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A powdered rock is a mixture of calcium carbonate and magnesium carbonate only…

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A powdered rock is a mixture of calcium carbonate and magnesium carbonate only. A student heats $16.2\ \text{g}$ of it strongly in an open crucible until the mass stops changing. Both carbonates decompose, and carbon dioxide escapes: $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ and $\text{MgCO}_3 \rightarrow \text{MgO} + \text{CO}_2$ . No carbonate is left in the residue. The white residue of the two oxides has a mass of $8.72\ \text{g}$ . A friend says that the composition of the rock cannot be found from these two masses alone. Show that it can, using the law of conservation of mass and the equations. Atomic masses in $\text{g mol}^{-1}$ : C 12, O 16, Mg 24, Ca 40. Take the molar volume of a gas at STP as $22.7\ \text{L mol}^{-1}$ .

What mass of carbon dioxide is given off?
What mass of calcium carbonate is present in the mixture?
What is the mass per cent of magnesium carbonate in the mixture?
What volume of carbon dioxide, measured at STP, is given off?
The residue is then dissolved completely in $2.00\ \text{M}$ hydrochloric acid: $\text{CaO} + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O}$ , and in the same way $\text{MgO}$ gives $\text{MgCl}_2$ . What is the smallest volume of the acid that dissolves the whole residue?

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. Mixture $16.2\ \text{g}$ , residue $8.72\ \text{g}$ . Molar masses: $\text{CaCO}_3$ $100\ \text{g mol}^{-1}$ , $\text{MgCO}_3$ $84\ \text{g mol}^{-1}$ , $\text{CO}_2$ $44\ \text{g mol}^{-1}$ .

Step 1. Carbon dioxide. Mass is conserved, so $16.2\ \text{g} - 8.72\ \text{g} = 7.48\ \text{g}$ escaped. That is $\dfrac{7.48\ \text{g}}{44\ \text{g mol}^{-1}} = 0.17\ \text{mol}$ .

Step 2. Two unknowns, two facts. Let $x$ mol be $\text{CaCO}_3$ . Then $\text{MgCO}_3$ is $(0.17 - x)$ mol. The mass of the mixture gives $100x + 84(0.17 - x) = 16.2$ . So $16x = 1.92$ , and $x = 0.12\ \text{mol}$ .

Step 3. The masses. $\text{CaCO}_3$ : $0.12\ \text{mol} \times 100\ \text{g mol}^{-1} = 12.0\ \text{g}$ . $\text{MgCO}_3$ : $0.05\ \text{mol} \times 84\ \text{g mol}^{-1} = 4.2\ \text{g}$ , that is $\dfrac{4.2\ \text{g}}{16.2\ \text{g}} \times 100 = 25.9\%$ .

Step 4. Volume of carbon dioxide. $0.17\ \text{mol} \times 22.7\ \text{L mol}^{-1} = 3.86\ \text{L}$ .

Step 5. Acid for the residue. Each mole of either oxide needs $2$ mol of $\text{HCl}$ . The oxides total $0.17\ \text{mol}$ , so $0.34\ \text{mol}$ of $\text{HCl}$ is needed. Volume: $\dfrac{0.34\ \text{mol}}{2.00\ \text{mol L}^{-1}} = 170\ \text{mL}$ .

Answer: (a) $7.48\ \text{g}$ . (b) $12.0\ \text{g}$ . (c) $25.9\%$ . (d) $3.86\ \text{L}$ . (e) $170\ \text{mL}$ .

Check. Rebuild the residue: $0.12\ \text{mol}$ of CaO is $6.72\ \text{g}$ , and $0.05\ \text{mol}$ of MgO is $2.00\ \text{g}$ . The sum is $8.72\ \text{g}$ , the mass weighed.

Watch out. The acid in (e) depends only on the total moles of oxide, not on how they split. You did not need the split for (e) at all.

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