Empirical and Molecular Formulae – A Level Chemistry Revision

Empirical and Molecular Formulae

Welcome back to Beyond’s Science Blog! This A Level Chemistry post explores Empirical and Molecular Formulae.

Empirical formulae and molecular formulae are two different ways of representing the structure of chemical substances. Depending on the substance, the empirical and molecular formulae may be the same, or they may be different. This blog explains how to find the empirical and molecular formula of a chemical. To see this blog in a downloadable PDF format with a wider variety of questions and answers, click here.


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Chemical formulae represent the composition of substances. The formula of carbon dioxide, for example, is CO2. This tells us that one molecule of carbon dioxide contains twice as many oxygen atoms as carbon atoms.

Chemical formulae in balanced equations can help us to understand chemical reactions on an atomic scale. For example, the following equation tells us that one mole of carbon atoms reacts with two moles of oxygen atoms (one mole of O2 molecules contains two moles of O atoms) to make one mole of carbon dioxide molecules.

C(s) + O2(g) β†’ CO2(g)

Chemical substances can be represented by different types of formulae including empirical formulae and molecular formulae.

An empirical formula is the simplest whole-number ratio of atoms of each element in a compound.

A molecular formula is the actual number of atoms of each element in a molecule of a compound.

The empirical formula and molecular formula of a substance can be the same or different, as shown by the examples below:

Chemical SubstanceMolecular FormulaEmpirical Formula
waterH2OH2O
methaneCH4CH4
ethaneC2H6CH3
ethanoic acidC2H4O2CH2O
glucoseC6H12O6CH2O

Methods for determining the empirical and molecular formulae of substances are shown on the following pages.


Determining an Empirical Formula

Elemental compositions of compounds can be determined experimentally. For example, an experiment could show that an oxide of iron consists of 70% iron by mass. This means that 70% of the compound’s mass is iron atoms and 30% of the compound’s mass is oxygen atoms. The empirical formula of this compound can be deduced following three steps:

1. Find the mass of each element in the compound.

You may be given the mass of each element or the percentage by mass of
each element in the compound.

If percentage by mass values are given, calculate the mass of each element that would be present in 100 g of the compound (100 g is a nice round number, but you could use any amount for the mass of the compound because the percentage, and therefore the ratio, of each element will remain the same).

Example:

ironoxygen
70%
\frac{70}{100} x 100g
mass = 70g
30%
\frac{30}{100} x 100g
mass = 30g

2. Find the number of moles of each element in the compound.

Divide the mass of each element by its relative atomic mass (Ar).

IronOxygen
Ar = 55.8
moles = \frac{70}{55.8} = 1.254…
Ar = 16.0
moles = \frac{30}{16.0} = 1.875

3. Find the simplest whole number ratio.

Divide each number of moles by the smallest number of moles.

If the simplest ratio calculated does not contain whole numbers, multiply all
numbers by the same value to convert the ratio to whole numbers.

IronOxygen
\frac{1.254...}{1.254...} = 1
simplest whole number ratio = 1 x 2 = 2
\frac{1.875}{1.254...} = 1.5
simplest whole number ratio = 1.5 x 2 = 3

The empirical formula is Fe2O3.


Determining a Molecular Formula

If the relative molecular mass (Mr) and the empirical formula of a compound are known, we can deduce the molecular formula of the compound.

For example, an oxide of iron with a Mr of 159.6 g mol-1 has the empirical formula Fe2O3. The molecular formula of this compound can be deduced as follows:

1. Calculate the mass of one empirical formula unit.

Find the sum of the relative atomic masses (Ar) of each atom in the
empirical formula.

For example, the relative mass of one Fe2O3 unit:
Relative mass Fe2O3 = (55.8 Γ— 2) + (16.0 Γ— 3)
Relative mass Fe2O3 = 159.6

2. Calculate the number of empirical formula units that are present in one molecule of the compound.

You can do this by dividing the relative formula mass (Mr) by the relative mass of one empirical formula unit you just calculated.

