Relative Atomic Mass and Relative Formula Mass

This lesson covers relative atomic mass ( $A_r$ ) and relative formula mass ( $M_r$ ) as required by the AQA GCSE Combined Science Trilogy specification (8464). You need to understand what these quantities mean, how to read $A_r$ values from the periodic table, and how to calculate the $M_r$ of any compound from its formula. Mastering these calculations is essential — nearly every other topic in quantitative chemistry builds directly on them.

What Is Relative Atomic Mass ( $A_r$ )?

The relative atomic mass ( $A_r$ ) of an element is the average mass of one atom of that element compared to one-twelfth the mass of a carbon-12 atom.

Key facts:

It is a weighted average that accounts for the masses and abundances of all naturally occurring isotopes of the element.
$A_r$ has no units — it is a dimensionless ratio.
You read $A_r$ values directly from the periodic table (the larger number shown for each element).

Reference Table — Common $A_r$ Values

Element	Symbol	$A_r$
Hydrogen	H	1
Carbon	C	12
Nitrogen	N	14
Oxygen	O	16
Sodium	Na	23
Magnesium	Mg	24
Aluminium	Al	27
Silicon	Si	28
Phosphorus	P	31
Sulfur	S	32
Chlorine	Cl	35.5
Potassium	K	39
Calcium	Ca	40
Iron	Fe	56
Copper	Cu	63.5
Zinc	Zn	65
Bromine	Br	80
Silver	Ag	108

Exam Tip: You do NOT need to memorise $A_r$ values — they are given on the periodic table in the AQA exam. But knowing the common ones by heart saves valuable time in calculation-heavy questions.

Why Is $A_r$ a Weighted Average?

Most elements exist as a mixture of isotopes — atoms with the same number of protons but different numbers of neutrons. The $A_r$ reflects this mixture rather than any single isotope.

The formula is:

$A_r = \frac{(\text{mass}_1 \times \text{abundance}_1) + (\text{mass}_2 \times \text{abundance}_2) + \ldots}{\text{total abundance}}$

Worked Example 1 — Chlorine

Chlorine has two stable isotopes:

Isotope	Mass Number	Natural Abundance
$^{35}$ Cl	35	75%
$^{37}$ Cl	37	25%

$A_r = \frac{(35 \times 75) + (37 \times 25)}{100} = \frac{2625 + 925}{100} = \frac{3550}{100} = 35.5$

This is why the periodic table shows 35.5 for chlorine rather than a whole number.

Worked Example 2 — Copper

Isotope	Mass Number	Natural Abundance
$^{63}$ Cu	63	69%
$^{65}$ Cu	65	31%

$A_r = \frac{(63 \times 69) + (65 \times 31)}{100} = \frac{4347 + 2015}{100} = \frac{6362}{100} = 63.6$

The periodic table rounds this to 63.5.

Worked Example 3 — Boron

Boron has two isotopes: $^{10}$ B (20%) and $^{11}$ B (80%).

$A_r = \frac{(10 \times 20) + (11 \times 80)}{100} = \frac{200 + 880}{100} = 10.8$

Worked Example 4 — Working Backwards

An element has an $A_r$ of 24.3. It has two isotopes: mass 24 and mass 25. Find the percentage abundance of each isotope.

Let the abundance of mass 24 = $x\%$ . Then the abundance of mass 25 = $(100 - x)\%$ .

$24.3 = \frac{24x + 25(100 - x)}{100}$

$2430 = 24x + 2500 - 25x$

$2430 = 2500 - x$

$x = 70$

So the element is 70% mass 24 and 30% mass 25. This matches magnesium.

Common Exam Mistake: Students sometimes just average the two mass numbers (e.g. (35 + 37) / 2 = 36 for chlorine). This only works if the isotopes are equally abundant. You must always use the weighted-average formula.

What Is Relative Formula Mass ( $M_r$ )?

The relative formula mass ( $M_r$ ) of a substance is the sum of the relative atomic masses of all the atoms shown in its formula.

For covalent compounds it is sometimes called the relative molecular mass — but the calculation is identical.
Like $A_r$ , it has no units.

How to Calculate $M_r$ — Step by Step

flowchart TD
    A["Write out the chemical formula"] --> B["Count atoms of each element\n(watch brackets and subscripts)"]
    B --> C["Multiply each count by the element's Ar"]
    C --> D["Add all contributions together"]
    D --> E["Result = Mr"]
    style A fill:#3b82f6,color:#fff,stroke:#2563eb
    style E fill:#f59e0b,color:#000,stroke:#d97706

Worked Examples — Calculating $M_r$

Example 5 — Water ( $\text{H}_2\text{O}$ )

Element	Atoms	$A_r$	Contribution
H	2	1	2
O	1	16	16

$M_r = 2 + 16 = 18$

Example 6 — Carbon Dioxide ( $\text{CO}_2$ )

Element	Atoms	$A_r$	Contribution
C	1	12	12
O	2	16	32

$M_r = 12 + 32 = 44$

Example 7 — Calcium Carbonate ( $\text{CaCO}_3$ )

Element	Atoms	$A_r$	Contribution
Ca	1	40	40
C	1	12	12
O	3	16	48

$M_r = 40 + 12 + 48 = 100$

Example 8 — Magnesium Hydroxide ( $\text{Mg(OH)}_2$ )

The subscript 2 outside the bracket means there are two OH groups — so 2 oxygen atoms and 2 hydrogen atoms from the bracket.

Element	Atoms	$A_r$	Contribution
Mg	1	24	24
O	2	16	32
H	2	1	2

$M_r = 24 + 32 + 2 = 58$

Example 9 — Aluminium Sulfate ( $\text{Al}_2(\text{SO}_4)_3$ )

The subscript 3 outside the bracket multiplies everything inside by three: 3 sulfur atoms and 12 oxygen atoms.

