Isotopes: Count Particles and Calculate Relative Atomic Mass

Isotope questions combine two different kinds of numbers. A mass number counts particles in one nucleus. A percentage abundance describes how common an isotope is in a sample. Relative atomic mass is an average that depends on both isotope masses and their abundances. Keeping those roles separate makes the calculations much easier.

This lesson uses original relative atomic mass isotope calculations, including explicitly fictional datasets, plus a real chlorine illustration checked against NIST. Letters used for fictional elements are exercise labels, not proposed chemical symbols or claims about natural samples.

Read the notation before calculating

For a nuclide written as ^A_ZX, Z is the atomic number: the number of protons. A is the mass number: protons plus neutrons. Therefore neutrons = A − Z. A neutral atom has as many electrons as protons.

Isotopes have the same atomic number but different mass numbers, because their neutron counts differ. Ions differ in electron count from the neutral atom. Losing two electrons produces a 2+ ion; gaining one produces a 1− ion. Forming an ordinary ion does not change the nucleus or turn the atom into a different isotope.

Worked example 1: complete a particle-count table

Use the stated atomic and mass numbers for these real nuclides. Work out the nucleus first, then adjust the electron count for charge.

For chlorine-37, neutron count = 37 − 17 = 20. Its negative charge requires one additional electron, giving 18. For magnesium-24, neutron count = 24 − 12 = 12, and a 2+ charge means two electrons have been lost: 12 − 2 = 10.

The two chlorine entries refer to different isotopes, even though the listed species also have different charges. Changing chlorine-37 from a neutral atom to chloride would alter its electron count but leave it chlorine-37. Check charge by subtracting electrons from protons: 17 − 18 = −1 and 12 − 10 = +2.

Mass number, isotopic mass and the weighted mean

A mass number is an integer particle count. A measured isotopic mass is not generally an exact integer number of unified atomic mass units, u. In school calculations, a question may tell you to use mass numbers as approximate isotope masses. Follow that stated approximation rather than silently replacing more precise masses.

Relative atomic mass, A_r, compares the average atomic mass with one twelfth of the mass of a carbon-12 atom. It is a ratio and has no unit. An average atomic mass expressed in u has the same numerical value, but the quantity and its units should be described correctly.

Weighted mean = sum of (relative isotope mass × fractional abundance). Fractions must total 1. For percentage abundances, multiply each mass by its percentage, add the products and divide by 100. For a count or ratio, divide by the total count or total ratio parts instead.

Worked example 2: a fictional two-isotope sample

Fictional element J has two isotopes. In this constructed sample, use approximate relative isotope masses 46 and 48, with abundances 25% and 75% respectively. These values are a simplified calculation model, not measured natural isotope data.

A_r = [(46 × 25) + (48 × 75)] ÷ 100 = (1,150 + 3,600) ÷ 100 = 47.50. Alternatively, 46 × 0.25 + 48 × 0.75 = 11.50 + 36.00 = 47.50.

The answer lies between 46 and 48 and is closer to 48, the more abundant isotope. An ordinary mean, (46 + 48) ÷ 2 = 47, incorrectly gives the isotopes equal weight. It would be appropriate only for a 50:50 mixture.

A useful mental picture is 100 atoms: 25 contribute the lower approximate mass and 75 the higher one. The average does not imply that any individual J atom has mass number 47.5. Averaging a mixture does not create a new isotope with a fractional neutron count.

Worked example 3: reverse the mean to find abundance

A second fictional element, L, is modelled with approximate relative isotope masses 70 and 74. A constructed sample has A_r = 71.20 and contains only those two isotopes. Find their percentage abundances.

1. Let x be the fraction of the mass-70 isotope. The other fraction is 1 − x.
2. Write the mean: 70x + 74(1 − x) = 71.20.
3. Expand: 70x + 74 − 74x = 71.20.
4. Rearrange: 4x = 2.80, so x = 0.70.
5. Convert fractions to percentages: mass-70 isotope 70%; mass-74 isotope 30%.

Substitute back: 70 × 0.70 + 74 × 0.30 = 49 + 22.20 = 71.20. The sample is dominated by the lighter isotope, consistent with a mean nearer 70.

A mean of 75 would be impossible for a mixture containing only these two specified masses. A calculated fraction greater than 1 or below 0 is a warning to check the algebra, the values or the assumption that there are only two isotopes. Never “fix” an impossible result by removing its sign.

Worked example 4: a sourced chlorine illustration

NIST lists the following relative isotope masses and representative isotopic composition for chlorine. The table shows the central values; the source also reports uncertainties. Use them to illustrate the weighted-mean method, not to claim every chlorine sample has exactly this composition.

Calculated A_r = 34.968852682 × 0.7576 + 36.965902602 × 0.2424 = 35.452937582608, or approximately 35.45. The extra digits show the arithmetic before rounding; they are not a claim of equivalent experimental precision.

Using mass numbers 35 and 37 instead would give 35 × 0.7576 + 37 × 0.2424 = 35.4848. That difference illustrates why a mass number is not identical to a measured relative isotope mass. NIST also gives a standard atomic-weight interval of [35.446, 35.457], reflecting the need to distinguish a representative calculation from a universal exact value.

A reliable checking routine

Label every column before using a calculator. Are the abundances percentages, fractions or ratio parts? Percentages near 100 in total should not be used as fractions without dividing by 100. Conversely, fractions already summing to 1 should not be divided by 100 again.

Keep intermediate values unrounded where possible. Then check that the final average is inside the range of the supplied isotope masses. For a two-isotope mixture, it should lie nearer the more abundant isotope. If a table lists only some isotopes, do not assume the missing percentage is zero unless the question makes that assumption explicit.

Three practice tasks with solved answers

Practice 1: isotope or ion?

Calcium-42 has atomic number 20. Find protons, neutrons and electrons in its 2+ ion. Does losing the two electrons change its mass number?

Answer: Protons = 20; neutrons = 42 − 20 = 22; electrons = 20 − 2 = 18. The mass number remains 42 because the proton and neutron counts are unchanged.

Practice 2: a three-isotope average

A fictional sample uses approximate relative masses 50, 52 and 54 with abundances 20%, 50% and 30%. Calculate its relative atomic mass.

Answer: 50 × 0.20 + 52 × 0.50 + 54 × 0.30 = 10 + 26 + 16.2 = 52.2. The fractions total 1, and the answer lies between 50 and 54.

Practice 3: reverse another mean

A fictional sample contains only isotopes with approximate relative masses 90 and 94. Its mean is 93. Find the percentage of the lighter isotope.

Answer: Let its fraction be x: 90x + 94(1 − x) = 93. Thus 4x = 1 and x = 0.25, or 25%. Check: 90 × 0.25 + 94 × 0.75 = 22.5 + 70.5 = 93.

Sources and further reading

Scientific background was checked against these educational references. The examples, tables, diagrams and practice answers on this page are original learning material.

• OpenStax Chemistry 2e: atomic structure and symbolism
• NIST: atomic weights and isotopic compositions for chlorine
• IUPAC Gold Book: relative atomic mass

Continue your revision

You can complete every question above without signing in. For a wider resource collection, visit the KCSE Chemistry topic-tests page. Its resource content currently requires login; check membership access and the listed contents before joining. This lesson does not promise a particular downloadable file.