A formula that gives the simplest whole number ration of atoms in a compound found from an experiment. Empirical Formula means ‘from experiment’.

Molecular formula is the actual ration for a molecule

For example for Benzene:

– Actual Formula: C6H6

– Empirical Formula: CH

Examples

  • Determine the Empirical Formula of methane given that 6.0g of Methane can be decomposed into 4.5g of carbon and 1.5g of hydrogen. Work out the empirical formula.

Carbon

Hydrogen

Mass

4.5g

1.5g

Ar

12

1

Moles

4.5g / 12 = 0.375

1.5g /1 = 11.5

Divide by smallest

0.375 / 0.375

1.5 / 0.375

Ratio

1

4

= CH4

Empirical Formula from percentages

Empirical formula can be used using percentages, just using the percentage instead of it’s mass.

Examples

  • The composition of a compound is 40% sulphur and 60% oxygen by weight. What is its empirical formula?

Sulphur

Oxygen

Mass

40%

60%

Ar

32.1

16.0

Moles

40 / 32.1 = 1.25

60% / 16.0 = 3.75

Divide by smallest

1.25 / 1.25

3.25 / 1.25

Ratio

1

3

= SO3

  • Calculate the percentage of iron present in Fe2O3

2Fe 2 x 55.8 x100 = 69.92%

Fe2O2 x 55.8 + 48

  • Percentage composition of 81.8% carbon, 6.1% hydrogen and 12.1% oxygen.

Carbon

Oxygen

Hydrogen

Mass

81.8%

12.1%

6.1%

Ar

12.0

16.0

1

Moles

81.8 / 12.0 = 6.16

12.1% / 16.0 = 0.75625

6.1 / 1 = 6.1

Divide by smallest

6.16 / 0.75625 

 0.75625/ 0.75625

6.1/0.75625

Ratio

9

1

8

= C9H8O

  • Blue Sulphate crystals yield the following percentage composition, Cu – 25%, S – 12.8%, O – 25.6% and water crystallisation – 36.1%. Find the empirical formula of the crystals

Copper

Sulphur

Oxygen

H2O

Mass

25.0%

12.1%

25.6%

36.1%

Ar

63.5

32.1

16.0

18

Moles

25.0 / 63.5 = 0.39

12.1% / 32.1 = 0.37

25.6 / 16.0 = 1.60

36.1 / 18 =2.01

Divide by smallest

0.39 / 0.37 

0.37 / 0.37

1.60 / 0.37

2.01/ 0.37

Ratio

1

1

4

5

= CuSO4 • 5H2O

  • A sugar solution contains 40% carbon, 6.7% hydrogen and 53.3% oxygen.

Carbon

Oxygen

Hydrogen

Mass

40.0%

53.3%

6.7%

Ar

12.0

16.0

1

Moles

40.0 / 12.0 = 3.3

53.3% / 16.0 = 3.3

6.7 / 1 = 6.7

Divide by smallest

3.3 / 3.3 

3.3 / 3.3

3.3 / 3.3

Ratio

1

1

2

= CH2O

  • 24% Carbon and 76% fluorine

Carbon

Fluorine

Mass

24%

76%

Ar

12

19

Moles

24 / 12 = 2

76% / 19 =4

Divide by smallest

2 / 2 

1.5 / 0.375

Ratio

1

2

Calculating molecular formula from empirical formula

This can be deduced from the empirical formula if the Mr is known (as the Mr can come from mass spectrometry)

In mass spectrometry the molecular ion is the last peak on the graph. The Mr is this value.

21% in this example graph is the Molecular Ion.

A compound has an empirical formula of ClCH2 and a molecular weight of 98.96 g/mol. What is its molecular weight formula

Mass of empirical unit?

Mass of Chlorine + Mass of Carbon + Mass of Hydrogen2

35.5 + 12 + 2 = 49.50 g/mol

Empirical Formula = Mass of Molecular unit = 98.96

Mass of Empirical Formula 49.50

= 2.00 empirical units per molecular unit

It takes two empirical units to make a molecular unit, so the molecular formula is Cl2C2H4