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.
Empirical = information gained by means of observation, experience, or experiment.

Example:

Chemical Substances	Molecular Formula	Empirical 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

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