$latex $text{number of empirical formula units}=\frac{M_r}{\text{relative mass of one empirical formula unit}}&bg=fcfbfb&s=3$

For Fe2O3:

number of empirical formula units = \frac{159.6}{159.6} = 1

Therefore, there is one empirical formula unit in one molecule of Fe2O3 and the molecular formula is also Fe2O3.


Example Question 1

An experiment shows that a hydrocarbon contains 1.20 g of carbon (C) and 0.30 g of hydrogen (H).

a. Deduce the empirical formula of the hydrocarbon.

CarbonHydrogen
mass =1.20g0.30g
Ar =12.01.0
moles =\frac{1.20}{12.0} = 0.10\frac{0.30}{1.0} = 0.30
simplest ratio =\frac{0.10}{0.10} = 1\frac{0.30}{0.10} = 3
simplest whole number ratio =13

The empirical formula of this compound is CH3.

b. The same hydrocarbon has a relative molecular mass (Mr) of 60.0 g mol-1. Determine the molecular formula of the hydrocarbon.

relative mass of one CH3 unit = (12.0 Γ— 1) + (1.0 Γ— 3) = 15.0

number of empirical formula units in one molecule = \frac{60.0}{15} = 4

CH3 x 4 β†’ C4H12

The molecular formula of this compound is C4H12.


Example Question 2

The complete combustion of an unknown hydrocarbon in excess oxygen produced 11.0 g of carbon dioxide (CO2) and 4.50 g of water (H2O).

a. Determine the empirical formula of the hydrocarbon.

Mr of CO2 = (12.0 x 1) + (16.0 x 2) = 44.0

moles of CO2 produced = \frac{11.0}{44.0} = 0.25 moles

So, there were 0.25 moles of C in the original hydrocarbon that underwent complete combustion.

Mr of H2O = (1.0 x 2) + (16.0 x 1) = 18.0

moles of H2O produced = \frac{4.50}{18.0} = 0.25 moles

Each mole of H2O molecules contains two moles of H atoms.

So, there were 0.5 moles of H in the original hydrocarbon that underwent complete
combustion.

CarbonHydrogen
moles =0.250.5
simplest ratio =x4x4
simplest whole number ratio =12

The empirical formula of this compound is CH2.


Practice Question

2.40 g of phosphorus (P) reacts completely with excess oxygen (O) to form 5.50 g of phosphorus oxide.

a. Determine the empirical formula of the phosphorus oxide.

b. Analysis by mass spectrometry showed that the relative molecular mass (Mr) of the phosphorus oxide is 284.0. Determine the molecular formula of the phosphorus oxide.

2. An organic compound was found to contain 66.7% carbon (C), 11.1% hydrogen (H)
and 22.2% oxygen (O) by mass.

a. Determine the empirical formula of the organic compound.

b. Analysis by mass spectrometry showed that the relative molecular mass (Mr) of the organic compound was 72.0. Determine the molecular formula of the organic compound.


Answers

2.40 g of phosphorus (P) reacts completely with excess oxygen (O) to form 5.50 g of phosphorus oxide.

a. Determine the empirical formula of the phosphorus oxide.

phosphorusoxygen
mass =2.40g5.50 – 2.40 = 3.10g
Ar =31.016.0
moles =\boldsymbol{\frac{2.40}{31.0}} = 0.077419…\boldsymbol{\frac{3.10}{16.0}} = 0.19375
simplest ratio =\boldsymbol{\frac{0.077419...}{0.077419...}} = 1.0\boldsymbol{\frac{0.19375}{0.077419...}} = 2.5
simplest whole number ratio =1.0 x 2 = 22.5 x 2 = 5

empirical formula = P2O5

b. Analysis by mass spectrometry showed that the relative molecular mass (Mr) of the phosphorus oxide is 284.0. Determine the molecular formula of the phosphorus oxide.

mass of one P2O5 unit = (31.0 x 2) + (16.0 x 5) = 142.0

empirical formula units in 1 molecule = \boldsymbol{\frac{284.0}{142.0}} = 2

molecular formula = P4O10

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