Relative Atomic Mass from Isotopic Composition

A chlorine atom in nature is not always the same mass. Some chlorine atoms contain 18 neutrons, while others contain 20. Yet the periodic table gives chlorine a relative atomic mass of about 35.45.

So what is a chlorine atom with a mass of 35.45? There isn’t one.

That number is an average, but not the ordinary kind where every value counts equally. The more common isotopes must contribute more to the average than the rare ones. The skill is learning how to calculate that weighted relative atomic mass from isotope masses and natural abundances.

Before reading further, predict this: an element has two isotopes with masses 20 and 22. If 90% of its atoms are the lighter isotope, will its relative atomic mass be closer to 20, 21, or 22?

It should be closer to 20. A weighted average gets pulled towards the value that occurs more often.

01Start with the atoms, not the formula

Imagine collecting 100 atoms of an element.

Suppose:

• 80 atoms have an isotopic mass of 10
• 20 atoms have an isotopic mass of 11

You could find their average mass by adding the contribution from all 100 atoms:

\[
\frac{(80)(10)+(20)(11)}{100}
=
\frac{1020}{100}
=
10.2
\]

The relative atomic mass is therefore 10.2.

Notice what happened. We didn’t simply calculate:

\[
\frac{10+11}{2}=10.5
\]

That calculation would wrongly assume the two isotopes are equally common.

This is the central idea:

An isotope contributes to relative atomic mass according to both its mass and how common it is.

If you need a refresher on what makes isotopes different in the first place, see Stable and Unstable Isotopes Explained for HSC Chemistry.

02What relative atomic mass actually means

The relative atomic mass, \(A_r\), of an element is the weighted mean mass of its naturally occurring atoms relative to one-twelfth of the mass of a carbon-12 atom.

That definition contains two important ideas.

First, it is a weighted mean because different isotopes can have different natural abundances.

Second, it is relative. Relative atomic mass is a ratio, so it has no unit.

For most HSC calculations, the important equation is:

\[
A_r=\sum (\text{isotopic mass})(\text{fractional abundance})
\]

Here:

• \(A_r\) is the relative atomic mass
• isotopic mass is the relative mass of a particular isotope
• fractional abundance is the proportion of atoms that are that isotope, written as a decimal

For example, 72% becomes:

\[
72\% = 0.72
\]

If abundances are given as percentages, you can also calculate:

\[
A_r=
\frac{\sum(\text{isotopic mass})(\text{percentage abundance})}{100}
\]

Both methods are equivalent.

03Why the answer is usually not a whole number

Students sometimes look at a periodic table and wonder why atomic masses such as 24.305 or 63.546 appear when atoms contain whole numbers of protons and neutrons.

There are two reasons.

One is the weighted averaging of different isotopes.

The other is that the actual isotopic masses themselves are not generally exact whole numbers. A magnesium-24 atom, for example, has an isotopic mass close to 24, but not exactly 24.

In many HSC questions, you may be given mass numbers such as 24, 25, and 26 and use them as approximate isotope masses. If more precise isotopic masses are supplied, use those instead.

The question determines the precision of your model.

04A useful picture of weighted averages

Suppose two people are sitting on opposite ends of a seesaw. One isotope is on the left, and another isotope is on the right.

The relative atomic mass sits somewhere between them.

But abundance acts a little like extra influence. If one isotope represents 95% of the atoms, the average gets dragged strongly towards that isotope’s mass.

The analogy isn’t exact. No actual force or seesaw is involved. It is simply a way to picture why the average lies closer to the more abundant isotope.

This gives you a powerful error check:

Your relative atomic mass must lie between the smallest and largest isotope masses, and it should usually be closer to the mass of the more abundant isotope.

If an element contains only isotopes of mass 35 and 37, an answer of 38.1 is impossible.

So is an answer of 21.

06The most common mistake: averaging the isotope masses directly

Suppose an element is 99% isotope X-10 and 1% isotope X-11.

A student might calculate:

\[
\frac{10+11}{2}=10.5
\]

Why does that feel reasonable? Because when we hear “average”, we often think “add the values and divide by how many values there are”.

But that method answers the wrong question. It finds the average of two isotope masses, treating the isotopes as equally common.

The actual calculation is:

\[
A_r=(10)(0.99)+(11)(0.01)=10.01
\]

That makes physical sense. If 99 out of every 100 atoms are the lighter isotope, the average should be extremely close to 10.

A useful decision rule is:

08When an abundance is missing

You often don’t need every abundance to be stated explicitly.

