How do chemists count atoms they cannot see? How do you weigh out exactly the same number of particles of two completely different substances? The answer is the mole — chemistry’s most powerful counting tool. Once you understand the mole, calculations that once looked impossible will start to feel logical and manageable.
Every term below is examinable. Learn them in order because each one builds on the last.
Atomic Mass Unit (a.m.u.)
The unit used to express the mass of atoms and molecules. One atomic mass unit is defined as one-twelfth of the mass of a carbon-12 atom. Because atoms are unimaginably small, using grams to measure a single atom would be impractical — the atomic mass unit gives chemists a workable scale.
Why it matters: Every relative atomic mass and relative molecular mass value on your periodic table is measured in atomic mass units.
Relative Atomic Mass (Ar)
The average mass of one atom of an element compared to one-twelfth of the mass of a carbon-12 atom. It has no units because it is a ratio — a comparison, not an actual weight.
Formula: Ar = (Average mass of one atom of the element) ÷ (1/12 × mass of one carbon-12 atom)
Real-life example: The relative atomic mass of oxygen is 16. This means one oxygen atom is 16 times heavier than one-twelfth of a carbon-12 atom.
Exam note: When an element has isotopes, the relative atomic mass is the weighted average of all isotope masses based on their natural abundance. This is why chlorine has a relative atomic mass of 35.5 — it is a weighted average of chlorine-35 and chlorine-37.
Relative Molecular Mass (Mr)
The sum of the relative atomic masses of all the atoms in one molecule of a substance. Like relative atomic mass, it has no units.
Formula: Mr = sum of all Ar values of atoms in the molecule
Worked example: Mr of water (H₂O) = (2 × 1) + (1 × 16) = 2 + 16 = 18 Mr of carbon dioxide (CO₂) = (1 × 12) + (2 × 16) = 12 + 32 = 44 Mr of sulphuric acid (H₂SO₄) = (2 × 1) + (1 × 32) + (4 × 16) = 2 + 32 + 64 = 98
Exam tip: For ionic compounds such as sodium chloride (NaCl), the correct term is relative formula mass — not relative molecular mass — because ionic compounds do not exist as individual molecules. However, the calculation method is identical.
The Mole
The SI unit for amount of substance. One mole of any substance contains exactly 6.02 × 10²³ particles — whether those particles are atoms, molecules, ions or electrons. This number is called the Avogadro constant.
Think of it this way: Just as a dozen always means 12 of anything, a mole always means 6.02 × 10²³ of anything — regardless of what the substance is.
Real-life example: One mole of water contains 6.02 × 10²³ water molecules. One mole of sodium contains 6.02 × 10²³ sodium atoms. The number is always the same — only the mass changes.
Avogadro’s Constant (Avogadro’s Number)
The number of particles contained in one mole of any substance. Its value is 6.02 × 10²³ mol⁻¹.
Named after: Italian scientist Amedeo Avogadro, who first proposed that equal volumes of gases at the same temperature and pressure contain equal numbers of particles.
Significance: The Avogadro constant is the bridge between the atomic scale — where we count individual particles — and the laboratory scale — where we measure grams and litres. Without it, chemistry calculations would be impossible.
Molar Mass
The mass of one mole of a substance expressed in grams per mole (g/mol). Numerically, the molar mass of a substance equals its relative atomic mass or relative molecular mass — the only difference is the unit.
Examples:
- Molar mass of carbon (C) = 12 g/mol
- Molar mass of water (H₂O) = 18 g/mol
- Molar mass of sodium chloride (NaCl) = 23 + 35.5 = 58.5 g/mol
The key insight: Molar mass tells you the mass in grams of exactly 6.02 × 10²³ particles of that substance. Weigh out 18 g of water and you have exactly one mole — exactly 6.02 × 10²³ water molecules.
The Mole Formula (Amount of Substance)
The central formula that connects mass, molar mass and number of moles.
Formula: n = m ÷ M
Where:
- n = number of moles (mol)
- m = mass of the substance (g)
- M = molar mass (g/mol)
Worked example 1 — finding moles from mass: How many moles are in 44 g of carbon dioxide (CO₂)? Mr of CO₂ = 44 g/mol n = 44 ÷ 44 = 1 mole
Worked example 2 — finding mass from moles: What is the mass of 2 moles of water (H₂O)? Mr of H₂O = 18 g/mol m = n × M = 2 × 18 = 36 g
Worked example 3 — finding molar mass from mass and moles: A substance has a mass of 10 g and contains 0.5 moles. What is its molar mass? M = m ÷ n = 10 ÷ 0.5 = 20 g/mol
Molar Volume of a Gas
At standard temperature and pressure (STP — 0°C and 1 atm), one mole of any gas occupies 22.4 litres (22,400 cm³). At room temperature and pressure (RTP — 25°C and 1 atm), this value is 24 litres (24,000 cm³).
This is the same for every gas regardless of what the gas is — because at the same temperature and pressure, equal moles of all gases occupy equal volumes.
Formula: n = V ÷ 22.4 (at STP) or n = V ÷ 24 (at RTP)
Worked example: What volume does 2 moles of oxygen gas occupy at STP? V = n × 22.4 = 2 × 22.4 = 44.8 litres
Exam note: some exam questions usually specify which value to use. If no condition is stated, use 22.4 dm³ at STP. papers most commonly use 22.4 dm³.
Empirical Formula
The simplest whole-number ratio of atoms of each element in a compound. The empirical formula shows the ratio — not the actual number — of atoms present.
Example: Glucose has the molecular formula C₆H₁₂O₆. Its empirical formula is CH₂O because the ratio of carbon : hydrogen : oxygen is 1 : 2 : 1.
How to find the empirical formula from percentage composition:
- Write down the percentage of each element
- Divide each percentage by the element’s relative atomic mass to get the mole ratio
- Divide all values by the smallest value to get the simplest ratio
- If the ratio is not a whole number, multiply through to make it whole
Worked example: A compound contains 40% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Find the empirical formula.
- C: 40 ÷ 12 = 3.33
- H: 6.7 ÷ 1 = 6.7
- O: 53.3 ÷ 16 = 3.33
- Divide by smallest (3.33): C = 1, H = 2, O = 1
- Empirical formula = CH₂O
Molecular Formula
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.
Formula: Molecular formula = (Empirical formula) × n Where n = Mr of compound ÷ Mr of empirical formula
Worked example: The empirical formula of a compound is CH₂O and its relative molecular mass is 180. Mr of CH₂O = 12 + 2 + 16 = 30 n = 180 ÷ 30 = 6 Molecular formula = C₆H₁₂O₆ (glucose)
Percentage Composition by Mass
The percentage by mass of each element in a compound. Used to identify compounds, analyse purity and find empirical formulas.
Formula: % composition = (Total Ar of element in formula ÷ Mr of compound) × 100
Worked example: Find the percentage of nitrogen in ammonia (NH₃). Mr of NH₃ = 14 + (3 × 1) = 17 % N = (14 ÷ 17) × 100 = 82.4%
Real-life example: Fertiliser manufacturers calculate percentage composition to determine how much nitrogen a fertiliser delivers to crops per kilogram.
Mole Ratio (Stoichiometry)
The ratio of moles of reactants and products as shown by the coefficients in a balanced chemical equation. Mole ratios are used to calculate how much product forms from a given amount of reactant, or how much reactant is needed to produce a desired amount of product.
Example: N₂ + 3H₂ → 2NH₃ The mole ratio is: 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.
Worked example: How many moles of ammonia are produced when 5 moles of hydrogen react completely? Ratio H₂ : NH₃ = 3 : 2 Moles of NH₃ = (5 × 2) ÷ 3 = 3.33 moles
Limiting Reactant (Limiting Reagent)
The reactant that is completely used up first in a chemical reaction. The limiting reactant determines how much product is formed because the reaction cannot continue once it is exhausted.
