Moles, Mass & Mr Calculations Revision Notes | A Level AQA Chemistry
The mass, moles and Mr triangle shows one relationship in three forms: n = m ÷ Mr, m = n × Mr and Mr = m ÷ n. Free AQA notes and 14 questions answered.
Updated 8 Mar 2026 · Revise Right Team
Moles, Mass & Mr Calculations is part of Amount of substance in A Level AQA Chemistry.
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What this lesson covers
The mass, moles and Mr triangle shows one relationship in three forms: n = m ÷ Mr, m = n × Mr and Mr = m ÷ n. Free AQA notes and 14 questions answered.
Calculate the amount in moles of a substance given its mass and relative molecular mass
Calculate the mass of a substance given the amount in moles and its relative molecular mass
Calculate the relative molecular mass of a substance given its mass and amount in moles
Apply the moles-mass-Mr triangle to multi-step problems with correct units throughout
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Use this lesson to stay inside Amount of substance while connecting the detail back to the wider Physical chemistry topic.
Mole (mol): The SI unit of amount of substance. One mole contains exactly 6.022 × 10²³ particles (Avogadro's number). Used when counting atoms, molecules, or ions in a sample.
Relative Molecular Mass (Mr): A dimensionless ratio comparing the average mass of one molecule of a substance to one-twelfth the mass of a carbon-12 atom. Calculated by summing the relative atomic masses (Ar) of all atoms in the molecular formula.
Amount of Substance (n): The quantity of a chemical species measured in moles. Calculated from mass and Mr using n = mass ÷ Mr. Expressed in mol.
Revision notes
Read the core explanations from Moles, Mass & Mr Calculations before testing yourself from memory.
1. Why Chemists Count in Moles
Atoms are far too small to count individually. A single grain of salt contains roughly 1018 sodium ions — a number so large it is meaningless in a laboratory context. Chemists needed a practical unit that connects the invisible world of atoms to the measurable world of grams on a balance.
That unit is the mole. One mole of any substance contains 6.022 × 10²³ particles — Avogadro's number. The power of this idea is that one mole of carbon-12 atoms has a mass of exactly 12 g, one mole of water molecules has a mass of 18 g, and so on. The mole bridges atomic mass and laboratory mass.
Every calculation in this topic flows from one relationship: the amount in moles depends on the mass of the sample and the relative molecular mass of the substance. Get this triangle right and the rest follows.
2. The Core Relationship: Mass, Mr, and Moles
The formula rearranges into three forms depending on which quantity you need to find. Cover the quantity you want with your finger and the remaining two show the operation: $n = m \div M_r$, $m = n \times M_r$, and $M_r = m \div n$. All three are equally important at A Level.
3. Calculating Mr from a Chemical Formula
Before you can use the moles equation, you need the Mr of the substance. This means reading the molecular formula carefully and summing the relative atomic masses (Ar) of every atom present, accounting for subscripts and brackets.
The Ar values you need are always provided in the AQA data booklet during exams, so you do not need to memorise them. What you do need is the method for combining them correctly — particularly when brackets appear in a formula.
Worked example: Calculating Mr for Calcium Nitrate, Ca(NO₃)₂
Calculate the relative molecular mass of calcium nitrate, Ca(NO₃)₂. Use Ar: Ca = 40, N = 14, O = 16.
Ca: 1 atom
N: 2 atoms (bracket × 2)
O: 6 atoms (3 × 2 from bracket) — The subscript 2 outside the bracket multiplies everything inside. There are 3 oxygen atoms per NO₃ group, so 3 × 2 = 6 oxygen atoms in total.
Ca: 1 × 40 = 40
N: 2 × 14 = 28
O: 6 × 16 = 96 — Each element's contribution to Mr is Ar × number of atoms of that element.
$M_r = 40 + 28 + 96 = 164$ — Add all contributions together. Mr has no units.
Answer: Mr of Ca(NO₃)₂ = 164
4. Finding Moles from Mass — Worked Examples
With Mr in hand, the moles calculation itself is a single division. The challenge in exam questions is not the arithmetic — it is reading the question carefully to extract the correct mass and identify the correct substance whose Mr you need.
