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 every atom inside a bracket" a common mistake in Relative masses of atoms and molecules?

Why it happens: The subscript outside the bracket is easy to overlook when working quickly. How to avoid it: In Ca(NO_3)_2 the subscript 2 multiplies the WHOLE NO_3 group, giving 2 N and 6 O atoms. Work out the bracket's mass first, then multiply. Source: 0620/42 — recurring error in examiner reports

Why is "Mishandling the dot in hydrated salts (e.g. only adding the hydrogens)" a common mistake in Relative masses of atoms and molecules?

Why it happens: Students treat · 5H_2O as loose atoms instead of 5 whole water molecules. How to avoid it: Add 5 × M_r(H_2O) = 5 × 18 = 90 to the M_r of the anhydrous salt.

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

Why it happens: Operations are carried out in the wrong order. How to avoid it: Always do mass of elementM_r FIRST, and only then multiply by 100.

Why is "Using the mass of a single atom instead of all atoms of that element" a common mistake in Relative masses of atoms and molecules?

Why it happens: Forgetting that an element can appear more than once in a formula. How to avoid it: Count every atom of the element. In NH_4NO_3 there are TWO nitrogens, so use 2 × 14 = 28.

Relative masses of atoms and molecules – Study Notes & Past Paper Style Questions | Cambridge IGCSE Chemistry 0620 Extended

Short Notes - Relative masses of atoms and molecules

Detailed Study Notes

Detailed notes on Stoichiometry for Cambridge IGCSE Chemistry, covering key concepts, explanations, examples, and exam-focused revision points.

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.

At a glance

$A_r$ : weighted average mass of an element's atoms compared to $\frac{1}{12}$ of $^{12}$ C.
$M_r$ : sum of $A_r$ values for atoms in a formula unit.
No units for $A_r$ or $M_r$ — they're relative, not absolute.
Percentage by mass of element X: $\dfrac{n \times A_r(\text{X})}{M_r} \times 100\%$ .
Use $A_r$ values from the Periodic Table (rounded to whole numbers usually).

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 $2 \times 18 + 96 = 132$ .

Compound	Atoms	Working	$M_r$
H₂O	2 H, 1 O	2 × 1 + 16	18
CO₂	1 C, 2 O	12 + 2 × 16	44
Ca(OH)₂	1 Ca, 2 O, 2 H	40 + 2 × 17	74
(NH₄)₂SO₄	2 N, 8 H, 1 S, 4 O	2 × 18 + 96	132

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

The subscript outside the bracket applies to every atom inside it — here both the N and all four H.

$A_r$ : 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 $\text{H}_{2}\text{SO}_{4}$ .

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

Compound	Element	Atoms × $A_r$	$M_r$	% by mass
NH₄NO₃	N	2 × 14 = 28	80	35%
H₂SO₄	S	1 × 32 = 32	98	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.

Quick recap

$A_r$ : per atom (weighted average of isotopes).
$M_r$ : sum of $A_r$ for the formula.
Brackets in formulae: subscript multiplies inside.
$\% = (n \times A_r)/M_r \times 100$ .
Use for fertiliser comparisons, drug dosages, etc.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Relative atomic mass ( $A_r$ ) — weighted average atomic mass relative to $\frac{1}{12}$ of $^{12}$ C.
Relative molecular mass ( $M_r$ ) — sum of $A_r$ values for atoms in a formula.
Percentage composition — mass fraction of each element in a compound, expressed as a percentage.

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.

Take this whole topic with you

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.

