Last updated: 7 October 2026
Chemical equations may look as though they are simply telling us how many substances react. For example, 2H₂ + O₂ → 2H₂O contains the numbers 2, 1 and 2. But these numbers do not mean 2 g of hydrogen reacting with 1 g of oxygen. They describe a ratio between enormous numbers of particles.
That creates an obvious practical problem: we cannot count atoms and molecules one by one before carrying out an experiment. Instead, we weigh substances, measure gas volumes and prepare solutions of known concentration. The mole lets us convert all these measurable quantities into one common chemical language.
Once a quantity has been converted into moles, the balanced chemical equation becomes extremely useful. It acts like a chemical recipe, telling us how much reactant is needed, how much product can form and, when two reactants are present, which one will run out first.
Try these five questions before reading the chapter. They are designed to reveal whether the foundation ideas are already secure.
- 5 diagnostic questions
- Calculations and equations where useful
- Instant explanation for every answer option
Relative masses give atoms a practical mass scale
An individual atom has an extraordinarily small mass. Writing atomic masses in grams would therefore produce inconvenient numbers containing many zeros. Chemists instead use a relative scale based on carbon-12.
Relative atomic mass, Aᵣ
The relative atomic mass, Aᵣ, of an element is the average mass of one atom of that element compared with ¹⁄₁₂ of the mass of one carbon-12 atom.
The word relative matters. Aᵣ is a ratio, so it has no unit.
Many elements occur naturally as mixtures of isotopes. Their Aᵣ values are therefore weighted averages and need not be whole numbers. Chlorine, for example, has an Aᵣ of about 35.5 because naturally occurring chlorine contains atoms of more than one isotope.
The atomic number tells you the number of protons in an atom. Aᵣ tells you about the atom's relative mass. They describe completely different properties.
Relative molecular mass, Mᵣ
The relative molecular mass, Mᵣ, of a molecule is obtained by adding the Aᵣ values of all the atoms shown in its molecular formula.
Calculate the Mᵣ of sulfur dioxide, SO₂.
The formula contains one sulfur atom and two oxygen atoms, so the Aᵣ of oxygen must be counted twice.
Relative formula mass for ionic compounds
Ionic compounds such as sodium chloride do not consist of separate molecules. Their formulae show the simplest ratio of ions within a giant ionic lattice. The more accurate term is therefore relative formula mass.
The calculation is nevertheless carried out in the same way: add the Aᵣ values of every atom represented in the formula.
Calculate the relative formula mass of calcium nitrate, Ca(NO₃)₂.
The 2 outside the bracket multiplies everything inside the bracket. Ca(NO₃)₂ therefore represents one calcium atom, two nitrogen atoms and six oxygen atoms.
Aᵣ and Mᵣ: relative values with no units.
Molar mass: the mass of one mole of a substance, measured in g mol⁻¹.
The mole is a counting unit for particles
We already use special words for groups of objects. A dozen means 12 objects. A mole performs a similar job in Chemistry, except that the number involved is enormously larger.
The value 6.02 × 10²³ mol⁻¹ is known as the Avogadro constant, Nₐ.
The word particles must be interpreted carefully. Depending on the substance, the particles may be atoms, molecules, ions or formula units.
| Substance | Particle being counted | Particles in 1 mol |
|---|---|---|
| Helium, He | Atoms | 6.02 × 10²³ helium atoms |
| Oxygen, O₂ | Molecules | 6.02 × 10²³ O₂ molecules |
| Water, H₂O | Molecules | 6.02 × 10²³ H₂O molecules |
| Sodium chloride, NaCl | Formula units | 6.02 × 10²³ NaCl formula units |
| Sodium ions, Na⁺ | Ions | 6.02 × 10²³ Na⁺ ions |
A chemical formula tells you how many particles of each type are present
One mole of H₂O contains one mole of water molecules. But each water molecule contains two hydrogen atoms and one oxygen atom.
Therefore, one mole of H₂O contains:
- 2 mol of hydrogen atoms; and
- 1 mol of oxygen atoms.
Similarly, one mole of CaCl₂ contains:
- 1 mol of Ca²⁺ ions;
- 2 mol of Cl⁻ ions; and
- 3 mol of ions altogether.
One mole always means 6.02 × 10²³ specified particles, but the formula tells you whether those particles are atoms, molecules, ions or formula units.
