OCR AS Chemistry

Revision Notes

2.2.2 Determining Formulae

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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.

  • 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.

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 CuSO45H2O

  • 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•5H2OCuSO4 + 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




Exam Tip

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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