Real-life example: If you are making sandwiches and have 10 slices of bread but only 3 slices of cheese, the cheese is the limiting reactant — you can only make 3 sandwiches no matter how much bread you have.
How to identify the limiting reactant:
- Convert the mass of each reactant to moles
- Divide each by its coefficient in the balanced equation
- The reactant with the smaller value is the limiting reactant
Excess Reactant
The reactant that remains after the limiting reactant is completely used up. The excess reactant is present in greater quantity than required by the stoichiometry of the reaction.
Theoretical Yield
The maximum amount of product that could be obtained from a reaction if it proceeded perfectly with no losses. Calculated using the mole ratio from the balanced equation and the amount of the limiting reactant.
Percentage Yield
The percentage of the theoretical yield that is actually obtained in practice. In real laboratory and industrial reactions, the actual yield is almost always less than the theoretical yield due to side reactions, incomplete reactions, product loss during separation, and other practical factors.
Formula: % yield = (Actual yield ÷ Theoretical yield) × 100
Worked example: A reaction has a theoretical yield of 20 g of product. The actual yield obtained is 15 g. % yield = (15 ÷ 20) × 100 = 75%
Real-life example: Industrial chemists aim for the highest possible percentage yield to reduce waste and lower production costs.
Molar Concentration (Molarity)
The number of moles of solute dissolved in one litre (1 dm³) of solution. Measured in mol/dm³ or mol/L, sometimes written as M.
Formula: c = n ÷ V
Where:
- c = concentration (mol/dm³)
- n = number of moles of solute (mol)
- V = volume of solution (dm³)
Worked example: What is the concentration of a solution containing 4 moles of sodium hydroxide dissolved in 2 dm³ of solution? c = 4 ÷ 2 = 2 mol/dm³
Note: If volume is given in cm³, divide by 1000 to convert to dm³ before using the formula.
Common Exam Questions
Question 1: Calculate the number of moles in 9 g of water (H₂O). (Ar: H = 1, O = 16)
Model answer: Mr of H₂O = (2 × 1) + 16 = 18 g/mol n = m ÷ M = 9 ÷ 18 = 0.5 moles Therefore 9 g of water contains 0.5 moles of water molecules.
Question 2: A compound contains 75% carbon and 25% hydrogen by mass. Its relative molecular mass is 16. Determine its empirical formula and molecular formula. (Ar: C = 12, H = 1)
Model answer: Step 1 — Empirical formula: C: 75 ÷ 12 = 6.25 → divide by 6.25 → 1 H: 25 ÷ 1 = 25 → divide by 6.25 → 4 Empirical formula = CH₄
Step 2 — Molecular formula: Mr of CH₄ = 12 + (4 × 1) = 16 n = 16 ÷ 16 = 1 Molecular formula = CH₄ (methane)
Question 3: 25 cm³ of a sodium hydroxide solution contains 0.05 moles of NaOH. Calculate the molar concentration of the solution.
Model answer: Convert volume: 25 cm³ ÷ 1000 = 0.025 dm³ c = n ÷ V = 0.05 ÷ 0.025 = 2 mol/dm³ The concentration of the sodium hydroxide solution is 2 mol/dm³.
Question 4: Calculate the volume occupied by 3 moles of carbon dioxide gas at STP. (Molar volume at STP = 22.4 dm³/mol)
Model answer: V = n × 22.4 = 3 × 22.4 = 67.2 dm³ Three moles of carbon dioxide occupies 67.2 dm³ at STP.
Conclusion
The mole concept is the foundation of all quantitative chemistry — from calculating masses of reactants to preparing solutions of known concentration. Master these definitions and the three core formulas (n = m ÷ M, c = n ÷ V, and V = n × 22.4) and you will be able to handle every mole calculation your exam throws at you.
For more Chemistry revision, read our posts on [Chemical Formulae and Equations key terms] and Kinetic Theory and Diffusion key terms. Test yourself with our [Mole Concept quiz].