Relative Atomic Mass and Relative Formula Mass

This lesson covers relative atomic mass ($A_r$Ar​) and relative formula mass ($M_r$Mr​) as required by the AQA GCSE Combined Science Trilogy specification (8464). You need to understand what these quantities mean, how to read $A_r$Ar​ values from the periodic table, and how to calculate the $M_r$Mr​ of any compound from its formula.

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What Is Relative Atomic Mass ($A_r$Ar​)?

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

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

Key $A_r$Ar​ Values to Know

Element	Symbol	$A_r$Ar​
Hydrogen	H	1
Carbon	C	12
Nitrogen	N	14
Oxygen	O	16
Sodium	Na	23
Magnesium	Mg	24
Aluminium	Al	27
Sulfur	S	32
Chlorine	Cl	35.5
Potassium	K	39
Calcium	Ca	40
Iron	Fe	56
Copper	Cu	63.5

Exam Tip: You do NOT need to memorise $A_r$Ar​ values — they are given on the periodic table in the AQA exam. But knowing the common ones speeds up your work enormously.

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Why Is $A_r$Ar​ a Weighted Average?

Most elements exist as a mixture of isotopes — atoms with the same number of protons but different numbers of neutrons. The $A_r$Ar​ reflects this mixture.

Worked Example — Chlorine

Chlorine has two isotopes:

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

$A_r = \frac{(35 \times 75) + (37 \times 25)}{100} = \frac{2625 + 925}{100} = \frac{3550}{100} = 35.5$Ar​=100(35×75)+(37×25)​=1002625+925​=1003550​=35.5

This is why the periodic table shows 35.5 for chlorine, not a whole number.

Worked Example — Copper

Copper has two isotopes:

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

$A_r = \frac{(63 \times 69) + (65 \times 31)}{100} = \frac{4347 + 2015}{100} = \frac{6362}{100} = 63.6$Ar​=100(63×69)+(65×31)​=1004347+2015​=1006362​=63.6

The periodic table rounds this to 63.5.

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What Is Relative Formula Mass ($M_r$Mr​)?

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

• For molecules (covalent compounds), it is sometimes called the relative molecular mass — but the calculation is identical.
• Like $A_r$Ar​, it has no units.

How to Calculate $M_r$Mr​

flowchart TD
    A["Write out the chemical formula"] --> B["Count the number of atoms\nof each element"]
    B --> C["Multiply each count\nby the element’s Ar"]
    C --> D["Add up all the values"]
    D --> E["Result = Mr"]
    style A fill:#3b82f6,color:#fff,stroke:#2563eb
    style E fill:#f59e0b,color:#000,stroke:#d97706

1. Write out the formula.
2. Count the number of atoms of each element.
3. Multiply each count by the element's $A_r$Ar​.
4. Add up all the values.

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Worked Examples

Example 1 — Water ($\text{H}_2\text{O}$H2​O)

Element	Number of Atoms	$A_r$Ar​	Contribution
H	2	1	$2 \times 1 = 2$2×1=2
O	1	16	$1 \times 16 = 16$1×16=16

$M_r = 2 + 16 = 18$Mr​=2+16=18

Example 2 — Carbon Dioxide ($\text{CO}_2$CO2​)

Element	Number of Atoms	$A_r$Ar​	Contribution
C	1	12	$1 \times 12 = 12$1×12=12
O	2	16	$2 \times 16 = 32$2×16=32

$M_r = 12 + 32 = 44$Mr​=12+32=44

Example 3 — Calcium Carbonate ($\text{CaCO}_3$CaCO3​)

Element	Number of Atoms	$A_r$Ar​	Contribution
Ca	1	40	$1 \times 40 = 40$1×40=40
C	1	12	$1 \times 12 = 12$1×12=12
O	3	16	$3 \times 16 = 48$3×16=48

$M_r = 40 + 12 + 48 = 100$Mr​=40+12+48=100

Example 4 — Magnesium Hydroxide ($\text{Mg(OH)}_2$Mg(OH)2​)

Watch out for brackets! The subscript 2 outside the bracket means there are two OH groups.

Element	Number of Atoms	$A_r$Ar​	Contribution
Mg	1	24	$1 \times 24 = 24$1×24=24
O	2	16	$2 \times 16 = 32$2×16=32
H	2	1	$2 \times 1 = 2$2×1=2

$M_r = 24 + 32 + 2 = 58$Mr​=24+32+2=58

Example 5 — Sodium Carbonate ($\text{Na}_2\text{CO}_3$Na2​CO3​)

Element	Number of Atoms	$A_r$Ar​	Contribution
Na	2	23	$2 \times 23 = 46$2×23=46
C	1	12	$1 \times 12 = 12$1×12=12
O	3	16	$3 \times 16 = 48$3×16=48

$M_r = 46 + 12 + 48 = 106$Mr​=46+12+48=106

Example 6 — Aluminium Sulfate ($\text{Al}_2(\text{SO}_4)_3$Al2​(SO4​)3​)

The subscript 3 outside the bracket means there are three $\text{SO}_4$SO4​ groups.

