Copper(II) Sulphate Crystals for a Science Fair: Limiting Reactant, Yield and Water of Crystallisation
A Form 5 student plans to enter a science fair with copper(II) sulphate crystals she has prepared herself. She adds 2.00 g of black copper(II) oxide to 25.0 cm3 of 2.00 mol dm−3 sulphuric acid, warms the mixture, filters off any excess solid, then evaporates the filtrate and cools it to crystallise. The blue crystals are drained, dried overnight in an oven at 80 °C and weighed: 4.20 g. Her write-up must report a theoretical yield, a percentage yield, and some evidence that the product really is the pentahydrate CuSO4·5H2O. To obtain that evidence she heats 2.00 g of her crystals strongly in an open crucible to constant mass and obtains 1.44 g of a white solid. Her teacher warns that the oven-drying step may itself have altered the water content of the crystals. [Mr: CuO = 79.5; CuSO4 = 159.5; H2O = 18.0; CuSO4·5H2O = 249.5]
1. Identify the limiting reactant and calculate the maximum mass of CuSO4·5H2O the student could obtain.
Read the worked answer
Moles of each reactant
n(CuO) = 2.00 ÷ 79.5 = 0.02516 mol
n(H2SO4) = (25.0 ÷ 1000) × 2.00 = 0.0500 mol
Equation and ratio
CuO + H2SO4 → CuSO4 + H2O, so CuO and H2SO4 react in a 1 : 1 ratio.
Available ratio is 0.02516 : 0.0500, so the acid is in excess (about 0.0248 mol, roughly half of it, is left over) and copper(II) oxide is the limiting reactant.
Theoretical yield
All 0.02516 mol of CuO becomes CuSO4, and on crystallisation the salt carries five water molecules per formula unit, so n(CuSO4·5H2O) = 0.02516 mol.
Mass = 0.02516 × 249.5 = 6.28 g
A useful check: instead of scaling by the acid, the calculation must always follow the limiting reactant. Using the acid would wrongly predict 0.0500 × 249.5 = 12.5 g, which is more than the copper present could ever supply.
2. Calculate the percentage yield and give two likely reasons why it is below 100%, suggesting one improvement.
Read the worked answer
Percentage yield
= (actual mass ÷ theoretical mass) × 100
= (4.20 ÷ 6.28) × 100 = 66.9%
Why the yield is low - plausible reasons
- Mechanical loss during transfer. Some copper(II) sulphate stays dissolved in the filtrate and in the washings; crystals cling to the filter funnel, the beaker and the filter paper. Every transfer step loses a little product.
- Loss during evaporation and drying. Over-evaporating or heating too strongly can decompose or dehydrate the salt, and spitting during evaporation throws solution out of the dish.
- Incomplete reaction or recrystallisation losses. Not all the oxide may have reacted if the mixture was not warmed long enough.
Improvement
Rinse the filter paper, beaker and crystals with a minimum volume of cold distilled water and add the washings to the filtrate before evaporating; evaporate slowly and stop when crystals just begin to appear. Drying should be gentle, ideally in air or a desiccator, because 80 °C drives off water of crystallisation - which is exactly what the heating test is designed to detect.
3. Use the heating data to judge whether the 4.20 g of crystals are pure CuSO4·5H2O, and recommend what the student should change.
Read the worked answer
Data processing
Mass of crystals heated = 2.00 g; mass of anhydrous CuSO4 left = 1.44 g
Mass of water driven off = 2.00 − 1.44 = 0.56 g
n(H2O) = 0.56 ÷ 18.0 = 0.0311 mol
n(CuSO4) = 1.44 ÷ 159.5 = 0.00903 mol
Mole ratio H2O : CuSO4 = 0.0311 : 0.00903 = 3.45 : 1
Judgement
Pure CuSO4·5H2O would give exactly 5.00 : 1, and 2.00 g of it would leave 2.00 × 159.5 ÷ 249.5 = 1.28 g of anhydrous solid, not 1.44 g. A ratio of about 3.5 shows the crystals have already lost part of their water, so they are a partly dehydrated product - better described as an average hydrate near CuSO4·3.5H2O, or a mixture of the pentahydrate with a lower hydrate - not the pure pentahydrate she claims.
Recommendation
Dry the crystals between filter papers or in a desiccator at room temperature and store them in a stoppered container; avoid the 80 °C oven. Then repeat the crucible test, and report the ratio as the evidence of formula.