Lecture 4

Question 1

Give the influence of molecular formula on molecular ion cluster using C3H8O as an example

Answer 1

Every isotopologue of this compound has this formula – but the isotopologues vary
in their isotopic constitution. The Relative Molecular Mass of this formula
(sometimes called the “Formula Weight”) is 60.096 g mol−1, but recall that this is a
weighted average of all of the isotopologues; there are no individual molecules of
C3H8O with mass 60.096 amu.

Question 2

Give the influence of monoisotopic mass on molecular ion cluster

Answer 2

This is the accurately-determined mass of the lowest-mass isotopologue of C3H8O,
which contains only the lowest mass isotopes of each element: 12C3 1H8 16O

Question 3

Give the influence of relative intensity of M peak on molecular ion cluster

Answer 3

• This is the lowest-mass peak in the molecular ion cluster, which means that only the lowest-mass isotopologue of the molecular ion contributes to it: 12C3 1H8 16O
• In small molecules, depending on the formula, the M peak is often the tallest peak
  in the molecular ion cluster.
• The tallest peak in the entire mass spectrum (the “base peak”) is determined by
  which ion is most stable; this may or may not be the molecular ion.

Question 4

Give the influence of relative intensity of M+1 peak on molecular ion cluster

Answer 4

This peak is composed of all of the contributions of the isotopologues in which have mass ~61 amu: 12C2 13C1H8 16O, 12C3 1H7 2H16O, 12C3 1H8 17O etc. These isotopologues have different (but similar) masses (see p8)… so in this low-resolution mass spectrum, their m/z values are not resolved separately, and all of these isotopologues are lumped together on the peak at m/z 61.

Question 5

Give the influence of relative intensity of M+2 peak on molecular ion cluster

Answer 5

This peak is composed of all of the contributions of the isotopologues in which have mass ~62 amu. 12C13C2 1H8 16O, 12C2 13C1H7 2H16O, 12C3 1H8 18O etc. Again, the masses of each of these isotopologues are different to each other but not very, so all of these isotopologues are lumped together on the peak at m/z 62

Question 6

Method 1 for calculating peak intensities

Answer 6

We can use isotope abundance ratios to calculate relative peak intensities. This method works well enough for simple, small molecules, but becomes unwieldy with more complex examples.

Question 7

How can we improve the methods to become more accurate to real values for peak intensity?

Answer 7

1. we can get more accurate results if we use better approximations, e.g. for chlorine, using a 3.13:1 ratio
2. we can calculate and incorporate the contributions from more minor isotopes, more isotopologues

Question 8

Method 2 for calculating peak intensities

Answer 8

We can use the individual isotope abundance probabilities to calculate relative peak intensities. This method works for all examples, but this level of detail is excessive, given that some of the isotopologues are negligible.