IB DP Chemistry: SL复习笔记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:


  • 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 butanal4.1-Analytical-Techniques-Spec-1_Mass-Spectrometry


    • 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 CH3CH2CH2CHOion 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)


  • 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



Mass spectrum showing fragmentation of alkanes


  • 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 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