Locate the element chlorine on your copy of the periodic table. Notice that it appears to have an atomic number (Z) of 17 and a mass number (A) of 35.5.
How can this be? Surely the nucleus of the chlorine atom doesn’t contain 18.5 neutrons!!!
No! In fact there are two forms of chlorine atoms. They both contain 17electrons and 17 protons but one form (or isotope) of chlorine contains 18 neutrons while the other contains 20.
Isotopes are atoms of the same element with the same atomic number (Z) but different mass numbers (A).
Chemically the two chlorine isotopes, 37Cl and 35Cl are almost identical as they differ only in their respective masses.
Q6 (HARD QUESTION). What physical properties might you expect to be different for different isotopes of the same element?
Why 35.5?
The two forms of chlorine are not present in equal quantities.
In fact, in nature, only 25% of all chlorine is 37Cl, while 75% is 35Cl. Therefore the average mass of all chlorine is :
Average mass = (37 x 25/100) + (35 x 75/100) = 35.5
The number 35.5 is the relative atomic mass of chlorine.
Relative Atomic Mass : The average relative mass of the atoms of an element, taking into account natural abundance of isotopes, on a scale where 12C has a mass of 12.000.
Q7. Lithium has two isotopes with atomic masses 6 and 7. 5.9% of all Lithium is 7Li. Calculate the relative atomic mass of Li.
Q8. Indium (In) has two isotopes (of mass numbers 113 and 115) , 91% of all Indium is 115In . Calculate the relative atomic mass of Indium.
Q9. The table below gives information on the various isomers of Germanium. Use it to determine the Relative Atomic mass of Germanium.
Q10. The table below gives information on the various isomers of Palladium. Use it to determine the Relative Atomic mass of Palladium.
The Mass Spectrometer.
How is one able to detect the presence of isotopes? The Mass Spectrometer (shown below) allows us to discriminate between particles based on their masses. The operation of the mass spectrometer is described below.
A sample compound, for example a natural isotopic sample of atomic chlorine, is placed into a mass spectrometer.
How the mass spectrometer is able to differentiate between atoms of different mass.
One way of visualising the effect of a magnetic field on different ions is to envisage firing differently sized steel ball bearings close to a magnet.
Heavy ball bearings will only be slightly affected (if at all) by the magnetic field while the path of the light ball bearings will be strongly deflected by the magnetic field.
In the same way, at relatively light magnetic field strengths, ions of small size are able to be deflected sufficiently to pass through to the detector. However the magnetic field strength needs to be increased considerably to deflect heavy ions enough so that they are able to be detected.
During the mass spectrum of any sample, ionisation and acceleration take place continually. The magnetic field starts at its lowest setting before increasing gradually to its highest setting. The ions are detected by the sensitive screen as they pass through the mass spectrometer at the correct magnetic field strength. This very small detection is photomultiplied before being sent to a suitable detector.
What does a mass spectrum look like?
Double Ionisation.
Please note : Ionisation in a mass spectrometer is a precarious business. Atoms are passed into a field of fast moving electrons. Most of the time only one electron will be removed from the atoms at this point :
e.g.
Cl + e- ➔ Cl+ + 2e-
However, in some cases two electrons may be removed from the same atom.
Cl + e- ➔ Cl2+ + 3e-
This leaves us with a dipositive ion (2+) rather than a unipositive one (1+) and the mass to charge (m/z) ratio of these ions will be half of the mass to charge ratio of the (1+) ion.
Q11. A scientist wants to find the exact RAM of titanium. A sample of titanium was placed into a mass spectrometer the following results were obtained.:
- Using a periodic table, give the correct symbol for the isotope responsible for the peak at (m/z = 48). (1)
- In addition to these peaks, a very small peak was found at mass to charge ratio 24.
- Give the correct symbol for the ion causing this peak in the mass spectrometer. (1)
- Why was this peak not included in this data? (2)
- Calculate the exact RMM of titanium (2)
Q12. A scientist wants to find the exact RAM of Zinc. A sample of zinc was placed into a mass spectrometer the following results were obtained.:
- Using a periodic table, give the correct symbol for the isotope responsible for the peak at (m/z = 67). (1)
- In addition to these peaks, very small peaks were found at mass to charge ratio 32 and 33. Why were these peaks not included in this data? Give the correct symbol for the ion causing this peak in the mass spectrometer. (2)
- Calculate the exact RMM of zinc.
Mass Spectrometry of molecules.
