Isotopes

Look at a bottle labelled "chlorine, atomic mass 35.5 u" and something odd jumps out: no single atom can weigh a fraction of a unit — protons and neutrons come in whole numbers. The resolution is one of the most useful ideas in the whole "Structure of the Atom" chapter: elements are rarely made of one kind of atom. Chlorine gas is a mix of two slightly different atoms, both chemically identical, both called chlorine, but carrying a different number of neutrons in the nucleus. These twin atoms are called isotopes. This article defines them precisely using atomic number and mass number, explains why isotopes behave identically in reactions but differently in mass-dependent properties, walks the standard hydrogen/carbon/oxygen/chlorine examples, and drills the one calculation Class 9 exams ask for — turning isotope percentages into the decimal atomic mass printed on the periodic table.

The core insight

Isotopes are atoms of the same element with the same atomic number but different mass numbers, so they share chemical properties but differ in physical, mass-dependent ones, and their percentage-weighted mixture is what gives an element its non-integer average atomic mass.

The core idea

Definition. Isotopes are atoms of the same element — same atomic number $Z$ , meaning the same number of protons — that have different mass numbers $A$ , because they carry different numbers of neutrons. Since $A =$ protons $+$ neutrons, a change in neutron count changes $A$ without changing $Z$ , and since $Z$ alone decides which element an atom is, the atoms remain the same element throughout.

Why chemistry stays the same. An atom's chemical behaviour — how eagerly it bonds, what compounds it forms — is decided almost entirely by its electrons, specifically the outer ones. Since isotopes have identical proton counts, and a neutral atom has electrons equal to protons, isotopes of an element have identical electron arrangements. Same electron arrangement means identical chemical properties.

Why physical properties differ. Mass number counts protons plus neutrons, and neutrons contribute nothing to bonding but do contribute to mass. So isotopes of an element differ in mass-dependent physical properties — density, rate of diffusion, and similar bulk properties — even while every chemical reaction treats them identically.

The standard examples. Four cases recur across the syllabus and in exam questions:

Hydrogen has three isotopes: protium $^1_1\text{H}$ (0 neutrons), deuterium $^2_1\text{H}$ (1 neutron), and tritium $^3_1\text{H}$ (2 neutrons) — all with 1 proton.
Carbon: $^{12}_6\text{C}$ and $^{14}_6\text{C}$ , both with 6 protons.
Oxygen: $^{16}_8\text{O}$ , $^{17}_8\text{O}$ , $^{18}_8\text{O}$ , all with 8 protons.
Chlorine: $^{35}_{17}\text{Cl}$ and $^{37}_{17}\text{Cl}$ , both with 17 protons.

Figure

Three isotopes of hydrogen shown as nuclei with one proton each and zero one or two neutrons, labelled protium deuterium and tritium

Average atomic mass. Because a natural sample of an element is a mixture of its isotopes in fixed proportions, the atomic mass listed on the periodic table is not any single atom's mass — it is a percentage-weighted average across the mixture. For an element with two isotopes of mass $m_1$ and $m_2$ , present in abundances $p_1\%$ and $p_2\%$ (with $p_1 + p_2 = 100$ ):

\text{Average atomic mass} = \frac{m_1 \times p_1 + m_2 \times p_2}{100}

where: $m_1, m_2$ are the isotope masses in atomic mass units (u), and $p_1, p_2$ are their percentage abundances in a natural sample. This is the one formula Class 9 exams test, and it is the reason many atomic masses on the periodic table — chlorine's $35.5\ \text{u}$ among them — are not whole numbers, even though every individual atom's mass number is a whole number.

Derivation AWeighted-average atomic mass from two isotopes

The goal is to show, from first principles of what "average" means for a mixture, why the correct way to combine two isotope masses is a percentage-weighted sum divided by 100, and not a plain arithmetic mean of the two masses — a distinction that trips up students who average $35$ and $37$ to get $36$ regardless of abundance.

Step 1. Picture a sample of exactly 100 atoms of the element, since abundances are given as percentages of the whole. Of these, $p_1$ atoms have mass $m_1$ and $p_2$ atoms have mass $m_2$ , with $p_1 + p_2 = 100$ .

