Interpreting Mass Spectra (Edexcel International A Level Chemistry)

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Mass Spectroscopy - Interpretation

  • Mass spectroscopy is an analytical technique used to identify unknown compounds
  • The molecules in the small sample are bombarded with high energy electrons which can cause the molecule to lose an electron
  • This results in the formation of a positively charged molecular ion with one unpaired electron
    • One of the electrons in the pair has been removed by the beam of electrons

Interpreting Mass Spectra equation 1

  • The molecular ion can further fragment to form new ions, molecules, and radicals 

Analytical Techniques Fragmentation, downloadable AS & A Level Chemistry revision notes

Fragmentation of a molecule in mass spectroscopy

 

  • These fragmented ions are accelerated by an electric field
  • Based on their mass (m) to charge (z) ratio, the ion fragments are then separated by deflecting them into the detector
    • Most ions will only gain a charge of 1+ and therefore a ion with mass 12 and charge 1+ will have an m/z value of 12
    • It is, however, possible for a greater charge to occur. For example, an ion with mass 16 and charge 2+ will have a m/z value of 8

  • The smaller and more positively charged fragment ions will be detected first as they will get deflected the most and are more attracted to the negative pole of the magnet
  • Each fragment corresponds to a specific peak with a particular m/z value in the mass spectrum
  • The base peak is the peak corresponding to the most abundant ion
  • The m/z is sometimes referred to as the m/e ratio and it is almost always 1:1

Isotopes

  • Isotopes are different atoms of the same element that contain the same number of protons and electrons but a different number of neutrons.
    • These are atoms of the same elements but with different mass number
    • For example, Cl-35 and Cl-37 are isotopes as they are both atoms of the same element (chlorine, Cl) but have a different mass number (35 and 37 respectively)

  • Mass spectroscopy can be used to find the relative abundance of the isotopes experimentally
  • The relative abundance of an isotope is the proportion of one particular isotope in a mixture of isotopes found in nature
    • For example, the relative abundance of Cl-35 and Cl-37 is 75% and 25% respectively
    • This means that in nature, 75% of the chlorine atoms is the Cl-35 isotope and 25% is the Cl-37 isotope

  • The heights of the peaks in mass spectroscopy show the proportion of each isotope present

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

The peak heights show the relative abundance of the boron isotopes: boron-10 has a relative abundance of 19.9% and boron-11 has a relative abundance of 80.1%

Worked example

Calculating m/z ratio

In a sample of iron, the ions 54Fe2+ and 56Fe3+ are detected. Calculate their m/z ratio and determine which ion is deflected more inside the spectrometer.

Answer

    • 56Fe3+ has a smaller m/z ratio and will therefore be deflected more.
    • It also has the largest positive charge and will be more attracted to the negative pole of the magnet within the mass spectrometer.

Examiner Tip

A small m/z value corresponds to fragments that are either small or have a high positive charge or a combination of both.

Deducing molecular formulae

  • Each peak in the mass spectrum corresponds to a certain fragment with a particular m/z value
  • The peak with the highest m/z value is the molecular ion (M+) peak which gives information about the molecular mass of the compound
  • The molecular ion is the entire molecule that has lost one electron when bombarded with a beam of electrons

Interpreting Mass Spectra equation 1

  • The [M+1] peak is a smaller peak which is due to the natural abundance of the isotope carbon-13
  • The amount of naturally occurring C-13 is a little over 1%, so the [M+1] peak is very small
  • 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

Analysing mass spectra

Determine whether the following mass spectrum corresponds to but-1-ene or pent-1-ene:Analytical Techniques Spec 1_Mass Spectrometry, downloadable AS & A Level Chemistry revision notes

Answer

    • The mass spectrum corresponds to pent-1-ene as the molecular ion peak is at m/z = 70
      • The small peak at m/z = 71 is a C-13 peak, which does not count as the molecular ion peak
    • But-1-ene arises from the C4H8+ ion which has a molecular mass of 56
    • Pent-1-ene arises from the C5H10+ ion which has a molecular mass of 70

Mass Spectrum - Fragmentation

  • 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 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 Table 1_Identifying Molecules using Fragmentation, downloadable AS & A Level Chemistry revision notes

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

Mass spectrum showing the fragmentation of C10H22

Halogenoalkanes

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

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

Mass spectrum showing different isotopes of bromine in the molecular ion

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

Analytical Techniques Alcohols Spectrum_2, downloadable AS & A Level Chemistry revision notes

Mass spectrum showing the fragmentation patterns in propan-1-ol (alcohol)

Worked example

Ion fragmentation

Which of the following statements about the mass spectrum of CH3Br is correct?

A. There is one peak for the molecular ion with an m/e value of 44

B. There is one peak for the molecular ion with an m/e value of 95

C. The last two peaks have abundances in the ratio 3:1 and occur at m/e values of 94 and 96

D. The last two peaks are of equal size and occur at m/e values of 94 and 96

Answer

The correct answer is option D

    • Bromomethane (CH3Br) can produce 3 peaks
      • CH381Br → [CH381Br]+ + e at m/e 96
      • CH379Br → [CH379Br]+ + e at m/e 94
      • CH3Br → [CH3]+ + Br at m/e 15

    • The last two peaks (which correspond to the molecular ion peak) therefore are equal in size and occur at m/e values of 94 and 96

Analytical Techniques Answer Worked example - Ion fragmentation, downloadable AS & A Level Chemistry revision notes

Worked example

Alcohol fragmentation

Which alcohol is not likely to have a fragment ion at m/e at 43 in its mass spectrum?

A. (CH3)2CHCH2OH

B. CH3CH(OH)CH2CH2CH3

C. CH3CH2CH2CH2OH

D. CH3CH2CH(OH)CH3

Answer

The correct answer is option D

    • Because a line at m/e = 43 corresponds to an ion with a mass of 43 for example:
      • [CH3CH2CH2]+
      • [(CH3)2CH]+

    • 2-butanol is not likely to have a fragment at m/e = 43 as it does not have either of these fragments in its structure.

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Richard

Author: Richard

Expertise: Chemistry

Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.