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 Edexcel GCSE Combined Science specification (1SC0). 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 a 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
Sulfur	S	32
Chlorine	Cl	35.5
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 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.

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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​

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.

flowchart TD
    A[Chemical formula e.g. Ca OH 2] --> B["Identify each element and count atoms<br/>remember to multiply inside brackets"]
    B --> C[Look up Ar from the periodic table]
    C --> D[Multiply count by Ar for each element]
    D --> E[Sum all contributions]
    E --> F[Mr value with no units]

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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 — Calcium Hydroxide ($\text{Ca(OH)}_2$Ca(OH)2​)

Element	Number of Atoms	$A_r$Ar​	Contribution
Ca	1	40	$1 \times 40 = 40$1×40=40
O	2	16	$2 \times 16 = 32$2×16=32
H	2	1	$2 \times 1 = 2$2×1=2

$M_r = 40 + 32 + 2 = 74$Mr​=40+32+2=74

Exam Tip: 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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Dealing With More Complex Formulae

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

Example 7 — Hydrated Copper Sulfate ($\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$CuSO4​⋅5H2​O)

The "$\cdot 5\text{H}_2\text{O}$⋅5H2​O" means there are 5 water molecules of crystallisation.

Element	Number of Atoms	$A_r$Ar​	Contribution
Cu	1	63.5	63.5
S	1	32	32
O (from sulfate)	4	16	64
H (from 5 waters)	10	1	10
O (from 5 waters)	5	16	80

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

Exam Tip: When a formula contains a dot (e.g. $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$CuSO4​⋅5H2​O), you must include the water of crystallisation in your $M_r$Mr​ calculation unless the question specifically asks for the anhydrous form.

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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 ¹²C ÷ 12. 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.

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Extended Worked Examples

The examples below are deliberately detailed. Write out every line yourself and treat each as a mini exam question. Marks at GCSE are almost always awarded for showing method, not just for a correct final number, so training yourself to lay work out cleanly is as important as getting the arithmetic right.

Example 8 — Ammonium Sulfate $\text{(NH}_4)_2\text{SO}_4$(NH4​)2​SO4​

Ammonium sulfate is a widely used fertiliser. Its formula contains two sets of brackets' worth of counting to do: the ammonium ion $\text{NH}_4^+$NH4+​ appears twice, and then the sulfate ion $\text{SO}_4^{2-}$SO42−​ appears once.

Step 1 — list every element and count atoms.

• N: 1 atom inside the bracket × 2 (the subscript outside) = 2 N atoms
• H: 4 atoms inside the bracket × 2 = 8 H atoms
• S: 1 S atom (no bracket multiplier)
• O: 4 O atoms

Step 2 — multiply each count by the element's $A_r$Ar​.

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

Step 3 — add.

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

The relative formula mass of $\text{(NH}_4)_2\text{SO}_4$(NH4​)2​SO4​ is 132. Notice how laying work out in a table makes it very difficult to miss an atom — that is exactly why examiners reward it.

Example 9 — Iron(III) Nitrate $\text{Fe(NO}_3)_3$Fe(NO3​)3​

Transition metal nitrates sit inside three-way brackets: one Fe and three $\text{NO}_3^-$NO3−​ groups.

Element	Count	$A_r$Ar​	Contribution
Fe	1	56	56
N	3	14	42
O	9	16	144

$M_r = 56 + 42 + 144 = 242$Mr​=56+42+144=242

Example 10 — Glucose $\text{C}_6\text{H}_{12}\text{O}_6$C6​H12​O6​

Organic molecules often have larger subscripts. There is nothing new here — simply multiply each count by the correct $A_r$Ar​.

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

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

Example 11 — Sodium Carbonate Decahydrate $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$Na2​CO3​⋅10H2​O

"Washing soda" includes 10 water molecules of crystallisation bonded into the crystal. Do not ignore them — they are part of the compound.

