Empirical and molecular formula

What you'll learn

What empirical formula and molecular formula mean.
How to calculate an empirical formula from masses or percentages.
How to handle awkward ratios such as 1.5 or 1.33.
How to use empirical formula and relative molecular mass to find molecular formula.

Before formulae: masses, moles and atom ratios

An amount of substance, $n$ , tells you how many particles are present. It is measured in moles, mol.

In empirical formula questions, you compare how many atoms of each element are present. You cannot compare masses directly, because different atoms have different masses. For example, 12.0 g of carbon and 1.0 g of hydrogen both contain one mole of atoms, even though the masses are very different.

The key calculation is:

\frac{m}{M}

where $n$ is amount in mol, $m$ is mass in g, and $M$ is molar mass in g mol⁻¹. For an element, the numerical value of $M$ is usually its relative atomic mass, $A_r$ .

A subscript is the small number in a formula that shows how many atoms of that element are present. For example, in H₂O, the subscript 2 means two hydrogen atoms.

Key Idea

Moles reveal atom ratios

Empirical formula calculations are always mole-ratio calculations. Convert masses to moles first, then compare the mole amounts.

Empirical formula

Definition

Empirical formula

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

“Simplest” means the ratio has been reduced as far as possible. For example, the ratio 6:12:6 simplifies to 1:2:1.

Example

Simplifying a formula

Glucose has molecular formula C₆H₁₂O₆. Find its empirical formula.

Treat the subscripts as an atom ratio: C:H:O is 6:12:6.
Divide every part of the ratio by the highest common factor, 6:
$C:H:O=1:2:1\text{C:H:O} = 1 : 2 : 1$
Use the simplest ratio as the formula subscripts, so the empirical formula is CH₂O.

The calculation route

Composition by mass tells you the mass of each element in a compound. Percentage by mass tells you the mass of each element in every 100 g of compound.

The whole calculation is the same every time: turn mass information into amounts in moles, then turn mole amounts into the simplest whole-number ratio.

Flowchart showing how to calculate empirical and molecular formula from mass or percentage composition data

Calculating empirical formula from percentage by mass

If percentages are given, assume you have a 100 g sample. This makes the percentage numbers become masses in grams.

Tip

Percentage data

For percentage composition, assume 100 g of compound. So 40.0% carbon becomes 40.0 g carbon, 6.7% hydrogen becomes 6.7 g hydrogen, and so on.

Example

Calculating an empirical formula from percentages

A compound contains 52.2% carbon, 13.0% hydrogen and 34.8% oxygen by mass. Calculate its empirical formula. Use $A_r$ : C = 12.0, H = 1.0, O = 16.0.

Assume 100 g of compound, so the masses are carbon = 52.2 g, hydrogen = 13.0 g and oxygen = 34.8 g.
Convert each mass to moles by dividing by the relevant $A_r$ :
$mol\begin{aligned} n(\text{C}) &= \frac{52.2\ \text{g}}{12.0\ \text{g mol}^{-1}} = 4.35\ \text{mol} \\ n(\text{H}) &= \frac{13.0\ \text{g}}{1.0\ \text{g mol}^{-1}} = 13.0\ \text{mol} \\ n(\text{O}) &= \frac{34.8\ \text{g}}{16.0\ \text{g mol}^{-1}} = 2.175\ \text{mol} \end{aligned}$
Divide all mole amounts by the smallest mole amount, 2.175 mol:
$C:H:O=4.352.175:13.02.175:2.1752.175\text{C:H:O} = \frac{4.35}{2.175} : \frac{13.0}{2.175} : \frac{2.175}{2.175}$ $C:H:O=2.00:5.98:1.00\text{C:H:O} = 2.00 : 5.98 : 1.00$
Round the ratio to sensible whole numbers. Since 5.98 is very close to 6, the ratio is 2:6:1, so the empirical formula is C₂H₆O.

Handling awkward ratios

After dividing by the smallest mole value, you should get numbers close to whole numbers. Sometimes you get a simple fraction instead.

Common patterns:

1.5 means multiply every part of the ratio by 2.
1.33 means multiply every part of the ratio by 3.
1.25 means multiply every part of the ratio by 4.

