Hess's Law Virtual Lab: Enthalpy of Hydration of CuSO₄

Q: Can the class link open straight on the Lab tab?

Yes. On the Share page, pin Screen to "Calorimetry lab" (and Experiment to "Enthalpy of hydration of CuSO₄") in the link's starting values. Otherwise the first step of the method tells students to open the Lab tab.

Updated 2026-10-07

This Hess's law virtual lab runs the A-level enthalpy practical that finds the enthalpy change of hydration of copper(II) sulfate. That change cannot be measured directly, because anhydrous CuSO₄ dissolves as soon as it meets water. Instead, students dissolve anhydrous CuSO₄ and then CuSO₄·5H₂O in water in a polystyrene cup, record each temperature change, calculate both enthalpies of solution and combine them in a Hess cycle. They compare their answer with the data-book value and explain the gap. It fits one lesson and gives every group the same clean data set, with no copper sulfate waste.

Hess's law – enthalpy cycle builder

Curriculum links

AQA A-level Chemistry 3.1.4.2–3.1.4.3 (calorimetry and Hess's law) and the required practical on measuring an enthalpy change.
IB Chemistry Reactivity 1.1–1.2 (measuring enthalpy changes and energy cycles).
AP Chemistry Unit 6, topics 6.4 (calorimetry) and 6.9 (Hess's law).

Simulic is not affiliated with or endorsed by AQA, the IB, the College Board or any exam board.

Before the lab (5 min)

Ask students to commit to a prediction, on paper or as question 1 of the class link:

"You dissolve white anhydrous copper(II) sulfate in water, then blue hydrated copper(II) sulfate in a fresh cup of water. What happens to the temperature each time?"

Most students expect both to behave the same way. Don't correct them yet.

Method in the simulation

Open the Lab tab and keep Experiment on "Hydration of CuSO₄". Check that Lid is ticked.
Set Measure to "Reaction 1", Solid 4.00 g of anhydrous CuSO₄ and Solution 50 mL. Press Add solid and stir.
Watch the temperature–time graph. Wait until the run ends at t = 300 s, then press Record. The row shows ΔT, Q = m·c·ΔT and ΔH = −Q/n.
Set Measure to "Reaction 2", Solid 6.25 g of CuSO₄·5H₂O, 50 mL. Press Add solid and stir, wait, and press Record.
Use ΔH(hydration) = ΔH₁ − ΔH₂. Once both reactions are recorded, the box above the results table shows the Hess's law result next to the data-book value.
Untick Lid, press New run and repeat both reactions. Compare.

Reaction	Solid	Mass (g)	n (mol)	T₁ (°C)	T₂ (°C)	ΔT (°C)	Q (J)	ΔH (kJ/mol)
1	CuSO₄	4.00		20.0
2	CuSO₄·5H₂O	6.25		20.0

the Lab tab with Hydration of CuSO₄ and the lid on, after recording reaction 1 (4.00 g) and reaction 2 (6.25 g): the reaction 2 graph dipping to 18.7 °C, the results table with both rows and the Hess's law line giving −71.7 kJ/mol

Expected results

All values come from the simulation, which treats the solution as water (c = 4.18 J g⁻¹ K⁻¹) and lets the cup lose heat to the room at 20 °C.

Run	Highest or lowest T (°C)	ΔT (°C)	ΔH (kJ/mol)
Reaction 1, 4.00 g, lid on	27.3	+7.3	−60.9
Reaction 2, 6.25 g, lid on	18.7	−1.3	+10.9
Hess's law, lid on			−71.7 (data book −78.2, 8% off)
Hess's law, lid off			−63.4

Dissolving anhydrous CuSO₄ is exothermic; dissolving the hydrated salt is endothermic.
4.00 g and 6.25 g both contain about 0.025 mol of CuSO₄, so the two runs compare fairly.
Doubling the mass to 8.00 g doubles ΔT to +14.7 °C, but ΔH stays about −61 kJ/mol.
Without the lid, more heat is lost while the solid dissolves, so the result moves further from the data-book value.

Questions for students

(Prediction, asked again after the lab) What happens to the temperature when each salt dissolves?
Why are 4.00 g of CuSO₄ and 6.25 g of CuSO₄·5H₂O used?
In reaction 1 (4.00 g, 50 mL, lid on), what is ΔT?
From your two ΔH values, what is the enthalpy change of hydration of CuSO₄?
Your value is less negative than −78.2 kJ/mol. Explain why, and suggest one improvement.

Answers for teachers: (1) Anhydrous CuSO₄: the temperature rises; CuSO₄·5H₂O: it falls. (2) Both contain about 0.025 mol of CuSO₄, so the same amount dissolves in the same volume of water. (3) +7.3 °C (accept ±0.2). (4) About −71.7 kJ/mol (accept −73.7 to −69.7). (5) Heat is exchanged with the surroundings while the solid dissolves, so the measured ΔT values are too small; the cup's heat capacity is also ignored. Improve the insulation (keep the lid on) and extrapolate the cooling curve back to the time of mixing.

Common misconceptions

"ΔH₁ − ΔH₂ is the wrong way round." Follow the arrows. CuSO₄(s) → CuSO₄(aq) directly is ΔH₁. The other route goes through CuSO₄·5H₂O, then dissolves it (ΔH₂). So ΔH₁ = ΔH(hydration) + ΔH₂.
"A bigger temperature change means a bigger ΔH." ΔT depends on the mass and the volume of water. ΔH is per mole, so it stays about the same when the mass doubles.
"An endothermic dissolving does not need a lid." Heat flows in from the room, which makes the temperature drop too small, so the lid matters for both runs.

Extension

Switch Experiment to "Formation of MgO". React 0.20 g of Mg and 0.50 g of MgO with excess HCl, and add ΔfH(H₂O). The simulation gives −567.5 kJ/mol against the data-book −601.6 kJ/mol.
On the Hess cycle tab, follow the formation and combustion routes for the same reaction and see both give the same ΔrH.

FAQ

Can the class link open straight on the Lab tab?

Yes. On the Share page, pin Screen to "Calorimetry lab" (and Experiment to "Enthalpy of hydration of CuSO₄") in the link's starting values. Otherwise the first step of the method tells students to open the Lab tab.

Why is the result 8% off the data-book value?

The model loses heat to the room and ignores the cup, just like a real polystyrene cup. That makes the error worth discussing rather than a mistake to hide.

How does this differ from the calorimetry virtual lab?

The calorimetry virtual lab measures one enthalpy of neutralization directly. This lab measures two reactions and uses Hess's law to find a third that cannot be measured.