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GCSE and Advanced level Chemistry Calculations: Water of crystallization

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WATER of CRYSTALLISATION CALCULATIONS

hydrated and anhydrous salts (re-edit)

Doc Brown's Chemistry - GCSE/IGCSE/GCE (basic A level) O Level  Online Chemical Calculations

study examples carefully14. Other GCSE chemical calculations - WATER OF CRYSTALLISATION - method of determining and calculations


Keywords: Quantitative chemistry calculations How to determine the water of crystallisation in a salt like compound, hence determine the full formula of the hydrated salt. How to calculate the % water in a hydrated salt i.e. the percentage of water of crystallisation in a salt. How to deduce the number of molecules of water of crystallisation in a salt - fully worked out example calculations of water of crystallisation. Online practice exam chemistry CALCULATIONS and solved problems for KS4 Science GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB courses. These revision notes and practice questions on how to do water of crystallisation chemical calculations and worked examples should prove useful for the new AQA, Edexcel and OCR GCSE (91) chemistry science courses.

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soluble salt preparation from insoluble base-acid neutralisation14.4 Water of crystallisation in a crystallised salt

  • What is water of crystallisation?

    • Water of crystallization are the molecules of water that are incorporated into some salt crystals when they are crystallised out of water.

    • e.g. when blue crystals of copper(II) sulfate are crystallized out of water the actual formula of the crystals is ...

    • NOT simply CuSO4, but on crystallisation CuSO4.5H2O is formed from the aqueous solution,

    • because five water molecules are associated with each 'CuSO4'. in its 'hydrated' crystalline form.

  • Example 14.4.1: Reminders on calculating formula mass e.g. with MgSO4.7H2O

    • Relative atomic masses: Mg = 24, S = 32, O = 16 and H = 1

    • You need to add together the formula mass of MgSO4 plus the relative mass of seven water molecules.

    • Relative formula mass of water = (2 x 1) + 16 = 18

    • Relative formula mass of MgSO4 = 24 + 32 + (4 x 16) = 120

    • Relative mass of seven water molecules = 7 x 18 = 126

    • Relative formula mass of crystals = MgSO4 + (7 x H2O) = 120 + 126 = 246

  • Example 14.4.2 How to calculate the theoretical % of water in a hydrated salt

    • eg magnesium sulphate MgSO4.7H2O 'hydrated' salt crystals

    • Relative atomic masses: Mg = 24, S = 32, O = 16 and H = 1

    • Relative formula mass of crystals = 24 + 32 + (4 x 16) + {7 x (1 + 1 + 16)} = 246

    • Relative mass of seven water molecules = 7 x 18 = 126

    • so % water = 126 x 100 / 246 = 51.2%

  • Example 14.4.3 Determination and calculation of salt formula containing 'water of crystallisation'.

    • Some salts, when crystallised from aqueous solution, incorporate water molecules into the structure.

    • This is known as 'water of crystallisation', and the 'hydrated' form of the compound.

    • e.g. magnesium sulphate MgSO4.7H2O. The formula can be determined by a simple experiment (see the copper sulphate example below).

    • A known mass of the hydrated salt is gently heated in a crucible until no further water is driven off and the weight remains constant despite further heating.

      • The mass of the anhydrous salt left is measured.

      • The original mass of hydrated salt and the mass of the anhydrous salt residue can be worked out from the various weighings.

    • The % water of crystallisation and the formula and formula mass of the salt are calculated as follows:

      • Suppose 6.25g of blue hydrated copper(II) sulphate, CuSO4.xH2O, (x unknown) was gently heated in an evaporating dish until the mass of dish plus salt remaining was a constant leaving 4.00g of the anhydrous (dehydrated) salt.

        • (I'm just ignoring the actual masses of the evaporating dish before and after heating, but that maths is pretty simple!)

      • When the mass on subsequent weighings stays constant, you know all the water of crystallisation has driven off by the heat.

      • This is the white anhydrous copper(II) sulphate.

