Limiting reagent in an impure copper(II) oxide preparation
At a school senior chemistry practical, a student makes copper(II) sulfate crystals from copper(II) oxide that is contaminated with sand. The student weighs 5.00 g of the impure oxide, labelled as containing 8.0% inert sand by mass, and adds it to 40.0 mL of 1.50 mol L−1 sulfuric acid. The reaction is CuO(s) + H2SO4(aq) → CuSO4(aq) + H2O(l). After reaction, the mixture is filtered to remove sand and any unreacted solid. The filtrate is slowly evaporated and cooled, and blue crystals of CuSO4·5H2O are collected, dried and weighed. The mass obtained is 9.85 g. The teacher asks the student to decide whether the acid or the oxide was limiting, calculate the theoretical yield of hydrated crystals, and explain why the actual yield is below 100%. M(CuO) = 79.5 g mol−1; M(H2SO4) = 98.1 g mol−1; M(CuSO4·5H2O) = 249.7 g mol−1.
1. Show which reactant is limiting. Include amounts in moles and a clear comparison.
Read the worked answer
n(CuO) = m/M = (5.00 g × 0.920)/79.5 g mol−1 = 4.60/79.5 = 0.0579 mol. n(H2SO4) = cV = 1.50 mol L−1 × 0.0400 L = 0.0600 mol. The equation shows 1 mol CuO reacts with 1 mol H2SO4. The available H2SO4 (0.0600 mol) is greater than the CuO (0.0579 mol), so CuO is the limiting reagent. The sand is inert and is not included in the mole calculation; the 8.0% impurity only reduces the mass of CuO that can react.
2. Calculate the theoretical mass of CuSO4·5H2O and the percentage yield.
Read the worked answer
From the limiting CuO, the 1:1 reaction gives n(CuSO4·5H2O) = 0.0579 mol. Theoretical mass = n × M = 0.0579 mol × 249.7 g mol−1 = 14.5 g (14.44 g before rounding). Percentage yield = (actual/theoretical) × 100 = (9.85/14.44) × 100 = 68.2%.
3. The student says the sand must have lowered the yield because it used up some acid. Evaluate this statement and give two realistic reasons for the lower yield.
Read the worked answer
The statement is incorrect. Sand is an inert impurity: it does not react with sulfuric acid and does not change the moles of CuO available. It is removed by filtration before crystallisation. Two realistic reasons for the lower yield are:
- Product is lost during filtration, transfer between containers, and crystallisation because some CuSO4 remains dissolved in the mother liquor or on glassware.
- Not all crystals are recovered, or the crystals are heated too strongly during drying so water of crystallisation is lost or some decomposition occurs. Other acceptable points include soluble impurities interfering with complete crystallisation.