Hydrated salts
Hydrated salts
Hydrated salts and water of crystallisation
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A hydrated salt has water molecules in a fixed ratio within its crystal structure. This is water of crystallisation; it is not simply a salt dissolved in water.
- Hydrated salt
- A salt containing water molecules as part of its crystal structure.
- Water of crystallisation
- Water molecules present in a fixed ratio within the crystal structure of a salt.
- Anhydrous salt
- A salt containing no water of crystallisation.
In salt·xH₂O, x tells you how many water molecules occur per formula unit of salt. Heating can remove that water and leave an anhydrous salt.
- Find the mass of anhydrous salt left after heating.
- Find the mass of water lost by subtracting the final sample mass from the initial sample mass.
- Convert salt and water masses separately into moles.
- Divide moles of water by moles of salt to find x.
Check the stated chemistry: this method assumes that the loss is water and that the residue is the intended anhydrous salt.
n amount of the specified substance or particles, in mol; x number of water molecules per formula unit of the anhydrous salt; a whole-number ratio with no unit.
A hydrate from water percentage
MgSO4·xH2O contains 51.2% water by mass. M(MgSO4) = 120.4; M(H2O) = 18.0 g mol−1.
Determine x and write the formula of the hydrated salt.
Worked answer
| Quantity | MgSO4 | H2O |
|---|---|---|
| Mass / g | 48.8 | 51.2 |
| M / g mol−1 | 120.4 | 18.0 |
| Amount / mol | n = m/M n = 48.8/120.4 = 0.405316… | n = m/M n = 51.2/18.0 = 2.84444… |
| Salt : water | 1 | 7.01785… ≈ 7 |
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
(a) State what is meant by an anhydrous salt. [1 mark]
(b) State what is meant by water of crystallisation. [2 marks]
(c) Explain the meaning of the dot and the 7 in MgSO4·7H2O. [2 marks]
Total: 5 marks
Show answer & marking points
(a)
An anhydrous salt contains no water of crystallisation [1]
(b)
Water molecules incorporated into a crystal structure [1]
They occur in a fixed ratio to the salt [1]
(c)
The dot separates MgSO4 from its water of crystallisation [1]
There are seven H2O molecules per MgSO4 formula unit [1]
CoCl2·xH2O contains 45.4% water by mass. Determine x and the full formula. M(CoCl2) = 129.9 and M(H2O) = 18.0 g mol−1. (4 marks)
Total: 4 marks
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In 100 g, water mass = 45.4 g and anhydrous salt mass = 54.6 g. [1] n(water) = m/M
n(water) = 45.4/18.0 = 2.52222… mol; n(salt) = m/M
n(salt) = 54.6/129.9 = 0.420323… mol. [1] x = 2.52222…/0.420323… = 6.00067… ≈ 6. [1] CoCl2·6H2O. [1]
Hydrates by heating to constant mass
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The experiment measures a mass loss, so accurate weighing and controlled heating are essential.
- Weigh the dry empty crucible, then weigh it with the hydrated salt.
- Heat gently at first so the salt does not spit out of the crucible.
- Cool safely and weigh. Repeat heating, cooling and weighing until the mass is constant.
- Subtract the crucible mass to find the anhydrous salt mass.
- Subtract the final sample mass from the initial sample mass to find water lost, then convert both masses to moles.
Repeat heating, cooling and weighing until consecutive readings agree. This constant mass is evidence that all the water has been removed. Heat enough to remove the water without decomposing the salt, and cool before each weighing. Losing solid as well as water would spoil the calculated mole ratio.
Hydrated copper(II) sulfate
Crucible: 25.000 g. Crucible + hydrate: 27.496 g. After successive heatings: 26.622, 26.596, 26.596 g. M(CuSO4) = 159.6; M(H2O) = 18.0 g mol−1.
Each mass after heating includes the crucible. Use the final constant mass to determine x in CuSO₄·xH₂O.
Worked answer
| Quantity | CuSO4 | H2O lost |
|---|---|---|
| Mass / g | Mass of salt = mass with salt − mass of empty container 26.596 − 25.000 = 1.596 | Mass of water = mass before heating − mass after heating 27.496 − 26.596 = 0.900 |
| M / g mol−1 | 159.6 | 18.0 |
| Amount / mol | 0.0100 | 0.0500 |
| Salt : water | 1 | 5 |
Allow the crucible to cool before each weighing: a hot object creates air currents and buoyancy effects that can make the balance reading unreliable.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
FeSO4·xH2O is heated. Only water is lost. Empty crucible = 24.850 g; crucible + hydrate = 27.629 g. Successive heat–cool readings are 26.455, 26.371, 26.369 and 26.369 g. M(FeSO4) = 151.9 and M(H2O) = 18.0 g mol−1.
(a) Determine x and the full formula, explaining the final mass chosen and why heating, cooling and weighing are repeated. (8 marks)
(b) Explain why the solid is heated gently at first and cooled before weighing. (2 marks)
Total: 10 marks
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(a) Use 26.369 g, because two successive readings agree. [1] Anhydrous salt mass = 26.369 − 24.850 = 1.519 g. [1] Water mass = 27.629 − 26.369 = 1.260 g. [1] n(salt) = m/M
n(salt) = 1.519/151.9 = 0.0100 mol. [1] n(water) = m/M
n(water) = 1.260/18.0 = 0.0700 mol. [1] x = 7 [1]; FeSO4·7H2O. [1] Repeated heat–cool–weigh cycles check whether further measurable mass is lost, supporting complete dehydration under the stated water-only assumption. [1]
(b) Gentle initial heating reduces spitting and loss of solid. [1] Cooling avoids convection/buoyancy effects from a hot object giving an unreliable balance reading. [1]
Evaluate hydrate experiments
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To explain an experimental error, follow its effect on both quantities in x = n(water lost)/n(salt left). Saying 'x is inaccurate' does not explain the direction.
For example, incomplete removal of water makes the measured water loss too small and leaves the residue too heavy. Use the ratio to explain what that does to x.
Heat, cool and weigh the sample, then repeat until consecutive masses agree. This is evidence that no more water is being lost. Also check that the salt has not decomposed: a steady mass alone cannot identify the solid left behind.
| Error | Recorded anhydrous mass | Recorded water loss | Calculated x |
|---|---|---|---|
| Incomplete water removal | Too high | Too low | Too low |
| Solid spits out during heating | Too low | Too high | Too high |
| Salt reabsorbs water before weighing | Too high | Too low | Too low |
Heating is stopped too soon
A hydrate still contains some water at the final weighing.
Explain how this affects the calculated value of x in salt·xH₂O when the residue is assumed to be anhydrous.
Worked answer
The residue is recorded as anhydrous salt, so its calculated amount is too large. Too little mass is recorded as water lost, so the calculated water amount is too small.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
A student calculates x from the water lost and the anhydrous salt left when a hydrate is heated. For each error, state the effect on the calculated x and explain it.
(a) Some water remains after the final heating. (2 marks)
(b) Solid is lost by spitting and all mass loss is treated as water. (2 marks)
(c) The residue absorbs water before weighing. (2 marks)
(d) Explain why constant mass alone does not prove that no salt decomposed. (1 mark)
Total: 7 marks
Show answer & marking points
(a) x is too low [1]: water loss is underestimated and the mass treated as anhydrous salt is too large. [1]
(b) x is too high [1]: apparent water loss is too large and salt remaining is underestimated. [1]
(c) x is too low [1]: reabsorbed water makes the recorded residue too heavy and the apparent water loss too small. [1]
(d) A constant reading shows no further measurable mass change under the conditions; it does not establish the chemical identity of the residue. [1]