Converting between mass and moles
The molar mass of a substance is the mass of one mole of that substance. Its unit is g mol⁻¹.
The numerical value of the molar mass is the same as the relevant relative mass. For example, water has Mᵣ = 18 and a molar mass of 18 g mol⁻¹.
| Substance | Relative mass | Molar mass |
|---|---|---|
| Carbon, C | Aᵣ = 12 | 12 g mol⁻¹ |
| Water, H₂O | Mᵣ = 18 | 18 g mol⁻¹ |
| Carbon dioxide, CO₂ | Mᵣ = 44 | 44 g mol⁻¹ |
| Calcium carbonate, CaCO₃ | Relative formula mass = 100 | 100 g mol⁻¹ |
Worked example: mass to moles
Calculate the amount of Ca(NO₃)₂ present in 8.20 g of calcium nitrate.
The relative formula mass of Ca(NO₃)₂ is 164, so its molar mass is 164 g mol⁻¹.
Worked example: moles to mass
Calculate the mass of 0.250 mol of CO₂.
Converting between moles and number of particles
How many CO₂ molecules are present in 0.250 mol of CO₂?
If the particle number is known instead, reverse the calculation:
Percentage composition by mass
A chemical formula can also tell us what fraction of a compound's total mass comes from a particular element.
Worked example: percentage of oxygen in CaCO₃
Calculate the percentage by mass of oxygen in calcium carbonate, CaCO₃.
The three oxygen atoms contribute:
In CaCO₃, oxygen contributes 3 × 16 rather than just 16. A common error is to overlook subscripts when calculating percentage composition.
Empirical formulae show the simplest particle ratio
An empirical formula gives the simplest whole-number ratio of the atoms of each element in a compound.
For example, hydrogen peroxide has the molecular formula H₂O₂. The ratio 2:2 simplifies to 1:1, so its empirical formula is HO.
Empirical formula: simplest whole-number ratio of atoms.
Molecular formula: actual number of each type of atom in one molecule.
Why do we divide by Aᵣ?
A formula describes numbers of atoms, not their masses. Different atoms have different masses, so mass data cannot be used directly as an atomic ratio.
For instance, 12 g of carbon and 16 g of oxygen each represent one mole of atoms. Their mass ratio is 12:16, but their particle ratio is 1:1.
The standard empirical-formula method
- Write down the mass or percentage mass of each element.
- Divide each value by the element's Aᵣ to obtain moles.
- Divide every mole value by the smallest mole value.
- Convert the ratio into the smallest whole-number ratio.
- Use the ratio as the subscripts in the empirical formula.
Worked example: empirical formula from percentage composition
A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Determine its empirical formula.
When percentages are given, imagine a 100 g sample. The percentages then become masses in grams.
| Element | Mass / g | Mass ÷ Aᵣ | Divide by smallest |
|---|---|---|---|
| C | 40.0 | 40.0 ÷ 12 = 3.33 | 3.33 ÷ 3.33 = 1.00 |
| H | 6.7 | 6.7 ÷ 1 = 6.7 | 6.7 ÷ 3.33 ≈ 2.01 |
| O | 53.3 | 53.3 ÷ 16 = 3.33 | 3.33 ÷ 3.33 = 1.00 |
The ratio is approximately 1:2:1.
What if the ratio contains a fraction?
Values such as 1.50 should not simply be rounded to 2. Instead, multiply the entire ratio by a small whole number.
| Ratio obtained | Multiply all values by | Whole-number ratio |
|---|---|---|
| 1 : 1.50 | 2 | 2 : 3 |
| 1 : 1.33 | 3 | 3 : 4 |
| 1 : 1.25 | 4 | 4 : 5 |
| 1 : 1.67 | 3 | 3 : 5 |
Values such as 1.98 or 3.02 may reasonably be interpreted as 2 or 3 because small differences can arise from rounding or experimental error. A value such as 1.50 is clearly different and should not be rounded away.
Finding the molecular formula
A molecular formula is a whole-number multiple of its empirical formula.
Worked example
A compound has empirical formula CH₂O and Mᵣ = 180. Determine its molecular formula.
Multiply every subscript in CH₂O by 6:
A molecule cannot contain a fractional number of atoms. If the multiplier is not close to a whole number, recheck the empirical formula and the Mᵣ calculation.
