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Relative atomic, molecular and formula mass

Ar and Mr (formula mass): understand the carbon-12 comparison and calculate relative masses accurately using subscripts and brackets.

  • SEC G3 Pure Chemistry 2027

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Relative mass questions test careful reading of a formula. Use the supplied periodic-table values, count every atom and apply every subscript or bracket multiplier.

Relative mass is a comparison

Relative atomic mass (Aᵣ) is the weighted average mass of one atom of an element compared with 1/12 of the mass of one carbon-12 atom.

Relative molecular mass (Mᵣ) is the mass of one molecule compared with 1/12 of the mass of one carbon-12 atom.

For ionic compounds and other substances without separate molecules, such as silicon dioxide, use relative formula mass. Add the relative atomic masses in the formula ratio, rather than treating the solid as molecules.

All relative masses are ratios, so they have no units.

Read the formula before calculating

  • Aᵣ values are found in the periodic table and can be non-whole numbers because they represent an average for an element (see Elements & Isotopes).
  • Mᵣ (or relative formula mass) is found by summing Aᵣ × subscript for every element in the formula.
  • Use the Aᵣ values given in the question if they are provided (do not “correct” them).

From atomic masses to formula mass

Quick Recall (Aᵣ vs Mᵣ)
  • Aᵣ compares the average mass of the element’s atoms; Mᵣ compares molecular mass, and relative formula mass uses the formula ratio.
  • Relative masses have no units (they are ratios).
  • Multiply by subscripts and brackets, e.g. (NH₄)₂SO₄ contains 2 N and 8 H.

The carbon-12 reference

An individual atom is far too small to weigh on a school laboratory balance. Chemists express atomic masses on a convenient relative scale:

  • Carbon-12 is assigned a mass of exactly 12 (on this scale).
  • 1/12 of the mass of one carbon-12 atom is the reference mass. A relative mass is a mass divided by this reference mass, so the units cancel.

Why an element’s relative atomic mass can be a decimal

An element can contain more than one isotope, so its Aᵣ represents an abundance-weighted average on the carbon-12 scale. This explains why chlorine has Aᵣ = 35.5 even though no chlorine atom has a nucleon number of 35.5.

Illustrative chlorine isotope mixtureAn illustrative mixture contains 75% chlorine-35 and 25% chlorine-37. The more abundant lighter isotope makes the average closer to 35 than to 37.Illustrative chlorine isotope mixtureIsotopeAbundance (%)
These are rounded illustrative abundances. Approximating the isotope masses as 35 and 37 gives an average of 35.5, closer to the more abundant isotope's mass.
Data table
IsotopeChlorine
Cl-3575
Cl-3725
Relative masses have no units

Aᵣ and Mᵣ are comparison ratios. Do not attach g or g mol⁻¹ to them; those units belong to mass and molar mass, not relative mass.

Add the contributions from every atom

To find Mᵣ (or relative formula mass), add up the relative masses of every atom in the formula:

Mᵣ = ∑ (Aᵣ × subscript)

Check atom counts and units

  • Adding subscripts instead of multiplying (e.g., treating C₁₂H₂₂O₁₁ as 12 + 22 + 11).
  • Forgetting brackets (e.g., not multiplying correctly in (NH₄)₂SO₄).
  • Saying every chlorine atom has mass number 35.5; Aᵣ is an average for the element.
  • Mixing an element’s average relative atomic mass with a substance’s relative molecular or formula mass.

Set out the calculation clearly

Show each element’s contribution

Write the formula, write each element’s contribution as “subscript × Aᵣ”, then sum.

Use the question’s data

If the paper gives Aᵣ(Cu) = 63.5 (for example), use that value, not your memory.

Worked examples

Modelled example 1

Interpret a relative atomic mass

Core

Problem

The Periodic Table gives Aᵣ(Cl) = 35.5. A student says this means every chlorine atom has a nucleon number of 35.5. Explain the mistake and state what the value means.

Study the worked solution

  1. Identify what can be counted

    Method

    Reject a fractional nucleon count for one atom.

    Reason

    A nucleon number counts protons and neutrons, so it is a whole number for an individual atom.

    Working

    No single chlorine atom has 35.5 nucleons.

  2. Interpret the average

    Method

    Describe Aᵣ as the average for the element on the carbon-12 scale.

    Reason

    Natural chlorine contains isotopes in unequal abundances.

    Working

    35.5 is the abundance-weighted average relative mass of chlorine atoms.

  3. State the comparison

    Method

    Connect the value to the carbon-12 reference.

    Reason

    This completes the meaning of relative atomic mass.

    Working

    The average chlorine atom is 35.5 times as massive as 1/12 of one carbon-12 atom.

Guided practice 2

Relative formula mass with brackets

About 6 min

Problem

Calculate the relative formula mass of Al₂(SO₄)₃. Use Aᵣ(Al) = 27, Aᵣ(S) = 32 and Aᵣ(O) = 16.

