Isotopes
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Lesson overview
What isotopes are, why they behave alike chemically, calculating relative atomic mass from isotopic abundances, and reading simple mass spectra.
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Pacing
~40 seconds per MCQ (180 questions in 120 minutes, no negative marking).
How JAMB marks this
Fast dimensional/sign elimination and standard formula shortcuts — skip multi-line algebra when substitution tests options faster.
Watch out for
UTME distractors routinely include the intermediate value (e.g., forgetting to convert cm³ to dm³ or °C to Kelvin).
Lesson content
Overview
Look at chlorine in the periodic table and its relative atomic mass is 35.5. No single chlorine atom has half a neutron; the value is an average, because natural chlorine is a mixture of two kinds of atom with different masses. These different forms of the same element are called isotopes. In this lesson you will define isotopes, explain why they have the same chemical properties but different physical properties, calculate relative atomic masses from abundances and work backwards to abundances, and interpret the peaks a mass spectrometer produces. Expect at least one of these calculations in the UTME.
Key concept: isotopes and relative atomic mass
Isotopes are atoms of the same element that have the same number of protons (same atomic number) but different numbers of neutrons (different mass numbers). Examples: ¹H, ²H (deuterium) and ³H (tritium); ¹²C, ¹³C and ¹⁴C; ³⁵Cl and ³⁷Cl. Isotopy is the existence of an element in the form of isotopes. The relative atomic mass (Ar) of an element is the average mass of one atom of the element, taking account of the natural abundance of its isotopes, compared with one-twelfth of the mass of one atom of carbon-12. Isotopes have the same electron arrangement, so they have identical chemical properties. They differ in mass, so their physical properties (density, rate of diffusion, and nuclear stability) differ.
Calculating relative atomic mass
Because each isotope contributes to the average in proportion to how common it is, the relative atomic mass is a weighted average: multiply the mass of each isotope by its percentage abundance, add the products and divide by 100. A mass spectrometer measures the masses and relative abundances of isotopes. Atoms are ionised, accelerated, and deflected by a magnetic field; lighter ions are deflected more. The detector records a peak for each isotope: its position gives the mass and its height gives the relative abundance. Working backwards: if an element has two isotopes and you know Ar, let the abundance of one be x% and the other (100 − x)%, then solve the weighted-average equation for x. Diatomic molecules such as Cl₂ give more than one molecular peak, because each atom in the molecule can be either isotope. With ³⁵Cl (75%) and ³⁷Cl (25%), the probability of each combination is found by multiplying the fractions: ³⁵Cl–³⁵Cl, ³⁵Cl–³⁷Cl (which can occur two ways), and ³⁷Cl–³⁷Cl. The carbon-12 standard matters because it fixes the scale for every other element. One atom of carbon-12 is assigned a mass of exactly 12 units, so one atomic mass unit is one-twelfth of its mass. An element whose average atom is twice as heavy as a carbon-12 atom therefore has Ar = 24, which is magnesium. Before 1961 chemists used hydrogen or oxygen as the standard, which is why older books quote slightly different values. Uses of isotopes include carbon-14 for dating archaeological remains, cobalt-60 in cancer treatment, iodine-131 for thyroid disorders, and deuterium in heavy water for some nuclear reactors.
Relationships to know
Ar = Σ(isotopic mass × % abundance) ÷ 100 Two isotopes: Ar = [m₁x + m₂(100 − x)] ÷ 100 Neutrons = mass number − atomic number (same Z for all isotopes of an element) Probability of a diatomic combination = product of the fractional abundances (×2 for mixed pairs)
Worked example 1: relative atomic mass of chlorine
Natural chlorine contains 75% ³⁵Cl and 25% ³⁷Cl, as shown in the mass spectrum. Calculate the relative atomic mass of chlorine.
Active Worked Example · Predict Step 1
3 stepsWrite down the formula, rule, or first move you would make before revealing Step 1.
Worked example 2: finding abundances from Ar
Boron has two isotopes, ¹⁰B and ¹¹B, and a relative atomic mass of 10.8. Calculate the percentage abundance of each.
Active Worked Example · Predict Step 1
4 stepsWrite down the formula, rule, or first move you would make before revealing Step 1.
Worked example 3 (UTME standard): molecular peaks of Cl₂
Chlorine gas contains Cl₂ molecules made from ³⁵Cl (75%) and ³⁷Cl (25%). Predict the relative heights of the molecular peaks in its mass spectrum.
Active Worked Example · Predict Step 1
4 stepsWrite down the formula, rule, or first move you would make before revealing Step 1.
Common traps
• Isotopes differ in neutrons, never in protons; a different number of protons means a different element. • Do not average isotope masses without weighting by abundance. • Relative atomic mass has no units and need not be a whole number. • The mixed diatomic molecule is counted twice in probability questions. • Isotopes have the same chemical properties but can differ in radioactivity. • Check your answer lies between the lightest and heaviest isotope masses, and nearer the more abundant one.
Summary
• Isotopes: same atomic number, different mass numbers (different numbers of neutrons). • Same electron arrangement means identical chemical properties; physical properties differ. • Ar = Σ(mass × abundance) ÷ 100, compared with ¹/₁₂ of carbon-12. • A mass spectrometer shows one peak per isotope; height gives abundance. • Cl₂ gives peaks at 70, 72 and 74 in the ratio 9 : 6 : 1.
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