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O LevelChemistryCambridgeRevision Notes3. Stoichiometry3.3 The Mole & the Avogadro ConstantEmpirical & Molecular Formula

Empirical & Molecular Formula (Cambridge O Level Chemistry)

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Calculating Empirical & Molecular Formulae

Calculating Empirical Formula

  • The empirical formula is the simplest whole number ratio of the atoms of each element present in one molecule or formula unit of the compound
    • E.g. the empirical formula of ethanoic acid is CH2O
  • Organic molecules often have different empirical and molecular formulae
  • The formula of an ionic compound is always an empirical formula 

Worked example

A compound that contains 10 g of hydrogen and 80 g of oxygen.

What is its empirical formula of this compound?

Answer

  hydrogen oxygen
Write the mass of each element  10 g 80 g
Divide each mass by the relative atomic mass to find the number of moles 10/1 = 10 80/16 = 5
Find the molar ratio by dividing by the smallest number 10/5 = 2  5/5 = 1 

Empirical formula = H2O

Worked example

Substance X was analysed and found to contain 31.58% carbon, 5.26% hydrogen and 63.16% oxygen by mass.

What is the empirical formula of substance X?

Relative atomic masses, Ar:  C = 12;   H = 1;   O = 16

Answer

  carbon hydrogen oxygen
Convert % to g by assuming 100 g of substance is present 31.58 g 5.26 g 63.16 g
Divide each mass by the relative atomic mass to find the number of moles in 100 g 31.58/12 = 2.63 5.26/1 = 5.26 63.16/16 = 3.95
Find the molar ratio by dividing by the smallest number 2.63/2.63 = 1 5.26/2.63 = 2  3.95/2.63 = 1.5
Multiply all by 2 to obtain a whole number ratio 2 4 3

Empirical formula = C2H4O3

Examiner Tip

The molar ratio must be a whole number. If you don't get a whole number when calculating the ratio of atoms in an empirical formula, such as 1.5, multiply that and the other ratios to achieve whole numbers.

Calculating Molecular Formula

  • Molecular formula gives the actual numbers of atoms of each element present in the formula of the compound
  • To calculate the molecular formula:
    • Step 1: Find the relative formula mass of the empirical formula
    • Step 2: Use the following equation:

fraction numerator relative space formula space mass space of space molecular space formula over denominator relative space formula space mass space of space empirical space formula end fraction

    • Step 3: Multiply the number of each element present in the empirical formula by the number from step 2 to find the molecular formula

Table showing the Relationship between Empirical and Molecular Formula

Relationship between Empirical _ Molecular Formula table, IGCSE & GCSE Chemistry revision notes

Worked example

The empirical formula of X is C4H10S1

The relative formula mass (Mr ) of X is 180.

What is the molecular formula of X?

(Relative atomic mass, Ar:       Carbon : 12      Hydrogen : 1      Sulfur : 32 )

Answer

Step 1 - Calculate the relative formula mass of the empirical formula

Mr =  (12 x 4) + (1 x 10) + (32 x 1)   =   90

Step 2 - Divide relative formula mass of X by relative formula mass of empirical formula

180 / 90 = 2

Step 3 - Multiply each number of elements by 2

(C4 x 2) + (H10 x 2) + (S1 x 2)     

Molecular Formula of X = C8H20S2

Deducing formulae of hydrated salts

  • The formula of hydrated salts can be determined experimentally by weighing a sample of the hydrated salt, heating it until the water of crystallisation has been driven off, then reweighing the now anhydrous salt
  • From the results, you can determine the mass of anhydrous salt and the mass of the water of crystallisation
  • Applying a similar approach to deducing empirical formulae, the formula of the hydrated salt can be calculated

Worked example

11.25 g of hydrated copper sulfate, CuSO4.xH2O, is heated until it loses all of its water of crystallisation. It is reweighed and its mass is 7.19 g. What is the formula of the hydrated copper(II) sulfate?

Answer

  CuSO4 H2O
Deduce the mass of water of crystallisation and anhydrous salt 

= mass of salt after heating

= 7.19 g

mass of hydrated salt - mass of anhydrous salt:

11.25 - 7.19 = 4.06 g
Divide each mass by the relative formula mass to find the number of moles fraction numerator 7.19 over denominator 160 end fraction space equals space 0.045 fraction numerator 4.06 over denominator 18 end fraction space equals space 0.226
Find the molar ratio by dividing by the smallest number fraction numerator 0.045 over denominator 0.045 end fraction space equals space 1  

fraction numerator 0.226 over denominator 0.045 end fraction space equals space 5


Therefore the formula of hydrated copper(II) sulfate is CuSO4.5H2O

Examiner Tip

The specification is not clear about whether deducing the formula of hydrated salts is required, however, it is an application of deducing empirical formulae so it is worth knowing how to do this.

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    Caroline

    Author: Caroline

    Expertise: Physics Lead

    Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.