What is the relative atomic mass and how is it determined?

The relative atomic mass (Ar) is a measure of the mass of an atom compared to one-twelfth of the mass of a carbon-12 atom. It is a dimensionless quantity and is often used in stoichiometry to calculate the masses of atoms in chemical reactions. The relative atomic mass is determined using mass spectrometry, which measures the masses of isotopes and their relative abundances.

How do you calculate the relative molecular mass of a compound?

The relative molecular mass (Mr) of a compound is calculated by summing the relative atomic masses of all the atoms in its molecular formula. For example, to find the Mr of water (H2O), you add the relative atomic masses of two hydrogen atoms (2 x 1) and one oxygen atom (16), resulting in an Mr of 18. This calculation is crucial in stoichiometry for determining the proportions of substances in chemical reactions.

Why is the concept of relative masses important in stoichiometry?

The concept of relative masses is essential in stoichiometry because it allows chemists to quantify the amounts of reactants and products in a chemical reaction. By using relative atomic and molecular masses, chemists can convert between moles and grams, facilitating the calculation of reactant and product quantities needed or produced in a reaction, which is a key aspect of the Cambridge IGCSE Chemistry curriculum.

What is the difference between relative atomic mass and molar mass?

Relative atomic mass is a dimensionless number that represents the mass of an atom compared to carbon-12, while molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). Although both concepts are related, molar mass is used for practical calculations in stoichiometry, allowing conversion between the mass of a substance and the amount in moles.

How do isotopes affect the calculation of relative atomic mass?

Isotopes are atoms of the same element with different numbers of neutrons, resulting in different masses. The relative atomic mass of an element is a weighted average of the masses of its isotopes, based on their natural abundance. This means that elements with multiple isotopes will have a relative atomic mass that reflects the contribution of each isotope to the overall mass, which is important for accurate stoichiometric calculations in chemistry.

Why is "Forgetting to multiply through brackets" a common mistake in Relative masses of atoms and molecules?

Why it happens: Speed. How to avoid it: Ca(NO_3)_2: the subscript 2 multiplies EVERY atom inside the brackets. Source: 0620/42 — recurring

Why is "Stopping at moles without simplifying the ratio" a common mistake in Relative masses of atoms and molecules?

Why it happens: Forgetting the last step. How to avoid it: Always divide each mole value by the SMALLEST to get the simplest whole-number ratio.

Why is "Multiplying by 100 at the wrong step" a common mistake in Relative masses of atoms and molecules?

Why it happens: Misordering operations. How to avoid it: Compute mass of element ÷ M_r, THEN multiply by 100.

Why is "Using mass directly to find ratio in empirical formula" a common mistake in Relative masses of atoms and molecules?

Why it happens: Skipping the moles step. How to avoid it: Always convert mass to MOLES first (n = m/A_r).

StoichiometryCambridge IGCSE Chemistry 0620 Extended

Relative masses of atoms and molecules

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Relative Atomic and Molecular Mass — Cambridge IGCSE 0620 Chemistry Extended (2026)

Relative atomic mass ( $A_r$ ), relative molecular mass ( $M_r$ ), and percentage composition. The mass-bookkeeping skill that everything stoichiometric needs.

What you’ll learn

Mapped to the Cambridge IGCSE 0620 syllabus (2026-2028).

3.2 — Define relative atomic mass and relative molecular mass.
3.2 — Calculate $M_r$ for a given formula.
3.2 — Calculate percentage composition by mass.

$A_r$ and $M_r$

$A_r$ : per atom. $M_r$ : per molecule (or formula unit).

Relative atomic mass ( $A_r$ ). The weighted average mass of all the natural isotopes of an element, on a scale where $^{12}\text{C} = 12$ exactly. No units (it's a ratio).

$A_r$ values are on the Periodic Table. Common values:

H: 1
C: 12
N: 14
O: 16
Na: 23
Mg: 24
S: 32
Cl: 35.5
K: 39
Ca: 40
Fe: 56
Cu: 63.5

Relative molecular mass ( $M_r$ ). Sum of the $A_r$ values for all atoms in a formula unit.