Element	Number of Atoms	$A_r$Ar​	Contribution
Al	2	27	$2 \times 27 = 54$2×27=54
S	3	32	$3 \times 32 = 96$3×32=96
O	12	16	$12 \times 16 = 192$12×16=192

$M_r = 54 + 96 + 192 = 342$Mr​=54+96+192=342

Exam Tip (AQA 8464): Always lay your working out in a clear table or list. Show every step — you earn method marks even if you make an arithmetic error.

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Common Mistakes to Avoid

Mistake	Correction
Forgetting to multiply inside brackets	In $\text{Ca(OH)}_2$Ca(OH)2​, there are 2 O atoms and 2 H atoms, not 1 of each
Using mass number instead of $A_r$Ar​	Always use the periodic table value — $A_r$Ar​ may not be a whole number (e.g. Cl = 35.5)
Missing atoms in complex formulae	Count every atom carefully, especially with multiple brackets
Giving $M_r$Mr​ a unit	$M_r$Mr​ has no units — it is a dimensionless ratio

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Summary

• Relative atomic mass ($A_r$Ar​) is the average mass of an atom of an element relative to one-twelfth the mass of a carbon-12 atom. It has no units and is found on the periodic table.
• $A_r$Ar​ is a weighted average because elements have different isotopes.
• Relative formula mass ($M_r$Mr​) is the sum of the $A_r$Ar​ values of all atoms in a formula.
• To calculate $M_r$Mr​: count atoms (including inside brackets), multiply by $A_r$Ar​, and add up.
• Always show your working clearly in a table or step-by-step list.
• This topic appears on both Foundation and Higher tier AQA papers.

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Extended Worked Examples — $A_r$Ar​ and $M_r$Mr​ in Action

The rest of this lesson shows more step-by-step working on $A_r$Ar​ and $M_r$Mr​ calculations. Every single calculation in GCSE quantitative chemistry — moles, reacting masses, concentrations, percentage yield — starts with $M_r$Mr​, so you must be confident here.

Worked Example 7 — Ammonium Nitrate ($\text{NH}_4\text{NO}_3$NH4​NO3​)

Ammonium nitrate is a fertiliser and a tricky $M_r$Mr​ example because nitrogen appears twice in the formula.

Element	Number of Atoms	$A_r$Ar​	Contribution
N	2 (one in $\text{NH}_4$NH4​, one in $\text{NO}_3$NO3​)	14	$2 \times 14 = 28$2×14=28
H	4	1	$4 \times 1 = 4$4×1=4
O	3	16	$3 \times 16 = 48$3×16=48

$M_r = 28 + 4 + 48 = 80$Mr​=28+4+48=80

Common slip: Students who rush often write "N = 1" and get $M_r$Mr​ = 66. Always count atoms of each element across the whole formula, including the same element appearing in different parts.

Worked Example 8 — Copper(II) Sulfate Crystals ($\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$CuSO4​⋅5H2​O)

The dot-formula shows hydrated copper sulfate — five water molecules are bound into each formula unit.

Element	Count	$A_r$Ar​	Contribution
Cu	1	63.5	$63.5$63.5
S	1	32	$32$32
O (from $\text{SO}_4$SO4​)	4	16	$64$64
O (from $5\text{H}_2\text{O}$5H2​O)	5	16	$80$80
H (from $5\text{H}_2\text{O}$5H2​O)	10	1	$10$10

$M_r = 63.5 + 32 + 64 + 80 + 10 = 249.5$Mr​=63.5+32+64+80+10=249.5

Notice the anhydrous salt $\text{CuSO}_4$CuSO4​ has $M_r = 159.5$Mr​=159.5. The water of crystallisation contributes $249.5 - 159.5 = 90$249.5−159.5=90, which equals $5 \times 18$5×18 (five $\text{H}_2\text{O}$H2​O) — a neat check on your arithmetic.

Worked Example 9 — Calcium Nitrate ($\text{Ca(NO}_3)_2$Ca(NO3​)2​)

The bracket with subscript 2 means two whole $\text{NO}_3$NO3​ groups. You multiply everything inside the bracket by 2.

Element	Number of Atoms	$A_r$Ar​	Contribution
Ca	1	40	$40$40
N	2	14	$28$28
O	6	16	$96$96

$M_r = 40 + 28 + 96 = 164$Mr​=40+28+96=164

Tip: Read $\text{Ca(NO}_3)_2$Ca(NO3​)2​ out loud as "calcium, two lots of NO-three". You will then automatically double both the N and the O.