Step 2. Find the total mass of the sample by adding the mass contributed by each group: the $p_1$ atoms of mass $m_1$ contribute $m_1 \times p_1$ , and the $p_2$ atoms of mass $m_2$ contribute $m_2 \times p_2$ . Total mass of the 100 atoms is $m_1 p_1 + m_2 p_2$ .

Step 3. An average mass per atom is total mass divided by number of atoms, and here the number of atoms is exactly $100$ by construction. So the average atomic mass is $\dfrac{m_1 p_1 + m_2 p_2}{100}$ — which is the formula above.

Step 4. Notice this collapses to a plain mean only in the special case $p_1 = p_2 = 50$ , since then $\dfrac{50 m_1 + 50 m_2}{100} = \dfrac{m_1 + m_2}{2}$ . Whenever abundances are unequal (the usual case), the plain mean is wrong and gives a biased answer that ignores which isotope actually dominates the sample.

Worked example

Question: Naturally occurring chlorine is a mixture of two isotopes: $^{35}_{17}\text{Cl}$ (mass $35\ \text{u}$ ), present as $75\%$ of chlorine atoms, and $^{37}_{17}\text{Cl}$ (mass $37\ \text{u}$ ), present as $25\%$ of chlorine atoms. Calculate the average atomic mass of chlorine, and check that a plain (unweighted) mean would have given a different, wrong answer.

Step 1. Apply the weighted-average formula with $m_1 = 35$ , $p_1 = 75$ , $m_2 = 37$ , $p_2 = 25$ :

\text{Average atomic mass} = \frac{(35 \times 75) + (37 \times 25)}{100}

Step 2. Compute each product: $35 \times 75 = 2625$ , and $37 \times 25 = 925$ .

Step 3. Add the products and divide by 100: $2625 + 925 = 3550$ , so $\dfrac{3550}{100} = 35.5\ \text{u}$ .

Step 4. Compare with the plain mean: $\dfrac{35 + 37}{2} = 36\ \text{u}$ , which is close but not correct — it silently assumes a 50:50 mixture, whereas chlorine's actual mixture is 75:25, so the lighter isotope pulls the true average down to $35.5\ \text{u}$ . The average atomic mass of chlorine is $35.5\ \text{u}$ , exactly the value printed on the periodic table.

Common traps

Confusing isotopes (same element, same $Z$ , different $A$ ) with isobars (different elements, different $Z$ , same $A$ ) — the two definitions are near-opposites and are commonly swapped in exams.
Assuming isotopes have different chemical properties because their masses differ — chemistry depends on electron arrangement, which is identical for isotopes, so only mass-related physical properties differ.
Averaging the two isotope masses directly (e.g. $\frac{35+37}{2}=36$ ) instead of weighting by percentage abundance — correct only when abundances happen to be equal.
Thinking one isotope is the "real" element and the rest are impurities — natural samples of chlorine, oxygen, carbon, and many other elements are genuine mixtures of multiple isotopes, none more "real" than the other.

Ask the tutor: if you want to try the weighted-average calculation with three isotopes instead of two, or check your own working on a fresh isotope pair, ask the AI tutor to walk through it with you.

If isotopes react identically, why do chemists ever care which isotope they have?

Within ordinary Class 9 chemistry, they usually don't need to — reactions, bonding, and compound formulas all follow from the electron arrangement, which is the same for every isotope. The mass difference only shows up in mass-dependent physical measurements, like the exact density of a pure isotopic sample, or in calculating the average atomic mass of a natural mixture, which is the one place isotopes enter your syllabus calculations directly.

Why isn't the average atomic mass just the average of the mass numbers, like (35+37)/2 for chlorine?

Because a plain average silently assumes the two isotopes are equally common, which is rarely true. Chlorine's natural mixture is 75% of the lighter isotope and only 25% of the heavier one, so the true average sits closer to 35 than to 36. You must weight each isotope's mass by how often it actually occurs, using the formula $(m_1 p_1 + m_2 p_2)/100$ , not a simple midpoint.

Are isotopes radioactive, and is that part of this topic?

Not necessarily, and no. Some isotopes are stable (like the ones in every example here) and some are radioactive (like tritium and carbon-14), but radioactivity itself is a separate topic not covered at Class 9 under Structure of the Atom. For this chapter, treat isotopes purely as atoms with the same atomic number and different mass numbers, and focus on the definition, standard examples, and the weighted-average calculation.

Isotopes

The core idea

Worked example

Common traps

Before this

Where this leads