Copper(II) sulfate preparation: who is limiting, and where did the yield go?
An upper-sixth student prepares hydrated copper(II) sulfate from copper(II) oxide and dilute sulfuric acid. The student weighs 2.00 g of copper(II) oxide (Mr = 79.5) into a beaker, adds 25.0 cm3 of 1.50 mol dm-3 sulfuric acid, warms the mixture and stirs. A little black solid remains, so the mixture is filtered. The filtrate is evaporated to half volume, cooled in ice, and the crystals are filtered under suction, rinsed quickly with a little cold water and left to dry in air. The product, CuSO4.5H2O (Mr = 249.7), weighs 3.61 g. In the write-up the student states that copper(II) oxide was the limiting reactant and that the yield was about 90 per cent, and blames the rest on 'experimental error'. The teacher wants the limiting reactant confirmed by calculation, the true yield worked out, and the residual black solid accounted for.
1. Show by calculation which reactant is in excess, calculate the theoretical and percentage yield, and comment on the student's claim of 'about 90 per cent'.
Read the worked answer
n(CuO) = 2.00 / 79.5 = 2.52 x 10^-2 mol. n(H2SO4) = 0.0250 dm3 x 1.50 mol dm-3 = 3.75 x 10^-2 mol.
The equation is CuO + H2SO4 -> CuSO4 + H2O, a 1:1 ratio, so the acid is in excess by (3.75 - 2.52) x 10^-2 = 1.23 x 10^-2 mol. The student is right that CuO is limiting.
Theoretical n(CuSO4.5H2O) = 2.52 x 10^-2 mol, so theoretical mass = 2.52 x 10^-2 x 249.7 = 6.28 g.
Percentage yield = 3.61 / 6.28 x 100 = 57.5 per cent, not 90 per cent.
A likely source of the student's figure is using the anhydrous molar mass: 2.52 x 10^-2 x 159.6 = 4.02 g, and 3.61 / 4.02 = 89.8 per cent, which rounds to 90. The product is a pentahydrate, so the water of crystallisation must be included in the theoretical mass; comparing a hydrated mass with an anhydrous prediction inflates the yield.
2. The student saw 'a little black solid' left after warming. Reconcile this with your answer, and give two other plausible causes of the low yield together with a test that would distinguish between them.
Read the worked answer
If CuO really is limiting, none of it should survive, so the black solid signals that the reaction did not go to completion: the acid may be too dilute locally, the mixture may not have been warmed or stirred long enough, or each CuO particle may become coated with blue product, blocking further attack. A less likely possibility is that the black solid is an impurity in the oxide (carbon, or Cu2O) that never dissolves.
Other causes of a 57.5 per cent yield:
- Product lost in solution. Copper(II) sulfate is very soluble (about 30 g per 100 g of cold water), so rinsing the crystals with cold water, and discarding the mother liquor, removes real product.
- Mechanical losses and dehydration. Crystals lost in transfer or on the filter paper, or the hydrate partly dehydrated on a hot bench, both reduce the mass.
Test: add a few drops of the black solid to warm dilute sulfuric acid. If it dissolves to a blue solution it is unreacted CuO, confirming incomplete reaction; if it remains unchanged it is an inert impurity. Test for dissolved copper by adding excess aqueous ammonia to the filtrate and washings: a deep blue solution confirms Cu2+ and shows how much product was thrown away.
3. Calculate the mass of copper(II) oxide needed to make 5.00 g of CuSO4.5H2O if this method typically delivers only 60 per cent of the theoretical yield.
Read the worked answer
Work backwards from the required mass.
n(product required) = 5.00 / 249.7 = 2.00 x 10^-2 mol.
Because only 60 per cent of the theoretical amount is obtained, the theoretical amount must be 2.00 x 10^-2 / 0.60 = 3.34 x 10^-2 mol.
From the 1:1 equation, n(CuO) needed = 3.34 x 10^-2 mol, so mass = 3.34 x 10^-2 x 79.5 = 2.65 g.
The acid must also be sufficient: 3.34 x 10^-2 mol of H2SO4 is needed, which is 22.3 cm3 of 1.50 mol dm-3, so about 25 cm3 with a small excess is appropriate. This is a planning calculation, not a licence to use a 100 per cent yield assumption: the whole point is that the yield factor must be applied before the stoichiometry is used.