Mass Spectrometry (Oxford AQA International A Level Chemistry)

Revision Note

Richard Boole

Written by: Richard Boole

Reviewed by: Stewart Hird

Mass Spectrometry

  • Mass spectrometry (MS) is an analytical technique used to identify unknown compounds

  • When a molecule is analysed in a mass spectrometer, the vaporised sample is ionised which creates molecular ions, MOLECULE+•:

MOLECULE bold rightwards arrow with bold Electron bold space bold bombardment on top MOLECULE+• + e

  • The peak with the highest mass to charge ratio or m/z value is the molecular ion (M+) peak

  • This gives information about the molecular mass of the compound

    • The m/z value of the molecular ion peak is 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 the number of atoms in the molecule

    • More carbon atoms = a larger [M+1] peak

    • For example, the [M+1] peak for a hexane ion will be higher / bigger than the [M+1] peak for an ethane ion

Fragmentation

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

Fragmentation of a molecule in mass spectroscopy

Diagram showing different ways that a molecule can split during mass spectroscopy
The same molecule can produce several different fragments in mass spectroscopy
  • The relative abundances of the detected ions form a mass spectrum

Mass spectrum of pent-1-ene

The mass spectrum of pent-1-ene
The small peak at m/z = 70 corresponds to the molecular mass of pent-1-ene, with an [M+1] peak at m/z = 71
  • The fragments in a mass spectrum are like a molecular fingerprint that can be used by a computer to identify the compound from a database

  • Fragments appear due to the formation of characteristic fragments or the loss of small molecules

Table of common fragments and small molecules

Fragment

m/z

Small molecule

m/z

CH3+

15

H2O

18

C2H5+

29

CO

28

C3H7+

43

CO2

44

Worked Example

Which alcohol is not likely to have a peak at m/z = 43 in its mass spectrum?

  1. 2-methylpropan-1-ol, (CH3)2CHCH2OH

  2. Pentan-2-ol, CH3CH(OH)CH2CH2CH3

  3. Butan-1-ol, CH3CH2CH2CH2OH

  4. Butan-2-ol, CH3CH2CH(OH)CH3

Answer:

  • A line at m/z = 43 corresponds to an ion with a mass of 43 for example:

    • [CH3CH2CH2]+

    • [(CH3)2CH]+

  • Butan-2-ol does not have either of these C3H7+ ions within its structure so it cannot produce a mass spectrum peak at m/z = 43

Low resolution mass spectrometry

  • Low resolution mass spectrometry typically reports mass to 2 decimal places

    • This uses more accurate values for atomic mass than are shown on the periodic table

    • For example:

      • Hydrogen = 1.00

      • Carbon = 12.00

      • Oxygen = 16.00

  • Low resolution mass spectrometry is suitable for general analysis or identifying simple compounds

  • However, it does not work for molecules with similar molecular masses

High resolution mass spectrometry

  • High resolution mass spectrometry typically reports mass to 4 or 5 decimal places

    • The values for atomic mass are far more precise than the periodic table

    • For example:

      • Hydrogen = 1.00783

      • Carbon = 12.0000

      • Oxygen = 15.99491

  • Therefore, a more precise molecular mass can be calculated

  • This means that it is possible to:

    • Determine the molecular formula of a compound

    • Distinguish between molecules with similar molecular masses

  • However, it does not work for isomers because they have an identical molecular mass

Comparing low and high resolution mass spectrometry

Molecule

Low resolution MS

High resolution MS

Propane, C3H8

(3 x 12.00)

+ (8 x 1.00)

=

44.00

(3 x 12.0000)

+ (8 x 1.00783)

=

44.06264

Ethanal, CH3CHO

(2 x 12.00)

+ (4 x 1.00)

+ 16.00

=

44.00

(2 x 12.00)

+ (4 x 1.00783)

+ 15.99491

=

44.02623

Worked Example

Fluoroethyne, CHCF, and carbon dioxide, CO2, both have a relative molecular mass of 44.00.

Explain how mass spectrometry can be used to distinguish between these compounds. Show your working.

  • Ar (H) = 1.00783

  • Ar (C) = 12.0000

  • Ar (O) = 15.99491

  • Ar (F) = 18.9984

Answer:

  • High resolution mass spectrometry will distinguish between the compounds because it it more accurate

  • Mr (CHCF) = (2 x 12.0000) + 1.00783 + 18.9984 = 44.00623

  • Mr (CO2) = 12.00 + (2 x 15.99491) = 43.98982

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Richard Boole

Author: Richard Boole

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.

Stewart Hird

Author: Stewart Hird

Expertise: Chemistry Lead

Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.