18.1 - Mass of Water in Hydrated Crystals
The experiment
Aim
To find the mass of water of crystallisation in a sample of hydrated crystals, and from it the formula of the hydrate.
Background
- The crystals of a hydrated salt contain water of crystallisation, which is chemically combined into the crystals and adds to their mass.
- Heating the crystals drives this water off as vapour and leaves the anhydrous salt.
- You can't count the water molecules in the crystals, but you can weigh the crystals before heating and the anhydrous salt after.
- So in this practical you heat hydrated crystals to constant mass, and use the masses to work out how many water molecules the formula holds.
Method
- Weigh the empty basin - Weigh an evaporating basin or a crucible and record its mass to two decimal places.
- Add the crystals - Add between 1.5 g and 2.0 g of hydrated iron(II) sulfate crystals, spread thinly across the bottom, and weigh the basin again.
- Heat gently - Heat the basin on a gauze and tripod with a Bunsen burner for a couple of minutes. The pale green crystals turn to an off-white powder as the water leaves. Hydrated copper(II) sulfate, used in the same way, turns from blue to white.
- Cool and reweigh - Let the basin cool completely on a heatproof mat, then weigh it.
- Heat to constant mass - Heat, cool, and weigh the basin again, and keep going until two readings in a row give the same mass.
Safety
- Eye protection - The solid can spit as it heats.
- Ventilation - Overheating iron(II) sulfate decomposes it and gives off sulfur dioxide, a toxic gas, so work in a well-ventilated room and stop once the mass is constant.
- Hot basin - Move it with tongs.
- Iron(II) sulfate is an irritant - Wash off any crystals that touch your skin, and wash your hands afterwards.
Worked example - Formula of a hydrated salt
A student heated a sample of hydrated iron(II) sulfate, FeSO4·nH2O, in an evaporating basin to constant mass.
| Reading | Mass |
|---|---|
| Empty evaporating basin | 20.00 g |
| Basin and hydrated crystals, before heating | 22.78 g |
| Basin and contents, after heating to constant mass | 21.52 g |
Find the value of n, and the percentage of water in the hydrated salt.
Step 1 - Find the mass of hydrated salt used
22.78 - 20.00 = 2.78 g
Step 2 - Find the mass of anhydrous salt left behind
21.52 - 20.00 = 1.52 g
Step 3 - Find the mass of water driven off
2.78 - 1.52 = 1.26 g
Step 4 - Convert the mass of water to moles
Divide the mass by the relative formula mass of water.
of H2O = (2 × 1) + 16 = 18
1.26 ÷ 18 = 0.0700 mol
Step 5 - Convert the mass of anhydrous salt to moles
of FeSO4 = 56 + 32 + (4 × 16) = 152
1.52 ÷ 152 = 0.0100 mol
Step 6 - Find n and write the formula
One mole of the anhydrous salt is combined with seven moles of water, so the hydrated salt is FeSO4·7H2O.
Step 7 - Find the percentage of water
(1.26 ÷ 2.78) × 100 = 45.3%
Nearly half the mass of the crystals was water of crystallisation.
Step 8 - Check the answer makes sense
Here n came out as exactly 7. Real weighings rarely divide exactly, so a value such as 6.6 or 7.4 would still be rounded to 7. A value far from any whole number would suggest the heating stopped too early or the salt began to decompose.
Common exam questions
Question 1
Why is the basin heated more than once, to constant mass?
A single heating might leave some water in the crystals. Getting the same mass twice shows no more water is being lost, so all the water of crystallisation has gone.
Question 2
Why must the crystals be heated gently rather than strongly?
Strong heating decomposes iron(II) sulfate into iron(III) oxide and sulfur dioxide, instead of just driving off the water. A solid that turns brown or orange shows this has happened.
Question 3
Why is the sample spread thinly across the bottom of the basin rather than left in a pile?
So heat reaches all of the sample evenly and it all loses its water. Crystals at the bottom of a pile can stay hydrated.
Question 4
A student's calculated value of n came out higher than the true value. Suggest why.
Some of the salt decomposed, so extra mass was lost as well as the water. The calculation counts that extra loss as water, so n comes out too high.
Question 5
A different student's calculated value of n came out lower than the true value. Suggest why.
The crystals weren't heated to constant mass, so some water was still in the "anhydrous" salt. Less mass was recorded as water, so n comes out too low.