Thermal Decomposition Practical
CIE A-Level Chemistry 9701 · Hydrated Aluminium Sulfate Al₂(SO₄)₃·xH₂O
In this experiment, you determine the value of x in hydrated aluminium sulfate, Al₂(SO₄)₃·xH₂O (FB 1). A known mass of the salt (1.80–2.00 g) is placed in a crucible and heated until constant mass is achieved, driving off all water of crystallisation as steam. The mass of water lost and the mass of anhydrous residue are each converted to moles, and their ratio gives x as the nearest integer. Two separate heatings confirm that constant mass has been reached.
👇 Click a section below to explore
01 — experiment design
The core reaction
Heating drives off water of crystallisation, leaving the anhydrous salt.
Al₂(SO₄)₃·xH₂O(s) → Al₂(SO₄)₃(s) + xH₂O(g) Key ratio x = mol H₂O lost ÷ mol Al₂(SO₄)₃ produced — both calculated from masses recorded during the experiment.
Why heat twice?
Heating to constant mass confirms all water has been expelled. After the second heating, the mass must be within ±0.05 g of the reading after the first heating.
1
First strong heating (≈5 min, lid removed after 2 min) — removes most water of crystallisation.
2
Cool, reweigh. Then a second 2-min heating, cool, reweigh. Compare the two "after" masses.
3
If the two residue masses agree within +0.02 and −0.05 g, constant mass is confirmed and the lower value is used.
Procedural details to know
Crucible + lid (first 2 min)Prevents the solid spitting or frothing out of the crucible as water is released rapidly. Lid is removed after 2 min so steam can escape.
Pipeclay triangleSupports the crucible over a Bunsen flame, allowing direct high-temperature heating.
Mass range 1.80–2.00 gToo little → large percentage error in mass lost. Too much → incomplete decomposition in the time allowed.
Cool before weighingA hot crucible causes convection currents above the balance pan, producing a falsely low mass reading.
Appearance before vs after heating
The hydrated salt (FB 1) is crystalline / finely divided. After complete dehydration, the residue becomes lumpy / crusty or develops a hard "skin" — it is no longer free-flowing powder.
02 — data presentation (part a)
✕ Incomplete headings
| Measurement | Value |
|---|---|
| Crucible mass | 18.4 |
| Mass with FB1 | 20.3 |
| After 1st heating | 19.96 |
| After 2nd heating | 19.95 |
✕
No units — every heading must include / g.
✕
Vague labels — "Crucible mass" must be "(mass of) empty crucible". Lid must be mentioned where applicable.
✕
Inconsistent dp — 18.4 and 20.3 have only 1 dp. All four raw weighings must share the same number of decimal places.
✕
Derived masses absent — mass of FB 1 used, mass of residue, and mass of water lost must be explicitly calculated and shown.
✓ All seven mark-scheme headings
| Measurement | Value |
|---|---|
| (mass of) empty crucible / g | 18.40 |
| (mass of) crucible + FB 1 (hydrated aluminium sulfate) / g | 20.32 |
| (mass of) crucible and residue / contents, after 1st heating / g | 19.96 |
| (mass of) crucible and residue / contents, after 2nd heating / g | 19.95 |
| (mass of) FB 1 used / g | 1.92 |
| (mass of) residue / g | 1.55 |
| (mass of) water / mass loss / g | 0.37 |
✓
Seven unambiguous headings, all with / g units. Derived masses explicitly stated.
✓
All four raw weighings to the same number of decimal places (2 dp shown; 3 dp equally accepted).
Balance precision±0.01 g (2 dp) or ±0.001 g (3 dp)
All four weighings must sharethe same number of dp
Mass of FB 1 must be in range1.80 g – 2.00 g
2nd heating must be within+0.02 and −0.05 g of 1st heating
✕ Inconsistent decimal places
| Measurement | Value |
|---|---|
| Empty crucible / g | 18.4 |
| Crucible + FB 1 / g | 20.324 |
| After 1st heating / g | 19.96 |
| After 2nd heating / g | 19.95 |
✕
18.4 is 1 dp; 20.324 is 3 dp; the others 2 dp — inconsistent. All four must match.
✓ Consistent 2 dp throughout
| Measurement | Value |
|---|---|
| Empty crucible / g | 18.40 |
| Crucible + FB 1 / g | 20.32 |
| After 1st heating / g | 19.96 |
| After 2nd heating / g | 19.95 |
The second heating result must fall within the tolerance band of the first heating. This confirms all water of crystallisation has been expelled.
✕ Outside tolerance — further heating needed
| Measurement | Value |
|---|---|
| After 1st heating / g | 19.96 |
| After 2nd heating / g | 19.83 |
| Difference | 0.13 g |
✕
0.13 g exceeds the −0.05 g tolerance. Constant mass is not yet reached — heat again.
