Determination of Formulae

This lesson covers: 

  1. The difference between empirical and molecular formulae
  2. Calculating empirical formula from experimental data
  3. Determining molecular formula when given molecular mass
  4. Writing formulae for ionic compounds

Empirical vs. molecular formula

The empirical formula gives the simplest whole number ratio of atoms of each element present in a compound.

The molecular formula gives the actual number and type of atoms of each element in a molecule.

For example, glucose has:

  • An empirical formula of CH2O as this gives the simplest ratio.
  • A molecular formula of C6H12O6 as this shows the actual numbers of atoms in a glucose molecule.


So the molecular formula is always a whole number multiple of the empirical formula.

Calculating empirical formula

The empirical formula of a compound can be determined from experimental data including:

  • Percentage composition by mass
  • Masses of products from combustion reactions


Percentage composition

Using the percentage mass contribution of each element, one can calculate the empirical formula as follows:

  1. Assume a total sample size of 100 g - This simplifies the calculation as percentage values directly translate to grams.
  2. Calculate the moles of each element based on its percentage mass.
  3. Divide the mole values by the smallest mole number among them to simplify the ratio to its lowest form.
  4. Construct the empirical formula from this simplified mole ratio.

Combustion reactions

Empirical formulas can also be derived from the masses of combustion products:

  1. Determine moles of products using their mass and molar mass.
  2. Infer moles of each element in the original compound from these product moles.
  3. Simplify to the lowest mole ratio.
  4. Derive the empirical formula from this ratio.

Worked example 1 - Calculating empirical formula from percentage composition

Calculate the empirical formula for a compound with 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.


Step 1: Assume 100 g of the compound

Assuming a 100 g sample makes the percentages equal to grams:

C: 40.0 g

H: 6.7 g

O: 53.3 g


Step 2: Convert mass to moles

C: n =Ar ​m​=12.040.0​=3.33 mol

H: n =Ar ​m​=1.06.7​=6.7 mol

O: n =Ar ​m​=16.053.3​=3.33 mol


Step 3: Determine the simplest mole ratio

Divide all mole quantities by the smallest number of moles calculated:

C: n =3.333.33​=1

H: n =3.336.7​≈2

O: n =3.333.33​=1


Step 4: Write the empirical formula

The empirical formula is CH2O.

Worked example 2 - Calculating the empirical formula from combustion data

A hydrocarbon sample weighing 2.20 g was completely combusted in oxygen, producing 3.30 g of CO2 and 1.35 g of H2O.

Determine the empirical formula of the hydrocarbon.


Step 1: Calculate moles of CO2 produced

moles of CO2​ =Mr ​mass​=44.03.30​=0.075 mol


Step 2: Calculate moles of carbon in the sample

Each molecule of CO2 contains 1 atom of carbon.

Moles of C atoms = Moles of CO2 = 0.075 mol


Step 3: Calculate moles of H2O produced

moles of H2​O =Mr ​mass​=18.01.35​=0.075 mol


Step 4: Calculate moles of hydrogen in the sample

Each molecule of H2O contains 2 atoms of hydrogen.

Moles of H atoms = 2 x moles of H2O = 2 x 0.075 = 0.150 mol


Step 5: Determine the simplest C : H mole ratio

C : H = 0.075 : 0.150 = 1 : 2


Step 6: Write the empirical formula

The empirical formula is CH2.

Worked example 3 - Determining the molecular formula from empirical formula

A compound has an empirical formula of CH and a molar mass of 78.0 g mol-1.

Determine its molecular formula.

Step 1: Calculate empirical formula mass

C: 12.0 g mol-1

H: 1.0 g mol-1

Empirical formula mass = 12.0 + 1.0 = 13.0 g mol-1


Step 2: Divide molecular mass by empirical formula mass

empirical formula massmolar mass​=13.078.0​=6.0


Step 3: Determine the number of empirical units per molecule

The compound contains 6 empirical units per molecule.


Step 4: Write the molecular formula

The molecular formula is C6H6.

Formulae for ionic compounds

  • Ionic compounds consist of a lattice of alternating positive and negative ions.
  • Some ionic compounds also contain water of crystallisation - water molecules lodged within the lattice framework.
  • An ionic compound without water of crystalisation is termed anhydrous, while one with water of crystallisation is hydrated.
  • Anhydrous and hydrated forms of the same compound are often different colours, for example, the anhydrous form of copper sulfate is white whereas the hydrated form is blue.


Hydrated compounds are denoted with a dot separating the salt from the water molecules, for example, CuSO4•5H2O for copper sulfate pentahydrate.


The formula of a hydrated ionic compound can be experimentally determined from the mass loss on heating using the following steps:

  1. Heat the hydrated salt to remove water, leaving anhydrous salt
  2. Calculate moles of water lost and anhydrous salt remaining
  3. Find the mole ratio of salt to water.
  4. Express this ratio in the formula as a whole number.

Worked example 4 - Calculating the molecular formula of a hydrated ionic salt

Calculate the formula of a hydrated magnesium sulfate, MgSO4•xH2O, given that heating 24.6 g of the hydrated salt leaves 12.0 g of anhydrous MgSO4.


Step 1: Calculate mass of water lost

Mass of water lost = Initial mass - Final mass = 24.6 - 12.0 = 12.6 g


Step 2: Calculate moles of anhydrous MgSO4

moles of MgSO4​=Mr ​mass​=120.312.0​≈0.100 mol


Step 3: Calculate moles of water lost

moles of H2​O =Mr ​mass​=18.012.6​=0.700 mol


Step 4: Determine the simplest H2O : MgSO4 mole ratio

H2O : MgSO4 = 0.700 : 0.100 = 7 : 1

x = 7


Step 5: Write the molecular formula

The molecular formula is MgSO4•7H2O.