Question

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

• A. 60
• B. 84
• C. 75
• D. 96

Answer 1

The Correct Option is B

To determine the percentage purity of magnesium carbonate in the sample, we need to follow these steps:

1. Use the chemical reaction equation for the decomposition of magnesium carbonate (\text{MgCO}_3):

   \text{MgCO}_3 \rightarrow \text{MgO} + \text{CO}_2

   This equation tells us that one mole of magnesium carbonate decomposes to produce one mole of magnesium oxide and one mole of carbon dioxide.
2. Calculate the molar masses of the compounds involved:
   • Molar mass of \text{MgCO}_3 (Magnesium Carbonate): 24.31 \,g/mol + 12.01 \,g/mol + (3 \times 16.00) \,g/mol = 84.32 \,g/mol
   • Molar mass of \text{MgO} (Magnesium Oxide): 24.31 \,g/mol + 16.00 \,g/mol = 40.31 \,g/mol
3. Determine how much magnesium carbonate is needed to produce the given amount of magnesium oxide in the sample:
   • According to the stoichiometry of the reaction, 84.32 \,g of \text{MgCO}_3 gives 40.31 \,g of \text{MgO}.
   • From the mass proportion: \frac{84.32 \,g}{40.31 \,g} = \frac{x \,g}{8 \,g} where x is the actual mass of \text{MgCO}_3 in the sample that would give 8.0 g \text{MgO}:
   • Solving for x: x = \frac{84.32 \times 8.0}{40.31} \approx 16.72 \,g
   Hence, 16.72 g of pure \text{MgCO}_3 would produce 8.0 g of \text{MgO}.
4. Calculate the percentage purity of the magnesium carbonate in the sample:
   • Start with the original sample weight: 20.0 \,g
   • The percentage purity is given by: \text{Percentage Purity} = \left(\frac{\text{Mass of Pure } \text{MgCO}_3}{\text{Total mass of sample}}\right) \times 100 = \left(\frac{16.72}{20.0}\right) \times 100 \approx 83.6\%, which can be approximated to 84% when rounded.

Thus, the percentage purity of magnesium carbonate in the sample is 84%.