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ASChemistryCambridge (CIE)Revision Notes2. Atoms, Molecules & StoichiometryFormulasEmpirical & Molecular Formulae

Empirical & Molecular Formulae (Cambridge (CIE) AS Chemistry)

Revision Note

Richard Boole

Last updated

Empirical & Molecular Formulae

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

    • Eg. the molecular formula of ethanoic acid is C2H4O2

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

    • Eg. the empirical formula of ethanoic acid is CH2O

  • Organic molecules often have different empirical and molecular formulae

  • Simple inorganic molecules however have often similar empirical and molecular formulae

  • Ionic compounds always have similar empirical and molecular formulae

Empirical & Molecular Formulae Calculations

Empirical formula

  • The 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

Calculating empirical formula from mass

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

Answer:

Elements

Hydrogen

Oxygen

Mass of each element (g)

10 

80

Atomic mass

1.0

16.0

Moles = mass / Ar

begin mathsize 14px style fraction numerator 10 over denominator 1.0 end fraction end style = 10

fraction numerator 80 over denominator 16.0 end fraction = 5

Ratio (divide by smallest value)

10 over 5 = 2

5 over 5 = 1

  • So, the empirical formula of the compound is H2O

Worked Example

Calculating empirical formula from %

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

Answer:

Elements

Carbon

Hydrogen

Mass of each element (%)

85.7 

14.3

Atomic mass

12.0

1.0

Moles = mass / Ar

fraction numerator 85.7 over denominator 12.0 end fraction = 7.14

fraction numerator 14.3 over denominator 1.0 end fraction = 14.3

Ratio (divide by smallest value)

begin mathsize 14px style fraction numerator 7.14 over denominator 7.14 end fraction end style = 1

fraction numerator 14.3 over denominator 7.14 end fraction = 2.00

  • So, the empirical formula of the compound is CH2 

Molecular formula

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

  • The molecular formula can be found by dividing the relative formula 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 formula mass of X is 180.

What is the molecular formula of X?

Relative atomic mass   Carbon: 12   Hydrogen:1   Sulfur:32

Answer:

  • Step 1: Calculate the relative formula mass of the empirical formula

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

    • Relative formula mass = (12 x 4) + (1 x 10) + (32 x 1)

    • Relative formula mass = 90

  • Step 2: Divide the relative formula mass of X by the relative formula mass of the empirical formula

    • Ratio between Mr of X and the Mr of the empirical formula = 180/90

    • 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) + (S1 x 2) = (C8) + (H20) + (S2)

    • The molecular formula of X is C8H20S2

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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.