If an element has only two naturally occurring isotopes and one has an abundance of 73%, the other must have:

\[
100\%-73\%=27\%
\]

For three or more isotopes, the same principle applies. Natural abundances must add to 100%, assuming the question includes all isotopes present in the sample.

Be careful with that assumption. If a question says that three listed isotopes are the only naturally occurring isotopes, you can safely subtract from 100%. If the list is incomplete, you cannot.

That is the sort of hidden assumption that becomes important in harder questions.

09Relative abundance is not always given as a percentage

A mass spectrum may show isotope peaks using relative intensity instead.

Suppose two isotope peaks have relative intensities of 3 and 1. That does not mean 3% and 1%.

It means their abundance ratio is:

\[
3:1
\]

There are four total parts, so their fractional abundances are:

\[
\frac{3}{4}=0.75
\]

and

\[
\frac{1}{4}=0.25
\]

If their isotope masses were 34 and 36, the weighted relative atomic mass would be:

\[
A_r=(34)(0.75)+(36)(0.25)=34.5
\]

The important idea is not “always divide by 100”. It is “turn each abundance into its fraction of the total”.

10You can also work backwards

Sometimes the isotope masses and relative atomic mass are known, but an abundance is missing.

Suppose an element has two isotopes of mass 79 and 81, and its relative atomic mass is 79.6.

Let the fraction of the mass-79 isotope be \(x\).

Because there are only two isotopes, the fraction of the mass-81 isotope must be \(1-x\).

Then:

\[
79x+81(1-x)=79.6
\]

Expand:

\[
79x+81-81x=79.6
\]

\[
-2x=-1.4
\]

\[
x=0.70
\]

So the abundances are:

• mass-79 isotope: 70%
• mass-81 isotope: 30%

Again, the result passes the physical check. The average of 79.6 is closer to 79, so the mass-79 isotope should be more abundant.

11Relative atomic mass versus mass number

These quantities are easy to mix up.

For example, chlorine-35 has a mass number of 35 because its nucleus contains 35 nucleons in total.

That does not mean chlorine’s relative atomic mass must be 35. Chlorine occurs naturally as more than one isotope, so its periodic-table value is a weighted mean.

The electron arrangement is a separate idea. Changing the number of neutrons produces a different isotope, while changing the number of electrons produces an ion. If that distinction needs refreshing, Electronic Configuration for HSC Chemistry: Atoms and Ions is the useful next step.

12Questions and solutions

Question 1

An element has two isotopes with masses 50.0 and 52.0. Their natural abundances are 84.0% and 16.0%, respectively. Calculate the relative atomic mass.

Solution 1

The relative atomic mass is \(\boxed{50.32}\).

Relative atomic mass is a weighted mean, so each isotope mass must be multiplied by its fractional abundance.

Convert the percentages:

\[
84.0\%=0.840,\qquad16.0\%=0.160
\]

Then:

\[
\begin{aligned}
A_r
&=(50.0)(0.840)+(52.0)(0.160)\\
&=42.0+8.32\\
&=50.32
\end{aligned}
\]

The answer is closer to 50.0 because the mass-50 isotope is much more abundant.

Simply averaging 50.0 and 52.0 to obtain 51.0 would incorrectly treat the two isotopes as equally abundant.

Question 2

An element contains three isotopes with the following composition:

Calculate its relative atomic mass.

Solution 2

The relative atomic mass is approximately \(\boxed{70.03}\).

Convert the abundances to fractions:

\[
0.600,\qquad0.250,\qquad0.150
\]

They total 1.000, so the abundance data are internally consistent.

Now calculate the weighted mean:

\[
\begin{aligned}
A_r
&=(68.93)(0.600)+(70.92)(0.250)+(71.92)(0.150)\\
&=41.358+17.730+10.788\\
&=69.876
\end{aligned}
\]

Therefore:

\[
\boxed{A_r=69.88}
\]

The average is closer to 68.93 than to either heavier isotope because the lightest isotope represents 60% of the sample.

Question 3

An element has exactly two naturally occurring isotopes. One isotope has a mass of 106.9 and an abundance of 51.8%. The second isotope has a mass of 108.9.

Calculate the relative atomic mass.

Solution 3

The relative atomic mass is approximately \(\boxed{107.86}\).