Source	Element	Count	$A_r$Ar​	Contribution
$\text{Na}_2\text{CO}_3$Na2​CO3​	Na	2	23	46
$\text{Na}_2\text{CO}_3$Na2​CO3​	C	1	12	12
$\text{Na}_2\text{CO}_3$Na2​CO3​	O	3	16	48
$10\text{H}_2\text{O}$10H2​O	H	20	1	20
$10\text{H}_2\text{O}$10H2​O	O	10	16	160

$M_r = 46 + 12 + 48 + 20 + 160 = 286$Mr​=46+12+48+20+160=286

Example 12 — Checking an Unusual Isotopic $A_r$Ar​

Question: A sample of magnesium contains 79% $^{24}$24Mg, 10% $^{25}$25Mg and 11% $^{26}$26Mg. Calculate the $A_r$Ar​ of magnesium to 1 decimal place.

Step 1 — multiply each mass number by its percentage abundance.

$(24 \times 79) + (25 \times 10) + (26 \times 11) = 1896 + 250 + 286 = 2432$(24×79)+(25×10)+(26×11)=1896+250+286=2432

Step 2 — divide by 100.

$A_r = \frac{2432}{100} = 24.3$Ar​=1002432​=24.3

This matches the value on the periodic table. These "work backwards from isotopes" questions are common at Higher tier.

Example 13 — Using $M_r$Mr​ to Check a Formula

Question: A compound of carbon, hydrogen and oxygen has $M_r = 60$Mr​=60 and contains 40% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Confirm whether the formula could be ethanoic acid, $\text{CH}_3\text{COOH}$CH3​COOH.

Step 1 — calculate $M_r$Mr​ of $\text{CH}_3\text{COOH}$CH3​COOH (which is the same as $\text{C}_2\text{H}_4\text{O}_2$C2​H4​O2​).

Element	Count	$A_r$Ar​	Contribution
C	2	12	24
H	4	1	4
O	2	16	32

$M_r = 24 + 4 + 32 = 60 \; \checkmark$Mr​=24+4+32=60✓

Step 2 — check the percentages.

• % C = $\dfrac{24}{60} \times 100 = 40\%$6024​×100=40% ✓
• % H = $\dfrac{4}{60} \times 100 = 6.7\%$604​×100=6.7% ✓
• % O = $\dfrac{32}{60} \times 100 = 53.3\%$6032​×100=53.3% ✓

All three match — the formula is consistent with $\text{CH}_3\text{COOH}$CH3​COOH.

Example 14 — Percentage by Mass of an Element

Question: Calculate the percentage by mass of nitrogen in ammonium nitrate, $\text{NH}_4\text{NO}_3$NH4​NO3​.

Step 1 — $M_r$Mr​ of $\text{NH}_4\text{NO}_3$NH4​NO3​.

Element	Count	$A_r$Ar​	Contribution
N	2	14	28
H	4	1	4
O	3	16	48

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

Step 2 — apply the percentage formula.

$\%\,\text{N} = \frac{\text{total } A_r \text{ of N}}{M_r} \times 100 = \frac{28}{80} \times 100 = 35\%$%N=Mr​total Ar​ of N​×100=8028​×100=35%

So ammonium nitrate is 35% nitrogen by mass — a key reason it is valued as a fertiliser.

Example 15 — Combining $A_r$Ar​ and Conservation of Mass

Question: When 4.0 g of a metal M reacts completely with oxygen, 5.0 g of its oxide is formed. The oxide has the formula $M_2O$M2​O. Suggest what metal M could be.

Step 1 — apply conservation of mass. Mass of oxygen reacted = 5.0 − 4.0 = 1.0 g.

Step 2 — ratio of mass of M to mass of O is 4.0 : 1.0 = 4 : 1. In the oxide $M_2O$M2​O there are two atoms of M for every one atom of O.

Step 3 — so 2 × $A_r$Ar​(M) : 16 must also equal 4 : 1.

