Copper Sulfate Crystals That Went Wrong
A student prepares copper(II) sulfate crystals. She measures 25 cm³ of 1.0 mol/dm³ sulfuric acid into a beaker, warms it gently, and adds 4.0 g of black copper(II) oxide a spatula at a time, stirring. The black solid disappears at first; she keeps adding until some black solid remains at the bottom. She filters the mixture, collects the blue filtrate, and evaporates it in an evaporating basin over a roaring Bunsen flame until a dry solid is left. The product is a pale greenish-white powder with a mass of 2.6 g, not blue crystals. She is told her maximum possible yield of hydrated copper(II) sulfate was much higher.
(Ar: Cu = 63.5, O = 16, H = 1, S = 32)
1. Explain why the student added copper(II) oxide until some remained, and use a mole calculation to show that the acid was the limiting reactant.
Read the worked answer
Adding excess copper(II) oxide ensures all the acid is used up, so the filtrate contains only copper(II) sulfate and water; any unreacted black oxide is removed by filtration. This is how a pure soluble salt is made from an insoluble base.
Calculation: moles of H₂SO₄ = concentration × volume = 1.0 mol/dm³ × 0.025 dm³ = 0.025 mol. The equation is CuO + H₂SO₄ → CuSO₄ + H₂O, a 1:1 ratio, so 0.025 mol of CuO is needed. Mr of CuO = 63.5 + 16 = 79.5, so mass needed = 0.025 × 79.5 = 1.99 g.
The student added 4.0 g, which is about twice the required amount, so copper(II) oxide is in excess and sulfuric acid is limiting. The maximum amount of copper(II) sulfate is therefore fixed by the 0.025 mol of acid.
2. Explain why the product was a greenish-white powder rather than blue crystals, and describe how the method should be changed.
Read the worked answer
Blue crystals are hydrated copper(II) sulfate, CuSO₄·5H₂O. Heating strongly over a roaring flame drives off the five molecules of water of crystallisation, leaving white or pale grey-green anhydrous copper(II) sulfate. Any impurity or partial decomposition can give the greenish tinge.
The method should be changed to:
- Filter off the excess black copper(II) oxide while the solution is warm.
- Heat the filtrate gently in an evaporating basin until the solution is saturated - stop when crystals just begin to form at the edge, or when about half the water has gone.
- Remove from the heat and leave to cool slowly, so large blue crystals form.
- Filter off the crystals, press between filter papers and dry them in air or a warm oven, not over a flame.
The key judgement is that evaporation should stop before dryness; strong heating destroys the desired product.
3. Calculate the maximum mass of hydrated copper(II) sulfate the student could have obtained, and comment on her 2.6 g yield.
Read the worked answer
Maximum moles of CuSO₄ = moles of acid = 0.025 mol (1:1 ratio from CuO + H₂SO₄). Mr of CuSO₄·5H₂O = 63.5 + 32 + (4 × 16) + (5 × 18) = 249.5. Maximum mass = 0.025 × 249.5 = 6.24 g, so about 6.2 g.
Her 2.6 g is only about 42% of this. Part of the loss is normal: some solution stays on the filter paper and in the beaker, and some crystals are lost during filtration. But the main issue is that the product is not the pentahydrate, so the comparison is unfair - 0.025 mol of anhydrous CuSO₄ would be 0.025 × 159.5 = 3.99 g, and 2.6 g is about 65% of that. A fair comparison would need the product dried gently and weighed as the hydrate, or the calculation done for the anhydrous salt.