Worked. $M_r$ of $\text{H}_{2}\text{O}$ .

$2 \times 1 + 1 \times 16 = 18$ .

Worked. $M_r$ of $\text{CO}_{2}$ .

$12 + 2 \times 16 = 44$ .

Worked. $M_r$ of $\text{Ca(OH)}_{2}$ .

Ca: 40. OH: $16 + 1 = 17$ . So $40 + 2 \times 17 = 74$ .

Mr is just the Ar of every atom in the formula added together.

Worked. $M_r$ of $(\text{NH}_{4})_{2}\text{SO}_{4}$ .

NH₄: $14 + 4 \times 1 = 18$ . SO₄: $32 + 4 \times 16 = 96$ . So $2 \times 18 + 96 = 132$ .

Tip. Take care with brackets. Subscript outside multiplies everything inside.

$A_r$ : average isotope mass, no units.
$M_r$ : sum of $A_r$ for atoms in formula.
Brackets: subscript multiplies everything inside.
Look up $A_r$ on Periodic Table.

Percentage composition by mass

Find what fraction of a compound's mass is a particular element.

Formula. $\%\text{ of element X} = \dfrac{(\text{number of atoms of X}) \times A_r(\text{X})}{M_r} \times 100\%.$

Worked. Percentage of nitrogen in ammonium nitrate $\text{NH}_{4}\text{NO}_{3}$ .

$M_r$ : $14 + 4 + 14 + 48 = 80$ .
N atoms: 2 (one in NH₄, one in NO₃).
$\% \text{N} = (2 \times 14)/80 \times 100 = 35\%$ .

Worked. Percentage of sulfur in sulfuric acid $\text{H}_{2}\text{SO}_{4}$ .

$M_r$ : $2 + 32 + 64 = 98$ .
$\% \text{S} = 32/98 \times 100 \approx 32.7\%$ .

Percentage composition splits the total Mr into each element's share.

Use. Percentage composition matters for fertilisers (you want high N percent), drug doses, food labels, and quality control.

Worked. A farmer wants high-nitrogen fertiliser. Compare ammonium nitrate ( $\%$ N = 35%) and ammonium phosphate $(\text{NH}_{4})_{3}\text{PO}_{4}$ .

$M_r$ of $(\text{NH}_{4})_{3}\text{PO}_{4}$ : $3 \times 18 + 31 + 64 = 149$ .
$\%$ N = $(3 \times 14)/149 \times 100 \approx 28.2\%$ .
Ammonium nitrate is the better N source by mass.

$\% = (n \times A_r)/M_r \times 100$ .
Useful for fertilisers, drugs, food.
Always sum atom counts FIRST.

How it’s examined

$M_r$ and percentage composition are foundational and appear most years on Paper 2 (2-3 marks: calculate $M_r$ , percentage of element) and Paper 4 (4-6 marks: combined with stoichiometry). Examiner reports flag bracket-handling errors in formulae and missed atoms when counting elements.

Sources: Cambridge IGCSE Chemistry 0620 syllabus 2026-2028 (3.2); 0620/22 May/Jun 2024 — Q7 ($M_r$, percentage by mass); 0620 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

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Step-by-step worked examples — Relative masses of atoms and molecules

Step-by-step solutions to past-paper-style questions on relative masses of atoms and molecules, written exactly the way a tutor would explain them at the board.

