AQA · A-Level · Chemistry
Course version: 7405 · Version from Sept 2016
Students should be able to: interpret simple mass spectra of elements…
Full specification wording
Mass number and isotopes: Students should be able to: interpret simple mass spectra of elements calculate relative atomic mass from isotopic abundance, limited to mononuclear ions.
Reading a spectrum of an element means listing each isotopic mass with its abundance, then taking an abundance-weighted mean.
Checked against the AQA 7405 specification (opens in a new tab) on .
- Point 7 of 7
Study notes
From the checked teaching notes for this specification point.
The idea
Reading a spectrum of an element means listing each isotopic mass with its abundance, then taking an abundance-weighted mean. Multiply each relative isotopic mass by its abundance, add the products, and divide by the total abundance, which is 100 only when the data are percentages. The same expression solves in reverse: if the relative atomic mass and one isotopic mass are known, the unknown mass can be found. For rhenium with peaks at 185 and an unknown mass in abundances 10 and 17, setting 186.3 equal to (185 × 10 + x × 17) ÷ 27 gives 187.1. Only mononuclear singly charged ions are examined, so each peak corresponds to one atom.
What earns the marks
- A correct weighted-mean expression with each isotopic mass multiplied by its abundance.
- Dividing by the total abundance rather than by the number of isotopes.
- The correct value of the relative atomic mass or of the unknown isotopic mass.
- An answer quoted to an appropriate number of significant figures rather than rounded to a whole number.
Common mistakes and how to put them right
Dividing by the number of isotopes instead of by the sum of the abundances.
Assuming the abundances add to 100 when they are relative rather than percentage values.
Rounding the relative atomic mass to a whole number so it no longer matches the Periodic Table value.
Multiplying abundance by atomic number rather than by relative isotopic mass.
Exam tips
- Write the full fraction before substituting; the expression itself carries a mark even if the arithmetic fails.
- Add the abundances first and write the total down so you cannot forget to divide by it.
- Sense-check the answer: it must lie between the smallest and largest isotopic masses, nearer the most abundant.
Try the lesson
The start of the Mass number and isotopes: Students should be able to: interpret simple mass spectra of elements calculate relative atomic mass from isotopic abundance, limited to mononuclear ions. lesson, made from checked teaching notes for this specification point.
Have a go first
A simple mass spectrum of an element shows two peaks, at m/z = 10 and m/z = 11. What do these peaks represent?
Listen
Reading a spectrum of an element means listing each isotopic mass with its abundance, then taking an abundance-weighted mean. Multiply each relative isotopic mass by its abundance, add the products, and divide by the total abundance, which is 100 only when the data are percentages. The same expression solves in reverse: if the relative atomic mass and one isotopic mass are known, the unknown mass can be found. For rhenium with peaks at 185 and an unknown mass in abundances 10 and 17, setting 186.3 equal to (185 × 10 + x × 17) ÷ 27 gives 187.1. Only mononuclear singly charged ions are examined, so each peak corresponds to one atom. A strong answer earns marks for these points. a correct weighted-mean expression with each isotopic mass multiplied by its abundance. dividing by the total abundance rather than by the number of isotopes. the correct value of the relative atomic mass or of the unknown isotopic mass. an answer quoted to an appropriate number of significant figures rather than rounded to a whole number. Exam tip: Write the full fraction before substituting; the expression itself carries a mark even if the arithmetic fails.
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