1Mr of a simple molecule
Getting started• Mr
Question
Calculate the relative molecular mass, $M_{r}$ , of water, $\text{H}_{2}\text{O}$ . ( $A_{r}$ : H = 1, O = 16.)
Step-by-step solution
1. Step 1
  Multiply each $A_{r}$ by the number of those atoms in the formula. There are 2 H atoms and 1 O atom.
  $M_{r} = (2 \times 1) + (1 \times 16)$
2. Step 2
  Add the contributions together.
  $M_{r} = 2 + 16 = 18$
Answer
$M_{r}(\text{H}_{2}\text{O}) = 18$
Examiner tip
$M_{r}$ has no units — it is a RATIO of masses on the carbon-12 scale.
2Mr of carbon dioxide
Getting started• Mr
Question
Calculate the relative molecular mass of carbon dioxide, $\text{CO}_{2}$ . ( $A_{r}$ : C = 12, O = 16.)
Step-by-step solution
1. Step 1
  One C atom plus two O atoms.
  $M_{r} = (1 \times 12) + (2 \times 16)$
2. Step 2
  Add up.
  $M_{r} = 12 + 32 = 44$
Answer
$M_{r}(\text{CO}_{2}) = 44$
Examiner tip
The subscript 2 belongs only to oxygen here — there is just one carbon atom.
3Mr of sulfuric acid
Building confidence• Mr
Question
Calculate the relative molecular mass of sulfuric acid, $\text{H}_{2}\text{SO}_{4}$ . ( $A_{r}$ : H = 1, S = 32, O = 16.)
Step-by-step solution
1. Step 1
  Identify how many of each atom: 2 H, 1 S, 4 O.
2. Step 2
  Multiply each $A_{r}$ by its count.
  $M_{r} = (2 \times 1) + (1 \times 32) + (4 \times 16)$
3. Step 3
  Add: $2 + 32 + 64$ .
  $M_{r} = 2 + 32 + 64 = 98$
Answer
$M_{r}(\text{H}_{2}\text{SO}_{4}) = 98$
Examiner tip
Lay the calculation out term by term — most lost marks here come from miscounting the four oxygens ( $4 \times 16 = 64$ ).
4Relative formula mass with brackets
Building confidence• Adapted from 0620/42 May/Jun 2023 Q3• Mr, brackets, ionic
Question
Calculate the relative formula mass of calcium nitrate, $\text{Ca(NO}_{3})_{2}$ . ( $A_{r}$ : Ca = 40, N = 14, O = 16.)
Step-by-step solution
1. Step 1
  Work out the mass of ONE nitrate group, $\text{NO}_{3}$ , first.
  $M_{r}(\text{NO}_{3}) = 14 + (3 \times 16) = 62$
2. Step 2
  The subscript 2 outside the bracket multiplies the WHOLE nitrate group, so there are two of them.
  $2 \times 62 = 124$
3. Step 3
  Add the one calcium atom.
  $M_{r} = 40 + 124 = 164$
Answer
$M_{r}(\text{Ca(NO}_{3})_{2}) = 164$
Examiner tip
The bracket subscript multiplies EVERY atom inside it (so $2 \times 3 = 6$ oxygens in total). Forgetting this is the single most common error.
5Mr of a hydrated salt
Stretch• Adapted from 0620/41 Oct/Nov 2022 Q4• Mr, water of crystallisation
Question
Calculate the relative formula mass of hydrated copper(II) sulfate, $\text{CuSO}_{4}\cdot 5\text{H}_{2}\text{O}$ . ( $A_{r}$ : Cu = 64, S = 32, O = 16, H = 1.)
Step-by-step solution
1. Step 1
  Find $M_{r}$ of the anhydrous part, $\text{CuSO}_{4}$ .
  $M_{r}(\text{CuSO}_{4}) = 64 + 32 + (4 \times 16) = 160$
2. Step 2
  Find the mass of the water of crystallisation. The dot means '+ 5 water molecules', and $M_{r}(\text{H}_{2}\text{O}) = 18$ .
  $5 \times 18 = 90$
3. Step 3
  Add the two parts together.
  $M_{r} = 160 + 90 = 250$
Answer
$M_{r}(\text{CuSO}_{4}\cdot 5\text{H}_{2}\text{O}) = 250$
Examiner tip
The dot ( $\cdot$ ) is an ADDITION of complete water molecules — multiply the whole $18$ by 5, not just the hydrogens.
6Percentage by mass of an element
Stretch• Adapted from 0620/42 May/Jun 2024 Q9• %, composition
Question
Calculate the percentage by mass of nitrogen in ammonium nitrate, $\text{NH}_{4}\text{NO}_{3}$ . ( $A_{r}$ : N = 14, H = 1, O = 16.)
Step-by-step solution
1. Step 1
  Calculate $M_{r}$ of the whole compound: 2 N, 4 H, 3 O.
  $M_{r} = (2 \times 14) + (4 \times 1) + (3 \times 16) = 28 + 4 + 48 = 80$
2. Step 2
  Find the total mass of the element of interest. There are 2 nitrogen atoms.
  $\text{mass of N} = 2 \times 14 = 28$
3. Step 3
  Divide the element mass by $M_{r}$ , THEN multiply by 100.
  $\%\,\text{N} = \dfrac{28}{80} \times 100 = 35\%$
Answer
$\%\,\text{N} = 35\%$
Examiner tip
Use the TOTAL mass of that element (here $2 \times 14$ , not just one N). Do the division first, then $\times 100$ .

