Empirical Formula
- 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