`1.00 g` of a hydrated salt contains `0.2014 g` of iron, `0.1153 g` fo sulphur, `0.2301 g` of oxygen and `0.4532 g` of water of crystallisation. Find its empirical formula. `(Fe = 56, S = 32, O = 16)`

Explanation

To find the empirical formula of the hydrated salt, we will follow these steps:

### Step 1: Calculate the number of moles of each component.
We will use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

1. **Iron (Fe)**:
- Mass = 0.2014 g
- Molar mass = 56 g/mol
\[ \text{Moles of Fe} = \frac{0.2014}{56} = 0.00359 \text{ moles} \]

2. **Sulfur (S)**:
- Mass = 0.1153 g
- Molar mass = 32 g/mol
\[ \text{Moles of S} = \frac{0.1153}{32} = 0.00360 \text{ moles} \]

3. **Oxygen (O)**:
- Mass = 0.2301 g
- Molar mass = 16 g/mol
\[ \text{Moles of O} = \frac{0.2301}{16} = 0.01438 \text{ moles} \]

4. **Water (H₂O)**:
- Mass = 0.4532 g
- Molar mass = 18 g/mol
\[ \text{Moles of H₂O} = \frac{0.4532}{18} = 0.02518 \text{ moles} \]

### Step 2: Determine the simplest ratio of the moles.
Now, we will divide each of the calculated moles by the smallest number of moles obtained.

- Smallest number of moles = 0.00359 (for Fe)

1. **For Fe**:
\[ \frac{0.00359}{0.00359} = 1 \]

2. **For S**:
\[ \frac{0.00360}{0.00359} \approx 1 \]

3. **For O**:
\[ \frac{0.01438}{0.00359} \approx 4 \]

4. **For H₂O**:
\[ \frac{0.02518}{0.00359} \approx 7 \]

### Step 3: Write the empirical formula.
From the ratios calculated, we can write the empirical formula:
- Fe: 1
- S: 1
- O: 4
- H₂O: 7

Thus, the empirical formula of the hydrated salt is:
\[ \text{Fe}_1\text{S}_1\text{O}_4\cdot 7\text{H}_2\text{O} \quad \text{or simply} \quad \text{FeSO}_4\cdot 7\text{H}_2\text{O} \]