1Calculate relative formula mass
Core• Mr
Question
Find $M_{r}$ of $\text{Ca(NO}_{3})_{2}$ . ( $A_{r}$ : Ca = 40, N = 14, O = 16.)
Step-by-step solution
1. Step 1
  Inside the brackets first.
  $M_{r}(\text{NO}_{3}) = 14 + 3 \times 16 = 62$
2. Step 2
  Multiply bracketed group by 2, then add Ca.
  $40 + 2 \times 62 = 164$
Answer
$M_{r} = 164$
2Percentage by mass of an element
Extended• Adapted from 0620/42 May/Jun 2024 Q9• %
Question
Find the percentage by mass of N in ammonium nitrate, $\text{NH}_{4}\text{NO}_{3}$ .
Step-by-step solution
1. Step 1
  $M_{r}$ first.
  $14 + 4 + 14 + 48 = 80$
2. Step 2
  $2 \times 14 = 28\,$ from N atoms.
  $\%\,\text{N} = \dfrac{28}{80} \times 100 = 35\%$
Answer
$35\%$
3Empirical formula from masses
Extended• empirical
Question
A compound contains $1.20\,\text{g}$ C and $0.20\,\text{g}$ H. Find its empirical formula. ( $A_{r}$ : C = 12, H = 1.)
Step-by-step solution
1. Step 1
  Convert masses to moles.
  $n_{C} = \dfrac{1.20}{12} = 0.100;\ n_{H} = \dfrac{0.20}{1} = 0.200$
2. Step 2
  Divide by the smallest.
  $C : H = \dfrac{0.100}{0.100} : \dfrac{0.200}{0.100} = 1 : 2$
Answer
$\text{CH}_{2}$
4Molecular formula from empirical
Extended• molecular formula
Question
A compound has empirical formula $\text{CH}_{2}$ and $M_{r} = 56$ . Find its molecular formula.
Step-by-step solution
1. Step 1
  Empirical $M_{r} = 12 + 2 = 14$ .
2. Step 2
  Ratio.
  $\dfrac{56}{14} = 4 \Rightarrow (\text{CH}_{2})_{4}$
Answer
$\text{C}_{4}\text{H}_{8}$

Key Formulae — Relative masses of atoms and molecules

The formulae you need to memorise for relative masses of atoms and molecules on the Cambridge IGCSE 0620 paper, with every variable defined in plain English and a note on when to use it.

Relative formula mass
$M_{r} = \Sigma (\text{number of atoms} \times A_{r})$
When to use
Sum each element's contribution.
Percentage by mass
$\%\,\text{element} = \dfrac{n \times A_{r}}{M_{r}} \times 100$
$n$
number of atoms of that element in the formula
When to use
Composition by mass questions.

Key Definitions and Keywords — Relative masses of atoms and molecules

Definitions to memorise and the exact keywords mark schemes credit for relative masses of atoms and molecules answers — sharpened from recent examiner reports for the 2026 0620 sitting.

Relative atomic mass ( $A_{r}$ )
Examiner keyword
Average mass of an atom of the element on a scale where ${}^{12}\text{C} = 12$ .
Relative formula (molecular) mass ( $M_{r}$ )
Examiner keyword
Sum of the relative atomic masses of all atoms in the formula.
Empirical formula
Examiner keyword
Simplest whole-number ratio of atoms in a compound.
Molecular formula
Examiner keyword
Actual number of each type of atom in one molecule (a whole-number multiple of the empirical formula).

Common Mistakes and Misconceptions — Relative masses of atoms and molecules

The traps other students keep falling into on relative masses of atoms and molecules questions — taken from recent Cambridge IGCSE 0620 examiner reports and mark schemes — and how to avoid them.

✕Forgetting to multiply through brackets
0620/42 — recurring
Why it happens
Speed.
How to avoid it
$\text{Ca(NO}_{3})_{2}$ : the subscript 2 multiplies EVERY atom inside the brackets.
✕Stopping at moles without simplifying the ratio
Why it happens
Forgetting the last step.
How to avoid it
Always divide each mole value by the SMALLEST to get the simplest whole-number ratio.
✕Multiplying by 100 at the wrong step
Why it happens
Misordering operations.
How to avoid it
Compute mass of element ÷ $M_{r}$ , THEN multiply by 100.
✕Using mass directly to find ratio in empirical formula
Why it happens
Skipping the moles step.
How to avoid it
Always convert mass to MOLES first ( $n = m/A_{r}$ ).

Practice questions

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