}} \times 100 $

$ \% N = \frac{\frac{700}{760} \times \frac{50}{1000} \times 2 \times 14}{0.0821 \times 300} \times \frac{1000}{292} \times 100 $

$ \% N = 18\% $

The percentage of nitrogen in the organic compound is 18%.