Question

Question 3.3.2 [5 marks]
A compound consists of 24.8% C, 2.0% H, and 73.1% Cℓ. Calculate its empirical
formula.

Answer 1

To determine the empirical formula of a compound based on its percentage composition, follow these steps:

1. **Convert the percentages to grams**: Assume you have 100 grams of the compound. This means you have:
- Carbon (C): 24.8 grams
- Hydrogen (H): 2.0 grams
- Chlorine (Cl): 73.1 grams

2. **Convert grams to moles**: Use the molar masses of each element to convert grams to moles.

- Molar mass of Carbon (C) = 12.01 g/mol
- Molar mass of Hydrogen (H) = 1.008 g/mol
- Molar mass of Chlorine (Cl) = 35.45 g/mol

Now calculate moles for each element:
\[
\text{Moles of C} = \frac{24.8 \, \text{g}}{12.01 \, \text{g/mol}} \approx 2.07 \, \text{mol}
\]
\[
\text{Moles of H} = \frac{2.0 \, \text{g}}{1.008 \, \text{g/mol}} \approx 1.98 \, \text{mol}
\]
\[
\text{Moles of Cl} = \frac{73.1 \, \text{g}}{35.45 \, \text{g/mol}} \approx 2.06 \, \text{mol}
\]

3. **Divide by the smallest number of moles**: Find the smallest number of moles calculated:

- Smallest moles is approximately 1.98 (for H).

Now divide the moles of each element by the smallest number of moles:
\[
\text{Ratio of C} = \frac{2.07}{1.98} \approx 1.04
\]
\[
\text{Ratio of H} = \frac{1.98}{1.98} = 1
\]
\[
\text{Ratio of Cl} = \frac{2.06}{1.98} \approx 1.04
\]

4. **Get the simplest whole number ratio**: The ratios are approximately:
- C: 1.04
- H: 1
- Cl: 1.04

Round to the nearest whole numbers:
- C: 1
- H: 1
- Cl: 1

5. **Write the empirical formula**: The empirical formula is \( C_1H_1Cl_1 \) or simply \( CHCl \).

Thus, the empirical formula of the compound is **CHCl**.