Limiting Reactant and Yield in Copper(II) Sulfate Crystallisation
A student is making hydrated copper(II) sulfate crystals by reacting copper(II) oxide with dilute sulfuric acid. She adds 2.00 g of copper(II) oxide to 50.0 cm3 of 1.00 mol/dm3 sulfuric acid, warms and stirs the mixture, then filters off the excess solid. She evaporates the filtrate slowly and obtains 4.40 g of blue crystals. The equation is CuO + H2SO4 → CuSO4 + H2O. The hydrated crystals are CuSO4·5H2O. She claims the yield is about 50% because she started with 2.00 g of copper oxide and got only 4.40 g of crystals. A classmate says the acid is the limiting reactant and that adding more acid would increase the yield. The teacher asks for a mole-based check and a comment on the classmate's claim. Molar masses: CuO = 79.5 g/mol; H2SO4 = 98.0 g/mol; CuSO4·5H2O = 249.5 g/mol.
1. Determine which reactant is limiting and calculate the theoretical yield of hydrated copper(II) sulfate crystals.
Read the worked answer
Calculate the amount of each reactant:
- Copper(II) oxide: n = m / M = 2.00 / 79.5 = 0.02516 mol
- Sulfuric acid: n = c × V = 1.00 × (50.0 / 1000) = 0.0500 mol
The equation shows a 1:1 mole ratio, so 0.02516 mol of CuO needs only 0.02516 mol of H2SO4. There is 0.0500 mol of acid available, so the acid is in excess and CuO is limiting.
The amount of hydrated crystals expected is therefore also 0.02516 mol, because each mole of CuO gives one mole of CuSO4 and hence one mole of CuSO4·5H2O.
Theoretical mass = n × M = 0.02516 × 249.5 = 6.28 g.
2. Calculate the percentage yield and comment on the student's claim that the yield is about 50%.
Read the worked answer
Percentage yield = (actual mass / theoretical mass) × 100
= (4.40 / 6.28) × 100 = 70.1%, approximately 70%.
The student's 50% figure is not a valid percentage yield. She compared the mass of product obtained with the mass of copper(II) oxide started with. Those are masses of different substances and have no direct 1:1 relationship. Percentage yield must compare the actual mass of product with the theoretical mass of that same product, calculated from the limiting reactant. A yield below 100% is expected because some product is lost during filtration, transfer and crystallisation.
3. Evaluate the classmate's claim that adding more acid would increase the yield.
Read the worked answer
The claim is incorrect. The acid is already in excess: 0.0500 mol of H2SO4 is available but only 0.02516 mol is needed to react with all the copper(II) oxide. Adding more acid cannot increase the theoretical yield because the amount of copper(II) oxide is what limits the reaction. The extra acid would remain unused and could make crystallisation more difficult.
To increase the yield, the student would need to ensure that all the copper(II) oxide reacts, for example by warming and stirring thoroughly, and then reduce transfer losses during filtration and crystallisation. Adding more CuO would not help either while some solid remains unreacted, because the acid would then become limiting. The 70% yield is most likely caused by practical losses, not by insufficient acid.