Relative Masses and Moles

Summary: The mole is the unit of amount of substance. One mole contains 6.02 x 10^23 particles (Avogadro’s constant). Relative atomic mass (Ar) is found on the Periodic Table. Relative molecular/formula mass (Mr) is the sum of Ar values. The central equation is moles = mass / Mr. Tags: igcse chemistry stoichiometry Created: 2026-07-14 Last Updated: 2026-07-16

Relative Atomic Mass (Ar)

Definition: The relative atomic mass (Ar) of an element is the average mass of all the isotopes of that element, compared to 1/12 of the mass of a carbon-12 atom.

How to find Ar on the Periodic Table:

• The Ar is the larger number shown with each element on the Periodic Table
• It is usually written above or below the element symbol
• Ar values are often not whole numbers because they are weighted averages of isotopes (e.g., chlorine = 35.5)

Examples of common Ar values:

Relative Molecular / Formula Mass (Mr)

Definition: The relative molecular mass (or relative formula mass) is the sum of the relative atomic masses of all the atoms in a formula.

• Relative molecular mass is used for covalent compounds (molecules), e.g., H2O, CO2
• Relative formula mass is used for ionic compounds (no discrete molecules), e.g., NaCl, CaCO3
• In IGCSE, both are given the symbol Mr and calculated the same way

Method for calculating Mr:

1. Write down the formula of the compound
2. Count how many atoms of each element are present (look at subscripts; multiply brackets)
3. Multiply each count by the Ar of that element
4. Add all the values together

Example 1: Water, H2O

H: 2 atoms x Ar(H) = 2 x 1 = 2
O: 1 atom  x Ar(O) = 1 x 16 = 16
Mr of H2O = 2 + 16 = 18

Example 2: Carbon dioxide, CO2

C: 1 atom  x Ar(C) = 1 x 12 = 12
O: 2 atoms x Ar(O) = 2 x 16 = 32
Mr of CO2 = 12 + 32 = 44

Example 3: Calcium carbonate, CaCO3

Ca: 1 atom x Ar(Ca) = 1 x 40 = 40
C:  1 atom x Ar(C)  = 1 x 12 = 12
O:  3 atoms x Ar(O) = 3 x 16 = 48
Mr of CaCO3 = 40 + 12 + 48 = 100

Example 4: Sulfuric acid, H2SO4

H: 2 atoms x Ar(H) = 2 x 1  = 2
S: 1 atom  x Ar(S) = 1 x 32 = 32
O: 4 atoms x Ar(O) = 4 x 16 = 64
Mr of H2SO4 = 2 + 32 + 64 = 98

Example 5: Magnesium hydroxide, Mg(OH)2

Mg: 1 atom  x Ar(Mg) = 1 x 24 = 24
O:  2 atoms x Ar(O)  = 2 x 16 = 32   (bracket subscript multiplies everything inside)
H:  2 atoms x Ar(H)  = 2 x 1  = 2
Mr of Mg(OH)2 = 24 + 32 + 2 = 58

When a formula contains brackets with a subscript outside, e.g., Ca(OH)2, Mg(OH)2, Al2(SO4)3 — multiply everything inside the bracket by the subscript outside.

The Mole

Definition: One mole is the amount of substance that contains 6.02 x 10^23 particles (atoms, molecules, ions, or formula units).

This number, 6.02 x 10^23, is called Avogadro’s constant.

Why the mole is useful:

• One mole of any substance has a mass in grams equal to its Mr
• This means: 1 mole of carbon atoms (Ar = 12) has a mass of exactly 12 g
• 1 mole of water molecules (Mr = 18) has a mass of exactly 18 g
• 1 mole of CaCO3 (Mr = 100) has a mass of exactly 100 g

Molar mass = the mass of one mole of a substance, in g/mol. Numerically, it is equal to Mr.

The Moles Equation

The central equation in IGCSE Chemistry stoichiometry:

moles = mass / Mr

or written symbolically:

n = m / Mr

Where:

• n = number of moles (mol)
• m = mass of substance (g)
• Mr = relative molecular/formula mass (g/mol)

Rearranging the equation:

Memory aid — cover up the quantity to find:

      m
    -----
    n | Mr

Cover m → n x Mr; Cover n → m / Mr; Cover Mr → m / n

Example: Calculating Moles Calculate the number of moles in 8.0 g of sulfur (S). [Ar(S) = 32]

n = m / Mr
n = 8.0 / 32
n = 0.25 mol

Example: Calculating Moles (Compound) Calculate the number of moles in 22 g of carbon dioxide (CO2). [Ar: C = 12, O = 16]

