How to Calculate Moles in Chemistry (Without Losing Marks)

Quick answer: A mole is a counting number — 6.02 x 10^23 particles — and nearly every mole question comes down to four formulas: moles = mass / Mr, moles = volume in dm3 / 24 for a gas at room temperature and pressure, moles = concentration x volume in dm3 for a solution, and number of particles = moles x 6.02 x 10^23.

I do IGCSE and A-Levels, and chemistry is not my strongest subject — English is. The first time I sat a quantitative chemistry paper I dropped four marks on one question, and not one of them was for not knowing the chemistry. I divided a volume by 24 while it was still in cm3. The mole is not a hard idea. It is an idea that punishes sloppy units, which is a completely different problem with a completely different fix.

What a mole actually is

A mole is a fixed quantity of particles: 6.02 x 10^23 of them, in the same way a dozen is 12 of anything. Avogadro's constant was chosen so the periodic table becomes usable in a lab — the relative atomic mass of any element, written in grams, is one mole of it. So Mr (relative formula mass) and molar mass are the same number wearing different units: water's Mr is 18, so its molar mass is 18 g/mol.

Be strict about one thing: a mole is always a mole of something. Before you touch a calculator, write down what the moles are of. This matters most with ions — one mole of MgCl2 contains one mole of Mg2+ and two moles of Cl-, so three moles of ions in total. Examiners love that gap.

The four formulas that cover nearly every question

Choose your formula from the units the question hands you, not from the chapter title.

• Given a mass in grams: moles = mass / Mr
• Given a volume of gas: moles = volume in dm3 / 24 (at RTP)
• Given a solution with a concentration in mol/dm3: moles = concentration x volume in dm3
• Asked for a number of atoms, molecules or ions: number = moles x 6.02 x 10^23

Every reacting-mass question is then the same three-step sandwich. Convert what you are given into moles. Use the balanced equation to swap to moles of the substance you want. Convert those moles back out into grams, dm3 or mol/dm3. The chemistry lives only in the middle step, and that step is worthless if the equation is not balanced first — fix that before you go near mole arithmetic, on the chemistry hub if you need it.

Worked example: 5.0 g of limestone to a volume of gas

Calcium carbonate reacts with excess hydrochloric acid:

CaCO3 + 2HCl -> CaCl2 + H2O + CO2

Calculate the mass and volume of carbon dioxide made from 5.0 g of CaCO3 at RTP.

1. Mr of CaCO3 = 40 + 12 + (3 x 16) = 100
2. Moles of CaCO3 = 5.0 / 100 = 0.050 mol
3. Ratio from the equation is 1 : 1, so moles of CO2 = 0.050 mol
4. Mr of CO2 = 12 + 32 = 44, so mass of CO2 = 0.050 x 44 = 2.2 g
5. Volume of CO2 at RTP = 0.050 x 24 = 1.2 dm3, which is 1200 cm3

The words "excess hydrochloric acid" are doing real work there. They tell you the acid never runs out, so the carbonate controls the answer and you can ignore the acid completely.

Mark schemes here usually give a mark for the Mr, one for the moles, one for the ratio, and one for the final answer with units. A wrong Mr at step 1 still scores the rest on error carried forward — but only if every line is written down.

Gas volumes: use 24 dm3 per mole at RTP

For GCSE (AQA, Edexcel, OCR) and Cambridge IGCSE, the molar gas volume is 24 dm3 per mole at room temperature and pressure, roughly 20 degrees C and normal atmospheric pressure. That is also 24 000 cm3 per mole, which is the version worth memorising, because gas volumes in questions are so often printed in cm3.

The 22.4 dm3 per mole you may have seen is the molar volume at STP (0 degrees C), and it is the figure CBSE material quotes most often. AP Chemistry goes further again and expects the ideal gas equation, PV = nRT, for non-standard conditions rather than one fixed number. Check which conditions your own syllabus states before you memorise anything: using 22.4 in a paper that specified RTP loses the accuracy mark even when the method is perfect.

The unit trap is the real killer. If a gas volume is in cm3, divide by 1000 first, then divide by 24. Dividing 1200 cm3 straight by 24 gives 50, which is a thousand times too big.

