Mass Spectrometry (Edexcel AS Chemistry)

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Interpreting Mass Spectra

  • 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 numbers
  • Therefore, the mass of an element is given as relative atomic mass (Ar) by using the average mass of the isotopes
    • The relative atomic mass of an element can be calculated by using the relative abundance values

Ar = fraction numerator open parentheses r e l a t i v e space a b u n d a n c e subscript i s o t o p e space 1 end subscript space cross times m a s s subscript i s o t o p e space 1 end subscript close parentheses space plus open parentheses r e l a t i v e space a b u n d a n c e subscript i s o t o p e space 2 end subscript space cross times m a s s subscript i s o t o p e space 2 end subscript close parentheses space e t c over denominator 100 end fraction

    • The relative abundance of an isotope is either given or can be read off the mass spectrum

Worked example

Calculating relative atomic mass of oxygen
A sample of oxygen contains the following isotopes:What is the relative atomic mass, Ar, of oxygen in this sample, to 2dp? 

Answer

    • Arfraction numerator open parentheses 99.76 space cross times space 16 close parentheses space plus open parentheses 0.04 space cross times space 17 close parentheses space plus open parentheses 0.20 space cross times 18 close parentheses over denominator 100 end fraction
    • Ar = 16.0044
    • Ar = 16.00

Worked example

Calculating relative atomic mass of boron

Calculate the relative atomic mass of boron using its mass spectrum, to 1dp:Analytical Techniques Mass Spectrum Boron, downloadable AS & A Level Chemistry revision notes

Answer

    • Arfraction numerator left parenthesis 19.9 space cross times 10 right parenthesis space plus space left parenthesis 80.1 space cross times 11 right parenthesis over denominator 100 end fraction space equals space 10.801 space equals space 10.8

Examiner Tip

You can be expected to work with tables or graphs of data to calculate relative atomic mass

You can also be expected to do these calculations backwards to determine the abundance of one isotope given sufficient information

Predicting Mass Spectra

  • You can also predict how a mass spectrum might appear for a given compound, e.g. ethanol, CH3CH2OH
    • The methyl, CH3+, fragment has a mass of 15.0
    • The ethyl, CH3CH2+, fragment has a mass of 29.0
    • The base ion, CH2OH+, fragment has a mass of 31.0
    • The whole molecule has a mass of 46.0
  • Predicting mass spectra becomes more complex with the inclusion of halogen isotopes such as chlorine and bromine

Chlorine

Chlorine exists as two isotopes, 35Cl and 37Cl

  • A compound containing one chlorine atom will therefore have two molecular ion peaks due to the two different isotopes it can contain
    • 35Cl = M+ peak
    • 37Cl = [M+2] peak
    • The ratio of the peak heights is 3:1 (as the relative abundance of 35Cl is 3x greater than that of 37Cl)
  • A diatomic chlorine molecule or a compound containing two chlorine atoms will have three molecular ion peaks due to the different combinations of chlorine isotopes they can contain
    • 35Cl + 35Cl = M+ peak
    • 35Cl + 37Cl = [M+2] peak
      • There is an alternative of 37Cl + 35Cl doubling the [M+2] peak
    • 37Cl + 37Cl = [M+4] peak
    • The ratio of the peak heights is 9:6:1
      • This ratio can be deduced by using the probability of each chlorine atom being 35Cl or 37Cl
      • 35Cl + 35Cl = begin mathsize 16px style bevelled 3 over 4 space cross times bevelled 3 over 4 space equals space bevelled 9 over 16 end style
      • 35Cl + 37Cl = begin mathsize 16px style bevelled 3 over 4 space cross times bevelled 1 fourth space equals space bevelled 3 over 16 end style but this doubles for the 37Cl + 35Cl option, therefore, begin mathsize 16px style bevelled 6 over 16 end style
      • 37Cl + 37Cl = begin mathsize 16px style bevelled 1 fourth space cross times bevelled 1 fourth space equals space bevelled 1 over 16 end style

  • The presence of bromine or chlorine atoms in a compound gives rise to a [M+2] and possibly [M+4] peak 

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

Mass spectrum of compounds containing one chlorine atom (1) and two chlorine atoms (2)

Bromine

  • Bromine too exists as two isotopes, 79Br and 81Br
  • A compound containing one bromine atom will have two molecular ion peaks
    • 79Br = M+  peak
    • 81Br = [M+2] peak
    • The ratio of the peak heights is 1:1 (they are of similar heights as their relative abundance is the same!)

  • A diatomic molecule of bromine or a compound containing two bromine atoms will have three molecular ion peaks
    • 79Br + 79Br= M+ peak
    • 79Br+ 81Br = [M+2] peak
    • 81Br + 81Br= [M+4] peak
    • The ratio of the peak heights is 1:2:1

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

Mass spectrum of compounds containing one bromine atom

Mass Spectra & Mr

  • 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/z 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 butanalAnalytical 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/z = 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

Examiner Tip

A mass spectrum can give lots of information about fragments of the overall compound being analysed

Your specification states that this is not expected knowledge, you are only required to know the implications of the M+1 peak from a mass spectrum

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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.