It is quite possible to place compounds (rather than elements) into a mass spectrometer.
If the information provided by the mass spectrum is used correctly we can often determine the relative molecular mass of the molecules in the compound.
Molecules undergo the same four processes in a mass spectrometer that atoms do : Ionisation, acceleration, deflection, detection
This means that molecules often become ionised (through the removal of one electron) to form a molecularion.
M + e- ➔ M+ + 2e-
The molecularion can then be detected. The peak produced would give a mass to charge ratio equivalent to the relative molecular mass of the compound.
However the conditions within a ,mass spectrometer are such that molecules can be broken apart by passing through the field of fast moving electrons :
M + e- ➔ X + Y+ + 2e-
This random process is known as fragmentation and means that the mass spectrum of molecules is often complicated by having any number of peaks due to fragments in addition to the peak due to the molecularion.
For example, the mass spectrum of ethanol, C2H5OH is shown above .
Notice that despite there being a number of peaks in the spectrum it is relatively easy to locate the molecularion peak (and hence the relative molecular mass of the compound).
The molecularion will always be the peak with the highest mass to charge ratio.
Attempt Questions on Mass Spectra Analysis
Molecules containing isotopic elements..
We have seen that there are two isotopic forms of chlorine, 35Cl and 37Cl, and they occur in the ratio 3:1 respectively in nature.
Because of this, molecules that contain chlorine (chloroalkanes - for example) will form two molecularions that will always be in the ratio 3:1.
Q1. Draw mass spectra showing the molecularions you would expect for the following compounds :
Chloromethane (CH3Cl), Chloroethane(C2H5Cl) and Chloropropane (C3H7Cl).
Answer
Ans = All of these compounds should give mass spectra with a characteristic region around the molecularion with a large peak at, for example m/e = 50 (chloromethane) and other molecularion peak 2 mass units further (52) on but only 1/3rd of the size of the first peak.
Therefore chloroethane would have a large molecularion peak with a mass to charge ration of 64, with a smaller peak (one third of the size) at 66.
peak with a mass to charge ration of 78, with a smaller peak (one third of the size) at 80. Therefore chloropropane would have a large molecularion
Brominated Organic molecules.
Bromoalkanes can also often be identified by the shape of their molecularion peaks. Bromine has two isotopes 79Br (50.5%) and 81Br (49.5%).
As a result mass spectra of compounds containing one bromine atom per molecule consist of two molecularion peaks of near equal intensity two mass units apart.
Q2. What would the shapes of the molecularion peaks be for Bromomethane (CH3Br), Bromoethane (C2H5Br), Bromopropane (C3H7Br)?
Answer
Ans : Bromomethane would produce two molecularion peaks of near identical relative intensity with mass units 94 and 96 (see above)
Bromoethane would produce two molecularion peaks of near identical relative intensity with mass units 108 and 110
Bromopropane would produce two molecularion peaks of near identical relative intensity with mass units 122 and 124 (see above)
How would you use this information?
The examiner may give you information that suggests the presence of a halogen group within an organic molecule, you could use a mass spectrum of the compound (if one is provided) to determine if chlorine or bromine is present in the molecule, by looking for the presence of multiple molecularion peaks and their relative shapes.
Q3. Draw the low resolution mass spectra you would expect for the following compounds, clearly showing the molecularions that would form.
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CH2CHBr
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CH2CHCl
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CH2BrCH2Br
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CH2ClCH2CL
Ans :
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Should contain two peaks of equal size at 106 (containing 79Br) and 108 (containing 81Br)
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Should contain a large molecularion peak at 62 (containing 35Cl) and a smaller molecularion peak at 64 (containing 37Cl) that is one third of the size of the other peak.
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There should be three molecularion peaks in this spectrum, one at 186 (containing two 79Br atoms), one at 188 (containing one 79Br and one 81Br – there are two possibilities as to how this might arise) and one peak at 190 (containing two 81Br atoms). The heights of these peak should be roughly in the ratio (1:2:1) reflecting the probability of each occuring.
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There should be three molecularion peaks in this spectrum, one at 98 (containing two 35Cl atoms), one at 100 (containing one 35Cl and one 37Cl – there are two possibilities as to how this might arise) and one peak at 102 (containing two 37Cl atoms). The heights of these peak should be in the ratio (0.5625:0.375:0.0625) reflecting the probability of each occuring.
(0.75 x 0.75 = 0.5625) : ((0.75 X 0.25) X2 = 0.375), (0.25 X 0.25 = 0.0625)
