Copper(II) Sulfate Crystals: Yield and Purity Decisions
A candidate prepares hydrated copper(II) sulfate crystals from copper(II) carbonate and dilute sulfuric acid. 6.20 g of dry copper(II) carbonate powder is added in small portions to 40.0 cm³ of 2.00 mol/dm³ sulfuric acid, stirred until no more solid dissolves, and the mixture is filtered. The blue filtrate is evaporated to about half its volume, cooled in ice, and the crystals are filtered off, washed with a little ice-cold distilled water and pressed dry. The candidate obtains 8.90 g of blue crystals. A second candidate, using the same mass of carbonate and the same volume and concentration of acid, does not filter the mixture and evaporates it to dryness over a strong flame, obtaining 13.40 g of a pale blue-green solid. Mr: CuCO3 = 123.5; CuSO4·5H2O = 249.5; H2SO4 = 98.0.
1. Show by calculation which reactant is in excess, and calculate the maximum mass of hydrated copper(II) sulfate that could be obtained.
Read the worked answer
n(CuCO3) = 6.20 / 123.5 = 0.0502 mol. n(H2SO4) = c × V = 2.00 × 0.0400 = 0.0800 mol.
The equation is CuCO3 + H2SO4 → CuSO4 + CO2 + H2O, a 1:1 ratio, so 0.0502 mol of acid is needed and 0.0800 mol is present. Acid is in excess by 0.0298 mol; copper(II) carbonate is the limiting reactant.
Maximum n(CuSO4·5H2O) = 0.0502 mol, so m = 0.0502 × 249.5 = 12.5 g.
This also tells us that adding more acid cannot raise the yield above 12.5 g; any shortfall must come from handling losses or unreacted solid, not from lack of acid.
2. Calculate the percentage yield obtained by the first candidate and explain two reasons why it is below 100%, suggesting one improvement that does not involve changing the amounts of reactant.
Read the worked answer
Percentage yield = (8.90 / 12.5) × 100 = 71.1%.
Reasons for the shortfall:
- Copper(II) sulfate is soluble, so a significant fraction remains in the mother liquor. Cooling in ice reduces this loss but does not remove it.
- Washing with cold water removes some product along with the impurities.
- Small amounts are lost in transferring and filtering, and some solution is lost by spitting during evaporation.
Improvement without changing reactants: evaporate only until crystals just begin to form, cool thoroughly (ice or ice/salt), filter through the same filter paper, wash with the minimum volume of ice-cold distilled water, then recrystallise from the minimum volume of hot water. Recovering a second crop from the mother liquor would also raise the mass obtained, although it would be less pure.
3. The second candidate's 13.40 g is greater than the calculated maximum. Explain how this is possible and describe a test that would show the solid is not pure hydrated copper(II) sulfate.
Read the worked answer
13.40 g is greater than the theoretical maximum of 12.5 g, so the solid cannot be pure CuSO4·5H2O; the extra mass must come from another substance. The most likely impurity is unreacted copper(II) carbonate, because this candidate did not filter the mixture: any solid left after the acid was used up stays in the basin. Copper(II) carbonate is green, which fits the pale blue-green colour, whereas the pure hydrate is blue. Some excess sulfuric acid may also remain, since it is not volatile and is not removed simply by evaporating.
Test: add a few drops of dilute hydrochloric acid to a small sample. Effervescence, with the gas turning limewater milky, shows carbonate; a pure sample dissolves without effervescence. Heating a weighed sample gently would also show the water loss expected for the hydrate (5 × 18 / 249.5 = 36% of the mass). Strong heating is unsuitable for that check because it can also decompose the sulfate to black copper(II) oxide.