SPM Chemistry

Empirical Formula

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Empirical Formula

  1. The empirical formula of a substance is the chemical formula that gives the simplest whole-number ratio of atoms of each element in the substance.
  2. Empirical = information gained by means of observation, experience, or experiment.

Example:

Chemical SubstancesMolecular FormulaEmpirical Formula
Glucose
C6H12O6
CH2O
Water
H2O
H2O
Carbon Dioxide
CO2
CO2
Benzene
C6H6
CH
Butane
C4H8
CH2

Finding Empirical Formula

Steps to determine the empirical formula of a compound
STEP 1: Find the mass
STEP 2: Find the mole
STEP 3: Find the simplest ratio

Example:
In a chemical reaction, 4.23g of iron reacts completely with 1.80g of oxygen gas, producing iron oxide. Calculate the empirical formula of iron oxide. [Relative atomic mass: Iron = 56; Oxygen = 16]

Answer:

Element
Fe
O
Mass
 4.23g
 1.80g
Number of mole
4.23/56 =0.0755
1.80/16 =0.1125
Simple ratio
0.0755/0.0755 =1
0.1125/0.0755 =1.5
Ratio in round number
2
3

The empirical formula of iron oxide = Fe2O3

Example:
Determine the empirical formula of a compound which has a percentage of composition Mg: 20.2%, S: 26.6%, O: 53.2%. [Relative atomic mass: Mg = 24; S = 32; O = 16]

Answer

Element
Mg
S
O
Percentage
20.2%
26.6%
53.2%
Mass in 100g
20.2g
26.6g
53.2g
Number of mole
20.2/24 =0.8417mol
26.6/32 =0.8313mol
53.2/16 =3.325mol
Simple ratio
0.8417/0.8313 =1
0.8313/0.8313 =1
3.325/0.8313 =4

The empirical formula of the compound is MgSO4

Example:
From an experiment, a scientist found that a hydrocarbon contains 85.7% of carbon according to its mass. Find the empirical formula of the hydrocarbon. [Relative atomic mass: Carbon = 12; Hydrogen = 1]

Answer:

Element
C
H
Percentage
85.7%
14.3%
Mass in 100g
85.7g
14.3g
Number of mole
85.7/12 =7.142mol
14.3/1 =14.3mol
Simple ratio
7.142/7.142 =1
14.3/7.142 =2

 The empirical formula of the hydrocarbon = CH2