Calculate Relative Atomic Mass

An element is a mixture of isotopes — atoms with the same number of protons but different numbers of neutrons, so different masses. When you weigh a real sample you are weighing all of those isotopes together, in their natural proportions.

The relative atomic mass ($A_r$) is the weighted-mean (average) mass of the atoms of an element, measured relative to 1/12 the mass of a carbon-12 atom. Because it is a comparison of two masses, $A_r$ has no units.

"Weighted" is the key word: the more abundant an isotope is, the more it "pulls" the average towards its own mass.

Why this matters for the exam. $A_r$ is the number you read off the periodic table and use in every mole and reacting-mass calculation later in the course. The skill examined here is turning a list of isotope masses and abundances into that single weighted-mean number.

Every $A_r$ calculation uses one formula:

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

The denominator is 100 because the percentage abundances add up to 100%.

The four-step method (works every time):

1. Multiply each isotope's mass by its percentage abundance.
2. Add those products together.
3. Divide the total by 100.
4. Round sensibly (usually to 1 decimal place) and write the answer with no units.

Worked example — chlorine. ³⁵Cl is 75% abundant and ³⁷Cl is 25% abundant:

$A_r(\text{Cl}) = \dfrac{(35 \times 75) + (37 \times 25)}{100} = \dfrac{2625 + 925}{100} = \dfrac{3550}{100} = 35.5$

That is exactly the chlorine value on the periodic table.

Mark-scheme tip. Even though a calculator is allowed, you must write the multiply-and-add line (e.g. "(35 × 75) + (37 × 25)"). That line is the method mark — students who write only a final answer can lose it if the number is wrong, and so lose error-carried-forward credit too.

The method does not change when an element has three or more isotopes — you simply add more terms to the sum on top.

Worked example — magnesium. Mg has three isotopes: ²⁴Mg (79%), ²⁵Mg (10%) and ²⁶Mg (11%):

$A_r(\text{Mg}) = \dfrac{(24 \times 79) + (25 \times 10) + (26 \times 11)}{100}$

$= \dfrac{1896 + 250 + 286}{100} = \dfrac{2432}{100} = 24.32 \approx 24.3$

Check it makes sense. ²⁴Mg is by far the most common isotope, so the answer should sit close to 24 — and 24.3 does. This "does my answer sit between the lightest and heaviest isotope, near the most common one?" sanity check catches most arithmetic slips.

If abundances are given as decimals or fractions (e.g. 0.79 instead of 79%), one safe approach is to convert them to percentages first so the "÷ 100" still applies. Always make sure the abundances add up to 100% (or 1.00) before you start.

Mark-scheme tip. The most common slip with three isotopes is dropping a term or mis-pairing a mass with the wrong abundance. Lay the working out in a neat row of products so each mass is clearly multiplied by its own abundance.

Sometimes the exam gives you the Ar and asks you to work backwards to find a missing abundance. For a two-isotope element this is a short piece of algebra.

The trick: if one abundance is x%, the other must be (100 − x)%, because the two add to 100.

Worked example — neon. Neon has two isotopes, ²⁰Ne and ²²Ne, and $A_r(\text{Ne}) = 20.2$. Find the percentage abundance of each.

Let ²⁰Ne = x%, so ²²Ne = (100 − x)%:

$\dfrac{20x + 22(100 - x)}{100} = 20.2$

Multiply both sides by 100:

$20x + 2200 - 22x = 2020$

$-2x = 2020 - 2200 = -180$

$x = 90$

So ²⁰Ne = 90% and ²²Ne = (100 − 90) = 10%.

Check: (20 × 90 + 22 × 10) ÷ 100 = (1800 + 220) ÷ 100 = 2020 ÷ 100 = 20.2 ✓ — always substitute back to confirm.

Mark-scheme tip. Writing the two abundances as x and (100 − x) earns the first mark even before you finish the algebra — set it up clearly.

A single atom has a whole-number mass number (you cannot have part of a proton or neutron). Yet most $A_r$ values on the periodic table are not whole numbers — chlorine is 35.5, copper 63.5, magnesium 24.3.

The reason: $A_r$ is an average of a mixture of isotopes. Chlorine is roughly a 3 : 1 mixture of ³⁵Cl and ³⁷Cl, so its weighted mean (35.5) falls between 35 and 37 — closer to 35 because ³⁵Cl is more common. No single chlorine atom weighs 35.5; the average does.

So when an exam asks "explain why the relative atomic mass of chlorine is 35.5 and not a whole number", the marking points are:

• chlorine exists as (more than one) isotope (³⁵Cl and ³⁷Cl);
• $A_r$ is the weighted average / mean mass of these isotopes;
• so the value lies between the isotope masses and is not a whole number.

Reading a mass-spectrum-style bar chart. Some questions show the abundances as a bar chart (height = % abundance) instead of writing them in words. Read each bar's height off the axis, pair it with the mass on the horizontal axis, then use the same multiply-add-divide method. Treat the chart as just another way of giving you the abundances.