Moles in solutions: concentration and titration numbers

For solutions, moles = concentration (mol/dm3) x volume (dm3), and the volume is nearly always given in cm3.

Worked example: 25.0 cm3 of 0.100 mol/dm3 sodium hydroxide is exactly neutralised by 20.0 cm3 of hydrochloric acid, where NaOH + HCl -> NaCl + H2O.

1. Volume of NaOH = 25.0 / 1000 = 0.0250 dm3
2. Moles of NaOH = 0.100 x 0.0250 = 0.00250 mol
3. Ratio is 1 : 1, so moles of HCl = 0.00250 mol
4. Volume of HCl = 20.0 / 1000 = 0.0200 dm3
5. Concentration of HCl = 0.00250 / 0.0200 = 0.125 mol/dm3

One conversion is worth knowing: concentration in g/dm3 = concentration in mol/dm3 x Mr. So that acid is 0.125 x 36.5 = 4.56 g/dm3. Papers swap between the two units deliberately.

When two amounts are given: find the limiting reactant

If a question gives quantities for two reactants instead of saying "excess", one runs out first and decides the answer. Work out the moles of both, divide each by its coefficient in the balanced equation, and the smaller result is limiting.

Say 4.0 g of magnesium is added to 100 cm3 of 1.0 mol/dm3 hydrochloric acid, with Mg + 2HCl -> MgCl2 + H2.

• Moles of Mg = 4.0 / 24 = 0.167, divided by 1 = 0.167
• Moles of HCl = 1.0 x 0.100 = 0.100, divided by 2 = 0.050

The acid is limiting. Moles of H2 = 0.100 / 2 = 0.050 mol, so the volume of hydrogen at RTP = 0.050 x 24 = 1.2 dm3. Use the magnesium instead and you get 4 dm3 and lose every mark after the first.

The mistakes that cost me marks

• Not converting cm3 to dm3 first. Every factor-of-1000 error I have made started here.
• Rounding halfway through. Carry the full calculator value, round only at the end, usually to the significant figures of the data you were given.
• Assuming a 1 : 1 ratio because most examples are. Read the coefficients every time.
• Writing a bare number. Moles need "mol", masses need "g", concentrations need "mol/dm3", and units are often a separate mark.
• Using the Mr of the wrong substance. Label every line with the formula it belongs to.

To check your working rather than just your answer, paste the whole question into the step-by-step solver or ask for the reasoning in explain mode, then compare line by line. Where you diverge is the thing you actually need to relearn.

Test yourself

1. How many moles are there in 8.0 g of sodium hydroxide, NaOH (Mr = 40)?
2. What volume, in cm3, does 0.25 mol of carbon dioxide occupy at RTP?
3. 20.0 cm3 of hydrochloric acid is exactly neutralised by 0.00200 mol of sodium hydroxide in a 1 : 1 reaction. What is its concentration in mol/dm3?

Answers: 0.20 mol; 6000 cm3; 0.100 mol/dm3. For more of these with instant marking, generate a set on the quiz page.

FAQ

Is Mr the same thing as molar mass?

Numerically yes. Mr is a plain number with no units, found by adding the relative atomic masses in the formula. Molar mass is that same number in grams per mole. Water: Mr = 18, molar mass = 18 g/mol.

Do I have to memorise 6.02 x 10^23 and 24 dm3?

Learn 24 dm3 per mole at RTP, because it is not always printed on the paper. Avogadro's constant normally is given, either in the question or in the data booklet.

Why is my answer always a thousand times out?

Almost certainly cm3 versus dm3. 1 dm3 = 1000 cm3, so divide any cm3 figure by 1000 before it enters a mole formula, and make that a separate written line.

What is the difference between RTP and STP?

RTP is room temperature and pressure, about 20 degrees C, where one mole of gas occupies 24 dm3. STP is standard temperature and pressure, 0 degrees C, where it occupies 22.4 dm3. Use whichever your syllabus states — GCSE and Cambridge IGCSE use RTP, and 22.4 appears mainly in CBSE material.

In short: the mole is a counting unit, not something to be scared of. Learn the four formulas, always write down what the moles are of, convert cm3 to dm3 before anything else, and use 24 dm3 per mole at RTP unless your syllabus says otherwise. Most marks lost in this topic are lost to units, not to chemistry.