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:
1. Write the element
2. 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
3. Write the relative atomic mass of each element
4. Calculate the moles of each element
  - Moles = $\frac{mass}{A_{r}}$
5. 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
6. 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.

A_r (H) = 1 A_r (O) = 16

Answer:

1. Element	H	O
2. Value	10	80
3. Relative atomic mass	1	16
4. Moles = $\frac{mass}{A_{r}}$	$\frac{10}{1}$ = 10	$\frac{80}{16}$ = 5
5. Ratio (divide by smallest)	$\frac{10}{5}$ = 2	$\frac{5}{5}$ = 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.

A_r(H) = 1 A_r(C) = 12 A_r(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 = $\frac{mass}{A_{r}}$	$\frac{31.58}{12}$ = 2.63	$\frac{5.26}{1}$ = 5.26	$\frac{63.16}{16}$ = 3.95
5. Ratio (divide by smallest)	$\frac{2.63}{2.63}$ = 1	$\frac{5.26}{2.63}$ = 2	$\frac{3.95}{2.63}$ = 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:
1. Find the relative formula mass of the empirical formula
  - Add the relative atomic masses of all the atoms in the empirical formula
2. Use the following equation:
  - $\frac{relative formula mass of molecular formula}{relative formula mass of empirical formula}$
3. 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 (M_r) of X is 180.

Calculate the molecular formula of X.

A_r (C) = 12 A_r (H) = 1 A_r (S) = 32

Answer:

Calculate the relative formula mass of the empirical formula:
- M_r= (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
- (C_{4 x 2}) + (H_{10 x 2}) + (S_{1 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:
1. Write the salt and water
2. Write the value given for the salt and water
  - There are exam questions where you are required to calculate one of these values
3. Write the relative molecular / formula mass of the salt and water
4. Calculate the moles of the salt and water
  - Moles = $\frac{mass}{M_{r}}$
5. 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
6. 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.

A_r (Cu) = 63.5 A_r (S) = 32 A_r (O) = 16 A_r (H) = 1

Answer:

1. Salt and water	CuSO₄	H₂O
2. Value	7.19	11.25 - 7.19 = 4.06
3. M_r	63.5 + 32 + (16 x 4) = 159.5	(1 x 2) + 16 = 18
4. Moles = $\frac{mass}{M_{r}}$	$\frac{7.19}{159.5}$ = 0.045	$\frac{4.06}{18}$ = 0.226
5. Salt : water ratio	$\frac{0.045}{0.045}$ = 1	$\frac{0.226}{0.045}$ = 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.

Empirical & Molecular Formulae (Edexcel IGCSE Chemistry: Double Science)

Revision Note