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ASChemistryOCRRevision Notes2. Foundations in ChemistryAmount of SubstanceDetermining Formulae

Determining Formulae (OCR AS Chemistry A)

Revision Note

Richard Boole

Last updated

Defining Empirical & Molecular Formulae

  • The molecular formula shows the number and type of each atom in a molecule

    • E.g. the molecular formula of ethanoic acid is C2H4O2

  • The empirical formula shows the simplest whole number ratio of the elements present in one molecule of the compound

    • E.g. the empirical formula of ethanoic acid is CH2O

Worked Example

Deducing molecular & empirical formulae

Deduce the molecular and empirical formula of the following compounds:

Answer

Calculating Empirical & Molecular Formulae

Empirical formula

  • Empirical formula is the simplest whole number ratio of the elements present in one molecule or formula unit of the compound

  • It is calculated from knowledge of the ratio of masses of each element in the compound

  • The empirical formula can be found by determining the mass of each element present in a sample of the compound

  • It can also be deduced from data that gives the percentage compositions by mass of the elements in a compound

Worked Example

Empirical formula from mass Determine the empirical formula of a compound that contains 10 g of hydrogen and 80 g of oxygen.

Answer:

  • The above example shows how to calculate empirical formula from the mass of each element present in the compound

  • The example below shows how to calculate the empirical formula from percentage composition

Worked Example

Empirical formula from % Determine the empirical formula of a compound that contains 85.7% carbon and 14.3% hydrogen.

Answer:

Molecular formula

  • The molecular formula gives the exact numbers of atoms of each element present in the formula of the compound

  • The molecular formula can be found by dividing the relative molecular mass of the molecular formula by the relative formula mass of the empirical formula

  • Multiply the number of each element present in the empirical formula by this number to find the molecular formula

Worked Example

Calculating molecular formula The empirical formula of X is C4H10S and the relative molecular mass of X is 180.2 What is the molecular formula of X? (Ar data: C = 12.0, H = 1.0, S = 32.1)

Answer

Step 1: Calculate relative mass of the empirical formula

  • Relative empirical mass = (C x 4) + (H x 10) + (S x 1)

    • Relative empirical mass = (12.0 x 4) + (1.0 x 10) + (32.1 x 1)

    • Relative empirical mass = 90.1

Step 2: Divide relative molecular mass of X by relative empirical mass

  • Ratio between Mr of X and the Mr of the empirical formula = 180.2 / 90.1

    • Ratio between Mr of X and the Mr of the empirical formula = 2

Step 3: Multiply each number of elements by 2

  • (C4 x 2) + (H10 x 2) + (S x 2)

    • (C8) + (H20) + (S2)

    • Molecular formula of X is C8H20S2

Hydrated salts & Water of Crystallisation

  • Water of crystallisation is when some compounds can form crystals which have water as part of their structure

  • A compound that contains water of crystallisation is called a hydrated compound

  • The water of crystallisation is separated from the main formula by a dot when writing the chemical formula of hydrated compounds

    • E.g. hydrated copper(II) sulfate is CuSO4∙5H2O

  • A compound which doesn’t contain water of crystallisation is called an anhydrous compound

    • E.g. anhydrous copper(II) sulfate is CuSO4

  • A compound can be hydrated to different degrees

    • E.g. cobalt(II) chloride can be hydrated by six or two water molecules

    • CoCl2 ∙6H2O or CoCl2 ∙2H2O

  • The conversion of anhydrous compounds to hydrated compounds is reversible by heating the hydrated salt:

Hydrated:        CuSO4•5H2O ⇌ CuSO4 + 5H2O        :Anhydrous

  • The degree of hydration can be calculated from experimental results:

    • The mass of the hydrated salt must be measured before heating

    • The salt is then heated until it reaches a constant mass

    • The two mass values can be used to calculate the number of moles of water in the hydrated salt - known as the water of crystallisation

Worked Example

Calculating water of crystallisation 10.0 g of hydrated copper sulfate are heated to a constant mass of 5.59 g. Calculate the formula of the original hydrated copper sulfate. (Mr data: CuSO4 = 159.6, H2O = 18.0) 

Answer

List the components 

CuSO4

H2O

Note the mass of each component

5.59 g

10 - 5.59 = 4.41 g

Divide the component mass by the components Mr

fraction numerator 5.59 over denominator 159.6 end fraction = 0.035

fraction numerator 4.41 over denominator 18.0 end fraction = 0.245

Divide by the lowest figure to obtain the ratio

fraction numerator 0.035 over denominator 0.035 end fraction= 1

fraction numerator 0.245 over denominator 0.035 end fraction= 7

Hydrated salt formula

CuSO4•7H2O

Examiner Tips and Tricks

A water of crystallisation calculation can be completed in a similar fashion to an empirical formula calculation

  • Instead of elements, you start with the salt and water

  • Instead of dividing by atomic masses, you divide by molecular / formula masses

  • The rest of the calculation works the same way as the empirical formula calculation

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    Richard Boole

    Author: Richard Boole

    Expertise: Chemistry

    Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.