The Mole and Mole Calculations
The mole is chemistry’s counting unit. Almost every calculation in the subject runs through it, and tutors gave the same instruction over and over: always find the moles first.
1. What a mole is
A mole is the amount of substance containing 6.02 × 10²³ particles — the Avogadro constant.
That number of particles of an element has a mass in grams equal to its relative atomic mass.
Molar mass is the mass of one mole, in g/mol. Numerically it equals the Ar (for elements) or the Mr (for compounds).
Calculating Mr: add up the Ar values of every atom in the formula.
- H₂O = (2 × 1) + 16 = 18
- NaOH = 23 + 16 + 1 = 40
- CaCO₃ = 40 + 12 + (3 × 16) = 100
- Ca(OH)₂ = 40 + 2(16 + 1) = 74 — note the brackets apply to both O and H
Use the correct Ar values from the periodic table. Recorded errors included a wrong molar mass for water and a wrong Mr for sodium hydroxide — both from arithmetic rather than method. Tutors recommended having a printed periodic table to hand while practising.
2. The three core equations
moles = mass ÷ molar mass (n = m / Mr) moles = concentration × volume in dm³ (n = c × V) moles of gas = volume in dm³ ÷ 24 (at room temperature and pressure)
The formula is moles = MASS ÷ MOLAR MASS. A recorded error stated it as “molar mass ÷ atomic mass”, which is meaningless. Learn the triangle: mass on top, moles × molar mass below.
Rearranged:
- mass = moles × molar mass
- concentration = moles ÷ volume
- volume of gas = moles × 24
Worked example. How many moles in 25 g of CaCO₃ (Mr = 100)?
- n = 25 ÷ 100 = 0.25 mol
Worked example. Mass of 0.5 mol of NaOH (Mr = 40)?
- m = 0.5 × 40 = 20 g
Worked example. Number of molecules in 0.25 mol?
- 0.25 × 6.02 × 10²³ = 1.51 × 10²³
3. Units — where most marks go
Concentration is in mol/dm³, so the VOLUME must be in dm³.
1 dm³ = 1000 cm³. To convert cm³ → dm³, DIVIDE by 1000.
Example: 25.0 cm³ = 0.025 dm³
Unit conversion was the single most frequent error in this topic, recorded three separate times — failing to convert to dm³, and converting the wrong way. Make it the first line of your working.
Divide by 1000 going from cm³ to dm³; multiply by 1000 going the other way. If your answer looks a million times out, this is why.
A shortcut worth knowing: for volumes in cm³, you can use n = (c × V) ÷ 1000 directly.
4. Concentration
concentration (mol/dm³) = moles ÷ volume (dm³)
Concentration in g/dm³:
concentration (g/dm³) = concentration (mol/dm³) × molar mass
Worked example. 5.85 g of NaCl (Mr = 58.5) dissolved in 250 cm³.
- n = 5.85 ÷ 58.5 = 0.1 mol
- V = 250 ÷ 1000 = 0.25 dm³
- c = 0.1 ÷ 0.25 = 0.4 mol/dm³
5. Gas volumes
At room temperature and pressure (r.t.p.), one mole of any gas occupies 24 dm³ (24 000 cm³).
Worked example. Volume of 0.5 mol of CO₂ at r.t.p.?
- V = 0.5 × 24 = 12 dm³
The molar gas volume is the same for ALL gases — it doesn’t depend on which gas it is.
6. The method that works every time
1. Write the BALANCED equation. 2. Find the MOLES of what you’re given. 3. Use the MOLE RATIO from the equation. 4. Convert back to the quantity asked for.
“Always find the number of moles first” — tutors said this in almost identical words at least three times. It is the single most useful habit in stoichiometry.
Then take the ratio of what you’re given to what you’re asked for — also tutors’ own phrasing.
Worked example. What mass of CO₂ forms when 25 g of CaCO₃ decomposes?
CaCO₃ → CaO + CO₂
- Moles of CaCO₃ = 25 ÷ 100 = 0.25 mol
- Ratio is 1 : 1, so moles of CO₂ = 0.25 mol
- Mr of CO₂ = 12 + 32 = 44
- Mass = 0.25 × 44 = 11 g
Don’t try to convert mass to mass directly. A recorded error attempted to find a mass without going through the mole ratio — masses do not scale with the equation coefficients, but moles do.
7. Mistakes that cost marks
Not converting cm³ to dm³.
Dividing when you should multiply in a unit conversion.
Misquoting the moles formula.
Wrong Mr — miscounting atoms or ignoring brackets.
Skipping the mole ratio and working mass to mass.
Forgetting to balance the equation first.
Not showing working.
Rounding too early in multi-step calculations.
Wrong units on the answer — mol, g, dm³, mol/dm³.
Frequently asked questions
What is a mole? The amount of substance containing 6.02 × 10²³ particles.
What is the Avogadro constant? 6.02 × 10²³ — the number of particles in one mole.
How do I calculate moles from mass? moles = mass ÷ molar mass.
How do I find molar mass? Add the Ar values of all the atoms in the formula.
How do I convert cm³ to dm³? Divide by 1000.
What is the formula for concentration? moles ÷ volume in dm³, giving mol/dm³.
What volume does one mole of gas occupy? 24 dm³ at r.t.p. — the same for any gas.
What’s the first step in any stoichiometry question? Find the moles.
Can I convert mass to mass directly? No — go via moles and the mole ratio.
How do I find the number of particles? moles × 6.02 × 10²³.
Quick revision checklist
- I know what a mole is and the value of the Avogadro constant
- I can calculate Mr, handling brackets correctly
- I know n = m / Mr and can rearrange it
- I know n = c × V with V in dm³
- I know 1 dm³ = 1000 cm³ and which way to convert
- I know one mole of gas is 24 dm³ at r.t.p.
- I balance the equation first
- I find the moles before anything else
- I use the mole ratio from the equation
- I never convert mass to mass directly
- I show all working and keep full accuracy
- I give the correct units
These notes cover the mole and mole calculations in the Cambridge IGCSE Chemistry (0620) syllabus and are written for Grade 9–11 / Year 10–11 students. They are based on teaching patterns observed across a large set of one-to-one IGCSE Chemistry lessons, with particular attention to the errors students make most often and the wording examiners reward. Always check the current syllabus and data booklet for your own exam series.
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