Working Backwards: Finding Isotope Abundances from Average Atomic Mass

Calculating relative atomic mass from isotope abundances is a straightforward weighted average. Exams often flip the problem around: “The relative atomic mass of an element is 63.55, and it has two isotopes, 63 and 65. What percentage of each is present?” This needs a little algebra, but the method is always the same. This article sets it out step by step, with several worked examples and a way to check every answer.

If you haven’t met the forward calculation yet, start with calculating relative atomic mass from isotope abundances.

The key idea

For an element with two isotopes, the abundances must add up to 100% (or to 1 as fractions). So if one isotope’s abundance is x, the other’s must be 100 − x.

Put these into the relative atomic mass formula:

Aᵣ = [m₁ × x + m₂ × (100 − x)] ÷ 100

Then solve for x.

Method: step by step

1. Let the abundance of the first isotope be x %.
2. Write the abundance of the second isotope as (100 − x) %.
3. Set up the weighted average equation equal to the given Aᵣ.
4. Multiply both sides by 100 to clear the fraction.
5. Expand the bracket, collect the x terms and solve.
6. Find the second abundance: 100 − x.
7. Check by substituting back into the forward calculation.

Worked example 1: copper

Copper has Aᵣ = 63.55 and two isotopes, copper-63 and copper-65. Find the percentage abundance of each.

Let the abundance of Cu-63 = x %. Then Cu-65 = (100 − x) %.

63.55 = [63x + 65(100 − x)] ÷ 100

Multiply by 100: 6355 = 63x + 6500 − 65x

Collect terms: 6355 − 6500 = 63x − 65x −145 = −2x x = 72.5

So Cu-63 is 72.5% and Cu-65 is 27.5%.

Check: (63 × 72.5 + 65 × 27.5) ÷ 100 = (4567.5 + 1787.5) ÷ 100 = 63.55 ✓

(The measured natural abundances are 69.15% and 30.85%; the difference comes from using mass numbers instead of exact isotopic masses in this simplified calculation. Exam questions are designed with whatever values they give you.)

Worked example 2: chlorine

Chlorine has Aᵣ = 35.5 with isotopes of mass 35 and 37.

35.5 = [35x + 37(100 − x)] ÷ 100 3550 = 35x + 3700 − 37x −150 = −2x x = 75

Cl-35: 75%; Cl-37: 25%. So chlorine atoms are in roughly a 3 : 1 ratio, which is why the M and M+2 peaks in the mass spectrum of a chlorine compound have heights in about a 3 : 1 ratio. See mass spectrometry.

Worked example 3: using fractions

Some students prefer fractions. Let the fraction of the lighter isotope be y; the heavier is 1 − y.

Boron has Aᵣ = 10.81 with isotopes of mass 10 and 11.

10.81 = 10y + 11(1 − y) 10.81 = 10y + 11 − 11y 10.81 − 11 = −y y = 0.19

B-10: 19%; B-11: 81%.

Check: 10 × 0.19 + 11 × 0.81 = 1.9 + 8.91 = 10.81 ✓

A shortcut for two isotopes

For two isotopes, there’s a neat shortcut. The abundance of the heavier isotope equals:

(Aᵣ − lighter mass) ÷ (heavier mass − lighter mass) × 100%

For copper: (63.55 − 63) ÷ (65 − 63) × 100 = 0.55 ÷ 2 × 100 = 27.5% Cu-65 ✓

This works because Aᵣ sits between the two masses, and its position along that gap reflects the proportions. If Aᵣ were exactly halfway, the abundances would be 50 : 50. It’s a good way to check the algebra, but show the full method in exams unless told otherwise.

Worked example 4: gallium

Gallium has Aᵣ = 69.72, with isotopes ⁶⁹Ga and ⁷¹Ga.

Shortcut for ⁷¹Ga: (69.72 − 69) ÷ (71 − 69) × 100 = 0.72 ÷ 2 × 100 = 36% So ⁶⁹Ga = 64%.

Algebra check: 69.72 = [69x + 71(100 − x)] ÷ 100 6972 = 69x + 7100 − 71x −128 = −2x x = 64 ✓

Three isotopes: when one abundance is given

With three isotopes, there are two unknowns, so the question must give you one abundance (or another relationship).

