Question

In Dumas' method for estimation of nitrogen 1 g of an organic compound gave 150 mL of nitrogen collected at 300 K 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)

Given ...

Answer 1

20

Answer 2

1. Calculate the pressure of dry nitrogen ($P_1$) by subtracting the aqueous tension from the total pressure: $P_1 = 900 \text{ mm Hg} - 15 \text{ mm Hg} = 885 \text{ mm Hg}$.

2. Convert the volume of nitrogen collected at the given conditions ($V_1 = 150 \text{ mL}$, $T_1 = 300 \text{ K}$, $P_1 = 885 \text{ mm Hg}$) to STP conditions ($P_2 = 760 \text{ mm Hg}$, $T_2 = 273 \text{ K}$) using the combined gas law: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$.

   $V_2 = \frac{P_1V_1T_2}{P_2T_1} = \frac{885 \text{ mm Hg} \times 150 \text{ mL} \times 273 \text{ K}}{760 \text{ mm Hg} \times 300 \text{ K}} = \frac{36230250}{228000} \text{ mL} \approx 158.9046 \text{ mL}$.

3. Calculate the mass of nitrogen corresponding to this volume at STP. At STP, 22400 mL of $N_2$ weighs 28 g.

   Mass of $N_2 = \frac{\text{Volume of } N_2 \text{ at STP}}{\text{Molar volume at STP}} \times \text{Molar mass of } N_2 = \frac{158.9046 \text{ mL}}{22400 \text{ mL/mol}} \times 28 \text{ g/mol}$.

   Mass of $N_2 = \frac{158.9046}{800} \text{ g} \approx 0.19863 \text{ g}$.

4. Calculate the percentage composition of nitrogen in the organic compound:

   Percentage of $N = \frac{\text{Mass of } N_2}{\text{Mass of organic compound}} \times 100$.

   Percentage of $N = \frac{0.19863 \text{ g}}{1 \text{ g}} \times 100 = 19.863 \%$.

5. Round the percentage to the nearest integer. The nearest integer to 19.863 % is 20 %.