Worked example: Calculate Moles from Mass
Calculate the amount, in moles, of sodium hydroxide (NaOH) in 8.00 g of the substance. Use Ar: Na = 23, O = 16, H = 1. [2 marks]
$M_r(\text{NaOH}) = 23 + 16 + 1 = 40$ — Sum the Ar values for each atom in the formula. [M1] for correct Mr.
$n = \frac{m}{M_r} = \frac{8.00}{40}$ — Substitute mass = 8.00 g and Mr = 40 into n = m ÷ Mr.
$n = 0.200 \text{ mol}$ — 8.00 ÷ 40 = 0.200. Always state the unit mol. [A1] for correct answer with unit.
Answer: n = 0.200 mol
Notice the answer is given to 3 significant figures, matching the precision of the data given (8.00 g). AQA mark schemes typically accept answers rounded to 2 or 3 significant figures unless the question specifies otherwise — but always match the precision of the data you are given.
Worked example: Calculate Mass from Moles
Calculate the mass of 0.250 mol of sulfuric acid, H₂SO₄. Use Ar: H = 1, S = 32, O = 16. [2 marks]
$m = n \times M_r = 0.250 \times 98$ — Use the rearrangement m = n × Mr. Substitute n = 0.250 mol and Mr = 98.
$m = 0.250 \times 98 = 24.5 \text{ g}$ — 0.250 × 98 = 24.5. Units are grams (g). [A1] for 24.5 g.
Answer: m = 24.5 g
5. Finding Mr from Mass and Moles
The third rearrangement — $M_r = m \div n$ — appears less often but is tested at A Level, particularly in questions involving experimental data. You might be given the mass of a sample and the amount in moles determined from another measurement, and asked to identify the substance.
Worked example: Determine Mr from Experimental Data
A student dissolves a sample of an unknown acid in water. The sample has a mass of 4.90 g and the amount of acid is found to be 0.0500 mol. Calculate the Mr of the acid and suggest its identity. [3 marks]
$M_r = \frac{m}{n}$ — We know mass and moles, so rearrange to find Mr. [M1] for correct rearrangement.
$M_r = \frac{4.90}{0.0500}$ — Mass = 4.90 g, n = 0.0500 mol. Take care with the decimal — 0.0500 is 5.00 × 10⁻².
This type of question rewards methodical working. Even if you cannot identify the substance, you still earn the method and accuracy marks for the correct Mr value — so always show your working fully.
6. Handling Units and Significant Figures
The moles equation requires mass in grams. If a question gives mass in kilograms or milligrams, convert first. This is one of the most reliable ways to drop a mark in an otherwise correct calculation.
Worked example: Unit Conversion Before Calculating Moles
Calculate the amount, in moles, of magnesium oxide (MgO) in 2.40 kg of the substance. Use Ar: Mg = 24, O = 16. [3 marks]
$m = 2.40 \text{ kg} \times 1000 = 2400 \text{ g}$ — The moles equation requires mass in grams. 1 kg = 1000 g, so 2.40 × 1000 = 2400 g. [M1] for correct conversion.
$M_r(\text{MgO}) = 24 + 16 = 40$ — One Mg atom and one O atom per formula unit.
$n = \frac{m}{M_r} = \frac{2400}{40}$ — Substitute the converted mass and the calculated Mr. [M1] for correct substitution.
$n = 2400 \div 40 = 60.0 \text{ mol}$ — 2400 ÷ 40 = 60.0. State the unit mol. [A1] for 60.0 mol.
Answer: n = 60.0 mol
Significant figures follow the data in the question. Here, 2.40 kg has 3 significant figures, so the answer is given as 60.0 mol rather than just 60. That trailing zero after the decimal point signals precision and is worth keeping.
7. Exam Technique and Common Mistakes
AQA mark schemes for moles calculations typically award one method mark for the correct Mr, one method mark for the correct substitution into the equation, and one accuracy mark for the final answer with the correct unit. Losing the unit 'mol' on the final answer costs the accuracy mark — even if the number is right.