Element	Atoms	$A_r$	Contribution
Al	2	27	54
S	3	32	96
O	12	16	192

$M_r = 54 + 96 + 192 = 342$

Example 10 — Ammonium Nitrate ( $\text{NH}_4\text{NO}_3$ )

Element	Atoms	$A_r$	Contribution
N	2	14	28
H	4	1	4
O	3	16	48

$M_r = 28 + 4 + 48 = 80$

Example 11 — Hydrated Copper Sulfate ( $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$ )

This is a higher-demand example. The "dot 5H₂O" means 5 water molecules are included in the formula.

Element	Atoms	$A_r$	Contribution
Cu	1	63.5	63.5
S	1	32	32
O	4 + 5 = 9	16	144
H	10	1	10

$M_r = 63.5 + 32 + 144 + 10 = 249.5$

Percentage Composition by Mass

You may be asked to find the percentage by mass of an element in a compound:

$\text{Percentage by mass} = \frac{A_r \times \text{number of atoms of that element}}{M_r} \times 100$

Worked Example 12 — % of Oxygen in $\text{CaCO}_3$

$M_r$ of CaCO $_3$ = 100.

$\text{\% O} = \frac{16 \times 3}{100} \times 100 = 48\%$

Worked Example 13 — % of Nitrogen in $\text{NH}_4\text{NO}_3$

$M_r$ of NH $_4$ NO $_3$ = 80.

$\text{\% N} = \frac{14 \times 2}{80} \times 100 = 35\%$

Worked Example 14 — Comparing Fertilisers

Which fertiliser has a higher percentage of nitrogen: ammonium nitrate (NH₄NO₃) or urea (CO(NH₂)₂)?

NH₄NO₃: $M_r$ = 80, N contribution = 28, so %N = (28/80) × 100 = 35.0%

CO(NH₂)₂: $M_r$ = 12 + 16 + 2(14 + 2) = 12 + 16 + 32 = 60, N contribution = 28, so %N = (28/60) × 100 = 46.7%

Urea has a higher percentage of nitrogen.

Empirical Formula from Percentage Composition

You should know how to work backwards from percentage composition to find the simplest whole-number ratio of atoms — the empirical formula.

Worked Example 15

A compound contains 40% calcium, 12% carbon, and 48% oxygen by mass. Find its empirical formula.

Element	%	÷ $A_r$	Ratio	Simplest ratio
Ca	40	40 ÷ 40 = 1	1	1
C	12	12 ÷ 12 = 1	1	1
O	48	48 ÷ 16 = 3	3	3

Empirical formula = CaCO $_3$ (calcium carbonate).

Common Exam Mistakes to Avoid

Mistake	Why It Happens	How to Fix It
Forgetting brackets when counting atoms	Students count Mg(OH) $_2$ as 1 O and 1 H	Always multiply every atom inside the bracket by the subscript outside
Using mass number instead of $A_r$	Confusing isotope mass with periodic table value	Always use the number from the periodic table
Averaging isotopes without weighting	Treating all isotopes as equally abundant	Use the weighted average formula every time
Giving $M_r$ with units (g)	Confusing $M_r$ with molar mass	$M_r$ is dimensionless — molar mass has units g/mol but is numerically the same
Rounding too early	Rounding $A_r$ values before finishing	Keep at least one decimal place throughout, round only at the end

Summary

flowchart LR
    subgraph Ar["Relative Atomic Mass"]
        A1["Weighted average of isotopes"]
        A2["No units"]
        A3["Read from periodic table"]
    end
    subgraph Mr["Relative Formula Mass"]
        M1["Sum of all Ar values in formula"]
        M2["No units"]
        M3["Count atoms carefully — watch brackets"]
    end
    Ar --> Mr
    style Ar fill:#e0f2fe,stroke:#0284c7
    style Mr fill:#fef3c7,stroke:#d97706

$A_r$ = weighted average mass of an atom compared to 1/12 of carbon-12.
$M_r$ = sum of all $A_r$ values in a formula.
Always count atoms carefully, especially when brackets or complex formulae are involved.
Percentage composition uses $A_r$ and $M_r$ together.
These calculations underpin everything in quantitative chemistry — the mole, reacting masses, concentration, and yield.

What Is Relative Atomic Mass (ArA_rAr​)?

Reference Table — Common ArA_rAr​ Values

Why Is ArA_rAr​ a Weighted Average?

Worked Example 1 — Chlorine

Worked Example 2 — Copper

Worked Example 3 — Boron

Worked Example 4 — Working Backwards

What Is Relative Formula Mass (MrM_rMr​)?

How to Calculate MrM_rMr​ — Step by Step

Worked Examples — Calculating MrM_rMr​

Example 5 — Water (H2O\text{H}_2\text{O}H2​O)

Example 6 — Carbon Dioxide (CO2\text{CO}_2CO2​)

Example 7 — Calcium Carbonate (CaCO3\text{CaCO}_3CaCO3​)

Example 8 — Magnesium Hydroxide (Mg(OH)2\text{Mg(OH)}_2Mg(OH)2​)

Example 9 — Aluminium Sulfate (Al2(SO4)3\text{Al}_2(\text{SO}_4)_3Al2​(SO4​)3​)

Example 10 — Ammonium Nitrate (NH4NO3\text{NH}_4\text{NO}_3NH4​NO3​)

Example 11 — Hydrated Copper Sulfate (CuSO4⋅5H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O}CuSO4​⋅5H2​O)