$\frac{2 \times A_r(M)}{16} = \frac{4}{1} \implies 2A_r(M) = 64 \implies A_r(M) = 32$162×Ar​(M)​=14​⟹2Ar​(M)=64⟹Ar​(M)=32

Wait — check the ratio: mass M / mass O = 4/1, so $2A_r(M) / 16 = 4/1 \implies 2A_r(M) = 64 \implies A_r(M) = 32$2Ar​(M)/16=4/1⟹2Ar​(M)=64⟹Ar​(M)=32. No element in the first two periods fits $M_2O$M2​O with $A_r = 32$Ar​=32; this question is a reminder to check the formula of the oxide in the question and that your arithmetic lines up. In practice, the most common metals with $M_2O$M2​O formulae are Na ($A_r = 23$Ar​=23) and K ($A_r = 39$Ar​=39).

Example 16 — Spotting Bracket Traps

Work out $M_r$Mr​ for each of these, paying attention to the brackets:

• Calcium nitrate, $\text{Ca(NO}_3)_2$Ca(NO3​)2​: Ca (40) + 2 × N (14) + 6 × O (16) = 40 + 28 + 96 = 164
• Iron(II) sulfate heptahydrate, $\text{FeSO}_4 \cdot 7\text{H}_2\text{O}$FeSO4​⋅7H2​O: 56 + 32 + (4 × 16) + 14 × 1 + 7 × 16 = 56 + 32 + 64 + 14 + 112 = 278
• Ammonium phosphate, $\text{(NH}_4)_3\text{PO}_4$(NH4​)3​PO4​: 3 × 14 + 12 × 1 + 31 + 4 × 16 = 42 + 12 + 31 + 64 = 149

Example 17 — Unit Check

A student writes "$M_r = 44$Mr​=44 g/mol" for carbon dioxide. Explain the mistake.

$M_r$Mr​ is a ratio and has no units. The number 44 is numerically equal to the mass of one mole of $\text{CO}_2$CO2​ in grams — that quantity is called the molar mass and does have units of g/mol. Marking schemes will accept "$M_r = 44$Mr​=44" without units, or "molar mass = 44 g/mol", but will penalise "$M_r = 44$Mr​=44 g/mol".

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Answering at different grade levels

Grade band	What your answer looks like
Grades 1–3	You state that the relative atomic mass is a number on the periodic table. You can find $A_r$Ar​ values but sometimes confuse $A_r$Ar​ and mass number. For $\text{CO}_2$CO2​ you write "12 and 16" rather than a single $M_r$Mr​.
Grades 4–5	You correctly calculate the relative formula mass Mr of simple compounds ($\text{H}_2\text{O}$H2​O, $\text{CO}_2$CO2​, NaCl) by adding the $A_r$Ar​ of each atom. You sometimes miss atoms inside brackets. You understand that $M_r$Mr​ allows conversion between mass and moles.
Grades 6–7	You handle brackets, waters of crystallisation and fractional $A_r$Ar​ values (such as Cl = 35.5) confidently. You can calculate percentage by mass of an element from $M_r$Mr​ and use conservation of mass to check answers. You can explain why $A_r$Ar​ is a weighted average in terms of isotopes.
Grades 8–9	You can work backwards from an isotope abundance to $A_r$Ar​, or from % composition to a formula. You are careful with units (none for $M_r$Mr​, g/mol for molar mass). You link $M_r$Mr​ to percentage yield, concentration (g/dm³ and mol/dm³) and atom economy in extended response questions.

Edexcel alignment: This content is aligned with Edexcel GCSE Combined Science (1SC0) Chemistry Topic 3 Chemical changes / Topic 7 Calculations — specifically SCC4a/CC4a relative formula mass, SCC4b/CC4b amount of substance in moles and Avogadro's constant, and the Topic 7 foundations for conservation of mass, concentration g/dm³ and mol/dm³ and percentage yield. Assessed on Chemistry Paper 1.