7. Converting Empirical Formulae into Molecular Formulae
The empirical formula is just the simplest ratio of atoms in a compound. 90% of the time,
the empirical formula and the molecular formula are the same. For example we could do an
experiment and find out that the empirical formula of magnesium oxide is MgO. The chemical
formula also happens to be MgO, so the empirical and chemical formulae are the same.
Occasionally, often for organic compounds (molecules), they are not the same. If we worked
out the empirical formula of a compound and found that it was CH2, the actual molecular
formula of the compound could be either C2H4, C3H6, C4H8, C5H10, C6H12, C7H14 or C8H16 because
these all have a one to two ratio of carbon to hydrogen atoms.
Manchester United and Manchester City played football last
night. Your friend tells you that the match was a draw, but they
didn’t know how many goals the teams scored. Knowing the
empirical formula of a compound, but not knowing its molecular
formula is a bit like knowing that a particular football match is a
draw, but not knowing the score. There is a way of working out
the molecular formula if you know the relative molecular mass
(RMM) of the compound. The RMM should be a multiple of the
relative mass of the fragment that corresponds to the empirical
formula.
To convert an empirical formula into a molecular formula, we first need to work out the
relative mass of the empirical formula fragment. Then we divide the RMM of the compound
by the relative mass of the fragment. This should be a simple, whole number. We then just
multiply the fragment’s formula by the whole number.
Example One An organic acid was analysed and found to have an empirical formula of CO2H
and an RMM of 90. Calculate the molecular formula of the organic acid.
Relative mass of the fragment CO2H = 12 + 32 + 1 = 45
90 / 45 = 2, therefore the molecular formula must be: 2 x CO2H = C2O4H2
Example Two A student analysed a sample of benzene and found that it contained 4.8g of
carbon and 0.4g of hydrogen. In another experiment, they found that the RMM was 72.
Work out the empirical formula and the molecular formula of benzene.
C H
mass 4.8g 0.4g
4.8 0.4
moles = 0.4 mol = 0.4 mol
12 1
ratio of moles 1:1
empirical formula CH
Relative mass of the fragment CH = 12 + 1 = 13
72 / 13 = 6, therefore the molecular formula must be: 6 x CH = C6H6

Questions 9. 2.17g of phosphorus was burned in air to
1. The empirical formula of lighter fuel is make 7.76g of phosphorus oxide. Calculate
C2H5 and the RMM is 58. Use this the empirical formula and then work out
information to calculate the molecular the molecular formula, given that it has an
formula. RMM of 284.
2. Calculate the molecular formula of a 10. Calculate the molecular formula of a
substance that has an empirical formula of hydrocarbon containing 85.7% carbon and
AlCl3 and an RMM of 267. 14.3% hydrogen by mass, and with an RMM
of 70.
3. A gas used to ripen bananas has an
empirical formula of CH2 and an RMM of
28. Work out the molecular formula of the
gas.
4. Plants produce glucose during
photosynthesis. The empirical formula of
glucose is CH2O. Given that the RMM of
glucose is 180, work out the molecular
formula.
5. A pale brown liquid contains 0.7g of
nitrogen and 1.6g of oxygen. This oxide of
nitrogen is found to have a RMM of 92.
First calculate the empirical formula and
then work out the molecular formula.
6. Gas from a canister in a plastics factory
was analysed. It was found to contain 150g
of carbon and 25g of hydrogen. Calculate
the empirical formula. Given that the RMM
was 42, work out the molecular formula.
7. A colourless liquid made up of 5.9g of
hydrogen and 94.1g of oxygen was found to
have a RMM of 34. First calculate the
empirical formula and then work out the
molecular formula.
8. A compound containing 0.304g of carbon
0.0127g of hydrogen and 1.013g of
bromine, was found to have an RMM of 315.
Calculate the empirical formula. Calculate Answers
1. C4H10 2. Al2Cl6 3. C2H4 4. C6H12O6 5. N2O4 6. C3H6 7.
the molecular formula. H202 8. C6H3Br3 9. P4O10 10. C5H5