Worked example 5: magnesium

Magnesium has Aᵣ = 24.31. Its isotopes are ²⁴Mg, ²⁵Mg and ²⁶Mg. The abundance of ²⁵Mg is 10.0%. Find the other two abundances.

Let ²⁴Mg = x %. Then ²⁶Mg = 100 − 10.0 − x = (90.0 − x) %.

24.31 = [24x + 25(10.0) + 26(90.0 − x)] ÷ 100 2431 = 24x + 250 + 2340 − 26x 2431 = 2590 − 2x 2x = 159 x = 79.5

²⁴Mg: 79.5%; ²⁵Mg: 10.0%; ²⁶Mg: 90.0 − 79.5 = 10.5%

Check: (24 × 79.5 + 25 × 10.0 + 26 × 10.5) ÷ 100 = (1908 + 250 + 273) ÷ 100 = 24.31 ✓

Worked example 6: a ratio is given

An element has Aᵣ = 28.11 and three isotopes of mass 28, 29 and 30. The abundances of the 29 and 30 isotopes are in the ratio 3 : 2. Find all three abundances.

Let the 29 isotope = 3k % and the 30 isotope = 2k %. Then the 28 isotope = (100 − 5k) %.

28.11 = [28(100 − 5k) + 29(3k) + 30(2k)] ÷ 100 2811 = 2800 − 140k + 87k + 60k 2811 = 2800 + 7k k = 11 ÷ 7 = 1.571

• mass 29: 3k = 4.71%
• mass 30: 2k = 3.14%
• mass 28: 100 − 5k = 92.14%

This is close to natural silicon (92.2%, 4.7%, 3.1%).

Checking your answer: three quick tests

1. Abundances add up to 100%.
2. The more abundant isotope should be the one Aᵣ is closer to. Copper’s 63.55 is nearer 63 than 65, so Cu-63 must be more abundant.
3. Substitute back into the forward formula; you should get the given Aᵣ.

If any test fails, recheck the signs when expanding brackets: this is where most errors happen.

Why this skill matters beyond exams

Working backwards from an average is exactly what scientists do when they interpret isotope data. Geologists use changes in the average mass of elements such as strontium or lead to identify where rocks and minerals formed; food scientists use shifts in carbon and oxygen isotope ratios to check whether honey, wine or fruit juice is what the label claims. The algebra is the same as in these examples, just with more precise masses.

Common mistakes

• Writing the second abundance as x again instead of (100 − x).
• Forgetting to multiply the whole of Aᵣ by 100 (or forgetting to divide by 100).
• Sign errors when expanding −65(100 − x) or collecting terms.
• Giving the abundance of the wrong isotope as the final answer; label clearly which isotope is x.
• Rounding intermediate values too early in three-isotope problems.

Practice questions

1. Antimony has Aᵣ = 121.8 and isotopes of mass 121 and 123. Find the abundance of each.
2. Lithium has Aᵣ = 6.94 and isotopes of mass 6 and 7. Find the abundance of each.
3. Rubidium has Aᵣ = 85.47 with isotopes of mass 85 and 87. Find the abundance of each.

Answers:

1. Sb-123 = (121.8 − 121) ÷ 2 × 100 = 40%; Sb-121 = 60%
2. Li-7 = (6.94 − 6) ÷ 1 × 100 = 94%; Li-6 = 6%
3. Rb-87 = (85.47 − 85) ÷ 2 × 100 = 23.5%; Rb-85 = 76.5%

Key takeaways

• For two isotopes, let one abundance be x and the other (100 − x), then solve the weighted average equation.
• For three isotopes, you need one abundance or a ratio to reduce to one unknown.
• The shortcut (Aᵣ − lighter) ÷ (heavier − lighter) gives the heavier isotope’s fraction.
• Always check by recalculating Aᵣ, and make sure the more abundant isotope is the one Aᵣ lies closer to.
• Small differences from real abundances come from using mass numbers instead of exact isotopic masses. See how to read and write isotope notation.