Final answer:
The percentage purity of magnesium carbonate in the sample, based on the mass of magnesium oxide formed upon decomposition, is calculated to be 84%. The correct answer is option (b) 84
Step-by-step explanation:
The percentage purity of magnesium carbonate in the sample can be calculated using the mass of the products formed when the compound is decomposed. The balanced chemical equation for the decomposition of magnesium carbonate (MgCO₃) is:
MgCO₃ → MgO + CO₂
From the question, 20.0 g of magnesium carbonate yields 8.0 g of magnesium oxide (MgO), meaning the rest of the mass is from the evolved carbon dioxide (CO₂). The molar mass of MgCO₃ is approximately 84 g/mol, and that of MgO is approximately 40 g/mol.
Using stoichiometry, we can find out the theoretical yield of MgO from pure magnesium carbonate:
(8.0 g / 40 g/mol) = 0.20 moles of MgO
For magnesium carbonate, the number of moles of MgCO₃ that would produce 0.20 moles of MgO is also 0.20 moles (from the balanced equation's 1:1 ratio).
0.20 moles * 84 g/mol = 16.8 g of MgCO₃
Therefore, the percentage purity of the magnesium carbonate is found by comparing the theoretical mass of pure MgCO₃ to the original sample mass:
(16.8 g / 20.0 g) * 100% = 84%
Hence, the correct answer is (b) 84