(i) Calculate the mass percentage of various elements present in magnesium sulphate, `MgSO_(4)` (ii) Calculate the percentage of cation in ammonium dichromate.

Explanation

Let's solve the problem step by step.

### Part (i): Calculate the mass percentage of various elements present in magnesium sulfate, \( \text{MgSO}_4 \).

1. **Determine the molar mass of \( \text{MgSO}_4 \)**:
- Magnesium (Mg): 1 atom × 24 g/mol = 24 g/mol
- Sulfur (S): 1 atom × 32 g/mol = 32 g/mol
- Oxygen (O): 4 atoms × 16 g/mol = 64 g/mol
- Total molar mass of \( \text{MgSO}_4 \) = 24 + 32 + 64 = 120 g/mol

2. **Calculate the mass percentage of Magnesium (Mg)**:
\[ \text{Mass \% of Mg} = \left( \frac{\text{Mass of Mg}}{\text{Molar mass of } \text{MgSO}_4} \right) \times 100 = \left( \frac{24}{120} \right) \times 100 = 20\% \]

3. **Calculate the mass percentage of Sulfur (S)**:
\[ \text{Mass \% of S} = \left( \frac{\text{Mass of S}}{\text{Molar mass of } \text{MgSO}_4} \right) \times 100 = \left( \frac{32}{120} \right) \times 100 \approx 26.67\% \]

4. **Calculate the mass percentage of Oxygen (O)**:
\[ \text{Mass \% of O} = \left( \frac{\text{Mass of O}}{\text{Molar mass of } \text{MgSO}_4} \right) \times 100 = \left( \frac{64}{120} \right) \times 100 \approx 53.33\% \]

### Summary of Part (i):
- Mass % of Mg = 20%
- Mass % of S = 26.67%
- Mass % of O = 53.33%

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### Part (ii): Calculate the percentage of cation in ammonium dichromate, \( \text{(NH}_4)_2\text{Cr}_2\text{O}_7 \).

1. **Determine the molar mass of ammonium dichromate**:
- Ammonium (NH₄): 2 × (1 × 14 + 4 × 1) = 2 × (14 + 4) = 2 × 18 = 36 g/mol
- Chromium (Cr): 2 atoms × 52 g/mol = 104 g/mol
- Oxygen (O): 7 atoms × 16 g/mol = 112 g/mol
- Total molar mass of \( \text{(NH}_4)_2\text{Cr}_2\text{O}_7 \) = 36 + 104 + 112 = 252 g/mol

2. **Calculate the mass of cations (from \( \text{NH}_4^+ \))**:
- Total mass of cations = Mass of 2 \( \text{NH}_4^+ \) = 36 g/mol

3. **Calculate the percentage of cation**:
\[ \text{Percentage of cation} = \left( \frac{\text{Mass of cation}}{\text{Molar mass of } \text{(NH}_4)_2\text{Cr}_2\text{O}_7} \right) \times 100 = \left( \frac{36}{252} \right) \times 100 \approx 14.29\% \]

### Summary of Part (ii):
- Percentage of cation in ammonium dichromate = 14.29%

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