1. Given data

• Mass of organic compound $= 292\,\text{mg} = 0.292\,\text{g}$
• Volume of $\mathrm{N_2}$ gas $= 50\,\text{mL} = 0.050\,\text{L}$
• Temperature $= 300\,\text{K}$
• Total pressure $= 715\,\text{mm Hg}$
• Aqueous tension at $300\,\text{K} = 15\,\text{mm Hg}$

So, pressure of dry nitrogen gas is $P_{\mathrm{N_2}} = 715 - 15 = 700\,\text{mm Hg}$

Convert into atm: $P = \frac{700}{760}\,\text{atm}$

2. Use ideal gas equation to find moles of nitrogen gas

$n = \frac{PV}{RT}$

Substituting, $n = \frac{\left(\frac{700}{760}\right)(0.050)}{(0.0821)(300)}$

$n \approx \frac{0.04605}{24.63} \approx 1.87 \times 10^{-3}\,\text{mol of } \mathrm{N_2}$

3. Mass of nitrogen present

Each mole of $\mathrm{N_2}$ has mass $28\,\text{g}$.

$m_{\mathrm{N}} = n \times 28$

$m_{\mathrm{N}} = 1.87 \times 10^{-3} \times 28$

$m_{\mathrm{N}} \approx 0.0524\,\text{g} = 52.4\,\text{mg}$

4. Percentage of nitrogen in compound

$\%N = \frac{\text{mass of nitrogen}}{\text{mass of compound}} \times 100$

$\%N = \frac{0.0524}{0.292} \times 100$

$\%N \approx 17.95\%$

Nearest integer: $\boxed{18}$

5. Comparison with stored answer

Stored correct answer = $18$

Our derived answer also is $18$, so it agrees.