      • The mass of anhydrous salt = 4.00g, mass of water (of crystallisation) driven off = 6.25-4.00 = 2.25g

      • The % water of crystallisation in the crystals  is 2.25 x 100 / 6.25 = 36%

      • [ Ar values: Cu=64, S=32, O=16, H=1 ]

      • The mass ratio of CuSO4 : H2O is 4.00 : 2.25 (or 64% : 36%, doesn't matter which mass ratio you use)

      • To convert from mass ratio to mole ratio, you divide by the masses molecular /f formula mass Mr of each 'species'

      • Mr CuSO4 = 64 + 32 + (4x18) = 160

      • and Mr H2O = 1+1+16 = 18

      • The mole ratio of CuSO4 : H2O is 4.00/160 : 2.25/18

      • which is 0.025 : 0.125 or 1 : 5

      • (divide through by 0.025 to get the simplest integer ratio)

      • so the formula of the hydrated salt is CuSO4.5H2O

      • The formula mass Mr can then be calculated as follows:

        • from the calculation a few lines above Mr CuSO4.5H2O = 160 + (5 x 18) = 250

  • There are some More sophisticated A Level problems involving titrations to determine the water of crystallisation in a salt like compound (See Q30)

  • Advanced level pre-university chemistry notes on the structure of hydrated salts


Practice questions involving water of crystallization

study the mole examples carefully You must be able to do calculations involving moles and mole ratios!

Q1 The formula for the blue hydrated crystals of copper(II) sulfate is CuSO4.5H2O.

Relative atomic masses: Cu = 63.5   S = 32.0   O = 16.0   H = 1.00

Calculate the % water of crystallisation in the crystals (to 3 sf).

Worked out ANSWERS to the water of crystallisation questions

 

Q2 The hydrated salt of cobalt(II) chloride has the formula CoCl2.xH2O.

From a heating-dehydration experiment it was found that the hydrated salt lost the equivalent of 21.7% water.

Relative atomic masses: Co = 59.0   Cl = 35.5   O = 16.0   H = 1.00

Deduce the value of x and hence the formula of hydrated cobalt(II) chloride.

Worked out ANSWERS to the water of crystallisation questions

 

Q3 An evaporating dish weighed 30.55 g. Pale green crystals of hydrated iron(II) sulfate (FeSO4.xH2O) were added until the dish and contents weighed 40.75 g. After gentle heating to constant weight, the dish and anhydrous salt weighed 36.12 g.

Relative atomic masses: Fe = 55.8   S = 32.0   O = 16.0   H = 1.00

Deduce the value of x and hence the formula of the hydrated iron(II) sulfate salt.

Worked out ANSWERS to the water of crystallisation questions

 

Q4 Crystals of Glauber's, a hydrated form of the salt sodium sulfate, has the formula Na2SO4.xH2O.

3.578 g of the salt was heated to dehydrate it, leaving a residue of 1.578 g

Deduce the formula of Glauber's salt.

Worked out ANSWERS to the water of crystallisation questions


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See also

14.1 % purity of a product and assay calculations

14.2a % reaction yield and theoretical yield calculations and why you can't actually get 100% yield in practice

14.2b atom economy calculations

14.3 dilution of solutions calculations 

14.5 how much of a reactant is needed? calculation of quantities required, limiting reactant quantities

Chemical & Pharmaceutical Industry Economics & Sustainability, Life Cycle Assessment, Recycling

Index of all my online chemical calculation notes and quizzes

All my GCSE/IGCSE/US grade 8-10 level Chemistry Revision notes

Advanced UK A/AS level, IB and US grade 11-12 pre-university chemistry

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OTHER CALCULATION PAGES

1. What is relative atomic mass, relative isotopic mass, calculating relative atomic mass

2. Calculating relative formula/molecular mass of a compound or element molecule

3. Law of Conservation of Mass and simple reacting mass calculations

4. Composition by percentage mass of elements in a compound

5. Empirical formula and formula mass of a compound from reacting masses (easy start, not using moles)

6a. Reacting mass ratio calculations of reactants and products from equations (NOT using moles) and brief mention of actual percent % yield and theoretical yield, atom economy and formula mass determination (see calculations section 14.)

6b. Reacting masses, concentration of solution and volumetric titration calculations (NOT using moles)

7. Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations (relating reacting masses and formula mass)

8. Using moles to calculate empirical formula and deduce molecular formula of a compound/molecule (starting with reacting masses or % composition)

9. Moles and the molar volume of a gas, Avogadro's Law

10. Reacting gas volume ratios, Avogadro's Law and Gay-Lussac's Law (ratio of gaseous reactants to products)

11. Molarity, volumes and solution concentrations (and diagrams of apparatus) (this page)

12. How to do acid-alkali titration calculations, diagrams of apparatus, details of procedures

13. Electrolysis products calculations (negative cathode and positive anode products)

14. Other calculations e.g. % purity, % percentage & theoretical yield, dilution of solutions (and diagrams of apparatus), water of crystallisation, quantity of reactants required, atom economy

14.1 % purity of a product 14.2a % reaction yield

14.2b atom economy 14.3 dilution of solutions

14.4 water of crystallisation calculation

14.5 how much of a reactant is needed? limiting reactant calculations

Energy transfers in physical/chemical changes, exothermic/endothermic reactions

Gas calculations involving PVT relationships, Boyle's and Charles Laws

Radioactivity and half-life calculations including dating materials

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 Website content Dr Phil Brown 2000+. All copyrights reserved on Doc Brown's chemistry revision notes, images, quizzes, worksheets etc. Copying of Doc Brown's website material is NOT permitted. Exam revision summaries and references to chemistry science course specifications are unofficial.

