Deildrin, an insecticide, contains `C,H,Cl and O.` Combustion of `29.72 mg` of dieldrin gave `41.21 mg CO_2 and 5.63 mg` of `H_2O` . In a separate analysis `25.31mg` of dieldein was converted into `57.13 mg` Ag Cl. What is the empirical formula of diedrin ?

Options

A. `C_(6)H_(4)Cl_(3)O`
B. `C_(8)H_(8)ClO`
C. `C_(12)H_(8)Cl_(6)O`
D. `C_(6)H_(4)Cl_(3)O_(2)`

Answer

To determine the empirical formula of Dieldrin, we will follow these steps:

### Step 1: Calculate the moles of Carbon from CO₂
Given the mass of CO₂ produced is 41.21 mg.

1. Convert mg to grams:
\[ 41.21 \, \text{mg} = 0.04121 \, \text{g} \]

2. Calculate the number of moles of CO₂:
\[ \text{Moles of CO₂} = \frac{\text{mass of CO₂}}{\text{molar mass of CO₂}} = \frac{0.04121 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.000936 \, \text{mol} \]

3. Since each mole of CO₂ contains 1 mole of Carbon:
\[ \text{Moles of C} = 0.000936 \, \text{mol} \]

### Step 2: Calculate the moles of Hydrogen from H₂O
Given the mass of H₂O produced is 5.63 mg.

1. Convert mg to grams:
\[ 5.63 \, \text{mg} = 0.00563 \, \text{g} \]

2. Calculate the number of moles of H₂O:
\[ \text{Moles of H₂O} = \frac{0.00563 \, \text{g}}{18.02 \, \text{g/mol}} \approx 0.000312 \, \text{mol} \]

3. Each mole of H₂O contains 2 moles of Hydrogen:
\[ \text{Moles of H} = 2 \times 0.000312 \approx 0.000624 \, \text{mol} \]

### Step 3: Calculate the moles of Chlorine from AgCl
Given the mass of AgCl produced is 57.13 mg.

1. Convert mg to grams:
\[ 57.13 \, \text{mg} = 0.05713 \, \text{g} \]

2. Calculate the number of moles of AgCl:
\[ \text{Moles of AgCl} = \frac{0.05713 \, \text{g}}{143.32 \, \text{g/mol}} \approx 0.000398 \, \text{mol} \]

3. Each mole of AgCl contains 1 mole of Chlorine:
\[ \text{Moles of Cl} = 0.000398 \, \text{mol} \]

### Step 4: Calculate the mass of each element
1. Mass of Carbon:
\[ \text{Mass of C} = \text{Moles of C} \times \text{molar mass of C} = 0.000936 \times 12.01 \approx 0.01123 \, \text{g} \]

2. Mass of Hydrogen:
\[ \text{Mass of H} = \text{Moles of H} \times \text{molar mass of H} = 0.000624 \times 1.008 \approx 0.000629 \, \text{g} \]

3. Mass of Chlorine:
\[ \text{Mass of Cl} = \text{Moles of Cl} \times \text{molar mass of Cl} = 0.000398 \times 35.45 \approx 0.01412 \, \text{g} \]

4. Total mass of Dieldrin is 29.72 mg (0.02972 g). To find the mass of Oxygen:
\[ \text{Mass of O} = 0.02972 - (\text{Mass of C} + \text{Mass of H} + \text{Mass of Cl}) \approx 0.02972 - (0.01123 + 0.000629 + 0.01412) \approx 0.00374 \, \text{g} \]

### Step 5: Calculate the moles of Oxygen
1. Calculate moles of Oxygen:
\[ \text{Moles of O} = \frac{0.00374 \, \text{g}}{16.00 \, \text{g/mol}} \approx 0.000234 \, \text{mol} \]

### Step 6: Find the ratio of moles
1. Normalize the moles by dividing each by the smallest number of moles calculated (0.000234):
- C:
\[ \frac{0.000936}{0.000234} \approx 4 \]
- H:
\[ \frac{0.000624}{0.000234} \approx 2.67 \approx 3 \]
- Cl:
\[ \frac{0.000398}{0.000234} \approx 1.70 \approx 2 \]
- O:
\[ \frac{0.000234}{0.000234} = 1 \]

### Step 7: Write the empirical formula
The empirical formula is:
\[ \text{C}_4\text{H}_3\text{Cl}_2\text{O} \]