What you'll be able to do: Determine the formula of a hydrate from heating data and evaluate whether a gravimetric result is trustworthy.
Best after: Empirical and Molecular Formulas, Percent Composition
Heating a hydrate drives off its water of crystallisation and leaves the anhydrous salt behind. Two balance readings are therefore enough to determine how many water molecules were bound to each formula unit. This lesson applies the empirical formula method to real laboratory data and asks the harder question of when that data can be believed.
These are recommended, not required. You can start this lesson at any time.
In CuSO₄ . 5 H₂O the dot is not multiplication in the usual sense, it records that five water molecules are built into each formula unit of the crystal.
Those waters count fully in the molar mass, so the hydrate (249.68 g/mol) is much heavier per mole than the anhydrous salt (159.61 g/mol). The colour change from blue to white on heating is the visible sign that the water has gone.
CuSO₄ . 5 H₂O
Mass of water driven off = mass of hydrate minus mass of residue. Mass of anhydrous salt = mass of residue. Every gravimetric hydrate question reduces to those two lines, so write them down before doing any mole arithmetic. Remember to subtract the crucible mass from both readings first.
Let m₀ be the empty crucible mass, m₁ crucible plus hydrate, and m₂ crucible plus dry residue. Then hydrate mass = m₁ − m₀, anhydrous-salt mass = m₂ − m₀, and apparent water mass = m₁ − m₂. The crucible cancels only in the last subtraction. For 30.00, 32.50 and 31.22 g, those quantities are 2.50, 1.22 and 1.28 g respectively. Label all three before choosing the molar masses.
mwater = mhydrate - m_residue
Convert the water mass to moles using 18.02 g/mol, convert the residue mass to moles using the molar mass of the anhydrous salt, then divide moles of water by moles of salt. That quotient is x. For the simple fixed-composition hydrates in this lesson, x is a small whole number. Decide whether a measured ratio supports that integer using the measurement uncertainty; a ratio such as 3.4 should trigger investigation rather than automatic rounding.
The ratio is x = [(m₁ − m₂)/18.02] / [(m₂ − m₀)/M(salt)]. The numerator counts water molecules and the denominator counts anhydrous formula units. Dividing their masses would not count waters because one water molecule and one salt formula unit have different masses. Use the anhydrous formula in the denominator even though the original sample was hydrated.
x = nwater / nanhydrous
Percent water = mass of water lost divided by hydrate mass, times 100. Compare it with the theoretical value calculated from the proposed formula. For CuSO₄ . 5 H₂O the theoretical figure is 5(18.02)/249.68 = 36.08 %.
Agreement within the experiment’s uncertainty supports the formula; the acceptable difference depends on balance precision, sample size and the procedure.
Incomplete heating leaves water behind, so the measured water loss is too small and x comes out low. Overheating that decomposes the salt itself removes extra mass, so x comes out high. Spattering loses solid and also inflates the apparent water loss. A hygroscopic residue that reabsorbs moisture while cooling in open air lowers the apparent loss. Good practice is heating to constant mass and cooling in a desiccator.
The analysis assumes the only lost material is water and that the residue has the stated anhydrous formula. Spattering violates the first assumption; decomposition can violate both. Incomplete drying and rehydration make the numerator smaller and the apparent salt denominator larger, pushing x down in both ways. Heat gently at first to avoid loss, use the specified temperature, and cool before weighing so a hot crucible does not distort balance readings. Constant mass is necessary evidence of a stable endpoint, but a decomposed residue can also reach constant mass.
This is the empirical formula method with the elements replaced by two components, water and the salt. If you can determine x here, you can determine any empirical formula from mass data, because the reasoning is identical.
A balance often reports crucible plus contents, not the sample alone. Subtract the empty crucible mass from both the initial and final readings; subtract the two sample masses to obtain the water lost.
Heat, cool in a dry environment and reweigh, then repeat until two successive masses agree within the experiment's tolerance. A stable mass supports completion of drying but does not prove the residue has not chemically decomposed.
Check the residue's appearance and the plausible hydrate ratio alongside the mass data. Moisture reabsorption during open-air cooling makes the final mass too high and the calculated water ratio too low.
An ionic compound whose crystal includes a fixed number of water molecules per formula unit.
The bound water represented after the dot in a hydrate formula; it counts in the molar mass.
For a well-behaved hydrate, mass lost on gentle heating equals the mass of water driven off.
For the simple fixed-composition hydrates here, use uncertainty and the proposed salt identity to assess a small integer x; do not force an unsupported noninteger result to the nearest whole number.
The method assumes the anhydrous salt is thermally stable at the temperature used. Salts that decompose break this assumption.
x = (mwater / 18.02) / (manhydrous / Manhydrous)
% H₂O = mwater / mhydrate x 100 %
A 4.98 g sample of hydrated copper(II) sulfate is heated to constant mass, leaving 3.18 g of white residue. Determine x in CuSO₄ . x H₂O.
Try it first: Which of the two masses is the water, and which is the salt?
0 of 4 steps revealed.
A student determines x = 3.4 for a hydrate expected to be the tetrahydrate. Give the most likely cause and its direction.
Try it first: Does an x that is too low mean too much or too little mass was lost?
0 of 3 steps revealed.
An empty crucible is 30.00 g; with hydrated MgSO₄ it is 32.50 g; after heating it is 31.22 g. Find x in MgSO₄·xH₂O. Use M(MgSO₄) = 120.37 g/mol.
Try it first: Remove the crucible mass from both measurements.
0 of 3 steps revealed.
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Dividing the residue mass by the molar mass of the hydrate. | The residue is the anhydrous salt, so using the heavier hydrate molar mass understates its moles and inflates x. | Match each mass to the substance it actually is before choosing a molar mass. |
| Using total crucible-plus-sample readings as the sample masses. | Both masses are then too large by the same amount, which distorts the mole ratio even though the difference is unaffected. | Subtract the empty crucible mass from every reading first. |
| Reporting x = 4.6 as the answer. | A hydrate has a fixed whole number of waters per formula unit, so a non-integer is evidence of experimental error. | Round to the nearest whole number and explain the deviation, or identify the error if it is large. |
| Explaining a deviation only as experimental error. | It gives no chemical information and does not account for whether x came out high or low. | Name the error and state which direction it pushes the measured mass loss. |
| Treating the residue mass as the water lost. | The two are swapped, so the calculated ratio is inverted and x is far from any sensible value. | Write mwater = mhydrate - m_residue explicitly before calculating. |
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After the lesson · Question bank
Work through new problems on your own. The lesson's quick checks and worked examples stay above if you need to review a step.
12 topic questions across 3 difficulty levels.
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Subtract the crucible mass, convert the apparent water loss and residue to their separate mole amounts, then assess the water-to-salt ratio. Inferring a hydrate requires a stable, correctly identified residue and evidence that only water was lost; constant mass and a plausible integer ratio support the assignment without proving it alone.
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