20.0 g of magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g of magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample?

Options

A. 60
B. 84
C. 75
D. 96

Explanation

To find the percentage purity of magnesium carbonate (MgCO₃) in the sample, we can follow these steps:

### Step 1: Write the decomposition reaction
The decomposition of magnesium carbonate can be represented by the following chemical equation:
\[ \text{MgCO}_3 (s) \rightarrow \text{MgO} (s) + \text{CO}_2 (g) \]

### Step 2: Calculate the molar masses
- Molar mass of magnesium carbonate (MgCO₃):
- Mg: 24 g/mol
- C: 12 g/mol
- O: 16 g/mol × 3 = 48 g/mol
- Total: \( 24 + 12 + 48 = 84 \, \text{g/mol} \)

- Molar mass of magnesium oxide (MgO):
- Mg: 24 g/mol
- O: 16 g/mol
- Total: \( 24 + 16 = 40 \, \text{g/mol} \)

- Molar mass of carbon dioxide (CO₂):
- C: 12 g/mol
- O: 16 g/mol × 2 = 32 g/mol
- Total: \( 12 + 32 = 44 \, \text{g/mol} \)

### Step 3: Relate the masses of products to the reactant
From the balanced equation, 1 mole of MgCO₃ produces 1 mole of MgO. Therefore, we can set up a ratio based on the molar masses:
- 84 g of MgCO₃ produces 40 g of MgO.

### Step 4: Calculate the amount of MgCO₃ that produced 8 g of MgO
Using the unitary method:
\[ \text{If } 40 \, \text{g of MgO} \text{ is produced from } 84 \, \text{g of MgCO}_3, \]
\[ \text{Then } 8 \, \text{g of MgO} \text{ is produced from } \left( \frac{84}{40} \times 8 \right) \, \text{g of MgCO}_3. \]

Calculating this gives:
\[ \text{Mass of MgCO}_3 = \frac{84}{40} \times 8 = 16.8 \, \text{g} \]

### Step 5: Calculate the percentage purity of MgCO₃
The percentage purity can be calculated using the formula:
\[ \text{Percentage Purity} = \left( \frac{\text{Mass of pure MgCO}_3}{\text{Total mass of sample}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage Purity} = \left( \frac{16.8 \, \text{g}}{20.0 \, \text{g}} \right) \times 100 = 84\% \]

### Final Answer
The percentage purity of magnesium carbonate in the sample is **84%**.

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