4 grams of copper chloride on analysis were found to contain 1.890 g of copper (Cu) and 2.110 g of chlorine (CI). What is the empirical formula of copper chloride?

Answer

To find the empirical formula of copper chloride from the given data, we will follow these steps:

### Step 1: Calculate the percentage composition of copper and chlorine in the copper chloride sample.

1. **Percentage of Copper (Cu)**:
\[ \text{Percentage of Cu} = \left( \frac{\text{mass of Cu}}{\text{mass of copper chloride}} \right) \times 100 = \left( \frac{1.890 \, \text{g}}{4 \, \text{g}} \right) \times 100 = 47.25\% \]

2. **Percentage of Chlorine (Cl)**:
\[ \text{Percentage of Cl} = \left( \frac{\text{mass of Cl}}{\text{mass of copper chloride}} \right) \times 100 = \left( \frac{2.110 \, \text{g}}{4 \, \text{g}} \right) \times 100 = 52.75\% \]

### Step 2: Convert the percentages to grams for a 100 g sample.

- For 100 g of copper chloride:
- Mass of Cu = 47.25 g
- Mass of Cl = 52.75 g

### Step 3: Calculate the number of moles of copper and chlorine.

1. **Moles of Copper (Cu)**:
\[ \text{Moles of Cu} = \frac{\text{mass of Cu}}{\text{molar mass of Cu}} = \frac{47.25 \, \text{g}}{63.5 \, \text{g/mol}} \approx 0.7437 \, \text{mol} \]

2. **Moles of Chlorine (Cl)**:
\[ \text{Moles of Cl} = \frac{\text{mass of Cl}}{\text{molar mass of Cl}} = \frac{52.75 \, \text{g}}{35.5 \, \text{g/mol}} \approx 1.4859 \, \text{mol} \]

### Step 4: Determine the simplest mole ratio.

- The ratio of moles of Cu to moles of Cl:
\[ \text{Ratio} = \frac{0.7437 \, \text{mol}}{0.7437} : \frac{1.4859 \, \text{mol}}{0.7437} \approx 1 : 2 \]

### Step 5: Write the empirical formula.

- Since the ratio of Cu to Cl is 1:2, the empirical formula of copper chloride is:
\[ \text{Empirical Formula} = \text{CuCl}_2 \]

### Final Answer:
The empirical formula of copper chloride is **CuCl₂**.

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