The two numbers a question gives you
Relative atomic mass
The weighted mean mass of an element's atoms, taking the abundance of each isotope into account, compared with one-twelfth of the mass of a carbon-12 atom.
Relative isotopic mass
The mass of one isotope compared with one-twelfth of the mass of a carbon-12 atom, which is equal to that isotope's mass number.
Isotopic abundance
The proportion of the atoms in a sample of an element that are a particular isotope, given as a percentage or a fraction.
- A question supplies the mass of each isotope and how common that isotope is.
- An isotope's relative isotopic mass is the same number as its mass number.
- Abundance may be given as a percentage, as a decimal fraction, or as a count of atoms detected.
- The more abundant isotope pulls the average towards its own mass.
- Relative atomic mass has no unit, because it compares one mass with another.

- The answer always lies between the smallest and the largest isotopic mass.
- It lands nearer the mass of the most abundant isotope.
The method: multiply, add, then divide by the total abundance
- Write each isotopic mass and its abundance side by side before calculating anything.
- Multiply each isotopic mass by its own abundance.
- Add all of those products together.
- Divide by the total abundance: Ar=(m1×a1)+(m2×a2)a1+a2A_r = \frac{(m_1 \times a_1) + (m_2 \times a_2)}{a_1 + a_2}Ar=a1+a2(m1×a1)+(m2×a2)
- The denominator is 100100100 when the abundances are percentages that add to 100100100.
- Extend the same pattern for three or more isotopes by adding one product per isotope.
- Use the total abundance as the denominator even when the percentages do not add to exactly 100100100.
- Keep one format throughout, so use percentages everywhere or decimals everywhere.
Worked example: weighting three isotopes
- An element has isotopes of mass 242424, 252525 and 262626, with abundances of 79%79\%79%, 10%10\%10% and 11%11\%11%.
- The abundances add to 100100100, so the denominator is 100100100.
- Work out the three products separately before adding, so that a slip in one does not hide inside the total.
- Find the relative atomic mass of the element.
- Multiply each isotopic mass by its abundance:
- 24×79=189624 \times 79 = 189624×79=1896
- 25×10=25025 \times 10 = 25025×10=250
- 26×11=28626 \times 11 = 28626×11=286
- Add the three products:
- 1896+250+286=24321896 + 250 + 286 = 24321896+250+286=2432
- Divide by the total abundance:
- Ar=2432100=24.32A_r = \dfrac{2432}{100} = 24.32Ar=1002432=24.32
Checking your answer
- Your answer must lie between the lowest and the highest isotopic mass in the data.
- It must sit nearer the mass of the most abundant isotope.
- Do not round the intermediate products, because rounding early shifts the final value.
- Do not take a plain average of the masses, because the isotopes are not equally abundant.
- Give ArA_rAr with no unit, and to the precision the question asks for.
- What is an isotope?
- How does isotopic abundance affect the value of the relative atomic mass?
- Why do you divide by 100100100 when the percentage abundances add to 100100100?
- Why is relative atomic mass written without a unit?
- An element is 60%60\%60% mass 696969 and 40%40\%40% mass 717171; what is its relative atomic mass?