In Dumas method for estimation of nitrogen, 0.50 g of an organic...

Given data:
Mass of organic compound ($w$) = $0.50$ g
Volume of nitrogen gas ($V_1$) = $70$ mL
Temperature ($T_1$) = $300$ K
Total pressure = $715$ mm Hg
Aqueous tension = $15$ mm Hg
Pressure of dry nitrogen ($P_1$) = $715 - 15 = 700$ mm Hg
Convert the volume of nitrogen to STP conditions ($P_0 = 760$ mm Hg, $T_0 = 273$ K):
Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_0 V_0}{T_0}$
$V_0 = \frac{P_1 V_1 T_0}{P_0 T_1} = \frac{700 \times 70 \times 273}{760 \times 300} \approx 58.68$ mL
At STP, $22400$ mL of $N_2$ weighs $28$ g.
Mass of $N_2$ collected = $\frac{28 \times 58.68}{22400}$ g
Percentage of nitrogen = $\frac{\text{Mass of } N_2}{\text{Mass of compound}} \times 100$
$\% N = \frac{28 \times 58.68}{22400 \times 0.50} \times 100$
$\% N = \frac{28 \times 58.68}{112} = \frac{58.68}{4} = 14.67 \%$
Rounding to the nearest integer, we get $15$.