Relative Atomic Mass

You may have noticed that the relative atomic masses printed in the periodic table are often not whole numbers — chlorine is given as 35.5, not 35 or 37. The reason is that an element is usually a mixture of isotopes with different masses, and its relative atomic mass is a weighted average that allows for how much of each isotope is present. This lesson, part of Topic C1 of your OCR Gateway Combined Science course, defines relative atomic mass and shows how to calculate it from isotopic abundances — including the reverse problem of finding an abundance from a known $A_r$Ar​.

By the end of this lesson you should be able to define relative atomic mass, explain why it is often not a whole number, calculate $A_r$Ar​ from the masses and abundances of isotopes, and work backwards to find an unknown abundance from a given $A_r$Ar​.

This lesson develops AO2 (carrying out the weighted-mean calculation for $A_r$Ar​ from isotope masses and abundances, and rearranging it to find an unknown abundance), alongside AO1 (defining relative atomic mass) and AO3 (explaining why the value is usually not a whole number).

---

What Is Relative Atomic Mass?

The relative atomic mass ($A_r$Ar​) of an element is the weighted mean (average) mass of its atoms, taking account of the relative abundance of each isotope, measured on a scale where an atom of carbon-12 has a mass of exactly 12.

Two ideas are doing the work in that definition:

• it is an average over all the atoms of the element, not the mass of a single atom;
• it is weighted — isotopes that are more abundant count for more in the average.

Because it is an average of isotopes with different masses, $A_r$Ar​ is often not a whole number. Chlorine, for example, is a mixture of about 75% chlorine-35 and 25% chlorine-37, so its average atomic mass works out to 35.5 — a value no single chlorine atom actually has, but the correct mean for the mixture.

The phrase "relative to carbon-12" is worth unpacking, because it is what makes these masses "relative" rather than measured in grams. The masses of individual atoms are far too small to weigh directly, so instead of quoting them in grams, chemists compare every atom against a chosen standard: an atom of carbon-12 is defined to have a mass of exactly 12 units. Every other atom's mass is then expressed as a number of these units. So when we say hydrogen has a relative atomic mass of about 1 and magnesium about 24, we mean a hydrogen atom is roughly one-twelfth the mass of a carbon-12 atom, and a magnesium atom is roughly twice its mass. Because it is a ratio of one mass to another, relative atomic mass has no units — it is just a number.

Exam Tip: A complete definition of $A_r$Ar​ mentions three things: it is the weighted mean mass of the atoms, it allows for the relative abundance of each isotope, and it is measured relative to carbon-12 (= 12). Saying just "the average mass" misses the weighting.

---

The Formula

To calculate the relative atomic mass from isotope data, use:

$A_r = \dfrac{\sum(\text{isotope mass} \times \% \text{ abundance})}{100}$Ar​=100∑(isotope mass×% abundance)​

In words: multiply each isotope's mass by its percentage abundance, add these together, then divide by 100. (If the abundances are given as fractions or as a ratio rather than percentages, divide by their total instead of by 100.)

The method has the same three steps every time:

1. Multiply each isotope's mass by its percentage abundance.
2. Add those products together.
3. Divide by 100 (the total percentage).

Exam Tip: Show the calculation as one clear line: $(\text{mass}_1 \times \%_1) + (\text{mass}_2 \times \%_2)$(mass1​×%1​)+(mass2​×%2​), all over 100. Writing the substitution out fully earns the method marks even if the final arithmetic slips.

---

Why a Weighted Average, Not a Simple One?

It is worth being clear about why we weight the average rather than just taking the middle of the isotope masses. Imagine a bag of marbles where most are light and only a few are heavy. If you found the average mass by simply adding the light and heavy values and halving, you would get a number that is far too high, because it ignores the fact that most of the marbles are light. To get the true average you have to take account of how many of each kind there are — and that is exactly what "weighting by abundance" means.

The same is true of atoms. Take chlorine: there are three times as many chlorine-35 atoms as chlorine-37 atoms (75% against 25%). A simple, unweighted average of 35 and 37 would give 36, but that would pretend the two isotopes are equally common. Because chlorine-35 is much more abundant, the true average is pulled down towards 35, giving 35.5. The weighting is therefore not a mathematical nicety — it is the whole point. The relative atomic mass tells you the average mass of a chlorine atom as you would actually find them in nature, in the proportions in which they occur.

This is also why the relative atomic mass always lies between the smallest and largest isotope masses, and never outside that range: an average of several numbers cannot be bigger than the largest or smaller than the smallest. If a calculation ever gives you a value outside the range of the isotope masses, you know you have made an arithmetic slip and should check your working.

Exam Tip: Use the position of the answer as a sanity check. The relative atomic mass must lie between the isotope masses and closer to the more abundant one. For chlorine that means between 35 and 37, nearer 35 — so 35.5 looks right, but a value like 38 would be impossible.

---

Worked Examples

Worked Example 1 — Chlorine

Chlorine is 75% chlorine-35 and 25% chlorine-37. Calculate the relative atomic mass of chlorine.

Step 1 — multiply each mass by its abundance: $(35 \times 75) + (37 \times 25) = 2625 + 925 = 3550$(35×75)+(37×25)=2625+925=3550.

Step 2 — divide by 100:

$A_r = \dfrac{3550}{100} = 35.5$Ar​=1003550​=35.5

Answer: the relative atomic mass of chlorine is 35.5. Notice it is closer to 35 than to 37, because chlorine-35 is the more abundant isotope — exactly what "weighted" means.

Worked Example 2 — Copper

Copper is 69% copper-63 and 31% copper-65. Calculate the relative atomic mass of copper.

Step 1 — multiply: $(63 \times 69) + (65 \times 31) = 4347 + 2015 = 6362$(63×69)+(65×31)=4347+2015=6362.

Step 2 — divide by 100:

$A_r = \dfrac{6362}{100} = 63.62 \approx 63.6$Ar​=1006362​=63.62≈63.6

Answer: the relative atomic mass of copper is about 63.6. (The periodic table lists copper as 63.5, worked out from more precise abundance figures than the rounded 69% and 31% used here.) The value lies between 63 and 65, closer to 63 because copper-63 is more abundant.

Worked Example 3 — A three-isotope element

An element has three isotopes: mass 24 (79%), mass 25 (10%) and mass 26 (11%). Calculate its relative atomic mass.

Step 1 — multiply each: $(24 \times 79) + (25 \times 10) + (26 \times 11) = 1896 + 250 + 286 = 2432$(24×79)+(25×10)+(26×11)=1896+250+286=2432.

Step 2 — divide by 100:

$A_r = \dfrac{2432}{100} = 24.32 \approx 24.3$Ar​=1002432​=24.32≈24.3

Answer: the relative atomic mass is about 24.3 (this is magnesium). The same method extends to any number of isotopes — multiply, add, divide.

Here is the same data laid out as a table, which is a tidy way to organise a calculation:

Isotope	Mass	Abundance / %	Mass × abundance
Mg-24	24	79	1896
Mg-25	25	10	250
Mg-26	26	11	286
Total		100	2432

$A_r = 2432 \div 100 = 24.3$Ar​=2432÷100=24.3.

---

Working Backwards: Finding an Abundance

Sometimes you are given the relative atomic mass and the masses of the two isotopes, and asked to find the percentage abundance of each. The trick is to call one abundance $x$x and the other $(100 - x)$(100−x), then solve.