Empirical formulae
The empirical formula gives the simplest ratio of the atoms within a compound. Some examples are given in the table below.
| Name | Molecular formula | Empirical formula |
|---|---|---|
| methane | CH4 | CH4 |
| cyclohexene | C6H10 | C3H5 |
| benzene | C6H6 | CH |
| ethanol | CH3CH2OH | C2H6O |
| ethanoic acid | CH3COOH | CH2O |
| propanal | CH3CH2CHO | C3H6O |
| butanoic acid | CH3CH2CH2COOH | C2H4O |
| ethyl ethanoate | CH3CH2OOCCH3 | C2H4O |
Points to note from this table:
- If the ratio in the formula cannot be simplified, the empirical formula of the compound is the same as the molecular formula.
- Different compounds can share the same empirical formula.
- The molecular formula can always be worked out by multiplying the empirical formula by a whole number.
Empirical formula calculation
Combustion data can be used for these types of problems. A known mass of the unknown compound is burned in excess oxygen and the masses of carbon dioxide and water produced are measured. Work through the following example.
Example
A 6.00 g sample of an organic compound that was known to contain carbon, hydrogen and oxygen was burned in excess oxygen. This produced 13.64 g of carbon dioxide and 5.58 g of water. Calculate the empirical formula of the compound.
Click on the headings below to reveal the stages in the calculation.
Calculate the mass of carbon and nitrogen in sample
mass of carbon in the CO2 = 13.64 × 
= 3.72 g
mass of hydrogen in the H2O = 5.58 × 
= 0.624 g
Calculate the mass of oxygen in sample
This is done by subtracting the masses of the carbon and hydrogen from the mass of the original sample. The remaining mass must be made up by oxygen.
mass of oxygen = 6.00 – 3.72 – 0.624 = 1.66 g
Calculate the empirical formula using the molar masses
The calculation should be set out in a table as seen here.
| molar mass (M) | (12.01) | (1.008) | (16.00) |
|---|---|---|---|
| elements | C | H | O |
| mass (m) | 3.72 | 0.624 | 1.66 |
 |
 |
 |
 |
| ratio of moles | 0.310 | 0.619 | 0.103 |
| divide by smallest value |
 = 3.0
|
 = 6.0
|
 = 1.0
|
| empirical formula | C3H6O |
The empirical formula is therefore C3H6O.
In this example it is known that the molecular formula of the substance is the same as the empirical formula.
In your notes name and draw the structures of two possible identities of the compound.
Click here to reveal possible answers
Propanal
Propanone
Describe a chemical test that could be used to confirm which of these two possibilities is the actual identity of the compound.
Hint: Think about the chemical properties of aldehydes and ketones.
Click here to reveal a possible answer
A small amount of acidified potassium permanganate solution is added to the sample of the unknown compound and the mixture warmed.
If there is no sign of a reaction, the compound is propanone, because ketones cannot be oxidised.
If there is a sign of a reaction (the permanganate changes from deep pink to colourless), the compound is likely to be propanal, because aldehydes can be oxidised.
The above example shows you how to use chemical knowledge combined with knowledge of the formula to work out the complete structure of the unknown compound. There are more examples like this for you to complete later.
