To solve the problem of calculating the percentage of nitrogen in the organic compound, we will follow these steps:
### Step 1: Gather the given data
- Mass of the organic compound (m) = 0.2046 g
- Volume of moist nitrogen (V1) = 30.4 cm³
- Temperature (T1) = 288 K
- Pressure (P1) = 732.7 mm Hg
- Aqueous tension at 288 K = 12.7 mm Hg
### Step 2: Calculate the pressure of dry nitrogen (P_dry)
To find the pressure of dry nitrogen, we subtract the aqueous tension from the total pressure:
\[ P_{\text{dry}} = P_1 - \text{Aqueous tension} \]
\[ P_{\text{dry}} = 732.7 \, \text{mm Hg} - 12.7 \, \text{mm Hg} = 720.0 \, \text{mm Hg} \]
### Step 3: Convert the volume of nitrogen to liters
Convert the volume from cm³ to liters:
\[ V_1 = 30.4 \, \text{cm}^3 = 30.4 \, \text{mL} = 0.0304 \, \text{L} \]
### Step 4: Use the ideal gas law to find the volume of nitrogen at NTP (V2)
Using the formula:
\[
\frac{P_1 \cdot V_1}{T_1} = \frac{P_2 \cdot V_2}{T_2}
\]
Where:
- \( P_2 = 760 \, \text{mm Hg} \) (NTP pressure)
- \( T_2 = 273 \, \text{K} \) (NTP temperature)
Rearranging the formula to solve for \( V_2 \):
\[
V_2 = \frac{P_1 \cdot V_1 \cdot T_2}{P_2 \cdot T_1}
\]
Substituting the known values:
\[
V_2 = \frac{720.0 \, \text{mm Hg} \cdot 0.0304 \, \text{L} \cdot 273 \, \text{K}}{760 \, \text{mm Hg} \cdot 288 \, \text{K}}
\]
Calculating \( V_2 \):
\[
V_2 = \frac{720.0 \cdot 0.0304 \cdot 273}{760 \cdot 288} \approx 0.02748 \, \text{L} = 27.48 \, \text{mL}
\]
### Step 5: Calculate the number of moles of nitrogen (n)
Using the molar volume of nitrogen at NTP (22.4 L/mol):
\[
n = \frac{V_2}{22.4 \, \text{L/mol}} = \frac{0.02748 \, \text{L}}{22.4 \, \text{L/mol}} \approx 0.001225 \, \text{mol}
\]
### Step 6: Calculate the mass of nitrogen in the compound
The molar mass of nitrogen (N₂) is approximately 28 g/mol:
\[
\text{Mass of nitrogen} = n \cdot \text{Molar mass of N}_2 = 0.001225 \, \text{mol} \cdot 28 \, \text{g/mol} \approx 0.0343 \, \text{g}
\]
### Step 7: Calculate the percentage of nitrogen in the compound
\[
\text{Percentage of nitrogen} = \left( \frac{\text{Mass of nitrogen}}{\text{Mass of organic compound}} \right) \times 100
\]
\[
\text{Percentage of nitrogen} = \left( \frac{0.0343 \, \text{g}}{0.2046 \, \text{g}} \right) \times 100 \approx 16.8\%
\]
### Final Answer:
The percentage of nitrogen in the organic compound is approximately **16.8%**.
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