To solve the problem of estimating the percentage of nitrogen in the organic compound using Duma's method, we will follow these steps:
### Step 1: Calculate the pressure of the nitrogen gas (P1)
The total pressure of the gas collected is given as 725 mm Hg, and the aqueous tension at 300 K is 25 mm Hg. The pressure of the nitrogen gas can be calculated by subtracting the aqueous tension from the total pressure.
\[
P_1 = P_{\text{total}} - P_{\text{aqueous tension}} = 725 \, \text{mm Hg} - 25 \, \text{mm Hg} = 700 \, \text{mm Hg}
\]
### Step 2: Use the ideal gas equation to find the volume of nitrogen at STP (V2)
We will use the formula derived from the ideal gas law:
\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]
Where:
- \(P_1 = 700 \, \text{mm Hg}\)
- \(V_1 = 40 \, \text{mL}\)
- \(T_1 = 300 \, \text{K}\)
- \(P_2 = 760 \, \text{mm Hg}\) (standard pressure)
- \(T_2 = 273 \, \text{K}\) (standard temperature)
Rearranging the equation to solve for \(V_2\):
\[
V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}
\]
Substituting the known values:
\[
V_2 = \frac{700 \, \text{mm Hg} \times 40 \, \text{mL} \times 273 \, \text{K}}{760 \, \text{mm Hg} \times 300 \, \text{K}}
\]
Calculating \(V_2\):
\[
V_2 = \frac{700 \times 40 \times 273}{760 \times 300} = 33.52 \, \text{mL}
\]
### Step 3: Calculate the mass of nitrogen (N2) at STP
We know that 1 mole of nitrogen gas (N2) occupies 22,400 mL and has a molar mass of 28 g. We can find the mass of nitrogen in the volume we calculated:
\[
\text{Mass of N}_2 = \left(\frac{28 \, \text{g}}{22400 \, \text{mL}}\right) \times V_2
\]
Substituting \(V_2\):
\[
\text{Mass of N}_2 = \left(\frac{28}{22400}\right) \times 33.52 = 0.0419 \, \text{g}
\]
### Step 4: Calculate the percentage of nitrogen in the organic compound
Now we can find the percentage of nitrogen in the organic compound using the formula:
\[
\text{Percentage of N} = \left(\frac{\text{Mass of N}_2}{\text{Mass of organic compound}}\right) \times 100
\]
Substituting the values:
\[
\text{Percentage of N} = \left(\frac{0.0419 \, \text{g}}{0.25 \, \text{g}}\right) \times 100 = 16.76\%
\]
### Final Answer
The percentage of nitrogen in the compound is **16.76%**.
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