For example, a ratio of 1:1.5 is not rounded to 1:2. Instead, multiply both parts by 2 to get 2:3.

Common Mistake

Rounding too early

Do not round mole values after the first calculation. Keep at least 3 significant figures until the final ratio, otherwise you may change the empirical formula.

Calculating empirical formula from experimental mass data

You may be asked about finding the empirical formula of a metal oxide. In this type of experiment, a metal is heated in air, reacts with oxygen, and forms a metal oxide.

The mass of oxygen is not measured directly. It is found by difference:

metal\text{mass of oxygen} = \text{mass of metal oxide} - \text{mass of metal}

In practical work, heating to constant mass means heating, cooling and reweighing until the mass stops changing. This helps show the reaction is complete.

Example

Finding a metal oxide formula

A sample of magnesium has mass 0.486 g. After heating in air to form magnesium oxide, the final mass is 0.806 g. Find the empirical formula of the oxide. Use $A_r$ : Mg = 24.3, O = 16.0.

Find the mass of oxygen by difference:
$g\text{mass of oxygen} = 0.806\ \text{g} - 0.486\ \text{g} = 0.320\ \text{g}$
Convert the masses of magnesium and oxygen to moles:
$mol\begin{aligned} n(\text{Mg}) &= \frac{0.486\ \text{g}}{24.3\ \text{g mol}^{-1}} = 0.0200\ \text{mol} \\ n(\text{O}) &= \frac{0.320\ \text{g}}{16.0\ \text{g mol}^{-1}} = 0.0200\ \text{mol} \end{aligned}$
Compare the mole amounts:
$Mg:O=0.0200:0.0200=1:1\text{Mg:O} = 0.0200 : 0.0200 = 1 : 1$
Use the ratio as the formula subscripts, so the empirical formula is MgO.

Common Mistake

Using the oxide mass as oxygen

The mass of the oxide is the total mass of metal plus oxygen. For a metal oxide question, calculate the oxygen mass by subtracting the original metal mass from the final oxide mass.

Molecular formula

Definition

Molecular formula

The molecular formula is the actual number of atoms of each element in one molecule of a compound.

The molecular formula is always a whole-number multiple of the empirical formula. Sometimes the multiplier is 1, meaning the empirical formula and molecular formula are the same.

Definition

Empirical formula mass

The empirical formula mass is the sum of the relative atomic masses in one empirical formula unit.

To find the multiplier:

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

where $k$ is the number you multiply every empirical formula subscript by.

Example

Finding a molecular formula

A compound has empirical formula CH₂O and relative molecular mass $M_r = 180.0$ . Find its molecular formula. Use $A_r$ : C = 12.0, H = 1.0, O = 16.0.

Calculate the empirical formula mass:
$mass=12.0+(2×1.0)+16.0=30.0\text{empirical formula mass} = 12.0 + (2 \times 1.0) + 16.0 = 30.0$
Find the multiplier by comparing $M_r$ with the empirical formula mass:
$\frac{180.0}{30.0} = 6$
Multiply every empirical formula subscript by 6: CH₂O becomes C₆H₁₂O₆.

Exam technique

In the exam

If data are percentages, assume 100 g; if data are masses, use them directly.
Convert masses to moles before comparing elements — never use mass ratios as atom ratios.
Divide all mole amounts by the smallest, then multiply the whole ratio if you get fractions like 1.5, 1.33 or 1.25.
For molecular formula, calculate empirical formula mass first, then use $\frac{M_r}{\text{empirical formula mass}}$ and multiply every subscript by $k$ .

Self review

Check yourself

Why must you convert masses to moles before finding an empirical formula?
A compound is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. What is its empirical formula?
A compound has empirical formula CH₂ and $M_r = 56.0$ . What is its molecular formula?

What you'll learn

What empirical formula and molecular formula mean.
How to calculate an empirical formula from masses or percentages.
How to handle awkward ratios such as 1.5 or 1.33.
How to use empirical formula and relative molecular mass to find molecular formula.

Before formulae: masses, moles and atom ratios

An amount of substance, $n$ , tells you how many particles are present. It is measured in moles, mol.