Worked Example 10 — Ammonium Sulfate ($(\text{NH}_4)_2\text{SO}_4$(NH4​)2​SO4​)

The subscript 2 applies to the whole $\text{NH}_4$NH4​ group.

Element	Count	$A_r$Ar​	Contribution
N	2	14	$28$28
H	8	1	$8$8
S	1	32	$32$32
O	4	16	$64$64

$M_r = 28 + 8 + 32 + 64 = 132$Mr​=28+8+32+64=132

Worked Example 11 — Glucose ($\text{C}_6\text{H}_{12}\text{O}_6$C6​H12​O6​)

Element	Count	$A_r$Ar​	Contribution
C	6	12	$72$72
H	12	1	$12$12
O	6	16	$96$96

$M_r = 72 + 12 + 96 = 180$Mr​=72+12+96=180

Worked Example 12 — Back-calculation from Mass

You can use $M_r$Mr​ to check if formulae are consistent with conservation of mass. For example, in $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$2Mg+O2​→2MgO:

• Reactant mass per equation: $(2 \times 24) + (2 \times 16) = 48 + 32 = 80$(2×24)+(2×16)=48+32=80.
• Product mass per equation: $2 \times (24 + 16) = 80$2×(24+16)=80.
• Total $M_r$Mr​ is conserved — as it must be for any balanced equation.

Worked Example 13 — Percentage by Mass

Question: What percentage of the mass of ammonium nitrate ($\text{NH}_4\text{NO}_3$NH4​NO3​, $M_r = 80$Mr​=80) is nitrogen?

$\%\text{N} = \frac{\text{mass of N in formula}}{M_r} \times 100 = \frac{2 \times 14}{80} \times 100 = \frac{28}{80} \times 100 = 35\%$%N=Mr​mass of N in formula​×100=802×14​×100=8028​×100=35%

This kind of question appears on fertiliser questions and only needs $A_r$Ar​ and $M_r$Mr​ — no moles required.

Worked Example 14 — Isotope $A_r$Ar​ with Decimals

Question: Bromine has two isotopes: $^{79}$79Br at 50.7% and $^{81}$81Br at 49.3%. Calculate $A_r$Ar​ to one decimal place.

$A_r = \frac{(79 \times 50.7) + (81 \times 49.3)}{100} = \frac{4005.3 + 3993.3}{100} = \frac{7998.6}{100} = 80.0 \text{ (1 d.p.)}$Ar​=100(79×50.7)+(81×49.3)​=1004005.3+3993.3​=1007998.6​=80.0 (1 d.p.)

This matches the periodic table value and shows why $A_r$Ar​ is a weighted average — even when isotopes are nearly equally abundant, the slight bias shifts $A_r$Ar​ away from the midpoint.

Answering at different grade levels

Quantitative questions on $A_r$Ar​ and $M_r$Mr​ appear on both Foundation and Higher papers. What changes is the depth of reasoning, the precision of the language, and the complexity of the formulae.

Grade 3–4 answer (Foundation). A minimally correct answer may say: "The relative formula mass is the atomic masses added up. For $\text{CO}_2$CO2​ it is $12 + 16 + 16 = 44$12+16+16=44." This earns marks for calculation but not for definition.

Grade 5 answer. At grade 5 the student uses the correct term relative formula mass ($M_r$Mr​), states it has no units, and lays out working in a table. They handle compounds with brackets (e.g. $\text{Mg(OH)}_2$Mg(OH)2​) correctly.

Grade 6–7 answer (Higher). A stronger answer explains that $A_r$Ar​ is a weighted mean of isotopic masses relative to one-twelfth of a carbon-12 atom, and uses this to justify why $A_r$Ar​ values like 35.5 (chlorine) are not whole numbers. The student can back-calculate isotope abundances if $A_r$Ar​ is given.

Grade 8–9 answer. The top-grade student links $M_r$Mr​ to conservation of mass, explicitly uses it in mole and concentration (g/dm³ and mol/dm³) calculations, and discusses the percentage yield or percentage by mass consequences. They use precise language: "Relative formula mass is a dimensionless quantity equal to the sum of the $A_r$Ar​ values of all atoms in the formula unit." They also recognise that $M_r$Mr​ is the numerical value of the molar mass in g/mol, which bridges to advanced mole work.

Practise writing definitions in precise full sentences — examiners award marks for accurate terminology, not just numerical answers.

AQA alignment: This content is aligned with AQA GCSE Combined Science: Trilogy (8464) specification section 5.3 Quantitative chemistry — specifically 5.3.1 Chemical measurements, conservation of mass and the quantitative interpretation of chemical equations, and 5.3.2.1 Relative formula mass. Assessed on Chemistry Paper 1.