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Relative Atomic Mass, Isotopic Abundance and Formula Mass

UNIT 2 CHEMISTRY · PEARSON AAQ · A1.1.2

Relative Atomic Mass, Isotopic Abundance and Formula Mass

FREE CHEMISTRY REVISION · TOPIC 2

Different isotopes. One weighted mean.

Calculate relative atomic mass, work backwards to isotopic abundances and build relative formula masses.

New Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · A1.1.2

What you will learn

  • Define relative atomic mass and distinguish it from mass number.
  • Calculate a weighted mean using percentage, fractional or ratio abundances.
  • Use relative atomic mass to determine unknown isotopic abundances.
  • Use relative atomic masses to calculate relative formula masses, including brackets.

Complete coverage of Unit 2 A1.1.2: both directions of the isotope-abundance calculation and relative formula mass. Verified against Pearson’s Extended Certificate specification, Issue 7 (May 2026), printed page 36. Isotope particle counts are covered in the previous topic.

1. What does relative atomic mass mean?

Relative atomic mass, Ar, is the weighted mean mass of an atom of an element compared with one-twelfth of the mass of a carbon-12 atom. The mean takes account of the relative numbers of each isotope in the sample. It is a ratio, so Ar has no units.

Relative isotopic mass compares the mass of an atom of a particular isotope with the same carbon-12 reference. Carbon-12 has relative isotopic mass exactly 12. Other isotopic masses are close to, but generally not exactly equal to, their mass numbers.

Mass number, A

Number of protons + neutrons in one nucleus. Always a whole number. Chlorine-35 has mass number 35.

Relative atomic mass, Ar

A weighted average for the element in a sample. Often a decimal because the sample contains a mixture of isotopes.

Relative abundance tells you the proportion of atoms that belong to each isotope. It can be expressed as a percentage, a fraction or a number ratio. These describe numbers of atoms, not percentages by mass. Isotopic composition can vary between samples, so use the data in the question.

2. Visualise a weighted mean

Imagine a simplified chlorine sample with three chlorine-35 atoms for every one chlorine-37 atom. That is 75% chlorine-35 and 25% chlorine-37. The more abundant isotope contributes more to the mean.

Three chlorine-35 atoms for each chlorine-37 atomFour circles represent three atoms of chlorine-35 and one atom of chlorine-37. A line from relative mass 35 to 37 marks the weighted mean at 35.5, one quarter of the way from 35.³⁵Cl³⁵Cl³⁵Cl³⁷Cl3 of 4 atoms: 75%1 of 4: 25%3537Mean 35.5
Illustrative 3:1 number ratio using mass numbers as approximate isotopic masses. The mean is closer to 35 because chlorine-35 is more abundant. Real isotopic data need not give exactly this value.

Ar = sum of (isotopic mass × abundance) ÷ sum of abundances

With percentages, divide by 100. With fractions totalling 1, divide by 1. With a ratio such as 3:1, divide by 4. Use relative isotopic masses when supplied. If only mass numbers are given, use them as the question’s approximation.

3. Calculate Ar from abundances

Worked example 1 · Two isotopes

A simplified chlorine sample contains 75.0% chlorine-35 and 25.0% chlorine-37. Use mass numbers as approximate isotopic masses.

  1. Check the total: 75.0 + 25.0 = 100.0%.
  2. Multiply each mass by its abundance: 35 × 75.0 = 2625; 37 × 25.0 = 925.
  3. Add the contributions: 2625 + 925 = 3550.
  4. Divide by 100: Ar = 3550 ÷ 100 = 35.5.
  5. Check: 35 < 35.5 < 37, and 35.5 is closer to the more abundant isotope, 35.

Worked example 2 · Three isotopes

A model sample has isotopes with relative isotopic masses 24.0, 25.0 and 26.0, and number abundances 80.0%, 10.0% and 10.0% respectively.

  1. Total abundance = 80.0 + 10.0 + 10.0 = 100.0%.
  2. Weighted contributions: 24.0 × 80.0 = 1920; 25.0 × 10.0 = 250; 26.0 × 10.0 = 260.
  3. Ar = (1920 + 250 + 260) ÷ 100 = 24.3.
  4. The mean lies between 24.0 and 26.0 and is near 24.0, the dominant isotope.

Worked example 3 · Ratios instead of percentages

Isotopes with relative isotopic masses 20.0 and 22.0 occur in a 9:1 number ratio.

Ar = [(20.0 × 9) + (22.0 × 1)] ÷ (9 + 1)
= 202 ÷ 10 = 20.2.

Equivalently, use fractions 0.90 and 0.10: (20.0 × 0.90) + (22.0 × 0.10) = 20.2. Do not divide by 100 again when using fractions.

4. Work backwards to isotopic abundance

For an element with only two isotopes, knowing both isotopic masses and Ar lets you find their abundances. Let x be the fraction of the lighter isotope; the heavier fraction is 1 − x.

Ar = mlightx + mheavy(1 − x)

Rearranging gives x = (mheavy − Ar) ÷ (mheavy − mlight). Multiply x by 100 to express it as a percentage.

Worked example 4 · Find both percentages

An element has only two isotopes, with relative isotopic masses 10.0 and 11.0. Its Ar is 10.8. Find the percentage of each isotope.

