What you'll learn
This guide covers the fundamental concept of the mole and how to perform calculations involving mass, molar mass and the number of particles. You'll learn how to calculate relative formula mass, use Avogadro's constant, and apply the mole concept to solve quantitative chemistry problems that appear regularly in AQA GCSE Chemistry papers.
Key terms and definitions
Mole (mol) — the unit for amount of substance; one mole of any substance contains 6.02 × 10²³ particles (atoms, molecules, ions or electrons)
Avogadro's constant — the number of particles in one mole of a substance, equal to 6.02 × 10²³ mol⁻¹
Relative atomic mass (Aᵣ) — the average mass of one atom of an element compared to 1/12th the mass of one carbon-12 atom; found on the periodic table
Relative formula mass (Mᵣ) — the sum of the relative atomic masses of all atoms in a formula; used for ionic compounds and simple molecules
Molar mass — the mass of one mole of a substance, measured in grams per mole (g/mol); numerically equal to the relative formula mass
Limiting reactant — the reactant that is completely used up in a chemical reaction, determining the maximum amount of product that can form
Molecular formula — shows the actual number of atoms of each element in one molecule of a substance
Empirical formula — the simplest whole number ratio of atoms of each element in a compound
Core concepts
Understanding the mole
The mole is a counting unit, similar to how a dozen represents 12 items. One mole represents 6.02 × 10²³ particles. This extremely large number allows chemists to count atoms, molecules and ions by weighing them.
Key points about the mole:
- One mole of any substance contains the same number of particles (Avogadro's constant)
- One mole of carbon-12 atoms has a mass of exactly 12 g
- One mole of different substances have different masses because atoms have different masses
- The mole links the microscopic world (atoms) to the macroscopic world (grams we can measure)
For example:
- 1 mole of sodium (Na) atoms = 23 g = 6.02 × 10²³ atoms
- 1 mole of water (H₂O) molecules = 18 g = 6.02 × 10²³ molecules
- 1 mole of sodium chloride (NaCl) = 58.5 g = 6.02 × 10²³ formula units
Calculating relative formula mass
To calculate the relative formula mass (Mᵣ), add up the relative atomic masses of all atoms in the chemical formula. Use the periodic table to find Aᵣ values.
Method:
- Write down the chemical formula
- List each element and count how many atoms are present
- Multiply each Aᵣ by the number of atoms of that element
- Add all values together
Example: Calculate Mᵣ of calcium carbonate (CaCO₃)
From the periodic table:
- Ca: Aᵣ = 40
- C: Aᵣ = 12
- O: Aᵣ = 16
Calculation:
- Ca: 1 × 40 = 40
- C: 1 × 12 = 12
- O: 3 × 16 = 48
- Mᵣ of CaCO₃ = 40 + 12 + 48 = 100
Example: Calculate Mᵣ of magnesium nitrate (Mg(NO₃)₂)
Be careful with brackets — multiply everything inside by the number outside.
From the periodic table:
- Mg: Aᵣ = 24
- N: Aᵣ = 14
- O: Aᵣ = 16
Calculation:
- Mg: 1 × 24 = 24
- N: 2 × 14 = 28 (the 2 outside brackets multiplies the N)
- O: 6 × 16 = 96 (2 × 3 = 6 oxygen atoms total)
- Mᵣ of Mg(NO₃)₂ = 24 + 28 + 96 = 148
Converting between mass and moles
The relationship between mass, moles and molar mass is fundamental to quantitative chemistry.
The key equation:
number of moles = mass (g) / molar mass (g/mol)
Or rearranged:
- mass (g) = number of moles × molar mass (g/mol)
- molar mass (g/mol) = mass (g) / number of moles
This equation appears on the AQA equation sheet as:
mass = Mr × moles
Remember: molar mass in g/mol is numerically equal to Mr (which has no units).
