Water of crystallisation
and associated calculations.
- Water of crystallisation, also known as water of hydration, refers to water molecules that are chemically bound to the ions or molecules in a crystalline structure of a compound.
- These water molecules are an integral part of the crystal lattice and are often present in a fixed stoichiometric ratio with the other components of the compound.
Common examples of hydrated salts along with their chemical formulae:
- Copper(II) sulfate pentahydrate (CuSO4·5H2O)
- Magnesium sulfate heptahydrate (MgSO4·7H2O)
- Sodium carbonate decahydrate (Na2CO3·10H2O) – washing soda
- Iron(II) sulfate heptahydrate (FeSO4·7H2O)
- Zinc sulfate heptahydrate (ZnSO4·7H2O)
- Potassium aluminum sulfate dodecahydrate (KAl(SO4)2·12H2O), commonly known as alum.
These salts contain a specific number of water molecules (as indicated by the subscripts) within their crystal structures, and the water of crystallization plays a significant role in their physical properties, such as solubility and colour. When these hydrated salts are heated, they lose their water of crystallisation and are transformed into anhydrous salts with different properties. When left in open air, they may lose some of their water of crystallisation, so for example, washing soda might become Na2CO3·7H2O
Finding the number of water molecules.
There are two very different approaches:
- Heating to constant mass
- Determination by titration, where the hydrated salt neutralises an acid, or
decolourises a purple solution of KMnO4
1. Heating to constant mass
WORKED EXAMPLE 1.
(a) Draw a labelled diagram of the setup required to determine the
value of x in the formula ZnSO4 . xH2O
(b) Give three measurements of mass required
(c) Outline the key steps required
Solution 1.
(a) Crucible, Gauze, Tripod, Heat or Bunsen burner
(b) mass of crucible
mass of crucible + hydrated zinc sulphate
mass of crucible + anhydrous zinc sulphate
(c) Heat, cool and weigh
Repeat until a constant mass is reached
WORKED EXAMPLE 2.
Hydrated calcium nitrate can be represented by the formula
Ca(NO3)2 . xH2O where x is an integer.
A 6.04 g sample of Ca(NO3)2 . xH2O contains 1.84 g of water of crystallisation.
Use this information to calculate a value for x.
Solution 2.
Hydrated calcium nitrate can be represented by the formula Ca(NO3)2.xH2O Ans ⇒
where x is an integer.
A 6.04 g sample of Ca(NO3)2.xH2O contains 1.84 g of water of crystallisation.
Use this information to calculate a value for x .
Show your working.
A sample of hydrated nickel sulfate (NiSO4.xH2O) with a mass of 2.287 g was heated to Ans ⇒
remove all water of crystallisation. The solid remaining had a mass of 1.344 g.
Calculate the value of the integer x.
Show your working.
4.20 g Ca(NO3)2
x = 4
Use the data below to calculate the number of moles of water of crystallisation in Ans ⇒
each mole of hydrated magnesium chloride.
A. 2 B. 4 C. 6 D 8
The answer is C.
- the mass of water of crystallisation
= 203.1 – 95.1 = 108 g per mole of salt - 108 g = 108 / 18 = 6 moles of water
- the formula is MgCl2.6H2O
A student carried out an experiment to determine the value of x in the formula of Ans ⇒
hydrated sodium bromide, NaBr.xH2O.
Hydrated sodium bromide is heated until all the water of crystallization is removed.
Anhydrous sodium bromide, NaBr, is formed.
The student was given the following instructions.
Weigh a sample of the hydrated sodium bromide crystals in a pre-weighed crucible.
Heat the crucible containing the sample to remove the water of crystallisation.
Allow the crucible to cool and then reweigh the crucible.
The student’s results are shown in the table below.
(a) Complete the table.
(b) Find the value of x to the nearest whole number.
(a) 2.49 and 0.98
(b) 
Question 1.
To find out the number of moles of water of crystallisation,
a student heated some hydrated magnesium sulfate in a crucible and recorded the results in the table below. Find the value of x in the formula of the hydrated salt: MgSO4 . xH2O
(Use these relative atomic masses: H = 1; O = 16; Mg = 24; S = 32)
The answers can be found here.
Question 2.
Some whitening toothpastes contain hydrated silica, SiO2 .2H2O,
which acts as an abrasive to remove stains and polish teeth.
Hydrated silica contains water of crystallisation.
(a) What is meant by the term water of crystallisation?
(b) Calculate the percentage of water of crystallisation present
in hydrated silica.
(Use these relative atomic masses: H = 1; O = 16; Si = 28)
The answers can be found here.
Question 3.
Magnesium chloride has healing effects on a wide range of diseases. The
hydrated form of the salt has the formula MgCl2 .nH2O.
(Relative atomic masses: H = 1; O = 16; Mg = 24; Cl = 35.5)
The following results were obtained in an experiment to determine the value of n in the formula.
Mass of empty crucible = 13.87 g
Mass of crucible and hydrated magnesium chloride = 15.90 g
Mass of crucible and anhydrous magnesium chloride = 14.82 g
Calculate the value of n in MgCl2 .nH2O.
The answers can be found here.
Question 4.
Calculate the percentage of water of crystallisation by mass in hydrated sodium sulfate, Na2SO4 .10H2O.
Use these relative atomic masses: Na = 23, S = 32, H = 1 and O = 16
The answers can be found here.
Question 5.
A student added 627 mg of hydrated sodium carbonate (Na2CO3 . xH2O)
to 200 cm 3 of 0.250 moldm -3 hydrochloric acid in a beaker and stirred the mixture.