Because there are exactly two isotopes, their abundances must total 100%. The second isotope therefore has an abundance of:

\[
100.0\%-51.8\%=48.2\%
\]

Convert both abundances to fractions:

\[
0.518,\qquad0.482
\]

Now calculate:

\[
\begin{aligned}
A_r
&=(106.9)(0.518)+(108.9)(0.482)\\
&=55.3742+52.4898\\
&=107.864
\end{aligned}
\]

Therefore:

\[
\boxed{A_r\approx107.86}
\]

The answer lies almost halfway between the two isotope masses because their abundances are fairly similar. It is slightly closer to 106.9 because that isotope is slightly more abundant.

Question 4

A mass spectrum for an element shows two isotope peaks at masses 84 and 86. Their relative intensities are 5 and 3, respectively.

Calculate the relative atomic mass of the element.

Solution 4

The relative atomic mass is \(\boxed{84.75}\).

The numbers 5 and 3 are a ratio, not percentages. Together they represent:

\[
5+3=8
\]

equal abundance parts.

The fractional abundances are therefore:

\[
\frac{5}{8}=0.625
\]

and

\[
\frac{3}{8}=0.375
\]

Now calculate the weighted mean:

\[
\begin{aligned}
A_r
&=(84)(0.625)+(86)(0.375)\\
&=52.5+32.25\\
&=84.75
\end{aligned}
\]

The tempting mistake is to treat 5 and 3 as 5% and 3%. They are relative intensities, so they must first be converted into fractions of their total.

Question 5

An element has exactly two naturally occurring isotopes with isotopic masses 112.0 and 115.0. Its measured relative atomic mass is 114.1.

Determine the percentage abundance of each isotope.

Solution 5

The isotope of mass 112.0 has an abundance of \(\boxed{30.0\%}\), and the isotope of mass 115.0 has an abundance of \(\boxed{70.0\%}\).

Let \(x\) be the fractional abundance of the mass-112 isotope.

Because there are only two isotopes, the fractional abundance of the mass-115 isotope is \(1-x\).

Use the weighted-average equation:

\[
112.0x+115.0(1-x)=114.1
\]

Expand and solve:

\[
\begin{aligned}
112.0x+115.0-115.0x&=114.1\\
-3.0x&=-0.9\\
x&=0.300
\end{aligned}
\]

Therefore:

\[
112.0:\quad0.300\times100\%=30.0\%
\]

and:

\[
115.0:\quad100.0\%-30.0\%=70.0\%
\]

The result also makes sense before doing the algebra. The relative atomic mass, 114.1, lies much closer to 115.0 than to 112.0, so the heavier isotope should be more abundant.

Question 6

A student is told that an element has three naturally occurring isotopes with masses 40, 42, and 44. Two measured abundances are 70% for mass 40 and 20% for mass 42.

The student calculates the abundance of the mass-44 isotope as 10%, then obtains:

\[
A_r=(40)(0.70)+(42)(0.20)+(44)(0.10)=40.8
\]

A second experiment later shows that the sample also contains a small amount of a fourth isotope of mass 43 that was omitted from the original data.

Explain why the student’s numerical calculation can no longer be accepted, even though the arithmetic itself is correct.

Solution 6

The calculation cannot be accepted because the assumption used to obtain the 10% abundance is no longer valid.

The student reasoned:

\[
100\%-70\%-20\%=10\%
\]

That would be correct only if the masses 40, 42, and 44 represented all isotopes present in the sample.

Once a mass-43 isotope is shown to be present, some of the remaining 10% abundance must belong to that isotope. The abundance assigned to mass 44 is therefore unknown.

For example, if mass 43 had an abundance of 2%, then mass 44 would have only:

\[
100\%-70\%-20\%-2\%=8\%
\]

The weighted mean would then be:

\[
\begin{aligned}
A_r
&=(40)(0.70)+(42)(0.20)+(43)(0.02)+(44)(0.08)\\
&=40.78
\end{aligned}
\]

The exact revised value depends on the abundance of the mass-43 isotope.

The important trap is that a correct piece of arithmetic can still produce an invalid scientific conclusion if the assumption behind the numbers is wrong.

13What this calculation leads to next

Weighted relative atomic mass connects the microscopic idea of isotopes to a number you use throughout chemistry.

Once you can move confidently between isotope masses, abundances, and \(A_r\), mass spectra become much easier to interpret. You can use peak positions to identify isotope masses, peak sizes to infer relative abundances, and both pieces of information together to explain why the periodic table gives an element the atomic mass that it does.

That same atomic structure then feeds into electron configuration, bonding, and periodic trends. The nucleus determines which element you have, isotopes change its neutron count and mass, and its electrons determine much of its chemistry.