Examples:
Calculate the number of moles in 50 g of calcium carbonate (CaCO₃):
- Mr of CaCO₃ = 100
- moles = mass / Mr = 50 / 100 = 0.5 mol
Calculate the mass of 0.25 moles of sodium hydroxide (NaOH):
- Mr of NaOH = 23 + 16 + 1 = 40
- mass = moles × Mr = 0.25 × 40 = 10 g
Using Avogadro's constant
Avogadro's constant links moles to the actual number of particles.
The equation:
number of particles = moles × Avogadro's constant
Or: number of particles = moles × 6.02 × 10²³
Example calculations:
How many molecules are in 0.5 moles of water?
- number of molecules = 0.5 × 6.02 × 10²³
- number of molecules = 3.01 × 10²³
How many moles is 1.204 × 10²⁴ atoms of copper?
- moles = number of particles / Avogadro's constant
- moles = 1.204 × 10²⁴ / 6.02 × 10²³
- moles = 2 mol
You can combine this with the mass equation:
How many atoms are in 4 g of calcium (Ar = 40)?
- First find moles: moles = 4 / 40 = 0.1 mol
- Then find atoms: atoms = 0.1 × 6.02 × 10²³ = 6.02 × 10²²
Empirical and molecular formulae
The empirical formula shows the simplest whole number ratio of atoms. The molecular formula shows the actual number of atoms in one molecule.
Method for finding empirical formula from masses or percentages:
- List each element with its mass or percentage
- Divide each by the element's Ar to get moles
- Divide all mole values by the smallest number of moles
- If needed, multiply all numbers to get whole number ratios
- Write the empirical formula
Finding molecular formula:
If you know the empirical formula and the Mr:
- Calculate Mr of empirical formula
- Divide actual Mr by empirical formula Mr
- Multiply all subscripts in empirical formula by this number
Example: Empirical formula = CH₂, Mr = 42
- Mr of CH₂ = 12 + 2 = 14
- 42 / 14 = 3
- Molecular formula = C₃H₆
Conservation of mass in reactions
Mass is conserved in chemical reactions — the total mass of reactants equals the total mass of products. This principle underpins all quantitative chemistry.
In a closed system, no substances can enter or leave, so you can directly measure that mass is conserved.
In a non-enclosed system, if a gas is produced and escapes, the measured mass appears to decrease. If a gas reactant is used from the air, the measured mass appears to increase. However, if you account for all substances including gases, mass is still conserved.
Reacting mass calculations use the balanced equation and mole ratios:
- Write the balanced equation
- Calculate moles of the known substance
- Use the balanced equation to find moles of required substance
- Convert moles back to mass if needed
Worked examples
Example 1: Multi-step mole calculation
Question: Calculate the mass of carbon dioxide produced when 10 g of calcium carbonate (CaCO₃) thermally decomposes. [5 marks]
The equation is: CaCO₃ → CaO + CO₂
Solution:
Step 1: Calculate Mr values [1 mark]
- Mr of CaCO₃ = 40 + 12 + (16 × 3) = 100
- Mr of CO₂ = 12 + (16 × 2) = 44
Step 2: Calculate moles of CaCO₃ [1 mark]
- moles = mass / Mr = 10 / 100 = 0.1 mol
Step 3: Use balanced equation to find moles of CO₂ [1 mark]
- From equation: 1 mole CaCO₃ produces 1 mole CO₂
- Therefore: 0.1 mol CaCO₃ produces 0.1 mol CO₂
Step 4: Calculate mass of CO₂ [1 mark]
- mass = moles × Mr = 0.1 × 44 = 4.4 g
Answer: 4.4 g [1 mark for correct answer with working]
Example 2: Limiting reactant problem
Question: 6.4 g of methane (CH₄) reacts with 16 g of oxygen (O₂). Calculate the maximum mass of carbon dioxide (CO₂) that can be produced. [6 marks]
The equation is: CH₄ + 2O₂ → CO₂ + 2H₂O
Solution:
Step 1: Calculate Mr values [1 mark]
