The empirical formula of the compound which has the following percentage composition.
Carbon=24.27% hydrogen =407% and chlorine=71.65% and molar mass 98.96 g, is found to be

Options

A. CHCl
B. `CH_2Cl`
C. `CHCl_2`
D. `C_2HCl`

Answer

To determine the empirical formula of the compound based on the given percentage composition, we will follow these steps:

### Step 1: Convert Percentage to Grams
Given the percentage composition:
- Carbon (C) = 24.27%
- Hydrogen (H) = 4.07%
- Chlorine (Cl) = 71.65%

Assuming we have 100 grams of the compound, we can directly convert the percentages to grams:
- Carbon = 24.27 g
- Hydrogen = 4.07 g
- Chlorine = 71.65 g

### Step 2: Calculate Moles of Each Element
Next, we will calculate the number of moles of each element using their atomic masses:
- Atomic mass of Carbon (C) = 12 g/mol
- Atomic mass of Hydrogen (H) = 1 g/mol
- Atomic mass of Chlorine (Cl) = 35.5 g/mol

Calculating the moles:
- Moles of Carbon = \( \frac{24.27 \, \text{g}}{12 \, \text{g/mol}} \approx 2.02 \, \text{mol} \)
- Moles of Hydrogen = \( \frac{4.07 \, \text{g}}{1 \, \text{g/mol}} = 4.07 \, \text{mol} \)
- Moles of Chlorine = \( \frac{71.65 \, \text{g}}{35.5 \, \text{g/mol}} \approx 2.02 \, \text{mol} \)

### Step 3: Divide by the Lowest Number of Moles
Now, we will divide the number of moles of each element by the smallest number of moles calculated:
- The smallest number of moles is 2.02 (for both Carbon and Chlorine).

Calculating the ratios:
- Ratio of Carbon = \( \frac{2.02}{2.02} = 1 \)
- Ratio of Hydrogen = \( \frac{4.07}{2.02} \approx 2 \)
- Ratio of Chlorine = \( \frac{2.02}{2.02} = 1 \)

### Step 4: Write the Empirical Formula
From the ratios calculated, we can write the empirical formula:
- The empirical formula is \( \text{C}_1\text{H}_2\text{Cl}_1 \) or simply \( \text{CH}_2\text{Cl} \).

### Conclusion
The empirical formula of the compound is \( \text{CH}_2\text{Cl} \).

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