The key calculation is:

\frac{m}{M}

where $n$ is amount in mol, $m$ is mass in g, and $M$ is molar mass in g mol⁻¹. For an element, the numerical value of $M$ is usually its relative atomic mass, $A_r$ .

A subscript is the small number in a formula that shows how many atoms of that element are present. For example, in H₂O, the subscript 2 means two hydrogen atoms.

Key Idea

Moles reveal atom ratios

Empirical formula calculations are always mole-ratio calculations. Convert masses to moles first, then compare the mole amounts.

Empirical formula

Definition

Empirical formula

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

“Simplest” means the ratio has been reduced as far as possible. For example, the ratio 6:12:6 simplifies to 1:2:1.

Example

Simplifying a formula

Glucose has molecular formula C₆H₁₂O₆. Find its empirical formula.

Treat the subscripts as an atom ratio: C:H:O is 6:12:6.
Divide every part of the ratio by the highest common factor, 6:
$C:H:O=1:2:1\text{C:H:O} = 1 : 2 : 1$
Use the simplest ratio as the formula subscripts, so the empirical formula is CH₂O.

The calculation route

Composition by mass tells you the mass of each element in a compound. Percentage by mass tells you the mass of each element in every 100 g of compound.

The whole calculation is the same every time: turn mass information into amounts in moles, then turn mole amounts into the simplest whole-number ratio.

Flowchart showing how to calculate empirical and molecular formula from mass or percentage composition data

Calculating empirical formula from percentage by mass

If percentages are given, assume you have a 100 g sample. This makes the percentage numbers become masses in grams.

Tip

Percentage data

For percentage composition, assume 100 g of compound. So 40.0% carbon becomes 40.0 g carbon, 6.7% hydrogen becomes 6.7 g hydrogen, and so on.

Example

Calculating an empirical formula from percentages

A compound contains 52.2% carbon, 13.0% hydrogen and 34.8% oxygen by mass. Calculate its empirical formula. Use $A_r$ : C = 12.0, H = 1.0, O = 16.0.

Assume 100 g of compound, so the masses are carbon = 52.2 g, hydrogen = 13.0 g and oxygen = 34.8 g.
Convert each mass to moles by dividing by the relevant $A_r$ :
$mol\begin{aligned} n(\text{C}) &= \frac{52.2\ \text{g}}{12.0\ \text{g mol}^{-1}} = 4.35\ \text{mol} \\ n(\text{H}) &= \frac{13.0\ \text{g}}{1.0\ \text{g mol}^{-1}} = 13.0\ \text{mol} \\ n(\text{O}) &= \frac{34.8\ \text{g}}{16.0\ \text{g mol}^{-1}} = 2.175\ \text{mol} \end{aligned}$
Divide all mole amounts by the smallest mole amount, 2.175 mol:
$C:H:O=4.352.175:13.02.175:2.1752.175\text{C:H:O} = \frac{4.35}{2.175} : \frac{13.0}{2.175} : \frac{2.175}{2.175}$ $C:H:O=2.00:5.98:1.00\text{C:H:O} = 2.00 : 5.98 : 1.00$
Round the ratio to sensible whole numbers. Since 5.98 is very close to 6, the ratio is 2:6:1, so the empirical formula is C₂H₆O.

Handling awkward ratios

After dividing by the smallest mole value, you should get numbers close to whole numbers. Sometimes you get a simple fraction instead.

Common patterns:

1.5 means multiply every part of the ratio by 2.
1.33 means multiply every part of the ratio by 3.
1.25 means multiply every part of the ratio by 4.

For example, a ratio of 1:1.5 is not rounded to 1:2. Instead, multiply both parts by 2 to get 2:3.

Common Mistake

Rounding too early

Do not round mole values after the first calculation. Keep at least 3 significant figures until the final ratio, otherwise you may change the empirical formula.

Calculating empirical formula from experimental mass data

You may be asked about finding the empirical formula of a metal oxide. In this type of experiment, a metal is heated in air, reacts with oxygen, and forms a metal oxide.

The mass of oxygen is not measured directly. It is found by difference:

metal\text{mass of oxygen} = \text{mass of metal oxide} - \text{mass of metal}

In practical work, heating to constant mass means heating, cooling and reweighing until the mass stops changing. This helps show the reaction is complete.

