Calculate the percentage of (i) `SO_(4)^(2-) (ii) H_(2)O` in pure crystals of Molar salt, viz., `FeSO_(4).(NH_(4))_(2)SO_(4).6H_(2)O.`

Explanation

To calculate the percentage of \( \text{SO}_4^{2-} \) and \( \text{H}_2\text{O} \) in the molar salt \( \text{FeSO}_4 \cdot (\text{NH}_4)_2\text{SO}_4 \cdot 6\text{H}_2\text{O} \), we will follow these steps:

### Step 1: Calculate the Molecular Mass of the Compound

1. **Identify the components**:
- Iron (Fe): 1 atom
- Sulfur (S): 3 atoms (1 from \( \text{FeSO}_4 \) and 2 from \( (\text{NH}_4)_2\text{SO}_4 \))
- Oxygen (O): 4 (from \( \text{FeSO}_4 \)) + 4 (from \( (\text{NH}_4)_2\text{SO}_4 \)) + 6 (from \( 6\text{H}_2\text{O} \)) = 4 + 4 + 6 = 14 atoms
- Nitrogen (N): 2 atoms (from \( (\text{NH}_4)_2\text{SO}_4 \))
- Hydrogen (H): 12 atoms (from \( 6\text{H}_2\text{O} \))

2. **Calculate the molar mass**:
- Molar mass of \( \text{Fe} = 56 \, \text{g/mol} \)
- Molar mass of \( \text{S} = 32 \, \text{g/mol} \) (3 atoms = \( 3 \times 32 = 96 \, \text{g/mol} \))
- Molar mass of \( \text{O} = 16 \, \text{g/mol} \) (14 atoms = \( 14 \times 16 = 224 \, \text{g/mol} \))
- Molar mass of \( \text{N} = 14 \, \text{g/mol} \) (2 atoms = \( 2 \times 14 = 28 \, \text{g/mol} \))
- Molar mass of \( \text{H} = 1 \, \text{g/mol} \) (12 atoms = \( 12 \times 1 = 12 \, \text{g/mol} \))

3. **Total Molar Mass**:
\[ \text{Total Molar Mass} = 56 + 96 + 224 + 28 + 12 = 416 \, \text{g/mol} \]

### Step 2: Calculate the Molar Mass of \( \text{SO}_4^{2-} \)

1. **Molar mass of \( \text{SO}_4^{2-} \)**:
- Molar mass of \( \text{S} = 32 \, \text{g/mol} \)
- Molar mass of \( \text{O} = 16 \, \text{g/mol} \) (4 atoms = \( 4 \times 16 = 64 \, \text{g/mol} \))

2. **Total Molar Mass of \( \text{SO}_4^{2-} \)**:
\[ \text{Molar Mass of } \text{SO}_4^{2-} = 32 + 64 = 96 \, \text{g/mol} \]

### Step 3: Calculate the Percentage of \( \text{SO}_4^{2-} \)

1. **Since there are 3 \( \text{SO}_4^{2-} \) ions in the compound**:
\[ \text{Total mass of } \text{SO}_4^{2-} = 3 \times 96 = 288 \, \text{g/mol} \]

2. **Percentage of \( \text{SO}_4^{2-} \)**:
\[ \text{Percentage of } \text{SO}_4^{2-} = \left( \frac{288}{416} \right) \times 100 \approx 69.23\% \]

### Step 4: Calculate the Molar Mass of \( \text{H}_2\text{O} \)

1. **Molar mass of \( \text{H}_2\text{O} \)**:
- Molar mass of \( \text{H} = 1 \, \text{g/mol} \) (2 atoms = \( 2 \times 1 = 2 \, \text{g/mol} \))
- Molar mass of \( \text{O} = 16 \, \text{g/mol} \)

2. **Total Molar Mass of \( \text{H}_2\text{O} \)**:
\[ \text{Molar Mass of } \text{H}_2\text{O} = 2 + 16 = 18 \, \text{g/mol} \]

### Step 5: Calculate the Total Mass of Water in the Compound

1. **Total mass of water**:
\[ \text{Total mass of } 6\text{H}_2\text{O} = 6 \times 18 = 108 \, \text{g/mol} \]

### Step 6: Calculate the Percentage of \( \text{H}_2\text{O} \)

1. **Percentage of \( \text{H}_2\text{O} \)**:
\[ \text{Percentage of } \text{H}_2\text{O} = \left( \frac{108}{416} \right) \times 100 \approx 25.00\% \]

### Final Results

- Percentage of \( \text{SO}_4^{2-} \): \( \approx 69.23\% \)
- Percentage of \( \text{H}_2\text{O} \): \( \approx 25.00\% \)