Empirical & Molecular Formulae (Edexcel IGCSE Chemistry (Modular))
Revision Note
Empirical & Molecular Formulae
The molecular formula is the formula that shows the number and type of each atom in a molecule
E.g. the molecular formula of ethanoic acid is C2H4O2
The empirical formula is the simplest whole number ratio of the atoms of each element present in one molecule or formula unit of the compound
E.g. the empirical formula of ethanoic acid is CH2O
Organic molecules often have different empirical and molecular formulae
The formula of an ionic compound is always an empirical formula
Calculating empirical and molecular formula
How to calculate empirical formulae
Empirical formula calculations are very methodical
Use a table and the following steps to complete an empirical formula calculation:
Write the element
Write the value given for each element
This may be given as a mass, in g, or as a percentage
There are exam questions where you are required to calculate the value of one of the elements
Write the relative atomic mass of each element
Calculate the moles of each element
Moles =
Calculate the ratio of elements
Divide all the moles by the smallest number of moles
If you get a ratio that does not have whole numbers, you multiply by an appropriate number to make all the values into whole numbers
Write the final empirical formula
Worked Example
A sample of a compound was found to contain 10 g of hydrogen and 80 g of oxygen.
Calculate the empirical formula of this compound.
Ar (H) = 1 Ar (O) = 16
Answer:
1. Element | H | O |
2. Value | 10 | 80 |
3. Relative atomic mass | 1 | 16 |
4. Moles = | = 10 | = 5 |
5. Ratio (divide by smallest) | = 2 | = 1 |
6. Answer | The empirical formula is H2O |
Worked Example
Carbohydrate X was analysed and found to contain 31.58% carbon and 5.26% hydrogen by mass.
Find the empirical formula of carbohydrate X.
Ar (H) = 1 Ar (C) = 12 Ar (O) = 16
Answer:
A carbohydrate contains carbon, hydrogen and oxygen
The percentages do not add up to 100%, which means that you need to calculate the percentage of oxygen needs to be calculated
Percentage of oxygen = 100 - 31.58 - 5.26 = 63.16%
1. Element | C | H | O |
2. Value | 31.58 | 5.26 | 63.16 |
3. Relative atomic mass | 12 | 1 | 16 |
4. Moles = | = 2.63 | = 5.26 | = 3.95 |
5. Ratio (divide by smallest) | = 1 | = 2 | = 1.5 |
5. Whole number ratio | 1 x 2 = 2 | 2 x 2 = 4 | 1.5 x 2 = 3 |
6. Answer | The empirical formula is C2H4O3 |
Examiner Tip
The molar ratio must be a whole number.
If you don't get a whole number when calculating the ratio of atoms in an empirical formula, such as 1.5, multiply that and the other ratios to achieve whole numbers.
How to calculate molecular formula
Molecular formula gives the actual numbers of atoms of each element present in the formula of the compound
Table showing the relationship between empirical and molecular formulae
Compound | Empirical formula | Molecular formula |
---|---|---|
Methane | CH4 | CH4 |
Ethane | CH3 | C2H6 |
Ethene | CH2 | C2H4 |
Benzene | CH | C6H6 |
To calculate the molecular formula:
Find the relative formula mass of the empirical formula
Add the relative atomic masses of all the atoms in the empirical formula
Use the following equation:
Multiply the number of each element present in the empirical formula by the number from step 2 to find the molecular formula
Worked Example
The empirical formula of X is C4H10S1
The relative formula mass (Mr ) of X is 180.
Calculate the molecular formula of X.
Ar (C) = 12 Ar (H) = 1 Ar (S) = 32
Answer:
Calculate the relative formula mass of the empirical formula:
Mr = (12 x 4) + (1 x 10) + (32 x 1) = 90
Divide the relative formula mass of X by the relative formula mass of empirical formula:
180 / 90 = 2
Multiply for the molecular formula:
The number of atoms of each elements should be multiplied by 2
(C4 x 2) + (H10 x 2) + (S1 x 2)
Molecular formula of X = C8H20S2
Deducing formulae of hydrated salts
A hydrated salt is a crystallised salt that contains water molecules as part of its structure
The formula of a hydrated salt shows the water molecules, e.g. CuSO4•2H2O
The • symbol shows that the water present is water of crystallisation
The formula of hydrated salts can be determined experimentally by:
Weighing a sample of the hydrated salt
Heating it until the water of crystallisation has been driven off
This is achieved by heating until a constant mass
Re-weighing the anhydrous salt
From the results, you can determine the mass of anhydrous salt and the mass of the water of crystallisation
Applying a similar approach to deducing empirical formulae, the formula of the hydrated salt can be calculated
How to calculate water of crystallisation
The steps for empirical formula can be adapted for hydrated salt / water of crystallisation calculations
Instead of writing elements, write the two components of a hydrated salt
The salt
Water
Instead of writing relative atomic mass, write the relative molecular / formula mass of the salt and water
Use a table and the following steps to complete the calculation:
Write the salt and water
Write the value given for the salt and water
There are exam questions where you are required to calculate one of these values
Write the relative molecular / formula mass of the salt and water
Calculate the moles of the salt and water
Moles =
Calculate the ratio salt : water
Divide all the moles by the smallest number of moles
The calculation should give a ratio of 1 salt : x water
Write the final hydrated salt formula
Worked Example
11.25 g of hydrated copper sulfate, CuSO4.xH2O, is heated until it loses all of its water of crystallisation.
It is re-weighed and its mass is 7.19 g.
Calculate the formula of the hydrated copper(II) sulfate.
Ar (Cu) = 63.5 Ar (S) = 3. Ar (O) = 16 Ar (H) = 1
Answer:
1. Salt and water | CuSO4 | H2O |
2. Value | 7.19 | 11.25 - 7.19 |
3. Mr | 63.5 + 32 + (16 x 4) | (1 x 2) + 16 |
4. Moles = | = 0.045 | = 0.226 |
5. Salt : water ratio | = 1 | = 5 |
6. Formula of hydrated salt | The formula is CuSO4•5H2O |
|
Examiner Tip
The specification is not clear about whether deducing the formula of hydrated salts is required.
However, it is an application of deducing empirical formulae so it is worth knowing how to do this.
Last updated:
You've read 0 of your 10 free revision notes
Unlock more, it's free!
Did this page help you?