After the reaction was complete, the resulting solution was transferred to a volumetric flask, made up to 250 cm 3 with deionised water and mixed thoroughly. Several 25.0 cm 3 portions of the resulting solution were titrated with 0.150 moldm -3 aqueous sodium hydroxide.
The mean titre was 26.60 cm 3 of aqueous sodium hydroxide.
Calculate the value of x in Na2CO3 . xH2O
Show your working. Give your answer as an integer
The answers can be found here.
2a. Determination by acid-base titration
WORKED EXAMPLE.
A sample of 4.64g of hydrated sodium carbonate, Na2CO3 . xH2O was dissolved in 1.00 dm3 of water.
25.0 cm 3 of this solution required 20.0 cm 3 of 0.0500 moldm -3 hydrochloric acid for neutralisation. Find the value of x.
Solution.
Moles of hydrochloric acid required for 25.0 cm 3.
n = c.v = 0.0500 x 0.0200 = 0.00100 mol of HCl
Equation: Na2CO3 + 2HCl → 2NaCl + CO2 + H2O
Moles of Na2CO3 in 25.0 cm 3 = 0.000500
Moles of Na2CO3 in 1000 cm 3 = 0.000500 x 1000 / 25 = 0.0200 mol
0.0200 mol = 4.64 g
1.00 mol = 4.64 / 0.02 = 232 = RFM of Na2CO3 . xH2O
Na2CO3 accounts for 106 of that.
The water of crystallisation then accounts for 232 – 106 = 126
126 / 18 = 7. So x = 7 and the formula is Na2CO3 . 7H2O
2b. Determination by redox-titration
WORKED EXAMPLE.
Calculate x in the formula from the formula FeSO4 .x H2O from the following data:
12.18 g of iron(II) sulfate crystals were made up to 500 cm 3, acidified with sulfuric acid. 25.0 cm 3 of this solution required 43.85 cm 3 of 0.0100 moldm -3 KMnO4 for complete oxidation.
Note that 1 mol of KMnO4 requires 5 mol of iron(II) in the titration.
NO FURTHER KNOWLEDGE OF THIS REDOX TITRATRATION IN REQUIRED.
Solution.
Moles of KMnO4 needed to react with 25.0 cm 3 of iron(II) solution.
n = c.v = 0.0100 x 0.04385 mol = 0.0004385 mol
Moles of iron(II) in the 25.0 cm 3 sample used in the titration.
= fives as many = 5 x 0.0004385 = 0.00219 mol iron(II)
Moles of iron(II) in the 500 cm 3 containing 12.18 g of hydrated salt.
= 0.00219 x 500/25 = 0.0439 mol
0.0439 mol = 12.18 g
1 mol = 12.18 / 0.0439 = 277.5 g
FeSO4 .x H2O has a formula mass of 277.5
FeSO4 contributes 151.9 to the RFM
That means the xH2O contributes 277.5 – 151.9 = 125.5
x = 125.5 / 18 = 6.97 = 7
The formula of the hydrated salt is FeSO4 .7 H2O
Question 6.
3.57 g of hydrated sodium carbonate was dissolved in water and the resulting solution was made up to 250 cm 3 in a volumetric flask.
A 25.0 cm 3 sample of this solution required 28.5 cm 3 of 0.100 moldm -3 hydrochloric acid to reach the end point.
Determine the number of water molecules of water of crystallisation.
The answers can be found here.
Question 7.
6.00 g of hydrated sodium carbonate was dissolved in water and the resulting solution was made up to 250 cm 3 in a volumetric flask.
A 25.0 cm 3 sample of this solution required 24.3 cm 3 of 0.200 moldm -3 sulfuric acid to reach the end point.
Determine the number of water molecules of water of crystallisation.
The answers can be found here.
Question 8.
Ammonium iron(II) sulfate crystals have the following formula:
(NH4)2SO4 . FeSO4 . nH2O.
In an experiment to find the value on n, 8.429 g of salt were dissolved and made up to 250 cm 3 of solution with distilled water and sulfuric acid.
A 25.0 cm 3 portion of this solution was titrated against 0.0150 moldm -3 KMnO4. 22.5 cm 3 was required. Calculate the value of n.
Note that 1 mol of KMnO4 requires 5 mol of iron(II) in the titration.
NO FURTHER KNOWLEDGE OF THIS REDOX TITRATRATION IN REQUIRED.
The answers can be found here.
Question 9. VERY DIFFICULT AND SHOULD BE DONE AT THE END OF GCE.
RFMs to use: HOOC—COOH.2H2O = 126; KOOC—COOK = 166.
The sample of solid ethanedioc acid (HOOC—COOH.2H2O) has been contaminated with potassium ethanedioate (KOOC—COOK . xH2O).
A 1.780 g sample of this mixture was made up to 250 cm 3 solution with distilled water.
A 25.0 cm 3 sample was titrated against 0.100 moldm -3 sodium hydroxide, requiring 17.35 cm 3.
Equation: HOOC—COOH + 2NaOH → NaOOC—COONa + 2H2O
THIS WILL GIVE US THE MOLES OF SOLID ETHANEDIOC ACID.
Another 25.0 cm 3 sample was acidified with sulfuric acid and titrated against 0.0200 moldm -3 KMnO4 solution, requiring 24.5 cm 3.
Equation: 2 mol of KMnO4 reacts with 5 mol of – OOC-COO –
THIS WILL GIVE US THE MOLES OF ETHANEDIOATE IONS
– AND WILL INCLUDE THOSE FROM THE CONTAMINANT.
Calculate the value of x.
The answers can be found here.