Determine the empirical formula of a compound having percentage composition as :
Iron = 20% , sulphur = 11.5% , oxygen = 23.1% and water molecules = 45.4% -(At. mass of Fe = 56, S = 32)

Explanation

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

### Step-by-Step Solution:

1. **Assume a Sample Size**:
Assume the total mass of the sample is 100 g. This simplifies the calculation since the percentage composition can be directly converted to grams.

2. **Calculate the Mass of Each Element**:
- Mass of Iron (Fe) = 20% of 100 g = 20 g
- Mass of Sulphur (S) = 11.5% of 100 g = 11.5 g
- Mass of Oxygen (O) = 23.1% of 100 g = 23.1 g
- Mass of Water (H₂O) = 45.4% of 100 g = 45.4 g

3. **Convert Mass to Moles**:
- Moles of Iron (Fe) = Mass / Atomic Mass = 20 g / 56 g/mol = 0.357 mol
- Moles of Sulphur (S) = Mass / Atomic Mass = 11.5 g / 32 g/mol = 0.359 mol
- Moles of Oxygen (O) = Mass / Atomic Mass = 23.1 g / 16 g/mol = 1.443 mol
- Moles of Water (H₂O) = Mass / Molar Mass = 45.4 g / 18 g/mol = 2.522 mol

4. **Determine the Smallest Number of Moles**:
Identify the smallest number of moles from the calculated values:
- Smallest number = 0.357 mol (for Fe)

5. **Calculate the Ratio of Moles**:
Divide the number of moles of each component by the smallest number of moles:
- Ratio of Fe = 0.357 / 0.357 = 1
- Ratio of S = 0.359 / 0.357 ≈ 1.005 ≈ 1
- Ratio of O = 1.443 / 0.357 ≈ 4.04 ≈ 4
- Ratio of H₂O = 2.522 / 0.357 ≈ 7.06 ≈ 7

6. **Write the Empirical Formula**:
Based on the ratios obtained:
- Fe: 1
- S: 1
- O: 4
- H₂O: 7

Therefore, the empirical formula of the compound is:
\[ \text{FeSO}_4 \cdot 7\text{H}_2\text{O} \]

### Final Answer:
The empirical formula of the compound is **FeSO₄·7H₂O**.