Analytical Flashcards
Mass spectrometry
- 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
- results in the formation of a + charged molecular ion with 1 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
- these fragmentation ions are accelerated by an electric field
molecules (electron bombardment)= molecule +. + e-
Deflection
- based on their mass (m) to charge (e) ratio, the fragments of ions are then separated by deflecting them into the detector
- 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
- the base peak is the peak corresponding to the most abundant ion
Isotopes
- 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
- mass spectrometry can be used to find the relative abundance of the isotopes experimentally
Relative abundance of an isotope
the proportion of one particular isotope in a mixture of isotopes found in nature
How to calculate m/e value ratio of isotopes
54 Fe2+ = 54/2=27
56 Fe3+ = 56/3=18.7=19
- since 19 is smaller than 27 and has a higher positive charge, 56 Fe3+ will be deflected more inside the spectrometer
Ar (relative atomic mass)
Ar= (relative abundance of isotope 1 x mass of isotope 1) + (relative abundance of isotope 2 x mass of isotope 2) /100
- divided by 100
Deducing molecular formula by using molecular peaks
- the peak with the highest m/e value is the molecular ion (M+) peak which gives information about the molecular mass of the compound
- 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 C atoms are present in that molecule, the more C atoms, the larger the [M+1] peak is
Determining number of C atoms using M+1 peak
- the [M+1] peak is caused by the presence of the carbon-13 isotope in the molecule
- carbon-13 makes up 1.1% of all carbon atoms, therefore the [M+1] peak is much smaller than the M peak as the isotope is less common (the ratio of C-13 to C-12 is 1:99)
- the greater the number of carbon atoms present in a molecule, the greater the height of the [M+1] peak
Formula for determining the number of C atoms using M+1peak
n= 100 x abundance of [M+1] /
1.1 x abundance of M+ ion
- if abundance is asked, substitute values into this formula
Chlorine
- exists as 2 isotopes: Cl-35 and Cl-37
- a compound containing 1 chlorine atom will have 2 molecular ion peaks due to the 2 different isotopes it can contain
- Cl-35 = M+ peak
- Cl-37 = [M+2] peak
- the ratio of the peak heights is 3:1 (as the relative abundance of Cl-35 is 3x greater than that of Cl-37
- a compound containing 2 chlorine atoms will have 3 molecular ion peaks due to the different combinations of chlorine isotopes they can contain
- Cl-35 + Cl-35 = M+ peak
- Cl-35 + Cl-37 = [M+2] peak
- Cl-37 + Cl-37 = [M+4] peak
- the ratio of the peak heights is 9:6:1
Bromine
- bromine too exists as 2 isotopes: Br-79 and Br-81
- a compound containing 1 bromine atom will have 2 molecular ion peaks
- Br-79 = M+ peak
- Br-81 = [M+2] peak
- the ratio of the peak height is 1:1 ( they are of similar heights as their relative abundance is the same)
- a compound containing 2 bromine atoms will have 3 molecular ion peaks
- Br-79 + Br-79 = M+ peak
- Br-79 + Br-81= [M+2] peak
- Br-81 + Br-81 = [M+4} peak
- the ratio of the peak heights is 1:2:1
Fragmentation
- molecules that break up when bombarded with high-energy electrons
- used to determine molecular structure
- fragmentation produces a positive fragment and a radical (only the positive charged fragment is detected)
For example, in the mass spectrum of ethanol (C₂H₅OH):
- The molecular ion (M⁺*) will appear at m/z = 46 (corresponding to the molecular mass of ethanol).
- Fragmentation may produce fragments such as C₂H₅⁺ (m/z = 29) and CH₃⁺ (m/z = 15).
Steps in Fragmentation:
- Ionization: The molecule (M) is bombarded with high-energy electrons, which can knock an electron out of the molecule, forming a positive ion (M⁺*), called the molecular ion or parent ion. This ion may be unstable.
Ionization: The molecule (M) is bombarded with high-energy electrons, which can knock an electron out of the molecule, forming a positive ion (M⁺*), called the molecular ion or parent ion. This ion may be unstable.
𝑀→𝑀+ + 𝑒−
- Fragmentation: The molecular ion (M⁺*) often breaks down into smaller charged fragments (usually cations) and neutral pieces. The fragmentation occurs because the high energy makes bonds within the molecule unstable.
Example of a fragmentation process:
𝑀+→𝐴+ + 𝐵
Where A⁺ is a fragment ion (detected) and B is a neutral fragment (not detected).
Common fragments
Alkanes- CH3+, C2H5, C3H7, C4H9, C5H11, C6H13
Halogenoalkanes- often have multiple peaks around the molecular ion peak, which 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, value 31 for CH2OH+ fragment
