Hydrated Copper(II) Sulphate: When Strong Heating Destroys the Evidence
A student is determining the value of x in CuSO₄·xH₂O. She weighs a clean, dry crucible with its lid: 23.46 g. She adds about 5 g of blue crystals and reweighs: 28.44 g, so the sample mass is 4.98 g. She heats gently with the lid partly open, cools in a desiccator and weighs. After a second gentle heating and cooling, the mass is constant at 26.64 g. Her teacher then asks her to heat a fresh 4.98 g sample strongly in a Bunsen flame for ten minutes. The residue weighs 1.59 g, is black, and a gas with a pungent smell is produced. The teacher says this strong-heating result must not be used to calculate x. (Mᵣ: CuSO₄ = 159.6, H₂O = 18.0, CuO = 79.5.)
1. Calculate x from the gentle-heating data. Show the mole calculation and explain how you decide the final value of x.
Read the worked answer
Mass of anhydrous salt: 26.64 − 23.46 = 3.18 g.
Moles of CuSO₄: 3.18 / 159.6 = 1.99 × 10⁻² mol.
Mass of water lost: 4.98 − 3.18 = 1.80 g.
Moles of H₂O: 1.80 / 18.0 = 0.100 mol.
Ratio: 0.100 / 1.99 × 10⁻² = 5.02.
The formula is CuSO₄·xH₂O, so x must be a whole number. 5.02 is within about 0.4% of 5, which is normal weighing error for a school balance. The correct deduction is x = 5, not x = 5.02. If the result had been 4.6 or 5.5, the method or the heating would need to be questioned before rounding.
2. Explain why the black residue from strong heating cannot be used to find x. Support your answer with an equation and a calculation.
Read the worked answer
At strong heat, copper(II) sulphate decomposes; the black residue is copper(II) oxide:
CuSO₄(s) → CuO(s) + SO₃(g)
The pungent gas is sulphur trioxide (which fumes in moist air as sulphuric acid mist).
Check the copper: n(CuO) = 1.59 / 79.5 = 0.0200 mol, which matches n(CuSO₄) = 0.0199 mol in the original sample. All the copper is still there, but the sulphate has been lost as SO₃.
If you wrongly assumed that the whole mass loss of 4.98 − 1.59 = 3.39 g was water:
n(H₂O) assumed = 3.39 / 18.0 = 0.188 mol ratio = 0.188 / 0.0199 = 9.4
A ratio of 9.4 is not a sensible hydrate formula; the extra mass loss is SO₃ (about 0.0200 × 80.1 = 1.60 g), not water. Therefore the strong-heating result cannot give x.
3. The student suggests keeping the lid fully closed during gentle heating 'to stop the crystals spitting'. Evaluate this suggestion.
Read the worked answer
Closing the lid completely would reduce spitting, but it would also trap the water vapour that must escape if the mass loss is to be measured. The hydrate would not be fully dehydrated, the final mass would be too high, and the calculated water loss - and therefore x - would be too low. There is also a practical risk: trapped vapour can build pressure and crack the crucible or lift the lid.
The correct technique is to heat gently at first with the lid partly open (or askew), then heat more strongly, cool in a desiccator, and reheat to constant mass. Spitting is better controlled by gentle initial heating and by not overfilling the crucible. The desiccator matters because anhydrous CuSO₄ is hygroscopic and would reabsorb water from the air before weighing.