Hydrate Analysis
● Ready
Initial Mass
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Current Mass
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Mass Lost (H₂O)
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Moles Salt
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Moles H₂O
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x in Formula
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Determining the Formula of a Hydrate
A hydrated salt contains a fixed ratio of water molecules in its crystal lattice, written as Salt·xH₂O. The goal is to find x.
- Step 1 — Weigh accurate sample of hydrate (mass = m₁)
- Step 2 — Heat strongly to drive off all water of crystallization
- Step 3 — Cool in desiccator (prevents reabsorption of moisture)
- Step 4 — Weigh anhydrous residue (mass = m₂)
- Step 5 — Mass of water lost = m₁ − m₂
- Step 6 — Calculate moles: n(salt) = m₂ ÷ M(salt), n(H₂O) = (m₁−m₂) ÷ 18.02
- Step 7 — Ratio x = n(H₂O) ÷ n(salt), round to nearest integer
The color change of CuSO₄ (blue → white) provides a visible confirmation that dehydration is complete.
IB Exam Strategies
- Why heat until constant mass? — Ensures all water has been removed. If mass keeps decreasing, more heating is needed
- Why use a desiccator? — Anhydrous salts are hygroscopic and reabsorb water from the air, giving falsely high final mass
- Overheating risk — If temperature is too high, the salt itself may decompose (e.g., CuSO₄ → CuO + SO₃ above ~650°C). This gives a falsely LOW final mass and too HIGH x value
- Calculation pattern — Always show: mass → moles → simplest ratio → x value
- Percentage water — % H₂O = (mass lost / initial mass) × 100. IB may ask for this too
Common Experimental Errors
| Error | Effect on x | Explanation |
|---|---|---|
| Incomplete heating | x too LOW | Not all water removed → mass loss underestimated |
| No desiccator used | x too LOW | Salt reabsorbs moisture → final mass too high |
| Overheating / decomposition | x too HIGH | Salt decomposes → extra mass lost beyond water |
| Spattering during heating | x too HIGH | Sample physically lost → mass loss overestimated |
| Sample not fully ground | x too LOW | Large crystals don't fully dehydrate in center |