A common exam trap is treating the identity step as exact when the calculated Mr is only close to one of the options. If your value is 48.0 and the nearest listed formula has Mr 46, you should state that the match is approximate and may reflect experimental uncertainty, rounding, or limited answer choices. The calculation itself is still valid, but the interpretation must acknowledge the mismatch.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-03-08.
The mass, moles and Mr triangle is a memory aid for one relationship with three forms: n = m ÷ Mr, m = n × Mr and Mr = m ÷ n. Covering the quantity needed leaves the operation using the other two, with mass in grams.
How do you calculate moles from mass?
Divide the mass in grams by the relative molecular mass: n = m ÷ Mr. For 8.00 g of sodium hydroxide, Mr(NaOH) = 23 + 16 + 1 = 40, so n = 8.00 ÷ 40 = 0.200 mol. Always state the unit mol, or the accuracy mark is lost.
How do you calculate mass from moles?
Multiply the amount in moles by the relative molecular mass: m = n × Mr. For 0.250 mol of sulfuric acid, H₂SO₄, Mr = (2 × 1) + 32 + (4 × 16) = 98, so m = 0.250 × 98 = 24.5 g.
How do you calculate the Mr of calcium nitrate?
The Mr of calcium nitrate, Ca(NO₃)₂, is 164. The subscript 2 outside the bracket multiplies everything inside, giving 1 Ca, 2 N and 6 O, so Mr = (1 × 40) + (2 × 14) + (6 × 16) = 40 + 28 + 96 = 164, with no units.
What is the Mr of magnesium oxide?
The Mr of magnesium oxide, MgO, is 40, from Mg = 24 plus O = 16. It is ionic, so it is a relative formula mass, but it uses the same symbol and the same calculation.
How do you work out Mr from mass and moles?
Rearrange the moles equation to Mr = m ÷ n. A 4.90 g sample of an unknown acid containing 0.0500 mol gives Mr = 4.90 ÷ 0.0500 = 98, which matches sulfuric acid, H₂SO₄. Show the working, as method and accuracy marks are awarded even if the identity is wrong.
Why must mass be in grams for moles calculations?
The moles equation n = m ÷ Mr requires mass in grams, so convert kilograms or milligrams first. For 2.40 kg of magnesium oxide, m = 2.40 × 1000 = 2400 g and Mr(MgO) = 40, so n = 2400 ÷ 40 = 60.0 mol, given to 3 significant figures to match the data.
How many moles are in 8.00 g of sodium hydroxide?
8.00 g of sodium hydroxide is 0.200 mol. The Mr of NaOH is 23 + 16 + 1 = 40, and n = m ÷ Mr = 8.00 ÷ 40 = 0.200 mol, given to 3 significant figures to match the data.
What is the mass of 0.250 mol of sulfuric acid?
0.250 mol of sulfuric acid has a mass of 24.5 g. The Mr of H₂SO₄ is (2 × 1) + 32 + (4 × 16) = 98, and m = n × Mr = 0.250 × 98 = 24.5 g.
How can Mr identify an unknown acid?
Mr is calculated from Mr = m ÷ n and compared with known Mr values. A 4.90 g sample containing 0.0500 mol of acid gives Mr = 4.90 ÷ 0.0500 = 98, which matches sulfuric acid, H₂SO₄.
What mass does one mole of carbon-12 have?
One mole of carbon-12 atoms has a mass of exactly 12 g, and one mole of water molecules has a mass of 18 g. The mole bridges atomic mass and laboratory mass, connecting the scale of atoms to grams on a balance.
How many significant figures should a moles answer have?
A moles answer should match the precision of the data given. Since 2.40 kg has 3 significant figures, 2400 ÷ 40 is given as 60.0 mol rather than 60; that trailing zero after the decimal point signals precision.
What if a calculated Mr does not exactly match any option?
If a calculated Mr is close to but not exactly one of the options, the closest match is chosen and the answer states that the match is approximate. The mismatch may reflect experimental uncertainty, rounding or limited answer choices.
Where are marks lost in moles calculations?
Marks in moles calculations are often lost when the unit mol is absent from the final answer, which costs the accuracy mark even when the number is right. Other common slips are using mass in kg or mg instead of grams and errors counting atoms inside brackets.