An impure sample of Magnesium carbonate () decomposes completely on strong heating to yield carbon d — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question
Question
An impure $20.0\text{ g}$ sample of Magnesium carbonate ($MgCO_3$) decomposes completely on strong heating to yield carbon dioxide gas and exactly $8.0\text{ g}$ of solid Magnesium oxide ($MgO$). What is the true percentage purity of the original $MgCO_3$ sample? (At. wt.: Mg = 24)
💡 Solution & Explanation
Step 1: Write the balanced decomposition equation: $MgCO_3(s) \rightarrow MgO(s) + CO_2(g)$. Step 2: Find moles of product obtained. Molar mass of $MgO = 24 + 16 = 40\text{ g/mol}$. Moles of $MgO = 8.0 / 40 = 0.2\text{ moles}$. Step 3: By stoichiometry, $1\text{ mole}$ of $MgO$ comes from exactly $1\text{ mole}$ of $MgCO_3$. Thus, moles of pure $MgCO_3$ reacted = $0.2\text{ moles}$. Step 4: Calculate mass of pure $MgCO_3$. Molar mass = $24 + 12 + 48 = 84\text{ g/mol}$. Pure mass = $0.2 \times 84 = 16.8\text{ g}$. Step 5: Percentage Purity = $(16.8 / 20.0) \times 100 = 84\%$.