Molar gas volume at room temperature and pressure
At the same temperature and pressure, equal numbers of gas particles occupy equal volumes. At room temperature and pressure, r.t.p., one mole of any gas occupies approximately:
Since 1 dm³ = 1000 cm³:
Worked example: gas volume to moles
Calculate the number of moles in 4.80 dm³ of oxygen at r.t.p.
Worked example: moles to gas volume
Calculate the volume occupied by 0.0750 mol of CO₂ at r.t.p.
Gas volume in dm³: use 24 dm³ mol⁻¹.
Gas volume in cm³: use 24 000 cm³ mol⁻¹ or convert the volume into dm³ first.
Concentration of solutions
Concentration describes how much solute is present in a particular volume of solution.
Two useful concentration units at this level are:
- g dm⁻³ for mass concentration; and
- mol dm⁻³ for molar concentration.
Mass concentration
A solution contains 5.00 g of NaCl in 250 cm³ of solution. Calculate its mass concentration.
Molar concentration
Molar concentration tells us the number of moles of solute present in 1 dm³ of solution.
Rearranging:
Always convert cm³ into dm³
Thus:
Worked example: finding moles in a solution
Calculate the amount of NaOH in 25.0 cm³ of 0.200 mol dm⁻³ NaOH.
Worked example: finding concentration
A solution contains 0.0300 mol of KNO₃ in 150 cm³. Calculate its concentration.
A concentration of mol dm⁻³ is defined per cubic decimetre. Substituting 25.0 instead of 0.0250 into n = cV would make the calculated amount 1000 times too large.
Stoichiometry: using a balanced equation as a mole recipe
Stoichiometry is the calculation of reacting quantities using a balanced chemical equation.
This equation tells us:
- 2 mol of H₂ react with 1 mol of O₂;
- 2 mol of H₂O are formed; and
- the mole ratio H₂ : O₂ : H₂O is 2 : 1 : 2.
It does not tell us that 2 g of hydrogen reacts with 1 g of oxygen.
| Substance | Moles | Molar mass / g mol⁻¹ | Mass / g |
|---|---|---|---|
| H₂ | 2 | 2 | 4 |
| O₂ | 1 | 32 | 32 |
| H₂O | 2 | 18 | 36 |
The corresponding mass relationship is therefore:
The standard stoichiometry route
- Write or identify the balanced chemical equation.
- Convert the given quantity into moles.
- Use the coefficient ratio to find the moles of the required substance.
- Convert those moles into the quantity requested.
Masses, gas volumes and solution volumes usually cannot be compared directly. Convert the known quantity into moles, apply the balanced-equation ratio, and then convert out of moles into the requested unit.
Worked example: reacting mass to gas volume
Magnesium reacts with excess hydrochloric acid. Calculate the volume of hydrogen produced at r.t.p. when 4.80 g of magnesium reacts completely.
First find the moles of magnesium:
Mg and H₂ occur in a 1:1 mole ratio.
Worked example: solution stoichiometry
25.0 cm³ of 0.100 mol dm⁻³ H₂SO₄ completely neutralises 20.0 cm³ of NaOH. Calculate the concentration of the NaOH.
The ratio H₂SO₄ : NaOH is 1:2.
The limiting reactant determines the maximum product
Sometimes a question gives the amounts of two reactants. In that case, one reactant may be completely consumed before the other.
The limiting reactant is the reactant that is used up first. Once it is gone, no more product can form. Any reactant left over is said to be in excess.
A reliable limiting-reactant method
- Convert each reactant amount into moles.
- Divide the moles of each reactant by its coefficient in the equation.
- The smaller value identifies the limiting reactant.
- Use the limiting reactant to calculate the product.
Worked example
A mixture contains 0.300 mol of H₂ and 0.200 mol of O₂. Determine the limiting reactant and the amount of H₂O formed.
| Reactant | Moles available | Coefficient | Moles ÷ coefficient |
|---|---|---|---|
| H₂ | 0.300 | 2 | 0.150 |
| O₂ | 0.200 | 1 | 0.200 |
Hydrogen gives the smaller value, so H₂ is the limiting reactant.
H₂ and H₂O have a 2:2, or 1:1, mole ratio.
Only 0.150 mol of O₂ is required:
Limiting reactant: runs out first and determines the maximum amount of product.
Excess reactant: more is present than required, so some remains after the reaction stops.
Percentage yield: how much product was actually obtained?
A stoichiometric calculation predicts the maximum amount of product that could form if everything happened perfectly. This is the theoretical yield.
