`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: List the masses of each component
We have the following masses from the question:
- Mass of Iron (Fe) = 0.2014 g
- Mass of Sulfur (S) = 0.1153 g
- Mass of Oxygen (O) = 0.2301 g
- Mass of Water (H₂O) = 0.4532 g

### Step 2: Calculate the number of moles of each component
To find the number of moles, we use the formula:
\[ \text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \]

- Moles of Iron (Fe):
\[ \text{Moles of Fe} = \frac{0.2014 \, \text{g}}{56 \, \text{g/mol}} = 0.0036 \, \text{mol} \]

- Moles of Sulfur (S):
\[ \text{Moles of S} = \frac{0.1153 \, \text{g}}{32 \, \text{g/mol}} = 0.0036 \, \text{mol} \]

- Moles of Oxygen (O):
\[ \text{Moles of O} = \frac{0.2301 \, \text{g}}{16 \, \text{g/mol}} = 0.0144 \, \text{mol} \]

- Moles of Water (H₂O):
\[ \text{Moles of H₂O} = \frac{0.4532 \, \text{g}}{18 \, \text{g/mol}} = 0.0252 \, \text{mol} \]

### Step 3: Determine the smallest number of moles
The smallest number of moles calculated is 0.0036 (for both Fe and S).

### Step 4: Calculate the ratio of moles
Now we divide each of the moles by the smallest number of moles (0.0036):

- Ratio for Iron (Fe):
\[ \frac{0.0036}{0.0036} = 1 \]

- Ratio for Sulfur (S):
\[ \frac{0.0036}{0.0036} = 1 \]

- Ratio for Oxygen (O):
\[ \frac{0.0144}{0.0036} \approx 4 \]

- Ratio for Water (H₂O):
\[ \frac{0.0252}{0.0036} \approx 7 \]

### Step 5: Write the empirical formula
From the ratios calculated, we can write the empirical formula:
- Iron (Fe) = 1
- Sulfur (S) = 1
- Oxygen (O) = 4
- Water (H₂O) = 7

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