Model Answers — Relative masses of atoms and molecules

High-scoring sample answers for relative masses of atoms and molecules on the Cambridge IGCSE 0620 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 4 structured style2 marks
Define the term relative atomic mass, $A_{r}$ . (2 marks)
Model answer
The relative atomic mass, $A_{r}$ , is the average mass of the atoms of an element, compared with $\tfrac{1}{12}$ of the mass of an atom of carbon-12 ( $^{12}\text{C}$ ). It is a weighted mean that takes the proportions of the different isotopes into account, so it has no units.
Why this scores
Two marks: (1) average mass of the atoms of the element; (2) on a scale where $^{12}\text{C}$ (or $\tfrac{1}{12}$ of a carbon-12 atom) is the reference. The word 'average' is essential because of isotopes.
Question 2
Paper 4 short-answer style1 mark
Calculate the relative molecular mass, $M_{r}$ , of water, $\text{H}_{2}\text{O}$ . ( $A_{r}$ : H = 1, O = 16.) (1 mark)
Model answer
$M_{r} = (2 \times 1) + 16 = 18$ .
Why this scores
One mark for the correct answer, $18$ . Always show the $(2 \times 1)$ for the two hydrogen atoms.
Question 3
Paper 4 structured style2 marks
Show that the relative molecular mass of sulfuric acid, $\text{H}_{2}\text{SO}_{4}$ , is 98. ( $A_{r}$ : H = 1, S = 32, O = 16.) (2 marks)
Model answer
Add the contribution of each element: hydrogen $2 \times 1 = 2$ , sulfur $1 \times 32 = 32$ , oxygen $4 \times 16 = 64$ . Total: $M_{r} = 2 + 32 + 64 = 98$ .
Why this scores
In a 'show that' question you must display the working. One mark for the correct contributions ( $2 + 32 + 64$ ) and one for reaching $98$ .
Question 4
Paper 4 structured style3 marks
Calculate the relative formula mass of calcium hydroxide, $\text{Ca(OH)}_{2}$ , and of calcium nitrate, $\text{Ca(NO}_{3})_{2}$ . ( $A_{r}$ : Ca = 40, O = 16, H = 1, N = 14.) (3 marks)
Model answer
For $\text{Ca(OH)}_{2}$ , one hydroxide group is $16 + 1 = 17$ , and the subscript 2 doubles it: $M_{r} = 40 + (2 \times 17) = 40 + 34 = 74$ . For $\text{Ca(NO}_{3})_{2}$ , one nitrate group is $14 + (3 \times 16) = 62$ , and the subscript 2 doubles it: $M_{r} = 40 + (2 \times 62) = 40 + 124 = 164$ .
Why this scores
Three marks: $\text{Ca(OH)}_{2} = 74$ ; correct nitrate group $= 62$ ; $\text{Ca(NO}_{3})_{2} = 164$ . The bracket subscript must multiply every atom inside it.
Question 5
Paper 4 (Extended) structured style4 marks
Hydrated copper(II) sulfate has the formula $\text{CuSO}_{4}\cdot 5\text{H}_{2}\text{O}$ . Calculate its relative formula mass. ( $A_{r}$ : Cu = 64, S = 32, O = 16, H = 1.) (4 marks)
Model answer
First find the anhydrous part: $M_{r}(\text{CuSO}_{4}) = 64 + 32 + (4 \times 16) = 64 + 32 + 64 = 160$ . Next find the water of crystallisation, where each $\text{H}_{2}\text{O}$ has $M_{r} = 18$ , so five of them give $5 \times 18 = 90$ . The dot in the formula means these are added on, so the total is $M_{r} = 160 + 90 = 250$ .
Why this scores
Four marks: $M_{r}(\text{CuSO}_{4}) = 160$ ; $M_{r}(\text{H}_{2}\text{O}) = 18$ ; $5 \times 18 = 90$ ; final total $250$ . The dot means '+ 5 whole water molecules'.
Question 6
Paper 4 (Extended) structured style5 marks
Ammonium nitrate, $\text{NH}_{4}\text{NO}_{3}$ , is used as a fertiliser because of its high nitrogen content. Calculate the percentage by mass of nitrogen in ammonium nitrate. ( $A_{r}$ : N = 14, H = 1, O = 16.) (5 marks)
Model answer
First calculate the relative formula mass of the whole compound. There are 2 nitrogen, 4 hydrogen and 3 oxygen atoms, so $M_{r} = (2 \times 14) + (4 \times 1) + (3 \times 16) = 28 + 4 + 48 = 80$ . Next find the total mass contributed by nitrogen: there are two nitrogen atoms, giving $2 \times 14 = 28$ . Finally, divide the mass of nitrogen by the total $M_{r}$ and multiply by 100: $\%\,\text{N} = \dfrac{28}{80} \times 100 = 35\%$ .
Why this scores
Five marks across: correct count of atoms; $M_{r} = 80$ ; mass of nitrogen $= 28$ ; correct formula $\tfrac{\text{mass of element}}{M_{r}} \times 100$ ; final answer $35\%$ . Use the TOTAL nitrogen mass ( $2 \times 14$ ), not one atom.

