Mass spectrometry (AQA A Level Chemistry): Revision Note

Interpreting a Mass Spectrum

• 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

  

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

• An electric field accelerates the fragmented ions

• 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 an 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 numbers

  • 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 are the Cl-35 isotope and 25% are the Cl-37 isotope

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

Deducing Molecular Formula

• 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

• 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 a 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

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 gives 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/z values of some of the common alkane fragments are given in the table below

The m/z Values of Common Fragments

Halogenoalkanes

• Halogenoalkanes often have multiple peaks around the molecular ion peak

• This is caused by the different isotopes of the halogens

Alcohols

• Alcohols often tend to lose a water molecule, giving rise to a peak at 18 units below the molecular ion

• Another common peak is found at m/z 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/z = 59

  • Loss of a water molecule to form a C₃H₆⁺ fragment with m/z = 42

  • Loss of a ^•C₂H₅ to form a CH₂OH⁺ fragment with m/z = 31

  • And the loss of ^•CH₂OH to form a C₂H₅⁺ fragment with m/z = 29