Worked out ANSWERS to the water of crystallisation questions

study the mole examples carefully You must be able to do calculations involving moles and mole ratios!

Q1 The formula for the blue hydrated crystals of copper(II) sulfate is CuSO4.5H2O.

Relative atomic masses: Cu = 63.5   S = 32.0   O = 16.0   H = 1.00

Calculate the % water of crystallisation in the crystals (to 3 sf).

Formula mass of hydrated salt = 63.5 + 32 + (4 x 16) + 5 x (1 + 1 + 16) = 249.5

Relative mass of 5H2O = 5 x 18 = 90

Therefore % water in crystals = 100 x 90 / 249.5 = 36.1 %

 

Q2 The hydrated salt of cobalt(II) chloride has the formula CoCl2.xH2O.

From a heating-dehydration experiment it was found that the hydrated salt lost the equivalent of 21.7% water.

Relative atomic masses: Co = 59.0   Cl = 35.5   O = 16.0   H = 1.00

Deduce the value of x and hence the formula of hydrated cobalt(II) chloride.

Formula mass of CoCl2 = 59.0 + (2 x 35.5) = 130

Formula mass of H2O = 1 + 1 + 16 = 18.

Imagine you have 100 g of the hydrated salt

Mass ratio of CoCl2 : H2O  is  78.3 : 21.7

study the mole examples carefullyMoles = mass / formula mass

So mole ratio of CoCl2 : H2O  is  78.3 / 130 : 21.7 / 18.0 = 0.602 : 1.206

Dividing through by 0.602 gives a mole ratio of 1 : 2.003

This is well within experimental error to be a 1 : 2 ratio.

Therefore the value of x = 2 and the formula of the cobalt salt is CoCl2.2H2O

 

Q3 An evaporating dish weighed 30.55 g. Pale green crystals of hydrated iron(II) sulfate (FeSO4.xH2O) were added until the dish and contents weighed 40.75 g. After gentle heating to constant weight, the dish and anhydrous salt weighed 36.12 g.

Relative atomic masses: Fe = 55.8   S = 32.0   O = 16.0   H = 1.00

Deduce the value of x and hence the formula of the hydrated iron(II) sulfate salt.

Mass of hydrated salt at the start = 40.75 - 30.55 = 10.20

Mass of dehydrated salt left after heating = 36.12 - 30.55 = 5.57 g

Therefore mass of water driven off on heating = 10.20 - 5.57 = 4.63 g

The mass ratio FeSO4 : H2O  is  5.57 : 4.63

Formula mass FeSO4 = 55.8 + 32 + (4 x16) = 151.8

Formula mass water = 1 + 1 + 16 = 18

study the mole examples carefullymoles = mass / formula mass

So, mole ratio FeSO4 : H2O  is  5.57 / 151.8 : 4.63 / 18 = 0.0367 :  0.257

Dividing through by 0.0367 gives a mole for FeSO4 : H2O   of  1: 7.003

Therefore within experimental error x = 7  and the formula of the salt is FeSO4.7H2O

 

Q4 Crystals of Glauber's, a hydrated form of the salt sodium sulfate, has the formula Na2SO4.xH2O.

3.578 g of the salt was heated to dehydrate it, leaving a residue of 1.578 g

Deduce the formula of Glauber's salt.

Relative atomic masses: Na = 23.0   S = 32.0   O = 16.0   H = 1.00

Mass of water driven off = 3.578 - 1.578 = 2.000 g

Mass ratio  Na2SO4 : H2O = 1.578 : 2.000

Formula masses: Na2SO4 = (2 x 23) + 32 + (4 x 16) = 142  and  H2O = 18

study the mole examples carefullyMole ratio for Na2SO4 : H2O = 1.578 / 142 : 2.000 / 18 = 0.0111 : 0.111

The molar ratio is 1 : 10 so the formula of Glauber's salt is Na2SO4.10H2O

Website content Dr Phil Brown 2000+.

All copyrights reserved on revision notes, images, quizzes, worksheets etc.

Copying of website material is NOT permitted. Exam revision summaries & references to science course specifications are unofficial.

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