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

Options

A. (a)`60`
B. (b)`84`
C. (c )`75`
D. (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 as:
\[ \text{MgCO}_3 \rightarrow \text{MgO} + \text{CO}_2 \]

### Step 2: Calculate the molar masses
- Molar mass of magnesium carbonate (MgCO3):
- Magnesium (Mg) = 24 g/mol
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol (3 O atoms)

\[ \text{Molar mass of MgCO}_3 = 24 + 12 + (3 \times 16) = 84 \text{ g/mol} \]

- Molar mass of magnesium oxide (MgO):
- Magnesium (Mg) = 24 g/mol
- Oxygen (O) = 16 g/mol

\[ \text{Molar mass of MgO} = 24 + 16 = 40 \text{ g/mol} \]

### Step 3: Establish the stoichiometric relationship
From the balanced reaction, we see that:
- 1 mole of MgCO3 produces 1 mole of MgO.

### Step 4: Calculate the mass of MgCO3 that produces 8 g of MgO
Using the molar masses:
- If 40 g of MgO is produced from 84 g of MgCO3, then for 8 g of MgO, we can set up a proportion:

\[ \frac{84 \text{ g MgCO}_3}{40 \text{ g MgO}} = \frac{x \text{ g MgCO}_3}{8 \text{ g MgO}} \]

Cross-multiplying gives:
\[ x = \frac{84 \times 8}{40} = 16.8 \text{ g MgCO}_3 \]

### Step 5: Calculate the percentage purity
Now, we know that 16.8 g of MgCO3 is present in the original 20 g sample. The percentage purity can be calculated as follows:

\[ \text{Percentage purity} = \left( \frac{\text{mass of pure MgCO}_3}{\text{total mass of sample}} \right) \times 100 \]

Substituting the values:
\[ \text{Percentage purity} = \left( \frac{16.8 \text{ g}}{20 \text{ g}} \right) \times 100 = 84\% \]

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

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