Atomic number, isotopes and isobars

Two whole numbers are all it takes to name any atom in the universe: how many protons sit in its nucleus, and how many nucleons — protons plus neutrons — it carries in total. Once Rutherford, Chadwick and their contemporaries had pinned down the proton, neutron and electron, chemists needed exactly this kind of precise labelling system, because it turned out that atoms of the very same element do not all weigh the same. Some carbon atoms are slightly heavier than others; some argon atoms weigh the same as some calcium atoms despite being completely different elements. Atomic number, mass number, isotopes, isobars and isotones are the vocabulary that makes sense of this. This article stays inside the ISC Class 11 Structure of Atom syllabus — pure definitions and arithmetic, no nuclear stability or radioactivity — and builds it into a fuller, worked-through picture with the weighted-average-mass calculation at its centre.

The core insight

An atom is fixed by two counts, atomic number $Z$ (protons) and mass number $A$ (protons plus neutrons), and comparing these counts across atoms sorts them into isotopes (same $Z$ ), isobars (same $A$ ) or isotones (same neutron count $n$ ).

The core idea

Atomic number ( $Z$ ) is the number of protons in an atom's nucleus. Because atoms are electrically neutral overall, $Z$ also equals the number of electrons in a neutral atom. No two different elements share the same $Z$ — it is the element's fingerprint, which is why the periodic table is ordered by $Z$ , not by mass.

Mass number ( $A$ ) is the total count of nucleons (protons and neutrons together) in the nucleus:

A = Z + n

where: $Z$ is the atomic number (proton count), $n$ is the number of neutrons, and $A$ is the mass number, always a whole number. An atom of element $X$ is written $^{A}_{Z}X$ , with $A$ as a left superscript and $Z$ as a left subscript. For a neutral atom, electrons $=$ protons $= Z$ . Ionisation changes only the electron count: a cation $X^{2+}$ has $Z-2$ electrons, an anion $X^{2-}$ has $Z+2$ electrons, but the nucleus — protons and neutrons alike — is untouched by gaining or losing electrons. Sorting atoms by which of these numbers they share gives three related families.

Isotopes share the same $Z$ but differ in $A$ (hence differ in $n$ ). They are the same element, sit in the same box of the periodic table, and have essentially identical chemical behaviour, because chemistry is governed by electron arrangement, which is fixed by $Z$ . Their physical properties — density, diffusion rate, and so on — differ slightly because they differ in mass. Hydrogen's three isotopes, protium $^1_1\text{H}$ , deuterium $^2_1\text{H}$ and tritium $^3_1\text{H}$ , all have $Z=1$ but $n = 0, 1, 2$ .

Isobars share the same $A$ but have different $Z$ . They are different elements that happen to weigh (in mass-number terms) the same, so they have different numbers of protons, electrons and neutrons, and unrelated chemistry. Argon-40 ( $^{40}_{18}\text{Ar}$ ) and calcium-40 ( $^{40}_{20}\text{Ca}$ ) both carry $A=40$ .

Isotones share the same neutron count $n$ across different elements. Silicon-30 ( $^{30}_{14}\text{Si}$ , $n=16$ ) and phosphorus-31 ( $^{31}_{15}\text{P}$ , $n=16$ ) are isotones. The figure below lines the three families up so the "same $Z$ ", "same $A$ ", "same $n$ " distinction is visible at a glance.

Figure

Three atom pairs illustrating isotopes with same atomic number, isobars with same mass number, and isotones with same neutron count

Finally, the atomic mass quoted on the periodic table (chlorine's famous 35.45, for instance) is not the mass of any single atom — it is the weighted average of the masses of an element's naturally occurring isotopes, weighted by how abundant each isotope is in nature. That weighted average is the one calculation this topic asks you to actually perform, and it is the subject of the derivation below.

Derivation AWeighted-average atomic mass from isotopic masses and abundances

The goal is to build, from nothing more than the everyday idea of a weighted average, the formula that turns a table of isotopic masses and their percentage abundances into the single decimal atomic mass printed on the periodic table — and to see why that number is almost never a whole number even though every individual atom's mass number is.

Step 1. Set up the idea of a weighted average. If a quantity takes value $m_1$ a fraction $f_1$ of the time and value $m_2$ a fraction $f_2$ of the time, its average is not $(m_1+m_2)/2$ unless $f_1=f_2$ ; it is $f_1 m_1 + f_2 m_2$ , giving more weight to whichever value occurs more often. Isotopic abundance is exactly this kind of situation: a sample of the element is a huge collection of atoms, some fraction of them one isotope, the rest another.

Step 2. Write the general formula for an element with isotopes of mass $m_1, m_2, \ldots$ and percentage abundances $p_1, p_2, \ldots$ (which must sum to $100\%$ ):

\bar{M} = \frac{p_1 m_1 + p_2 m_2 + \cdots}{100}

where: $\bar{M}$ is the average atomic mass (in u), $m_i$ is the mass of isotope $i$ , and $p_i$ is its percentage abundance. Dividing by $100$ converts each percentage into a fraction between $0$ and $1$ .

