A chemical formula is a ratio statement. In this lesson you will learn how composition data turns into an empirical formula, how relative molecular mass can scale that into a molecular formula, and how water of crystallisation is handled in hydrated salts. The common thread is simple but powerful: convert masses to amounts in moles, then compare mole ratios.

Formulae as ratios

The formula of a compound tells you which atoms or ions are present, and in what ratio. For simple molecular substances, the molecular formula gives the number and type of atoms in one molecule. For example, glucose has molecular formula C6H12O6, so one glucose molecule contains 6 carbon atoms, 12 hydrogen atoms and 6 oxygen atoms.

The empirical formula gives the simplest whole-number ratio of atoms of each element present. The empirical formula of glucose is CH2O, because the ratio 6:12:6 simplifies to 1:2:1.

You do not need to memorise formal definitions of these two terms for this specification point. You do need to use the terms accurately when interpreting formulae and calculation data.

Empirical formula is the simplest ratio. Molecular formula is the actual atom count in one molecule.

The two formulae may be the same. Water is H2O as both its empirical formula and its molecular formula because 2:1 is already the simplest ratio. Hydrogen peroxide has molecular formula H2O2 but empirical formula HO, because 2:2 simplifies to 1:1.

For ionic compounds, a formula such as MgCl2 is normally a ratio of ions in the giant lattice rather than the contents of one separate molecule. In this lesson, the same mole-ratio calculation method still works.

Empirical formula from composition data

Composition data tells you how much of each element is present in a compound. The masses are not the atom ratio yet, because atoms of different elements have different masses. A mass of 12.0 g of carbon is 1.00 mol of carbon atoms, but a mass of 16.0 g of oxygen is 1.00 mol of oxygen atoms.

To find the empirical formula:

Convert each mass to an amount in moles using amount = mass / A_r.
Divide every amount by the smallest amount.
Convert the ratio to the simplest whole-number ratio.
Write the empirical formula using that ratio.

Worked example 1: mass composition

A compound contains 2.40 g of carbon and 0.400 g of hydrogen. Use A_r values C = 12.0 and H = 1.0.

element	mass / g	divide by A_r	amount / mol	ratio
C	2.40	2.40 / 12.0	0.200	1
H	0.400	0.400 / 1.0	0.400	2

The mole ratio is C:H = 1:2, so the empirical formula is CH2.

Worked example 2: percentage composition

A compound has percentage composition by mass C = 40.0%, H = 6.7% and O = 53.3%. Treat this as a 100 g sample. The masses are then 40.0 g C, 6.7 g H and 53.3 g O.

Using A_r values C = 12.0, H = 1.0 and O = 16.0:

element	mass in 100 g / g	amount / mol	divide by smallest	ratio
C	40.0	40.0 / 12.0 = 3.33	3.33 / 3.33	1
H	6.7	6.7 / 1.0 = 6.7	6.7 / 3.33	2.01
O	53.3	53.3 / 16.0 = 3.33	3.33 / 3.33	1

The ratio is close to 1:2:1, so the empirical formula is CH2O.

If a ratio gives values such as 1:1.5, multiply all parts by 2 to give 2:3. If it gives 1:1.33, multiply all parts by 3 to give 3:4. Do this only after you have divided by the smallest amount.

Molecular formula from empirical formula

An empirical formula may be only a simplified version of the real molecular formula. To find the molecular formula, you need the empirical formula and the relative molecular mass, M_r.

Molecular formula multiplier

\text{multiplier} = \frac{M_r}{\text{empirical formula mass}}

The multiplier tells you how many empirical formula units are in one molecule. Multiply every subscript in the empirical formula by this number.

Worked example: from CH2O to a molecular formula

A compound has empirical formula CH2O and relative molecular mass 180. Use A_r values C = 12.0, H = 1.0 and O = 16.0.

First find the empirical formula mass:

(12.0) + (2 \times 1.0) + (16.0) = 30.0

Then find the multiplier:

\frac{180}{30.0} = 6

Multiply every atom count in CH2O by 6:

\text{molecular formula} = \text{C}_6\text{H}_{12}\text{O}_6

The multiplier should be a whole number. If it is not close to a whole number, check your empirical formula mass, your use of A_r values, and any rounding in earlier working.

Worked example: same empirical and molecular formula

A compound has empirical formula C2H6O and M_r = 46.0.

Empirical formula mass:

(2 \times 12.0) + (6 \times 1.0) + 16.0 = 46.0

Multiplier:

\frac{46.0}{46.0} = 1

So the molecular formula is also C2H6O.

Hydrated salt language

Some ionic solids form crystals that include a fixed ratio of water molecules in the crystal structure. This water is called water of crystallisation. The formula shows it after a dot.

For example:

\text{CuSO}_4 \cdot 5\text{H}_2\text{O}

This means there are 5 moles of water of crystallisation for every 1 mole of CuSO4 formula units in the solid crystal.

Hydrated

A hydrated salt contains water of crystallisation in its crystal structure.

Anhydrous

An anhydrous salt contains no water of crystallisation.

The dot is not a multiplication sign for one molecule joined to another molecule in solution. It is a formula convention showing the ratio of salt formula units to water molecules in the crystalline solid.

When a hydrated salt is heated, water of crystallisation can be driven off:

\text{hydrated salt} \rightarrow \text{anhydrous salt} + \text{water}

For copper(II) sulfate:

\text{CuSO}_4 \cdot 5\text{H}_2\text{O}(s) \rightarrow \text{CuSO}_4(s) + 5\text{H}_2\text{O}(g)

At this point, the important calculation idea is the mole ratio between the anhydrous salt and water:

\text{anhydrous salt} : \text{water} = 1 : x

So a hydrated salt formula can be written as:

\text{salt} \cdot x\text{H}_2\text{O}

Hydrated formula calculations

Hydrated-salt formula calculations use the same method as empirical formula calculations: turn masses into moles and compare ratios.

Worked example 1: from mass composition

A hydrated salt has formula MgSO4.xH2O. A sample contains 2.40 g of anhydrous MgSO4 and 2.52 g of water of crystallisation.

Use A_r values Mg = 24.3, S = 32.1, O = 16.0 and H = 1.0.

First calculate the formula mass of anhydrous MgSO4:

M_r(\text{MgSO}_4) = 24.3 + 32.1 + (4 \times 16.0) = 120.4

Now calculate amounts:

n(\text{MgSO}_4) = \frac{2.40}{120.4} = 0.0199 \text{ mol}

n(\text{H}_2\text{O}) = \frac{2.52}{18.0} = 0.140 \text{ mol}

Find the ratio by dividing by the smaller amount:

\text{MgSO}_4:\text{H}_2\text{O} = \frac{0.0199}{0.0199}:\frac{0.140}{0.0199} = 1:7.04

The experimental ratio is close to 1:7, so:

\text{formula} = \text{MgSO}_4 \cdot 7\text{H}_2\text{O}

Worked example 2: from percentage composition

A hydrated salt has formula CoCl2.xH2O. It contains 45.4% water by mass. Find x.

Use A_r values Co = 58.9, Cl = 35.5, H = 1.0 and O = 16.0.

Treat the percentages as a 100 g sample:

mass of water = 45.4 g
mass of anhydrous CoCl2 = 54.6 g

Formula mass of CoCl2:

58.9 + (2 \times 35.5) = 129.9

Amounts:

n(\text{CoCl}_2) = \frac{54.6}{129.9} = 0.420 \text{ mol}

n(\text{H}_2\text{O}) = \frac{45.4}{18.0} = 2.52 \text{ mol}

Ratio:

\text{CoCl}_2:\text{H}_2\text{O} = \frac{0.420}{0.420}:\frac{2.52}{0.420} = 1:6

So the formula is:

\text{CoCl}_2 \cdot 6\text{H}_2\text{O}

Formulae from experimental results

In practical work, the data may not be handed to you already split into anhydrous salt and water. You often need to work out the useful masses from balance readings.

For a hydrated salt heated to remove water of crystallisation:

\text{mass of hydrated salt} = \text{mass of container + hydrated salt} - \text{mass of container}

\text{mass of anhydrous salt} = \text{mass of container + anhydrous salt} - \text{mass of container}

\text{mass of water lost} = \text{mass of hydrated salt} - \text{mass of anhydrous salt}

Then use the normal mole-ratio method.

Worked example: experimental mass results

A student heats a hydrated sample of copper(II) sulfate, CuSO4.xH2O, until the mass is constant.

measurement	mass / g
crucible	32.18
crucible + hydrated salt	35.00
crucible + anhydrous salt	33.98

Use A_r values Cu = 63.5, S = 32.1, O = 16.0 and H = 1.0.

Find the masses:

\text{mass of hydrated salt} = 35.00 - 32.18 = 2.82 \text{ g}

\text{mass of anhydrous CuSO}_4 = 33.98 - 32.18 = 1.80 \text{ g}

\text{mass of water lost} = 2.82 - 1.80 = 1.02 \text{ g}

Find the amounts:

M_r(\text{CuSO}_4) = 63.5 + 32.1 + (4 \times 16.0) = 159.6

n(\text{CuSO}_4) = \frac{1.80}{159.6} = 0.0113 \text{ mol}

n(\text{H}_2\text{O}) = \frac{1.02}{18.0} = 0.0567 \text{ mol}

Find the ratio:

\text{CuSO}_4:\text{H}_2\text{O} = \frac{0.0113}{0.0113}:\frac{0.0567}{0.0113} = 1:5.02

So:

\text{CuSO}_4 \cdot 5\text{H}_2\text{O}

In real experiments, the ratio may be 5.02 rather than exactly 5. Keep enough figures during the calculation, then round to the nearest sensible whole-number value for x. If heating is incomplete, too much water remains in the solid and x may be underestimated. If the salt decomposes or solid is lost by spitting, the apparent mass of anhydrous salt becomes too low and x may be overestimated.

PAG1 links this topic to moles determination using measured masses and, in some activities, measured gas volumes. The practical skill is not a separate trick: the measured result must be converted into an amount in mol before it can be used in a ratio.

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2.1.3b-d - Empirical, molecular and hydrated formulae