✓ Within tolerance — constant mass confirmed
| Measurement | Value |
|---|---|
| After 1st heating / g | 19.96 |
| After 2nd heating / g | 19.95 |
| Difference | 0.01 g ✓ |
✓
0.01 g is well within tolerance. Use the lower (2nd heating) value as the final residue mass for calculation.
Which value? Always use the reading after the final heating as the residue mass — it is the closest approximation to the true anhydrous mass.
03 — calculations (part b) — interactive
1.92 g
1.55 g
Mass of water lost = mass FB 1 − mass residue= 0.37 g
b(i) — n(H₂O) = mass lost / 18.0= 0.02056 mol
b(ii) — n(Al₂(SO₄)₃) = mass residue / 342.3= 0.004529 mol
b(iii) — x = n(H₂O) / n(Al₂(SO₄)₃) → round to integer= 4.54 → x = 5
water lost
0.37
g
n(H₂O)
0.0206
mol
n(Al₂(SO₄)₃)
0.00453
mol
x (raw)
4.54
ratio
x (integer)
5
answer
Sig figs Give all intermediate values to 2–4 sf. The final answer for x must be the nearest integer. Show all steps — each sub-part carries its own mark.
Molar masses used
H₂O = 18.0 g mol⁻¹Used in b(i) to convert mass lost into moles of water.
Al₂(SO₄)₃ = 342.3 g mol⁻¹Al: 2×27 = 54 · S: 3×32 = 96 · O: 12×16 = 192 → total 342
Accuracy (Q) marks — mass ratio test: mass FB 1 / mass residue (to 2 d.p.)
IV Broader band — awarded if ratio is in range 1.56 – 2.12 inclusive
V Tighter band — awarded if ratio is in range 1.66 – 2.02 inclusive
How to check Divide mass of FB 1 ÷ mass of residue to 2 dp from your data. If it falls in the V band you earn both Q marks; if only in the IV band, one mark.
04 — error analysis (part d)
Balance (per weighing)
Each balance reading carries an uncertainty of ±0.01 g.
uncertainty = ±0.01 g per reading % error = (0.01 / mass) × 100 For residue mass ≈ 1.55 g:
= (0.01 / 1.55) × 100 ≈ 0.65%
Mass of water lost
Mass lost is derived from two readings — uncertainty doubles.
combined uncertainty = 0.01 + 0.01 = 0.02 g % error = (0.02 / mass loss) × 100 For mass loss ≈ 0.37 g:
= (0.02 / 0.37) × 100 ≈ 5.4%
Avoid Do not use the total mass of FB 1 (≈1.92 g) as the denominator when calculating % error in mass lost. The uncertain quantity is the small mass of water lost (≈0.37 g) — using the larger denominator gives an artificially small % error.
Why is the % error here larger than in a titration?
✓
The mass of water lost is small (≈0.37 g) relative to the balance uncertainty (±0.01 g), giving a large % error. In a titration the titre is much larger (≈16 cm³) relative to the burette uncertainty (±0.05 cm³), so the % contribution is far smaller.
✓
Using a more precise balance (e.g., 4 dp, ±0.0001 g) would substantially reduce the % error in mass measurement.
Incomplete decomposition — effect on x
✕
If not all water is driven off: mass of residue recorded is too high → mass of water lost calculated is too low → n(H₂O) too low → x calculated is too low (not too high).
Note This is precisely why the two-heating procedure exists — if the second mass is noticeably lower than the first, more heating is required before recording the final residue mass.
05 — short-answer Q&A (part c)
c(i) — How does the residue compare in appearance to the starting solid?
Before heating — FB 1crystalline / finely divided
After heating — residuelumpy / crusty / has a 'skin'
Mark Both descriptions are needed for the mark. The answer must contrast the state before AND after. "White powder becomes a hard lump" is acceptable phrasing.
c(ii) — Why is the crucible lid kept on for the first two minutes?
✓
To prevent the solid spitting / frothing out of the crucible. As the crystal structure breaks down rapidly, water is released quickly and can eject solid particles if the crucible is uncovered too soon.
Common error "To prevent water escaping" — this is wrong. Water must escape as steam; the lid prevents loss of solid. The lid is removed after 2 min precisely to allow steam to escape freely.
c(iii) — Anhydrous contamination causing x to be too high: is the student correct?
✕
The student is NOT correct.
Effect of anhydrous contaminationLess water present → mass lost is too low
n(H₂O) calculatedToo low
Residue massHigher than pure hydrate would give
Ratio mol water / mol residueLower than true x
Calculated xToo LOW — not too high
Full mark answer State the verdict AND give the reasoning: anhydrous salt contributes no water to the mass lost, so n(H₂O) is lower, and the ratio (= x) is lower — the opposite of what the student claimed.