# Empirical Formula

## 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 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]

 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]

 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]

 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