Apply the bracket to every atom inside it

Hints

Hint 1: expand the bracket

There are three sulfur atoms and twelve oxygen atoms.

Hint 2: write every contribution

Use 2(27) + 3[32 + 4(16)].

View solution step by step

  1. Expand the formula

    Method

    Apply the outside 3 to the whole sulfate group.

    Reason

    Al₂(SO₄)₃ contains 2 Al, 3 S and 12 O atoms.

    Working

    2(27) + 3(32) + 12(16).

  2. Add the contributions

    Method

    Calculate and sum all three element contributions.

    Reason

    Relative formula mass includes every atom in the formula unit.

    Working

    Relative formula mass = 54 + 96 + 192 = 342.

Common misconception 3

Relative molecular mass of sucrose

Find and correct the mistake

Learner response

For C₁₂H₂₂O₁₁, a student writes Mᵣ = 12 + 22 + 11 = 45. Given Aᵣ(C) = 12, Aᵣ(H) = 1, Aᵣ(O) = 16, locate the error and calculate the correct value.

Translate subscripts into atom contributions

Student's error

View solution step by step

  1. Locate the error

    Method

    Reject adding the atom counts alone.

    Reason

    A subscript must multiply the relative atomic mass of that atom type.

    Working

    Correct setup: 12(12) + 22(1) + 11(16).

  2. Sum mass contributions

    Method

    Add carbon, hydrogen and oxygen contributions.

    Reason

    Every atom in one molecule contributes to Mᵣ.

    Working

    144 + 22 + 176 = 342.

  3. State the result

    Working

    Mᵣ(C₁₂H₂₂O₁₁) = 342 with no unit.

Examiner practice 4

Relative formula mass of calcium carbonate

3 marks

Examination question

Calculate the relative formula mass of CaCO₃ using Aᵣ(Ca) = 40, Aᵣ(C) = 12, Aᵣ(O) = 16. [3 marks]

Show each elemental contribution

View solution step by step

  1. Read atom counts

    1 mark

    Method

    Identify one Ca, one C and three O atoms.

    Reason

    Only oxygen has an explicit subscript.

    Working

    Ca:1, C:1, O:3.

  2. Write contributions

    1 mark

    Method

    Multiply each count by Aᵣ.

    Reason

    The formula mass includes all atoms in one formula unit.

    Working

    40 + 12 + 3(16).

  3. Sum and report

    1 mark

    Working

    Relative formula mass = 40 + 12 + 48 = 100.

Challenge 5

Deduce an element from formula mass

Minimal support

Reverse calculation

A metal carbonate has formula XCO₃ and relative formula mass 100. Use Aᵣ(C) = 12 and Aᵣ(O) = 16 to find Aᵣ(X), then identify X from these supplied values: calcium 40, magnesium 24 and sodium 23.

Remove the known contributions

Element X

Hints

Hint 1: calculate the known group

Find the contribution from one C and three O atoms first.

Hint 2: use the whole formula mass

Subtract that known contribution from 100.

View solution step by step

  1. Calculate the carbonate contribution

    Method

    Add one carbon and three oxygen contributions.

    Reason

    These are the known atoms in XCO₃.

    Working

    12 + 3(16) = 60.

  2. Find the unknown contribution

    Method

    Subtract the known mass from the total.

    Reason

    The formula contains one X atom, so its contribution equals Aᵣ(X).

    Working

    Aᵣ(X) = 100-60 = 40.

  3. Identify the element

    Method

    Match 40 to the supplied element values.

    Reason

    Of the supplied metals, calcium has Aᵣ = 40.

    Working

    X is calcium, so the carbonate is CaCO₃.

Try it yourself

Mind stretcher 1: Bracket and unit checkExtension

Question: Calculate the relative formula mass of Ca(NO₃)₂ using Aᵣ(Ca) = 40, Aᵣ(N) = 14 and Aᵣ(O) = 16. Explain why the answer has no unit.

Show answer

The outer 2 multiplies the whole nitrate group, so the formula contains one Ca, two N and six O atoms.

relative formula mass = 40 + 2(14) + 6(16); = 164

The answer has no unit because relative formula mass is a comparison ratio to the carbon-12 standard.

Mind stretcher 2: Does a coefficient change molecular mass?Extension

A learner sees 2H₂O in an equation and writes, “The relative molecular mass of water is 36.” Explain the error and give the correct value, using Aᵣ(H) = 1 and Aᵣ(O) = 16.

Show answer

The coefficient 2 tells you the relative amount of water in the equation. It does not change the atoms in one molecule, so it does not change the relative molecular mass.

Mᵣ(H₂O) = 2(1) + 16 = 18

Two water molecules have twice the mass of one, but the relative molecular mass of the substance remains 18.

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