The amount actually collected in the laboratory is the actual yield.
Worked example
The theoretical yield of a reaction is 12.0 g. The experiment produces 9.60 g. Calculate the percentage yield.
Why might the yield be below 100%?
Common reasons include:
- the reaction may not go to completion;
- side reactions may form unwanted products;
- some product may remain dissolved;
- product may be lost during transfer, filtration or washing;
- some product may decompose before collection.
A measured yield above 100% usually suggests that the collected material contains impurities or water, or that there has been a measurement error.
Percentage purity: how much of the sample is actually useful?
An impure sample contains the required substance mixed with unwanted substances.
Rearranging:
Worked example: purity followed by stoichiometry
A 25.0 g limestone sample is 80.0% calcium carbonate. Calculate the maximum volume of CO₂ produced at r.t.p. when it reacts with excess hydrochloric acid.
First find the mass of pure CaCO₃:
CaCO₃ and CO₂ occur in a 1:1 mole ratio.
If a sample is only 80% pure, the whole sample mass cannot be treated as the reacting substance. Find the mass of the pure component first.
Percentage yield and percentage purity are different ideas
| Quantity | What is compared? | What does it tell us? |
|---|---|---|
| Percentage yield | Actual product ÷ theoretical product | How much of the expected product was obtained? |
| Percentage purity | Pure substance ÷ total sample | How much of the sample is the required substance? |
A substance may be very pure and still give a low percentage yield. Purity describes the composition of a sample, whereas yield describes the success of a reaction and product-collection process.
Choosing the correct route into moles
The easiest way to begin many calculations is to ask what type of information the question has given you.
| Information given | Convert into moles using | Important reminder |
|---|---|---|
| Mass | n = mass ÷ molar mass | Use mass in grams |
| Particle number | n = particles ÷ 6.02 × 10²³ | Identify the particle being counted |
| Gas volume at r.t.p. | n = volume ÷ 24 dm³ | Use 24 000 if volume is in cm³ |
| Solution concentration and volume | n = cV | V must be in dm³ |
Given quantity → moles of known substance → equation mole ratio → moles of required substance → required quantity.
Common mistakes in mole calculations
| Common mistake | Why it is wrong | What to do instead |
|---|---|---|
| Giving Aᵣ or Mᵣ a unit | Relative masses are ratios. | Write Aᵣ = 23 or Mᵣ = 44 with no unit. |
| Calling NaCl a molecule | NaCl forms a giant ionic lattice rather than separate molecules. | Refer to formula units and relative formula mass. |
| Ignoring brackets when finding relative mass | An outside subscript multiplies every atom inside the bracket. | Count the total number of each atom before adding Aᵣ values. |
| Using equation coefficients as mass ratios | Coefficients give mole ratios, not direct mass ratios. | Convert masses into moles first. |
| Using an unbalanced equation | The mole ratio will be incorrect. | Balance the equation before applying stoichiometry. |
| Using 24 with a gas volume in cm³ | 24 refers to dm³ per mole. | Convert to dm³ or divide cm³ by 24 000. |
| Using cm³ directly in n = cV | Concentration is expressed per dm³. | Divide cm³ by 1000 first. |
| Rounding an empirical ratio of 1.50 to 2 | 1.50 indicates a genuine fractional ratio. | Multiply the entire ratio by 2. |
| Choosing the limiting reactant simply because it has fewer moles | The equation may require unequal mole amounts. | Divide each amount by its equation coefficient. |
| Using the excess reactant to calculate product | The reaction stops once the limiting reactant has been consumed. | Base maximum product on the limiting reactant. |
| Reversing the percentage-yield formula | The theoretical yield belongs in the denominator. | Use actual ÷ theoretical × 100%. |
| Using the full mass of an impure sample | Only the pure reacting substance should enter the stoichiometry. | Calculate the mass of pure substance first. |
| Rounding too early | Repeated rounding can distort the final result. | Keep extra calculator digits and round at the end. |
Mole calculation checklist
Start correctly
- Write the correct chemical formula.
- Calculate Aᵣ, Mᵣ or molar mass carefully.
- Balance the chemical equation.
- Identify exactly what quantity is required.
Convert into moles
- Mass: divide by molar mass.
- Gas at r.t.p.: divide dm³ by 24.
- Solution: use n = cV with V in dm³.
- Particles: divide by 6.02 × 10²³.