Step 3. Apply it to chlorine, which occurs as $^{35}\text{Cl}$ (mass $34.97$ u, abundance $75.77\%$ ) and $^{37}\text{Cl}$ (mass $36.97$ u, abundance $24.23\%$ ):

\bar{M} = \left(34.97 \times \frac{75.77}{100}\right) + \left(36.97 \times \frac{24.23}{100}\right) = 26.494 + 8.960 = 35.45\ \text{u}

Step 4. Read the result. The answer, $35.45$ u, is exactly the value printed on the periodic table for chlorine, and it lies between the two isotopic masses ( $34.97$ and $36.97$ ) but much closer to $35$ than to $37$ , because the lighter isotope is roughly three times more abundant. This is why atomic masses on the periodic table are decimals rather than whole numbers — they are a weighted blend of whole-number-mass isotopes, not the mass of any single real atom.

Worked example

Question. For $^{35}_{17}\text{Cl}$ , state the number of protons, electrons and neutrons in one neutral atom, and then classify the following pairs as isotopes, isobars or isotones: (a) $^{14}_6\text{C}$ and $^{14}_7\text{N}$ , (b) $^{12}_6\text{C}$ and $^{14}_6\text{C}$ , (c) $^{3}_1\text{H}$ and $^{4}_2\text{He}$ .

Step 1. Read off $^{35}_{17}\text{Cl}$ : $Z=17$ , $A=35$ . Protons $= Z = 17$ . Electrons in the neutral atom $= 17$ . Neutrons $= A - Z = 35 - 17 = 18$ .

Step 2. Find $Z$ , $A$ and $n$ for each atom in pair (a). $^{14}_6\text{C}$ : $Z=6$ , $A=14$ , $n=14-6=8$ . $^{14}_7\text{N}$ : $Z=7$ , $A=14$ , $n=14-7=7$ . The mass number $14$ matches while $Z$ differs, so this pair is isobars.

Step 3. Pair (b): $^{12}_6\text{C}$ has $Z=6$ , $n=12-6=6$ ; $^{14}_6\text{C}$ has $Z=6$ , $n=14-6=8$ . The atomic number $6$ matches while $A$ differs, so this pair is isotopes.

Step 4. Pair (c): $^{3}_1\text{H}$ has $Z=1$ , $n=3-1=2$ ; $^{4}_2\text{He}$ has $Z=2$ , $n=4-2=2$ . Neither $Z$ nor $A$ matches, but the neutron count $2$ matches, so this pair is isotones.

Answer: $^{35}_{17}\text{Cl}$ has 17 protons, 17 electrons and 18 neutrons; pair (a) is isobars, pair (b) is isotopes, pair (c) is isotones.

Competitive stretch

Boron in nature is a blend of $^{10}\text{B}$ (mass $10.01$ u) and $^{11}\text{B}$ (mass $11.01$ u), with average atomic mass $10.81$ u. Using $\bar{M} = p_1 m_1 + p_2 m_2$ (as fractions, with $p_1+p_2=1$ ) and $\bar{M}=10.81$ : since $\bar M$ sits much closer to $11.01$ than to $10.01$ , the heavier isotope $^{11}\text{B}$ must be the dominant one — solving $10.01p_1 + 11.01(1-p_1) = 10.81$ gives $p_1 \approx 0.20$ , so $^{10}\text{B}$ is about $20\%$ abundant and $^{11}\text{B}$ about $80\%$ , matching the real measured values.

Common traps

Reversing the definitions: it is isotopes that share the same atomic number, and isobars that share the same mass number — not the other way round.
Assuming isotopes differ chemically because they differ in mass. Chemistry is decided by electron arrangement, fixed by $Z$ , so isotopes of an element behave almost identically in reactions.
Recomputing neutron count after ionisation. The nucleus (protons and neutrons) never changes when an atom loses or gains electrons; only the electron count changes.
Treating mass number $A$ (a whole number, protons plus neutrons) as the same thing as atomic mass (a decimal, the isotope-weighted average) — they are related but not identical, and confusing them derails the weighted-average calculation.

Ask the tutor: want extra practice classifying isotope/isobar/isotone pairs or working through a three-isotope weighted-average calculation?

Why is the periodic table mass of chlorine 35.45 when there is no such thing as an atom weighing 35.45?

Because that number is a weighted average across a huge natural sample of chlorine atoms, not the mass of one atom. About three quarters of chlorine atoms are $^{35}\text{Cl}$ (mass close to 35) and about a quarter are $^{37}\text{Cl}$ (mass close to 37). Blend those masses in that ratio and you land on 35.45 — close to 35 because the lighter isotope dominates, but never a value any single atom actually has.

If isotopes of an element have different masses, why do they react the same way chemically?

Chemical behaviour — which bonds an atom forms, how reactive it is — is decided almost entirely by the arrangement of electrons around the nucleus, and that arrangement is fixed by the atomic number $Z$ , which is identical for all isotopes of an element. The extra neutrons add mass to the nucleus but do not change $Z$ or the electron count, so they barely affect chemistry. They do cause small physical differences, such as slightly different diffusion rates or densities, because those depend directly on mass.

How do I quickly tell isotopes, isobars and isotones apart in an exam question?

Write down $Z$ , $A$ and $n=A-Z$ for both atoms first — that turns the question into simple number-matching. If $Z$ matches (and $A$ differs), it is isotopes. If $A$ matches (and $Z$ differs), it is isobars. If $n$ matches (and both $Z$ and $A$ differ), it is isotones. Doing this systematically avoids the common slip of eyeballing the mass numbers and guessing.

Atomic number, isotopes and isobars

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads