Topic 1 · page 9 of 10
Formulae and equations by experiment
To test a formula or an equation, predict what each candidate would give, measure what actually happens, and compare in moles. Then judge how far to trust the measurement.
Read first: Masses from equations, Core Practical 1: molar volume of a gas
Finding a formula: water of crystallisation 1.101.6
Many salts crystallise with water molecules built into the crystal, written after a dot: CuSO₄·5H₂O, MgSO₄·7H₂O. Heating drives the water off. Weighing before and after tells you how much water there was.
PracticalHeating a hydrated salt to constant mass
- Weigh an empty crucible, then the crucible with the hydrated salt.
- Heat gently at first, then more strongly.
- Cool in a desiccator, so the solid does not reabsorb water from the air, and weigh.
- Heat, cool and weigh again. Repeat until two weighings agree, which is constant mass.
Worked exampleFinding x in MgSO₄·xH₂O
2.46 g of hydrated magnesium sulfate is heated to constant mass, leaving 1.20 g of anhydrous MgSO₄.
- Mass of water lost = 2.46 − 1.20 = 1.26 g
- n(MgSO₄) = 1.20 ÷ 120.4 = 0.00997 mol
- n(H₂O) = 1.26 ÷ 18.0 = 0.0700 mol
- Ratio H₂O : MgSO₄ = 0.0700 ÷ 0.00997 = 7.02, so x = 7 and the formula is MgSO₄·7H₂O.
Confirming an equation 1.10
When a substance could react in more than one way, write each possible equation, predict the measurable result for each, and see which one the data match.
Worked exampleThermal decomposition of sodium hydrogencarbonate
1.68 g of NaHCO₃ (0.0200 mol) is heated to constant mass. The solid left weighs 1.09 g. Which equation is correct?
| Possible equation | Solid left | Predicted mass |
|---|---|---|
| 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂ | 0.0100 mol Na₂CO₃ (M 106.0) | 1.06 g |
| NaHCO₃ → NaOH + CO₂ | 0.0200 mol NaOH (M 40.0) | 0.80 g |
| 2NaHCO₃ → Na₂O + H₂O + 2CO₂ | 0.0100 mol Na₂O (M 62.0) | 0.62 g |
The measured 1.09 g is closest to 1.06 g, so the first equation is confirmed:
The result is slightly high, which suggests the solid had not quite reached constant mass.
Other measurements work the same way: the volume of gas given off (using molar volume), the mass of a precipitate, or the mass lost as a gas escapes from an open flask on a balance.
Evaluating the data 1.10
A confirmed formula is only as good as the data behind it. These are the ideas examiners expect you to use.
| Term | Meaning | Example |
|---|---|---|
| random error | scatter in both directions; reduced by repeats and averages | reading a gas syringe scale |
| systematic error | shifts every result the same way; not reduced by repeats | a gas leak; water left in a hydrated salt |
| percentage uncertainty | uncertainty ÷ measured value × 100 | ±0.01 g on 1.26 g is 0.8% |
| percentage difference | (accepted − measured) ÷ accepted × 100 | 23.4 vs 24.0 is 2.5% |
Ways to improve this kind of experiment:
- heat to constant mass, and cool in a desiccator;
- use a balance with more decimal places, or larger masses, so the percentage uncertainty falls;
- repeat until the results are concordant and take a mean.
Check yourself
- 2.50 g of hydrated copper(II) sulfate, CuSO₄·xH₂O, is heated to constant mass, leaving 1.60 g. Find x.
- Why is the anhydrous salt cooled in a desiccator before it is weighed?
- A student heats a hydrated salt only once and gets x = 4.6 when the true value is 5. Explain the result.
- 2.00 g of calcium carbonate is heated strongly and the solid left weighs 1.12 g. Show that this supports CaCO₃ → CaO + CO₂.
- A balance reads to ±0.005 g. A mass of water is found by difference as 0.90 g. Calculate the percentage uncertainty in the mass of water.
Answers cover these until you have tried the questions
- Water = 2.50 − 1.60 = 0.90 g, so n(H₂O) = 0.90 ÷ 18.0 = 0.0500 mol. n(CuSO₄) = 1.60 ÷ 159.6 = 0.01003 mol. Ratio = 0.0500 ÷ 0.01003 = 4.99, so x = 5.
- Hot anhydrous salts absorb water vapour from the air as they cool, which would make the final mass too high.
- The salt was not heated to constant mass. Some water remained, so the mass of water lost was too small, and x came out too low. This is a systematic error.
- n(CaCO₃) = 2.00 ÷ 100.1 = 0.01998 mol. The equation predicts 0.01998 mol of CaO, which is 0.01998 × 56.1 = 1.12 g. This matches the measurement.
- Two readings, each ±0.005 g, give ±0.01 g in total. 0.01 ÷ 0.90 × 100 = 1.1%