Preparing Copper(II) Sulfate Crystals: Excess Reactant and Yield
A student is preparing hydrated copper(II) sulfate crystals. They measure 50.0 cm³ of 1.00 mol/dm³ sulfuric acid into a beaker, warm it to about 60 °C, and add 5.00 g of copper(II) oxide powder in small portions with stirring. After the addition, a black solid remains at the bottom. The mixture is filtered while hot, and the blue filtrate is evaporated to about half its volume and left to crystallise. The student is disappointed: the crop of blue crystals weighs only 7.10 g, although their teacher predicted a higher mass. The student asks whether they should have added exactly the calculated mass of copper(II) oxide instead of an excess, and whether the leftover black solid affected the result. (Mᵣ: CuO = 79.5; CuSO₄·5H₂O = 249.5)
1. Show by calculation whether copper(II) oxide is in excess, and state the limiting reactant.
Read the worked answer
n(H₂SO₄) = c × V = 1.00 × 0.0500 = 0.0500 mol. n(CuO) = m / Mᵣ = 5.00 / 79.5 = 0.0629 mol. The equation is CuO + H₂SO₄ → CuSO₄ + H₂O, so 0.0500 mol of acid needs only 0.0500 mol of CuO. Since 0.0629 mol CuO is available, CuO is in excess and sulfuric acid is the limiting reactant. The leftover black solid is unreacted copper(II) oxide.
2. Explain why the student deliberately used an excess of copper(II) oxide and why the mixture is filtered before crystallisation.
Read the worked answer
An excess of copper(II) oxide ensures that all the sulfuric acid is neutralised. If exactly stoichiometric amounts were used, small weighing errors or incomplete reaction could leave acid in the filtrate, contaminating the crystals. Copper(II) oxide is insoluble, so the excess can be removed easily by filtration. Filtration also removes the black solid before evaporation, so it cannot contaminate the blue crystals. The excess solid does not increase the theoretical yield because the yield is fixed by the amount of limiting acid.
3. Calculate the maximum theoretical mass of CuSO₄·5H₂O and the percentage yield.
Read the worked answer
From the equation, n(CuSO₄) = n(H₂SO₄) = 0.0500 mol. The hydrated salt also contains 0.0500 mol of CuSO₄ units. m(CuSO₄·5H₂O) = n × Mᵣ = 0.0500 × 249.5 = 12.475 g ≈ 12.5 g. Percentage yield = (actual mass / theoretical mass) × 100 = (7.10 / 12.475) × 100 = 56.9% ≈ 57%. Losses are expected because some product remains in the mother liquor, some solution is retained during filtration, and some crystals are lost during transfer.