0.2046 g of an organic compound gave 30.4 mL of moist nitrogen measured at 288 K and 732.7 mm pressure. Calculate the percentage of nitrogen in the substance (Aqueous tension at 288 K is 12.7 mm).

Answer

To solve the problem step by step, we will follow the outlined procedure:

### Step 1: Identify the Given Data
- Mass of the organic compound (m) = 0.2046 g
- Volume of moist nitrogen (V1) = 30.4 mL
- Pressure of moist nitrogen (P1) = 732.7 mmHg
- Aqueous tension at 288 K = 12.7 mmHg
- Temperature (T1) = 288 K

### Step 2: Calculate the Effective Pressure of Nitrogen
The effective pressure of nitrogen (P_net) can be calculated by subtracting the aqueous tension from the total pressure:
\[ P_{net} = P_1 - P_{aqueous} \]
\[ P_{net} = 732.7 \, \text{mmHg} - 12.7 \, \text{mmHg} = 720.0 \, \text{mmHg} \]

### Step 3: Use the Ideal Gas Law to Find the Volume of Nitrogen at NTP
Using the ideal gas equation \( PV = nRT \), we can relate the conditions of the gas. We need to convert the volume of nitrogen to standard conditions (NTP: 0°C or 273 K, 1 atm or 760 mmHg).

Using the formula:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Where:
- \( P_2 = 760 \, \text{mmHg} \) (atmospheric pressure)
- \( T_2 = 273 \, \text{K} \)

Rearranging for \( V_2 \):
\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]
Substituting the values:
\[ V_2 = \frac{720.0 \, \text{mmHg} \times 30.4 \, \text{mL} \times 273 \, \text{K}}{760 \, \text{mmHg} \times 288 \, \text{K}} \]
Calculating \( V_2 \):
\[ V_2 \approx 27.3 \, \text{mL} \]

### Step 4: Calculate the Moles of Nitrogen
The molar volume of a gas at NTP is 22400 mL/mol. Therefore, the number of moles of nitrogen (n) can be calculated as:
\[ n = \frac{V_2}{22400} \]
Substituting the value of \( V_2 \):
\[ n = \frac{27.3 \, \text{mL}}{22400 \, \text{mL/mol}} \approx 0.00122 \, \text{mol} \]

### Step 5: Calculate the Mass of Nitrogen
The molar mass of nitrogen (N2) is approximately 28 g/mol. Therefore, the mass of nitrogen (m_N2) can be calculated as:
\[ m_{N2} = n \times \text{Molar mass of N2} \]
\[ m_{N2} = 0.00122 \, \text{mol} \times 28 \, \text{g/mol} \approx 0.03416 \, \text{g} \]

### Step 6: Calculate the Percentage of Nitrogen in the Organic Compound
The percentage of nitrogen in the organic compound can be calculated using the formula:
\[ \text{Percentage of N2} = \left( \frac{m_{N2}}{m} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage of N2} = \left( \frac{0.03416 \, \text{g}}{0.2046 \, \text{g}} \right) \times 100 \approx 16.68\% \]

### Final Answer
The percentage of nitrogen in the organic compound is approximately **16.68%**.

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