The document explains the concept of empirical formulas, which represent the simplest ratio of atoms in a compound, compared to molecular formulas that denote the actual number of each atom. It includes examples of calculating empirical formulas for various compounds and discusses the historical significance of empirical formulas in chemistry. Additionally, it provides practical exercises for determining empirical formulas based on given quantities of reactants.

1. 6. Empirical Formulae
What is an Empirical Formula?
You already know that the molecular formula of a
molecule represents the number of each kind of atom
in the molecule. So, for example, the molecular
formula of ethene is C2H4, because there are two
carbon atoms and four hydrogen atoms in the
molecule. But the empirical formula of a substance is
defined as the simplest ratio of atoms in a compound,
so the empirical formula of ethene is just CH2.
1. Write down the molecular and empirical formulae of the following molecules:
molecular
formula
empirical
formula
There are not very many cases of molecules in which the molecular formula and the empirical
formula are different. There is one in the table above.
2. Can you think of any more examples? (Clue: think of some hydrocarbons)
Example One What is the empirical formula of a compound containing 0.15 mol of zinc and
0.30 mol of chlorine?
Zn Cl
moles 0.15 mol 0.30 mol
ratio of moles 1 : 2
empirical formula ZnCl2
3. Write down the empirical formula of the compounds obtained when:
a) 0.2 mol of carbon and 0.8 mol of hydrogen are combined
b) 2 mol of oxygen and 2 mol of hydrogen are combined
c) 0.65 mol of oxygen and 1.30 mol of hydrogen are combined
d) 4.4 mol of barium and 8.8 mol of chlorine are combined
e) 1.8 mol of aluminium and 2.7 mol of oxygen are combined
Quantitative Analysis
You are probably wondering what empirical formulae are for. They are interesting from a
historical point of view. From 1860, when a list of relative atomic masses was agreed upon,
to about 1960, when NMR spectroscopy became widely available, chemists used empirical
formulae calculations to work out the formula of unknown substances. Modern chemists still

2. need to know how to do empirical formulae calculations, even if there are easier ways to
determine formulae experimentally.
The conventional Example 2 Thomas Edison weighed out 3.84g of copper and found that it
way to set out
empirical
reacted with 0.48g of oxygen. What was the empirical formula
formulae is as a of the oxide of copper that he was investigating? Notice that we have
divided by 16, which
table. The table
has a column for is the RAM of oxygen
atoms (not the RMM
each element
and we write the
Cu O of O2 molecules).
This is because we
symbol and then
the mass of each mass 3.83g 0.48g need to find out the
ratio of copper
one.
moles 3.83 0.48
= 0.0598 mol = 0.03 mol
atoms to oxygen
We then divide 64 16 atoms, in order to
obtain the empirical
the mass of each
elements by the ratio of moles 2 : 1 formula.
element’s RAM.
We are essentially empirical formula Cu2O Finally we have to write the two amounts as a
simple ratio. If you can’t ‘see’ the simple ratio
working out the
number of moles straight away, choose the smallest of the two
of each element, Don’t be put off if the ratio isn’t exact. It amounts and divide both amounts by this
because moles = is OK to round up or down, if your ratio smallest amount. This doesn’t always get you
mass / RAM. is close to a whole number. straight to the answer, but it often helps.
4. Find the empirical formulae of the d) 0.60g of magnesium was heated with
compounds formed when the following sulfur until further reaction occurred.
quantities react: The mass of magnesium sulfide formed
a) 4.32g of silver and 1.42g of chlorine was 1.40g.
b) 2.5g of calcium and 1.5g of carbon e) 10.0g of carbon was reacted with
c) 0.3g of carbon and 0.8g of oxygen hydrogen to give a compound that
d) 0.7g of iron and 3.0g of bromine weighed 12.5g.
e) 1.6g of calcium and 2.84g of chlorine f) 2.07g of sodium burnt completely in
f) 3.2g of oxygen and 5.5g of manganese oxygen to give 2.79g of sodium oxide.
g) 2.54g of iodine and 0.8g of oxygen g) 44.0g of manganese reacted completely
h) 2.1g of silicon and 2.66g of chlorine to give 88.8g of manganese oxide.
i) 5.0g of arsenic and 0.2g of hydrogen h) After reacting 1.5g of aluminium with
j) 4.00g of magnesium and 1.56g of fluorine gas, 4.67g of aluminium fluoride
nitrogen was produced.
i) 5.00g of chromium underwent complete
5. Calculate the empirical formulae of the combustion in oxygen to give an oxide
compounds formed in the following that weighed 7.31g
experiments: j) 1.54g of an oxide of nitrogen was found
a) 3.60g of magnesium was heated with to contain 0.40g of nitrogen.
bromine until further reaction
occurred. The mass of magnesium Answers
1. NH3, NH3, C3H8, C3H8, C2H6O, C2H6O, C4H10, C2H5
bromide formed was 27.60g.
b) 8.00g of a gaseous hydrocarbon was 3. a) CH4, b) HO, c) H2O, d) BaCl2, e) Al2O3
reduced to give 6.00g of carbon.
4. a) AgCl, b) CaC2, c) CO2, d) FeBr3, e) CaCl2, f) MnO2, g)
c) When 15.61g of an oxide of lead was I2O5, h) SiCl4, i) AsH3, j) Mg2N3
completely reduced, 14.49g of lead was
produced. 5. a) MgBr2, b) CH4, c) PbO, d) MgS, e) CH3, f) Na2O, g)
Mn2O7, h) AlF3, i) Cr2O3, j) N2O5