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
The average mass of the atoms of an element, on a scale where an atom of carbon-12 ( $^{12}\text{C}$ ) has a mass of exactly 12 (i.e. compared with $\tfrac{1}{12}$ of a carbon-12 atom). It is a weighted mean over the isotopes.
Relative molecular / formula mass ( $M_{r}$ )
Examiner keyword
The sum of the relative atomic masses of all the atoms shown in the formula of a substance. 'Relative molecular mass' is used for molecules; 'relative formula mass' is used for ionic compounds.
Percentage composition by mass
Examiner keyword
The percentage of the total mass of a compound contributed by a given element: $\%\,\text{element} = \dfrac{(\text{number of atoms}) \times A_{r}}{M_{r}} \times 100$ .
Water of crystallisation
Examiner keyword
Water molecules chemically bound into a crystal structure, shown after a dot in the formula (e.g. $\text{CuSO}_{4}\cdot 5\text{H}_{2}\text{O}$ ). Each $\text{H}_{2}\text{O}$ ( $M_{r} = 18$ ) is added to the $M_{r}$ of the salt.

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 every atom inside a bracket
0620/42 — recurring error in examiner reports
Why it happens
The subscript outside the bracket is easy to overlook when working quickly.
How to avoid it
In $\text{Ca(NO}_{3})_{2}$ the subscript 2 multiplies the WHOLE $\text{NO}_{3}$ group, giving 2 N and 6 O atoms. Work out the bracket's mass first, then multiply.
✕Mishandling the dot in hydrated salts (e.g. only adding the hydrogens)
Why it happens
Students treat $\cdot 5\text{H}_{2}\text{O}$ as loose atoms instead of 5 whole water molecules.
How to avoid it
Add $5 \times M_{r}(\text{H}_{2}\text{O}) = 5 \times 18 = 90$ to the $M_{r}$ of the anhydrous salt.
✕Multiplying by 100 at the wrong stage of a percentage calculation
Why it happens
Operations are carried out in the wrong order.
How to avoid it
Always do $\dfrac{\text{mass of element}}{M_{r}}$ FIRST, and only then multiply by 100.
✕Using the mass of a single atom instead of all atoms of that element
Why it happens
Forgetting that an element can appear more than once in a formula.
How to avoid it
Count every atom of the element. In $\text{NH}_{4}\text{NO}_{3}$ there are TWO nitrogens, so use $2 \times 14 = 28$ .

Relative masses of atoms and molecules — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Relative masses of atoms and molecules

Short Notes - Relative masses of atoms and molecules

Detailed Study Notes

At a glance

What you’ll learn

ArA_rAr​ and MrM_rMr​

Percentage composition by mass

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Mr of a simple molecule

2Mr of carbon dioxide

3Mr of sulfuric acid

4Relative formula mass with brackets

5Mr of a hydrated salt

6Percentage by mass of an element

Relative atomic mass (ArA_{r}Ar​)

Relative molecular / formula mass (MrM_{r}Mr​)

Percentage composition by mass

Water of crystallisation

✕Forgetting to multiply every atom inside a bracket

✕Mishandling the dot in hydrated salts (e.g. only adding the hydrogens)

✕Multiplying by 100 at the wrong stage of a percentage calculation

✕Using the mass of a single atom instead of all atoms of that element

Relative masses of atoms and molecules — frequently asked questions

Relative masses of atoms and molecules

Short Notes - Relative masses of atoms and molecules

Detailed Study Notes

At a glance

What you’ll learn

ArA_rAr​ and MrM_rMr​

Percentage composition by mass

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Mr of a simple molecule

2Mr of carbon dioxide

3Mr of sulfuric acid

4Relative formula mass with brackets

5Mr of a hydrated salt

6Percentage by mass of an element

Relative atomic mass (ArA_{r}Ar​)

Relative molecular / formula mass (MrM_{r}Mr​)

Percentage composition by mass

Water of crystallisation

✕Forgetting to multiply every atom inside a bracket

✕Mishandling the dot in hydrated salts (e.g. only adding the hydrogens)

✕Multiplying by 100 at the wrong stage of a percentage calculation

✕Using the mass of a single atom instead of all atoms of that element

Relative masses of atoms and molecules — frequently asked questions

$A_r$ and $M_r$

Relative atomic mass ( $A_{r}$ )

Relative molecular / formula mass ( $M_{r}$ )

$A_r$ and $M_r$

Relative atomic mass ( $A_{r}$ )

Relative molecular / formula mass ( $M_{r}$ )