S1.2.3 Mass Spectrometry (Higher Level Only)

Interpretation and Applications of Mass Spectra

1. You’re handed an unknown element sample and asked to identify it.
2. What steps would you take? How would you handle the challenge if the element has multiple isotopes?

This is where mass spectrometry becomes an invaluable tool.

Using Mass Spectra to Identify Isotopes and Their Relative Abundances

What Does a Mass Spectrum Display?

• In most cases, the charge (z) is +1, meaning the m/z value directly corresponds to the ion's mass.
• Each peak on the spectrum represents a specific isotope or molecular fragment.

Identifying Isotopes

• For elements, each peak corresponds to an isotope.
• The m/z value of the peak indicates the isotope’s mass number, while the peak height(or intensity) reflects its relative abundance.

Determining Relative Abundance

To calculate the relative abundance of isotopes:

1. Measure the height (or area) of each peak.
2. Divide each peak’s height by the total height of all peaks.
3. Multiply by 100 to express the result as a percentage.

Interpreting a Mass Spectrum Step-by-Step

To interpret a mass spectrum effectively:

1. Identify the molecular ion peak $M^{+}$: This peak gives the molecular mass of the compound.
2. Look for isotopic peaks: These highlight the presence of elements with multiple isotopes.
3. Analyze fragmentation patterns: Use the m/z values of the fragments to deduce the molecule’s structure.
4. Calculate relative atomic mass: For elements, use the isotopic abundances to calculate the weighted average.

Calculating the Relative Atomic Mass of a Diatomic Element from a Mass Spectrum

To calculate the relative atomic mass of a diatomic element like chlorine ($C{l}_2$) or bromine ($B{r}_2$) using a mass spectrum:

1. Identify the Isotopes Present:
   • Chlorine: ${}^{35}Cl$ (75% abundance) and ${}^{37}Cl$ (25% abundance).
   • Bromine: ${}^{79}Br$ (50%) and ${}^{81}Br$ (50%).
2. Determine All Possible Molecular Combinations:
   For $C{l}_2$:
   • ${}^{35}Cl{-}^{35}Cl$
   • ${}^{35}Cl{-}^{37}Cl$
   • ${}^{37}Cl{-}^{37}Cl$
3. Assign Masses to Each Combination:
   • ${}^{35}Cl{-}^{35}Cl$: Mass = $35+35=70$
   • ${}^{35}Cl{-}^{37}Cl$: Mass = $35+37=72$
   • ${}^{37}Cl{-}^{37}Cl$: Mass = $37+37=74$
4. Calculate the Relative Abundances Using Probabilities:
   • ${}^{35}Cl{-}^{35}Cl$: $0.75\times 0.75=0.5625$
   • ${}^{35}Cl{-}^{37}Cl$: $0.75\times 0.25\times 2=0.375$ (multiplied by 2 for both possible pairings)
   • ${}^{37}Cl{-}^{37}Cl$: $0.25\times 0.25=0.0625$
5. Normalize Abundances (Converting to Percentages):
   • ${}^{35}Cl{-}^{35}Cl$: $56.25$
   • ${}^{35}Cl{-}^{37}Cl$: $37.5$
   • ${}^{37}Cl{-}^{37}Cl$: $6.25$
6. Interpreting the Mass Spectrum:
   Peaks in the mass spectrum will appear at masses 70, 72, and 74 with heights proportional to the calculated percentages.

Limitations and Practical Considerations

While mass spectrometry is a powerful analytical method, it has some limitations:

• Operational complexity: The instrument requires regular calibration and maintenance.
• Fragmentation variability: Molecules may fragment unpredictably, making interpretation challenging.
• Isobaric interference: Species with the same m/z value (e.g., ${}^{40}Ar$ and ${}^{40}Ca$ can overlap, requiring additional techniques for distinction.

Reflection