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Why is chlorine 35.5 and not 35 or 37? Because a sample of chlorine is a mixture of isotopes, and the number in the periodic table is their weighted average. Exam questions ask for this average, or give the average and ask for the abundance. This post works both kinds with real isotope data, including a three-isotope case, and explains why some elements are now given as a range.
What does the syllabus ask?
The JEE (Main) 2026 syllabus lists "Atomic and molecular masses" and the mole concept in its first chemistry unit. The NCERT Class 11 chapter Some Basic Concepts of Chemistry, section 1.7.2, defines the average atomic mass using the natural abundance of each isotope. Its example for carbon:
(0.98892)(12 u) + (0.01108)(13.00335 u) + (2 × 10⁻¹²)(14.00317 u) = 12.011 u
Note that the abundances in its table are percentages (98.892%), and the calculation uses them as fractions. Check the syllabus for your exam year.
How do you calculate an average atomic mass?
Average atomic mass = Σ (fractional abundance × isotopic mass)
Example 1: chlorine. The NIST isotopic composition data for chlorine give:
³⁵Cl: mass 34.96885, abundance 0.7576
³⁷Cl: mass 36.96590, abundance 0.2424
34.96885 × 0.7576 = 26.4924 36.96590 × 0.2424 = 8.9605 Sum = 35.453
If you use the mass numbers 35 and 37 instead, you get 35.485, which is off in the second decimal place. For multiple-choice questions that give mass numbers, use them. When exact isotopic masses are given, use those.
Example 2: argon, three isotopes. NCERT exercise 1.32 gives ³⁶Ar at 35.96755 g mol⁻¹ (0.337%), ³⁸Ar at 37.96272 g mol⁻¹ (0.063%) and ⁴⁰Ar at 39.9624 g mol⁻¹ (99.600%).
0.00337 × 35.96755 = 0.1212 0.00063 × 37.96272 = 0.0239 0.99600 × 39.9624 = 39.8026 Sum = 39.948 g mol⁻¹
The abundances add to 100.000%, which is the first thing to check.
How do you find abundance from the average?
With two isotopes, let x be the fraction of one; the other is 1 − x.
Example 3: boron. Take the atomic mass as 10.81. The NIST boron data give ¹⁰B at 10.01294 and ¹¹B at 11.00931.
10.01294x + 11.00931(1 − x) = 10.81 11.00931 − 0.99637x = 10.81 x = 0.19931 ÷ 0.99637 = 0.200
So boron is about 20.0% ¹⁰B and 80.0% ¹¹B. NIST lists the representative ¹⁰B abundance as 0.199, so the answer is sensible.
Shortcut check. The average always lies between the isotope masses, closer to the more abundant one. Boron at 10.81 is closer to 11, so ¹¹B must be the majority.
Example 4: magnesium, three isotopes. The NIST magnesium data give ²⁴Mg 23.98504 (0.7899), ²⁵Mg 24.98584 (0.1000) and ²⁶Mg 25.98259 (0.1101).
18.9458 + 2.4986 + 2.8607 = 24.305
With three isotopes and only the average given, you cannot find all three abundances; a problem must give you one more piece of information, such as the ratio of two of them.
Why is chlorine's atomic weight given as a range?
The Commission on Isotopic Abundances and Atomic Weights gives chlorine's standard atomic weight as the interval [35.446, 35.457], in use since 2009. It explains that measurements since 1985 showed the isotope ratio of chlorine varies in natural and artificial materials. Environmental chloride samples alone range from about 35.450 to 35.455. One sample's average depends on where it came from, so a single "true" value would mislead. For exam work, use the value the question gives, usually 35.5.
Common mistakes
Using percentages as if they were fractions: 75.76 × 34.97 gives a number in the thousands.
Using mass numbers when the question gives exact isotopic masses, or the reverse.
Abundances that do not add to 100%.
Setting up a reverse problem with x for both isotopes.
Rounding too early. Keep four or five significant figures until the end.
For the mole calculations that use these masses, see our guide to mole concept errors and a checking routine. If you can set up the equation but keep slipping in the algebra, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning.
A practice routine
Redo the chlorine example with mass numbers and with exact masses, and note the difference.
Copper is listed as 63.546 with isotopes near 62.930 and 64.928. Find the fraction of ⁶³Cu. You should get about 0.69.
Make up a three-isotope problem with one abundance missing and solve it.
For each answer, apply the two checks in the callout before moving on.
Today, take the five most common elements in your syllabus and check that their table values sit nearer the isotope you expect to be most common.