Step 1: Calculate Mr of CO2
Mr = 12 + (2 x 16) = 44

Step 2: Calculate moles
n = m / Mr = 22 / 44 = 0.50 mol

Example: Calculating Mass What is the mass of 0.50 moles of sodium hydroxide (NaOH)? [Ar: Na = 23, O = 16, H = 1]

Step 1: Calculate Mr of NaOH
Mr = 23 + 16 + 1 = 40

Step 2: Calculate mass
m = n x Mr = 0.50 x 40 = 20 g

Example: Calculating Mass (Harder) What is the mass of 0.20 moles of aluminium oxide (Al2O3)? [Ar: Al = 27, O = 16]

Step 1: Calculate Mr of Al2O3
Mr = (2 x 27) + (3 x 16) = 54 + 48 = 102

Step 2: Calculate mass
m = n x Mr = 0.20 x 102 = 20.4 g

Example: Calculating Mr from Mass and Moles 4.6 g of an unknown gas contains 0.10 moles. Calculate its Mr.

Mr = m / n = 4.6 / 0.10 = 46
The gas could be NO2 (Mr = 14 + 32 = 46).

Example: Finding Number of Particles How many water molecules are present in 9.0 g of water? [Ar: H = 1, O = 16; Avogadro’s constant = 6.02 x 10^23]

Step 1: Mr of H2O = (2 x 1) + 16 = 18
Step 2: Moles of H2O = 9.0 / 18 = 0.50 mol
Step 3: Number of molecules = 0.50 x 6.02 x 10^23 = 3.01 x 10^23 molecules

Key Points

• Relative atomic mass (Ar) = the larger number on the Periodic Table
• Mr = sum of all Ar values in a formula (Mr has NO UNITS)
• 1 mole = 6.02 x 10^23 particles (Avogadro’s constant)
• Molar mass (g/mol) is numerically equal to Mr
• The central equation: moles = mass / Mr (n = m / Mr)
• Rearranged: mass = moles x Mr; Mr = mass / moles
• One mole of any substance has a mass in grams equal to its Mr

Key Concepts from Past Papers

• Relative atomic mass (Ar): the average mass of the isotopes of an element compared to 1/12 of the mass of a carbon-12 atom
• Relative molecular mass (Mr): the sum of the relative atomic masses of all the atoms in a molecule/formula
• Mole: the amount of substance containing 6.02 x 10^23 particles (Avogadro’s constant)
• Molar mass: the mass of one mole of a substance, in g/mol
• Mr is the sum of the Ar values of all atoms in the formula
• Moles = mass / Mr
• Mass of one mole = Mr in grams

Keywords from Past Papers

molecules, particles, concentration, movement, rate, reaction, lower, spread, koh, move, collide, mix, volume, amount, increases

Related Notes

• Empirical and Molecular Formulae
• Reacting Masses
• Concentration and Gas Volumes

Sources

• OpenStax Chemistry 2e — [Chapter 3: Composition of Substances and Solutions (The Mole)], Rice University (free, CC BY 4.0)
• BBC Bitesize GCSE Chemistry — [Moles and Masses], BBC (free educational resource)
• Cambridge IGCSE Chemistry 0620 — Syllabus Section 3: Stoichiometry (The Mole Concept), Cambridge Assessment International Education
• CK-12 Chemistry for High School — [Chapter 10: The Mole], CK-12 Foundation (free, CC BY-NC 3.0)

Past Paper Sources

• 0620/31 May/June 2015: Q22(b)(iv) (0m), Q44(a)(iii) (0m), Q55(c)(i) (0m) (+3 more)
• 0620/32 Feb/March 2017: Q33(b)(v) (2m)
• 0620/32 May/June 2018: Q11(b)(i) (1m)
• 0620/32 May/June 2020: Q66(b)(i) (1m)
• 0620/33 May/June 2016: Q66(d)(i) (1m)
• 0620/33 May/June 2017: Q66(b)(iv) (1m)
• 0620/33 May/June 2021: Q22(d)(ii) (3m)
• 0620/33 May/June 2022: Q33(b)(ii) (1m)
• 0620/33 May/June 2023: Q88(c)(v) (0m)
• 0620/33 May/June 2024: Q66(c)(ii) (1m)
• 0620/33 October/November 2015: Q77(b)(i) (2m), Q77(b)(ii) (0m)
• 0620/33 October/November 2018: Q22(a)(i) (2m)

Common Misconceptions