  1. Let x = fraction of the lighter isotope. Then 1 − x = fraction of the heavier isotope.
  2. Write the weighted mean: 10.8 = 10.0x + 11.0(1 − x).
  3. Expand: 10.8 = 10x + 11 − 11x = 11 − x.
  4. Rearrange: x = 11 − 10.8 = 0.20.
  5. Lighter isotope = 0.20 × 100 = 20%. Heavier isotope = 80%.
  6. Substitute back: (10.0 × 0.20) + (11.0 × 0.80) = 2.0 + 8.8 = 10.8.

Percentage alternative: if x is a percentage, use 100 − x for the other percentage and divide the whole weighted sum by 100. Do not mix percentage and fractional forms.

More than two isotopes? Ar alone is not enough to find all three abundances. You need additional independent information, such as a known abundance or ratio. Always state what your unknown represents.

5. From atomic mass to formula mass

Relative formula mass, Mr, is the sum of the relative atomic masses of all atoms shown in a chemical formula. It has no units. For a molecular substance it is also called relative molecular mass. For ionic substances such as NaCl, use relative formula mass: the formula gives a ratio of ions, not a discrete molecule.

Mr = sum of (number of each atom × its Ar)

A subscript multiplies the preceding element; a subscript outside brackets multiplies every atom inside. A coefficient before a formula in an equation changes the number of particles, not the Mr of the substance.

Worked example 5 · Calcium hydroxide, Ca(OH)2

Use Ar: Ca = 40.1, O = 16.0, H = 1.0.

Count every atom before calculating
ElementNumberContribution
Ca11 × 40.1 = 40.1
O22 × 16.0 = 32.0
H22 × 1.0 = 2.0

Mr = 40.1 + 32.0 + 2.0 = 74.1.

Worked example 6 · Aluminium sulfate, Al2(SO4)3

Use Ar: Al = 27.0, S = 32.1, O = 16.0.

  1. Count atoms: 2 Al; 3 S; 3 × 4 = 12 O.
  2. Mr = (2 × 27.0) + 3[32.1 + (4 × 16.0)].
  3. = 54.0 + 3(96.1) = 54.0 + 288.3 = 342.3.

Extension · Water of crystallisation

CuSO4·5H2O includes five water molecules per formula unit. Using Cu = 63.5, S = 32.1, O = 16.0 and H = 1.0:

Mr = 63.5 + 32.1 + (4 × 16.0) + 5[(2 × 1.0) + 16.0]
= 159.6 + 90.0 = 249.6.

The dot separates the water component; the 5 multiplies the whole H2O. This is an application of formula-mass calculation, not an additional isotope rule.

Link to later mole calculations: Mr = 74.1 is dimensionless; the corresponding molar mass is 74.1 g mol−1. Keep the quantity and its units distinct.

6. Common misconceptions and exam tips

  • “Just average the two masses.” Only when the isotopes have equal number abundances. Otherwise calculate a weighted mean.
  • “Ar is a mass number.” It averages a sample; no single atom needs to have that mass number. Never round Ar to invent an isotope.
  • “Every abundance question uses ÷ 100.” Divide by the total abundance in the form supplied: 100 for percentages, 1 for fractions, or the sum of ratio parts.
  • “The heavier isotope must be more abundant.” Abundance is independent of which isotope is heavier. Compare the mean with both masses.
  • “The 2 in Ca(OH)2 only multiplies H.” It multiplies both O and H inside the brackets.
  • “Mr is measured in grams.” Ar and Mr are ratios with no units. Molar mass uses g mol−1.

Exam method: show products, total and denominator. For reverse calculations, define the unknown, show the equation and check the percentages total 100%. Keep calculator digits until the final step, follow the question’s rounding instruction and use its supplied atomic masses.

7. Check your understanding

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Question 1 of 5 · Unit 2 A1.1.2

Which statement correctly defines relative atomic mass, Aᵣ?

Question 2 of 5 · Unit 2 A1.1.2

A sample has 20% isotope X-10 and 80% isotope X-11. Using these mass numbers as approximate isotopic masses, what is Aᵣ?

Question 3 of 5 · Unit 2 A1.1.2

An element has only isotopes of relative isotopic masses 63.0 and 65.0. Its Aᵣ is 63.6. What percentage is the lighter isotope?

Question 4 of 5 · Unit 2 A1.1.2

Two isotopes of relative isotopic masses 24.0 and 26.0 occur in a 3:1 number ratio. What is Aᵣ?

Question 5 of 5 · Unit 2 A1.1.2

Using Aᵣ values Ca = 40.1, O = 16.0 and H = 1.0, what is Mᵣ of Ca(OH)₂?

Your revision summary

  • Ar is a weighted mean relative to one-twelfth of carbon-12.
  • Multiply each isotopic mass by its number abundance, add and divide by the total abundance.
  • For two isotopes, use x and 1 − x (fractions), or x and 100 − x (percentages).
  • Check the mean lies between the isotope masses and closer to the more abundant one.
  • Mr adds every atom’s Ar; expand brackets carefully.
  • Neither Ar nor Mr has units.

Without notes, explain why a mean close to the lighter isotope implies a higher abundance of that isotope.

Specification checked: Pearson BTEC National AAQ Applied Science (Extended Certificate), Issue 7, May 2026, Unit 2 A1.1.2, printed page 36. Read the official Pearson specification (PDF).

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