Empirical formulas from experimental data: WASSCE Chemistry

How do chemists know that water is H₂O and rust contains Fe₂O₃? Not by looking at the molecules, but by measuring masses and turning them into a mole ratio. The simplest whole-number ratio of atoms in a compound is its empirical formula. This lesson shows how to find it from percentages and from experimental masses.

Learning objectives

By the end of this lesson you should be able to:

define the empirical formula of a compound
calculate an empirical formula from percentage composition
calculate an empirical formula from experimental masses
handle mole ratios that are not whole numbers
describe an experiment to find the formula of magnesium oxide

What an empirical formula is

Definition

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

Compound	Molecular formula	Empirical formula
Glucose	C₆H₁₂O₆	CH₂O
Ethane	C₂H₆	CH₃
Hydrogen peroxide	H₂O₂	HO
Benzene	C₆H₆	CH
Water	H₂O	H₂O

For ionic compounds such as NaCl and MgO, the formula is always an empirical formula, because there are no molecules.

The method

Step	Action
1	Write the mass or percentage of each element
2	Divide each by its relative atomic mass, giving moles
3	Divide every answer by the smallest one
4	If needed, multiply to get whole numbers
5	Write the formula

Worked example

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. Find its empirical formula. (C = 12, H = 1, O = 16)

	C	H	O
% (or mass in 100 g)	40.0	6.7	53.3
÷ Aᵣ	40.0 ÷ 12 = 3.33	6.7 ÷ 1 = 6.7	53.3 ÷ 16 = 3.33
÷ smallest (3.33)	1	2.0	1

Empirical formula: CH₂O.

Ratios that are not whole numbers

If the ratio comes out as about 1.5, multiply everything by 2; about 1.33 or 1.67, multiply by 3; about 1.25, multiply by 4. A value such as 1.98 or 2.03 is just experimental error: round it to 2.

Worked example

An oxide of iron contains 70% iron and 30% oxygen. Find its empirical formula. (Fe = 56, O = 16)

	Fe	O
%	70	30
÷ Aᵣ	70 ÷ 56 = 1.25	30 ÷ 16 = 1.875
÷ smallest (1.25)	1	1.5
× 2	2	3

Empirical formula: Fe₂O₃ (iron(III) oxide, the main part of rust and of iron ore).

From experimental masses

You can use masses directly instead of percentages.

Worked example

2.8 g of iron combines completely with 1.6 g of sulphur. Find the formula of the product. (Fe = 56, S = 32)

Fe: $2.8 \div 56 = 0.05$ mol
S: $1.6 \div 32 = 0.05$ mol
Ratio 1 : 1, so the formula is FeS.

Worked example

A hydrocarbon contains 85.7% carbon and 14.3% hydrogen.

C: $85.7 \div 12 = 7.14$ ; H: $14.3 \div 1 = 14.3$
Ratio: $7.14 \div 7.14 = 1$ ; $14.3 \div 7.14 = 2.0$

Empirical formula: CH₂ (the empirical formula of all alkenes).

Experiment: the formula of magnesium oxide

Magnesium heated in a crucible with a lid on a pipe-clay triangle; masses before and after heating give 0.24 g magnesium and 0.16 g oxygen, a 1 : 1 mole ratio

Figure 1. Finding the formula of magnesium oxide by heating magnesium. WASSCE Tutor diagram.

Method:

Weigh a clean, dry crucible and lid.
Add a coil of cleaned magnesium ribbon and weigh again.
Heat strongly, lifting the lid slightly from time to time to let air in while keeping the white smoke of magnesium oxide inside.
When the magnesium stops glowing, heat strongly for a few more minutes, then cool and weigh.
Repeat heating, cooling and weighing until the mass is constant (heating to constant mass), to be sure all the magnesium has reacted.

Worked example

Results: crucible + lid = 20.10 g; with magnesium = 20.34 g; with magnesium oxide = 20.50 g.

Mass of magnesium $= 20.34 - 20.10 = 0.24$ g → $0.24 \div 24 = 0.010$ mol
Mass of oxygen $= 20.50 - 20.34 = 0.16$ g → $0.16 \div 16 = 0.010$ mol
Ratio 1 : 1, so the formula is MgO.

Sources of error: some magnesium oxide smoke escapes when the lid is lifted (making the oxygen mass too low); not all the magnesium reacts; some magnesium reacts with nitrogen in the air to form magnesium nitride.

In Sierra Leone

Iron ore mined at Tonkolili is mainly haematite, Fe₂O₃. Using the empirical formula, 112 g of iron is combined with 48 g of oxygen in every 160 g of pure haematite, so pure ore is 70% iron by mass. Mining companies use exactly this kind of calculation to judge the grade of their ore.

Common mistake

Do not round a ratio such as 1 : 1.5 to 1 : 2. That would give FeO₂ instead of Fe₂O₃. Multiply to the nearest whole-number ratio instead.

Key points

The empirical formula is the simplest whole-number ratio of atoms in a compound.
Method: mass or % → ÷ Aᵣ → ÷ smallest → whole numbers.
Ratios near 1.5, 1.33 or 1.25 must be multiplied by 2, 3 or 4.
The magnesium oxide experiment needs heating to constant mass, lifting the lid to let in air.

WASSCE examination focus

WAEC tests this

Every year WASSCE asks for an empirical formula from percentage composition or masses. Set out the calculation in a table, show the ÷ Aᵣ and ÷ smallest steps, and watch for ratios that need multiplying. Practical questions may ask why the crucible is heated to constant mass.

Quick check

Define empirical formula.
A compound contains 52.2% C, 13.0% H and 34.8% O. Find its empirical formula.
0.48 g of magnesium forms 0.80 g of oxide. Find the formula of the oxide.
Why is the crucible heated to constant mass?

Answers:

The simplest whole-number ratio of the atoms of each element in a compound.
C 4.35, H 13.0, O 2.175 → 2 : 6 : 1, so C₂H₆O.
Mg 0.02 mol; O $0.32 \div 16 = 0.02$ mol; 1 : 1, so MgO.
To make sure all the magnesium has reacted.

Empirical formulas from experimental data

Learning objectives

What an empirical formula is

The method

Ratios that are not whole numbers

From experimental masses

Experiment: the formula of magnesium oxide

Key points

WASSCE examination focus

Quick check

Keep going with empirical and molecular formulas

More lessons in Empirical and Molecular Formulas