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Chemistry

1.18 - Empirical & Molecular Formulae

6 min

note 1 of 6

What an empirical formula is

An empirical formula is the simplest whole-number ratio of the atoms or ions in a compound.
It's also called the simplest formula.

Giant structures

For a giant structure, the empirical formula is the only formula it has, because there are no separate molecules:

Silicon dioxide is SiO₂ because silicon and oxygen are in a 1:2 ratio.
Sodium chloride is NaCl because sodium and chloride ions are in a 1:1 ratio.

Molecular substances

For substances made of separate molecules:
- Molecular formula - The actual number of atoms of each element in one molecule.
- Empirical formula - The simplest whole-number ratio of atoms in the compound.
Where the subscripts share no common factor, the two formulas are identical (e.g. water is H₂O, methane is CH₄).
Where the subscripts share a common factor, divide each subscript by the largest common factor to get the empirical formula:

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

Ball-and-stick models of ethane, benzene (a ring with alternating double bonds), and water beside their formulas C2H6, C6H6, and H2O, with arrows labelled divide by 2, divide by 6, and no common factor leading to CH3, CH, and H2O.

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Worked example - Empirical formula of ethane

Ethane has the molecular formula C₂H₆. Deduce its empirical formula.

Step 1 - Read the number of atoms of each element

Carbon: 2 atoms
Hydrogen: 6 atoms

Step 2 - Find the largest whole number that divides both counts

Both 2 and 6 divide exactly by 2.

Step 3 - Divide each count by that number and write as subscripts

Carbon: 2 ÷ 2 = 1
Hydrogen: 6 ÷ 2 = 3

A subscript of 1 isn't written, so the empirical formula of ethane is CH₃.

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Empirical formula from masses or percentages

An empirical formula can be calculated from the mass of each element in a sample.
Dividing each mass by that element's relative atomic mass gives a value proportional to the number of atoms present.

Formula for relative number of atoms

$Relative number of atoms = \frac{mass of that element}{its relative atomic mass}$

The four-step method

Divide the mass of each element by its relative atomic mass.
Divide each answer by the smallest of them.
If any numbers aren't whole, multiply all of them by the same factor so they all become whole (e.g. multiply by 2 for a value ending in 0.5, by 3 for 0.33).
Write the numbers as subscripts, leaving out any subscript of 1.

Using percentages

If composition is given as percentages, treat each percentage as a mass in grams in a 100 g sample.
If one element's mass or percentage is missing, subtract the known amounts from the total (100% or total mass).

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Worked example - Empirical formula from masses

A compound contains 5.4 g of aluminium and 4.8 g of oxygen. Deduce its empirical formula. Relative atomic masses: Al = 27, O = 16.

Step 1 - Divide each mass by its relative atomic mass

Aluminium: 5.4 ÷ 27 = 0.200
Oxygen: 4.8 ÷ 16 = 0.300

Step 2 - Divide each answer by the smallest

The smallest is 0.200.

Aluminium: 0.200 ÷ 0.200 = 1.00
Oxygen: 0.300 ÷ 0.200 = 1.50

Step 3 - Multiply so all become whole numbers

Multiply both by 2 to make them whole:

Aluminium: 1.00 × 2 = 2
Oxygen: 1.50 × 2 = 3

Step 4 - Write the numbers as subscripts

The empirical formula is Al₂O₃.

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From an empirical to a molecular formula

An empirical formula gives the ratio of atoms, not the number in a single molecule.
To find the molecular formula, you also need the compound's relative formula mass (or relative molecular mass).

Formula for n

$n = \frac{M_{r} of compound}{M_{r} of empirical formula}$

Where:

$n$ = Number of empirical formula units in one molecule
$M_{r}$ = Relative formula mass

The three-step method

Calculate the relative formula mass of the empirical formula.
Divide the compound's relative formula mass by it to get n.
Multiply every subscript in the empirical formula by n.

If $n = 1$ , the molecular formula and the empirical formula are the same.

The empirical formula CH2, worth 12 + 2 = 14, above rows of n = 2, 3, and 4 CH2 boxes, with arrows from totals 28, 42, and 56 to C2H4, C3H6, and C4H8.

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Worked example - Molecular formula from an Mr

A compound has the empirical formula CH₂O and a relative formula mass of 180. Deduce its molecular formula. Relative atomic masses: C = 12, H = 1, O = 16.

Step 1 - Work out the relative formula mass of the empirical formula

Relative formula mass of CH₂O = 12 + (2 × 1) + 16 = 30

Step 2 - Divide the compound's relative formula mass by it to get n

$n = 180 \div 30 = 6$

Step 3 - Multiply every subscript by n

Carbon: 1 × 6 = 6
Hydrogen: 2 × 6 = 12
Oxygen: 1 × 6 = 6

The molecular formula is C₆H₁₂O₆.

key terms

TermDefinition

Empirical formulaA formula that shows the simplest whole-number ratio of the atoms or ions of each element in a compound.

Molecular formulaA formula that shows the actual number of atoms of each element in one molecule.

end of notes

Key points to remember