- Mr of CH₄ = 12 + (1 × 4) = 16
- Mr of O₂ = 16 × 2 = 32
- Mr of CO₂ = 12 + (16 × 2) = 44
Step 2: Calculate moles of each reactant [1 mark]
- moles of CH₄ = 6.4 / 16 = 0.4 mol
- moles of O₂ = 16 / 32 = 0.5 mol
Step 3: Determine limiting reactant [2 marks]
- From equation: 1 mol CH₄ needs 2 mol O₂
- 0.4 mol CH₄ needs 0.4 × 2 = 0.8 mol O₂
- Only 0.5 mol O₂ available, so oxygen is the limiting reactant
Step 4: Calculate moles of CO₂ from limiting reactant [1 mark]
- From equation: 2 mol O₂ produces 1 mol CO₂
- 0.5 mol O₂ produces 0.5 / 2 = 0.25 mol CO₂
Step 5: Calculate mass of CO₂ [1 mark]
- mass = 0.25 × 44 = 11 g
Answer: 11 g
Example 3: Empirical formula calculation
Question: A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Calculate its empirical formula. (Ar: C = 12, H = 1) [4 marks]
Solution:
Step 1: Write element and percentage [given]
| Element | C | H |
|---|---|---|
| % | 85.7 | 14.3 |
Step 2: Divide by Ar to get moles [1 mark]
| Element | C | H |
|---|---|---|
| moles | 85.7/12 = 7.14 | 14.3/1 = 14.3 |
Step 3: Divide by smallest value [1 mark]
| Element | C | H |
|---|---|---|
| ratio | 7.14/7.14 = 1 | 14.3/7.14 = 2 |
Step 4: Write empirical formula [2 marks]
Answer: CH₂
Common mistakes and how to avoid them
Forgetting to multiply atoms inside brackets — In Ca(OH)₂, there are 2 oxygen atoms and 2 hydrogen atoms, not 1 of each. Always expand brackets correctly when calculating Mr.
Confusing Mr with molar mass units — Mr has no units (it's a ratio), but molar mass is in g/mol. However, they're numerically equal, so Mr of water = 18 and molar mass = 18 g/mol.
Rearranging the mole equation incorrectly — Use the triangle method or write the equation clearly. To find mass: multiply moles by Mr. To find moles: divide mass by Mr.
Not using the balanced equation — The mole ratio between substances comes from the balanced equation. In N₂ + 3H₂ → 2NH₃, one mole of nitrogen makes two moles of ammonia, not one.
Rounding too early — Keep full calculator values until the final answer, then round to an appropriate number of significant figures (usually 2 or 3 at GCSE).
Forgetting Avogadro's constant is per mole — It's 6.02 × 10²³ mol⁻¹, meaning this is the number in ONE mole. For 2 moles, you'd have 1.204 × 10²⁴ particles.
Exam technique for "Quantitative chemistry: the mole and molar mass"
Show all working clearly — Even if your final answer is wrong, you can gain method marks for correct steps. Write formulas, substitute values, and show calculations.
Use the correct equation — The AQA equation sheet provides key formulas. For mole calculations, identify whether you need "mass = Mr × moles" or the Avogadro's constant equation.
Command words matter — "Calculate" requires numerical working and an answer with units. "Determine" means use information to work something out. "State" needs a brief answer without explanation.
Check units in your answer — Mass in grams (g), moles in mol, Mr has no units. A common error is writing "mol" after an Mr value or forgetting "g" after a mass.
Quick revision summary
The mole is the unit for amount of substance; one mole contains 6.02 × 10²³ particles (Avogadro's constant). Calculate relative formula mass (Mr) by adding atomic masses from the periodic table. Use the equation: moles = mass / Mr to convert between mass and moles. Balance chemical equations to find mole ratios between substances in reactions. The empirical formula shows the simplest ratio of atoms; the molecular formula shows actual numbers. Mass is always conserved in chemical reactions — reactant mass equals product mass.