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This lesson covers relative atomic mass ($A_r$) and relative formula mass ($M_r$) as required by the Edexcel GCSE Combined Science specification (1SC0). 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 a compound from its formula.
The relative atomic mass ($A_r$) of an element is the average mass of one atom of the element compared to one-twelfth the mass of a carbon-12 atom.
| Element | Symbol | $A_r$ |
|---|---|---|
| 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 $A_r$ values — they are given on the periodic table in the exam. But knowing the common ones speeds up your work enormously.
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.
Chlorine has two 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, not a whole number.
The relative formula mass ($M_r$) of a substance is the sum of the relative atomic masses of all the atoms in its formula.
| Element | Number of Atoms | $A_r$ | Contribution |
|---|---|---|---|
| H | 2 | 1 | $2 \times 1 = 2$ |
| O | 1 | 16 | $1 \times 16 = 16$ |
$$M_r = 2 + 16 = 18$$
| Element | Number of Atoms | $A_r$ | Contribution |
|---|---|---|---|
| C | 1 | 12 | $1 \times 12 = 12$ |
| O | 2 | 16 | $2 \times 16 = 32$ |
$$M_r = 12 + 32 = 44$$
| Element | Number of Atoms | $A_r$ | Contribution |
|---|---|---|---|
| Ca | 1 | 40 | $1 \times 40 = 40$ |
| C | 1 | 12 | $1 \times 12 = 12$ |
| O | 3 | 16 | $3 \times 16 = 48$ |
$$M_r = 40 + 12 + 48 = 100$$
Watch out for brackets! The subscript 2 outside the bracket means there are two OH groups.
| Element | Number of Atoms | $A_r$ | Contribution |
|---|---|---|---|
| Mg | 1 | 24 | $1 \times 24 = 24$ |
| O | 2 | 16 | $2 \times 16 = 32$ |
| H | 2 | 1 | $2 \times 1 = 2$ |
$$M_r = 24 + 32 + 2 = 58$$
| Element | Number of Atoms | $A_r$ | Contribution |
|---|---|---|---|
| Ca | 1 | 40 | $1 \times 40 = 40$ |
| O | 2 | 16 | $2 \times 16 = 32$ |
| H | 2 | 1 | $2 \times 1 = 2$ |
$$M_r = 40 + 32 + 2 = 74$$
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.
The subscript 3 outside the bracket means there are three $\text{SO}_4$ groups.
| Element | Number of Atoms | $A_r$ | Contribution |
|---|---|---|---|
| Al | 2 | 27 | $2 \times 27 = 54$ |
| S | 3 | 32 | $3 \times 32 = 96$ |
| O | 12 | 16 | $12 \times 16 = 192$ |
$$M_r = 54 + 96 + 192 = 342$$
The "$\cdot 5\text{H}_2\text{O}$" means there are 5 water molecules of crystallisation.
| Element | Number of Atoms | $A_r$ | 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 |
$$M_r = 63.5 + 32 + 64 + 10 + 80 = 249.5$$
Exam Tip: When a formula contains a dot (e.g. $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$), you must include the water of crystallisation in your $M_r$ calculation unless the question specifically asks for the anhydrous form.
| Mistake | Correction |
|---|---|
| Forgetting to multiply inside brackets | In $\text{Ca(OH)}_2$, there are 2 O atoms and 2 H atoms, not 1 of each |
| Using mass number instead of $A_r$ | Always use the periodic table value — $A_r$ 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 $M_r$ a unit | $M_r$ has no units — it is a dimensionless ratio |