IB Chemistry SL

Revision Notes

11.1.2 Mass Spectrometry

Determining Molecular Mass

  • When a compound is analysed in a mass spectrometer, vaporised molecules are bombarded with a beam of high-speed electrons
  • These knock off an electron from some of the molecules, creating molecular ions:

Interpreting Mass Spectra equation 1

  • The relative abundances of the detected ions form a mass spectrum: a kind of molecular fingerprint that can be identified by computer using a spectral database
  • The peak with the highest m/e value is the molecular ion (M+) peak which gives information about the molecular mass of the compound
  • This value of m/z is equal to the relative molecular mass of the compound

The M+1 peak

  • The [M+1] peak is a smaller peak which is due to the natural abundance of the isotope carbon-13
  • The height of the [M+1] peak for a particular ion depends on how many carbon atoms are present in that molecule; The more carbon atoms, the larger the [M+1] peak is
    • For example, the height of the [M+1] peak for an hexane (containing six carbon atoms) ion will be greater than the height of the [M+1] peak of an ethane (containing two carbon atoms) ion

Worked Example

Determine whether the following mass spectrum belongs to propanal or butanal

Analytical Techniques Spec 1_Mass Spectrometry, downloadable AS & A Level Chemistry revision notes

Answer:

    • The mass spectrum corresponds to propanal as the molecular ion peak is at m/e = 58
    • Propanal arises from the CH3CH2CHO+ ion which has a molecular mass of 58
    • Butanal arises from the CH3CH2CH2CHO+ ion which has a molecular mass of 72

Fragmentation Patterns

  • The molecular ion peak can be used to identify the molecular mass of a compound
  • However, different compounds may have the same molecular mass
  • To further determine the structure of the unknown compound, fragmentation analysis is used
  • Fragments may appear due to the formation of characteristic fragments or the loss of small molecules
    • For example, a peak at 29 is due to the characteristic fragment C2H5+­­
    • Loss of small molecules give rise to peaks at 18 (H2O), 28 (CO), and 44 (CO2)

Alkanes

  • Simple alkanes are fragmented in mass spectroscopy by breaking the C-C bonds
  • M/e values of some of the common alkane fragments are given in the table below

m/e values of Fragments Table

Analytical Techniques - m_e values of fragments table, downloadable AS & A Level Chemistry revision notes

Analytical Techniques Alkane Spectrum, downloadable AS & A Level Chemistry revision notes

Mass spectrum showing fragmentation of alkanes

Halogenoalkanes

  • Halogenoalkanes have often multiple peaks around the molecular ion peak
  • This is caused by the fact that there are different isotopes of the halogens

Alcohols

  • Alcohols often tend to lose a water molecule giving rise to a peak at 18 below the molecular ion
  • Another common peak is found at m/e value 31 which corresponds to the CH2OH+­­ fragment
  • For example, the mass spectrum of propan-1-ol shows that the compound has fragmented in four different ways:
    • Loss of H to form a C3H7O+ fragment with m/e = 59
    • Loss of a water molecule to form a C3H6+ fragment with m/e = 42
    • Loss of a C2H5 to form a CH2OH+ fragment with m/e = 31
    • And the loss of CH2OH to form a C2H5+ fragment with m/e = 29

Exam Tip

A table of mass spectral fragments lost is included in the IB Chemistry Data Booklet Section 28 so you don’t need to learn all the likely fragments

Close

Join Save My Exams

Download all our Revision Notes as PDFs

Try a Free Sample of our revision notes as a printable PDF.

Join Now
Already a member?
Go to Top