Isotopes and Mass Spectrometry
This topic covers isotopes, which are atoms of the same element with different masses due to varying numbers of neutrons. You'll learn how to interpret mass spectra to identify an element's isotopes and their relative abundances.
Part of the ESAT Chemistry syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- Isotopes of an element have an identical number of protons but a different number of neutrons.
- This means isotopes share the same atomic number (Z) but have different mass numbers (A).
- A mass spectrometer separates ions based on their mass-to-charge (m/z) ratio, allowing for the identification and counting of different isotopes.
- In a mass spectrum, the x-axis shows the m/z ratio, which corresponds to the mass number for singly charged ions.
- The y-axis represents the relative abundance of each isotope; the height of each peak is proportional to the amount of that isotope present.
- The sum of the relative abundances of all isotopes is used as the total for calculating the percentage abundance of any single isotope.
Formulae
% Abundance (Isotope X) = (Abundance of X / Total Abundance of all isotopes) × 100 To convert the relative abundances from a mass spectrum into percentages for each isotope.
Definitions
- Isotopes
- Atoms of the same element (same number of protons) that have a different number of neutrons, resulting in different mass numbers.
- Mass Number (A)
- The total count of protons and neutrons in an atom's nucleus.
- Atomic Number (Z)
- The number of protons in an atom's nucleus, which uniquely defines the element.
Worked example
The mass spectrum of a sample of antimony (Sb) shows two peaks. The peak at m/z = 121 has a relative abundance of 135, and the peak at m/z = 123 has a relative abundance of 105. What is the percentage abundance of the antimony-121 isotope in the sample, to the nearest whole number?
- 1
Identify the relative abundance for each isotope:
Sb-121 has abundance 135, Sb-123 has abundance 105.
- 2
Calculate the total relative abundance by summing the individual values:
Total = 135 + 105 = 240 - 3
Set up the fraction for the percentage abundance of Sb-121:
(135 / 240) × 100.
- 4
Simplify the fraction.
Both numbers are divisible by 5:
(27 / 48).
Both are then divisible by 3:
(9 / 16).
- 5
Calculate the percentage:
(9 / 16) × 100.
This is (900 / 16).
Halve top and bottom repeatedly:
450/8 = 225/4 = 56.25% - 6
Round the result to the nearest whole number as requested:
56%.
Answer: 56%
Common mistakes
- ×Making arithmetic errors under pressure when summing peak heights or simplifying fractions.
- ×Confusing relative abundance with percentage. A peak at '100' on the y-axis is a reference point for the most abundant isotope, not its percentage, unless it's the only isotope present.
- ×Calculating the percentage of one isotope relative to another, instead of relative to the total sum of all isotopes.
No-calculator tips
- ✓When dealing with ratios, simplify them before doing any calculations. For abundances of 135 and 105, notice both end in 5, so divide by 5 to get 27:21. Then notice both are in the 3 times table, simplifying to 9:7.
- ✓Once you have a simple ratio like 9:7, the total 'parts' is 9 + 7 = 16. The percentage calculation is then (9/16) × 100, which is often easier to handle.
- ✓Memorise common fraction-to-percentage conversions, such as 1/8 = 12.5%, 1/4 = 25%, 3/4 = 75%, 1/5 = 20%, etc. This can speed up the final calculation.