Limiting reagent and crystal yield in copper(II) sulfate preparation
A technician must prepare copper(II) sulfate-5-water crystals from copper(II) oxide and 1.50 mol dm−3 sulfuric acid. The stock bottle contains 25.0 cm3 of acid, and 2.00 g of copper(II) oxide is available (Mr = 79.5). The planned method is to warm the acid, add the oxide in small portions until no further reaction occurs, filter off excess solid, concentrate the filtrate, and crystallise. After drying, the student obtains 4.10 g of blue crystals (Mr = 249.5). The technician suspects the acid, not the oxide, was limiting, and that some product was lost during filtration and crystallisation. Safety data require avoiding excess acid because the final crystals must not be acidic. Decide whether the preparation as described can use all the oxide, calculate the maximum mass of hydrated crystals, and judge whether 4.10 g is a credible yield.
1. Determine which reagent is limiting in the planned preparation and calculate the maximum mass of CuSO4·5H2O that could be obtained from 2.00 g of CuO and 25.0 cm3 of 1.50 mol dm−3 H2SO4.
Read the worked answer
- Amount of CuO: n = m/Mr = 2.00/79.5 = 2.52 × 10−2 mol.
- Amount of H2SO4: n = cV = 1.50 × (25.0/1000) = 3.75 × 10−2 mol.
- Equation: CuO + H2SO4 → CuSO4 + H2O, so the reacting ratio is 1:1.
- The CuO needs 2.52 × 10−2 mol of acid, but 3.75 × 10−2 mol is present. Therefore CuO is limiting and acid is in excess by about 1.23 × 10−2 mol.
- Maximum amount of hydrated product is therefore limited by CuO: 2.52 × 10−2 mol.
- Mr(CuSO4·5H2O) = 63.5 + 32.1 + (4 × 16.0) + (5 × 18.0) = 249.6 g mol−1.
- Maximum mass = 2.52 × 10−2 × 249.6 = 6.29 g, approximately 6.28 g using Mr = 249.5. The key judgement is that all the oxide can react because acid is in excess; the acid is not the limiting reagent.
2. The actual yield is 4.10 g. Calculate the percentage yield and judge whether it is credible. State one procedural change that would make the final crystals less likely to be acidic.
Read the worked answer
- Percentage yield = (4.10/6.28) × 100 = 65.3% using the theoretical mass from Mr = 249.5, or about 65.2% using 6.29 g.
- A yield of about 65% is credible for a school preparation because product is lost on the filter paper, in the mother liquor and during crystallisation. It is not evidence of purity, however.
- The planned method uses 25.0 cm3 of acid, but only 16.8 cm3 is needed to react with 2.00 g CuO: n = 2.52 × 10−2 mol, V = n/c = 2.52 × 10−2/1.50 = 1.68 × 10−2 dm3 = 16.8 cm3. The extra acid can leave the crystals acidic unless they are washed.
- Better procedure: use the calculated 16.8 cm3 of acid and add a slight excess of CuO, filter off the unused oxide, then crystallise. Alternatively, wash the crystals with a small volume of ice-cold distilled water and recrystallise. Cold water reduces product loss; washing with much warm water would dissolve too much product.