Example

Finding a metal oxide formula

A sample of magnesium has mass 0.486 g. After heating in air to form magnesium oxide, the final mass is 0.806 g. Find the empirical formula of the oxide. Use $A_r$ : Mg = 24.3, O = 16.0.

Find the mass of oxygen by difference:
$g\text{mass of oxygen} = 0.806\ \text{g} - 0.486\ \text{g} = 0.320\ \text{g}$
Convert the masses of magnesium and oxygen to moles:
$mol\begin{aligned} n(\text{Mg}) &= \frac{0.486\ \text{g}}{24.3\ \text{g mol}^{-1}} = 0.0200\ \text{mol} \\ n(\text{O}) &= \frac{0.320\ \text{g}}{16.0\ \text{g mol}^{-1}} = 0.0200\ \text{mol} \end{aligned}$
Compare the mole amounts:
$Mg:O=0.0200:0.0200=1:1\text{Mg:O} = 0.0200 : 0.0200 = 1 : 1$
Use the ratio as the formula subscripts, so the empirical formula is MgO.

Common Mistake

Using the oxide mass as oxygen

The mass of the oxide is the total mass of metal plus oxygen. For a metal oxide question, calculate the oxygen mass by subtracting the original metal mass from the final oxide mass.

Molecular formula

Definition

Molecular formula

The molecular formula is the actual number of atoms of each element in one molecule of a compound.

The molecular formula is always a whole-number multiple of the empirical formula. Sometimes the multiplier is 1, meaning the empirical formula and molecular formula are the same.

Definition

Empirical formula mass

The empirical formula mass is the sum of the relative atomic masses in one empirical formula unit.

To find the multiplier:

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

where $k$ is the number you multiply every empirical formula subscript by.

Example

Finding a molecular formula

A compound has empirical formula CH₂O and relative molecular mass $M_r = 180.0$ . Find its molecular formula. Use $A_r$ : C = 12.0, H = 1.0, O = 16.0.

Calculate the empirical formula mass:
$mass=12.0+(2×1.0)+16.0=30.0\text{empirical formula mass} = 12.0 + (2 \times 1.0) + 16.0 = 30.0$
Find the multiplier by comparing $M_r$ with the empirical formula mass:
$\frac{180.0}{30.0} = 6$
Multiply every empirical formula subscript by 6: CH₂O becomes C₆H₁₂O₆.

Exam technique

In the exam

If data are percentages, assume 100 g; if data are masses, use them directly.
Convert masses to moles before comparing elements — never use mass ratios as atom ratios.
Divide all mole amounts by the smallest, then multiply the whole ratio if you get fractions like 1.5, 1.33 or 1.25.
For molecular formula, calculate empirical formula mass first, then use $\frac{M_r}{\text{empirical formula mass}}$ and multiply every subscript by $k$ .

Self review

Check yourself

Why must you convert masses to moles before finding an empirical formula?
A compound is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. What is its empirical formula?
A compound has empirical formula CH₂ and $M_r = 56.0$ . What is its molecular formula?

What you'll learn

Before formulae: masses, moles and atom ratios

Empirical formula

The calculation route

Calculating empirical formula from percentage by mass

Handling awkward ratios

Calculating empirical formula from experimental mass data

Molecular formula

Review Empirical and molecular formula by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Empirical and molecular formula

What you'll learn

Before formulae: masses, moles and atom ratios

Empirical formula

The calculation route

Calculating empirical formula from percentage by mass

Handling awkward ratios

Calculating empirical formula from experimental mass data

Molecular formula

Review Empirical and molecular formula by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Empirical and molecular formula

What you'll learn

Before formulae: masses, moles and atom ratios

Empirical formula

The calculation route

Calculating empirical formula from percentage by mass

Handling awkward ratios

Calculating empirical formula from experimental mass data

Molecular formula

Review Empirical and molecular formula by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Empirical and molecular formula Revision Guide

Empirical and molecular formula

What you'll learn

Before formulae: masses, moles and atom ratios

Empirical formula

The calculation route

Calculating empirical formula from percentage by mass

Handling awkward ratios

Calculating empirical formula from experimental mass data

Molecular formula

Review Empirical and molecular formula by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Empirical and molecular formula Revision Guide