Use the equation
- Read the coefficients as a mole ratio.
- Apply the ratio to moles, not masses.
- Check for a limiting reactant if two reactants are given.
- Use the limiting reactant to find the maximum product.
Finish the calculation
- Convert the final moles into the requested quantity.
- Use the correct unit.
- Round only at the final stage.
- Ask whether the answer is chemically reasonable.
Now apply the ideas from the chapter. These questions involve calculations, equation ratios and common examination traps.
- 10 application questions
- Multi-step stoichiometry
- Instant explanation for every answer option
Frequently asked questions
Questions students often ask about the mole concept
These questions connect the calculation methods with the chemical ideas behind them.
Why do we need the mole if we can already measure mass?
A chemical equation compares numbers of particles, not masses. Equal masses of two different substances usually contain different numbers of particles because their molar masses are different. The mole converts measurable masses into amounts that can be compared using the balanced equation.
Are Mᵣ and molar mass the same?
They have the same numerical value but different meanings. Mᵣ is a relative mass and has no unit. Molar mass is the mass of one mole and is measured in g mol⁻¹. Water therefore has Mᵣ = 18 and molar mass 18 g mol⁻¹.
Does one mole of every substance have the same mass?
No. One mole of every substance contains the same number of specified particles, but the particles themselves have different masses. One mole of H₂ has a mass of 2 g, whereas one mole of CO₂ has a mass of 44 g.
Does one mole of NaCl contain one mole or two moles of ions?
One mole of NaCl contains one mole of NaCl formula units. Each formula unit contains one Na⁺ ion and one Cl⁻ ion. It therefore contains one mole of Na⁺ ions and one mole of Cl⁻ ions, giving two moles of ions altogether.
Why does one mole of every gas occupy about 24 dm³ at r.t.p.?
Gas particles are very widely spaced. At the same temperature and pressure, the volume depends mainly on the number of gas particles rather than the identity of the gas. Equal numbers of gas particles therefore occupy equal volumes under the same conditions.
Can I always use 24 dm³ mol⁻¹ for a gas?
No. The value 24 dm³ mol⁻¹ is the approximate molar gas volume at room temperature and pressure. Gas volume changes when temperature or pressure changes, so use this value only when the question states or implies r.t.p.
Why must volume be converted from cm³ to dm³ in n = cV?
Concentration in mol dm⁻³ means moles per cubic decimetre. The volume used with it must therefore be in dm³. Since 1000 cm³ = 1 dm³, divide a cm³ value by 1000 before using n = cV.
Why can I not compare reacting masses directly?
Balanced-equation coefficients represent mole ratios. Since different substances have different molar masses, their masses do not follow the same numerical ratio. Convert each relevant mass into moles before using the equation coefficients.
Is the reactant with fewer moles always the limiting reactant?
No. The equation coefficients must also be considered. For 2H₂ + O₂ → 2H₂O, twice as many moles of H₂ are required as O₂. Divide each available mole amount by its coefficient. The smaller result identifies the limiting reactant.
Why do we divide by the smallest mole value when finding an empirical formula?
The empirical formula requires the simplest ratio rather than the actual number of moles in the sample. Dividing every amount by the smallest value scales the numbers down while preserving their relative proportions.
Can percentage yield be greater than 100%?
A genuine chemical yield should not exceed the theoretical maximum. A measured value above 100% usually means the product contains water or another impurity, or that there has been a measurement error.
Can a sample be very pure but still give a low percentage yield?
Yes. Purity describes what fraction of a sample is the required substance. Yield describes how much product is obtained compared with the theoretical maximum. A pure reactant can still give a low yield if the reaction is incomplete or product is lost during separation.
How do I know which mole formula to use first?
Look at the information given. Use mass ÷ molar mass for a mass, volume ÷ 24 for a gas at r.t.p., n = cV for a solution, and particle number ÷ 6.02 × 10²³ when particles are given. If another substance is involved, use the balanced equation after obtaining moles.
The mole concept becomes much easier once you stop treating mass, gas volume, concentration and reacting quantities as separate calculation topics. They are different routes into and out of the same central quantity: moles. Convert the information given into moles, use the balanced-equation ratio when another substance is involved, and then convert the resulting moles into whatever the question asks for. Once that route becomes automatic, much of O-Level stoichiometry becomes one connected piece of Chemistry rather than a collection of formulas to memorise.