Hydrated salts
Hydrated salts
🔓 CRACK IT
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.
- Weigh the empty crucible and then the crucible containing hydrated salt.
- Heat gently at first to avoid spitting, then sufficiently to remove water without decomposing the salt.
- Cool and weigh; repeat heating, cooling and weighing to constant mass.
- Use the salt-left and water-lost mole amounts to calculate x.
m mass of the specified sample or substance, in g.
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.
Copper sulfate hydrate
2.496 g CuSO4·xH2O leaves 1.596 g CuSO4. Mr(CuSO4) = 159.6; Mr(H2O) = 18.0.
The hydrate is heated until all its water of crystallisation is removed. Determine x and calculate the relative formula mass of the hydrated salt.
Worked answer
| Quantity | CuSO4 | H2O lost |
|---|---|---|
| Mass / g | 1.596 | Mass of water = mass before heating − mass after heating 2.496 − 1.596 = 0.900 |
| Ar or Mr (no unit) | 159.6 | 18.0 |
| Amount / mol | n = m/Mᵣ n = 1.596/159.6 = 0.0100 | 0.0500 |
x = 0.0500/0.0100 = 5
Use constant mass
Crucible 19.840 g; with hydrated MgSO4 22.304 g. After heating: 21.068, 21.044, 21.044 g. Use Mr(MgSO4) = 120.4 and Mr(H2O) = 18.0.
Each mass after heating includes the crucible. Use the final constant mass to determine x in MgSO₄·xH₂O.
Worked answer
| Quantity | MgSO4 | H2O lost |
|---|---|---|
| Mass / g | 1.204 | Mass of water = mass before heating − mass after heating 22.304 − 21.044 = 1.260 |
| Ar or Mr (no unit) | 120.4 | 18.0 |
| Amount / mol | n = m/Mᵣ n = 1.204/120.4 = 0.0100 | 0.0700 |
x = 0.0700/0.0100
| Problem | Effect on residue and apparent water loss | Effect on x |
|---|---|---|
| Incomplete dehydration or water reabsorbed | Residue too heavy; apparent water loss too small | Too low |
| Solid lost by spitting | Residue too light; mass loss wrongly treated as water | Too high |
| Salt decomposes as well as losing water | Extra mass loss wrongly treated as water; wrong residue assumption | Usually too high under the assumed-salt calculation |
Why incomplete heating lowers x
1.722 g CaSO4·2H2O should leave 1.362 g CaSO4, but 0.090 g water remains. Mr(CaSO4) = 136.2; Mr(H2O) = 18.0.
Calculate the apparent value of x that would be obtained if the wet residue were incorrectly treated as anhydrous CaSO₄. State whether this value is higher or lower than the true value.
Worked answer
Water percentage in a hydrate
A CuSO4 hydrate contains 36.06% water. Mr(CuSO4) = 159.6; Mr(H2O) = 18.0.
Determine x in CuSO₄·xH₂O from this percentage by mass.
Worked answer
Use 100 g: 36.06 g water and 63.94 g anhydrous salt.
| Quantity | CuSO4 | H2O |
|---|---|---|
| Mass / g | 63.94 | 36.06 |
| Ar or Mr (no unit) | 159.6 | 18.0 |
| Amount / mol | n = m/Mᵣ n = 63.94/159.6 = 0.400626… | n = m/Mᵣ n = 36.06/18.0 = 2.00333… |
x = (36.06/18.0)/(63.94/159.6) = 5.0005…
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
4.76 g CoCl₂·xH₂O leaves 2.60 g anhydrous CoCl₂ after heating to constant mass. Use Mᵣ(CoCl₂) = 130.0 and Mᵣ(H₂O) = 18.0.
- Determine x and write the hydrate formula. (4 marks)
- Explain what water of crystallisation means. (1 mark)
- State the effect on calculated x if some water remains, and if solid is lost by spitting. (2 marks)
- Explain why a separate decomposition of the salt would make this method unreliable. (1 mark)
Total: 8 marks
Show answer & marking points
(a) Water lost = 4.76 − 2.60 = 2.16 g. [1] n(water) = m/Mᵣ
n(water) = 2.16/18.0 = 0.120 mol and n(CoCl₂) = m/Mᵣ
n(CoCl₂) = 2.60/130.0 = 0.0200 mol. [1] x = n(H₂O)/n(anhydrous salt)
x = 0.120/0.0200 = 6. [1] CoCl₂·6H₂O. [1]
(b) Water molecules incorporated in a fixed ratio in the salt’s crystal structure. [1]
(c) Water remaining: x too low. [1] Solid lost: x too high under the assumed-residue calculation. [1]
(d) The mass loss would not all be water and/or the residue would not be the assumed anhydrous salt. [1]
A student heats 5.72 g Na2CO3·xH2O. The table shows the mass after each heating. Mr(Na2CO3) = 106.0.
| Measurement | Mass after subtracting crucible / g |
|---|---|
| Before heating | 5.72 |
| After first heating | 2.46 |
| After second heating | 2.12 |
| After third heating | 2.12 |
(a) Select the valid final mass and justify your choice. (2 marks)
(b) Determine x and write the formula of the hydrate. (5 marks)
(c) If the first-heating mass were used, explain why the calculated x would be too low. (3 marks)
Use Mᵣ(H₂O) = 18.0.
Total: 10 marks
Show answer & marking points
(a) Use 2.12 g [1] because the second and third heating values agree, showing constant mass [1].
(b) m(H2O) = 5.72 − 2.12 = 3.60 g [1]. n(H₂O) = m/Mᵣ
n(H2O) = 3.60 ÷ 18.0 = 0.200 mol [1]. n(Na₂CO₃) = m/Mᵣ
n(Na2CO3) = 2.12 ÷ 106.0 = 0.0200 mol [1]. Ratio = 0.200 : 0.0200 = 10 : 1, so x = 10 [1]. Formula = Na2CO3·10H2O [1].
(c) After the first heating, some water remains in the sample [1]. The recorded residue is therefore too heavy and the apparent water loss is too small [1], making n(H2O) ÷ n(Na2CO3), and hence x, too low [1].
| Quantity | Na2CO3 | H2O lost |
|---|---|---|
| Mass / g | 2.12 | 3.60 |
| Ar or Mr (no unit) | 106.0 | 18.0 |
| Amount / mol | n = m/Mᵣ n = 2.12/106.0 = 0.0200 | n = m/Mᵣ n = 3.60/18.0 = 0.200 |