# Mass Spectrometry: A Level Chemistry

This study guide contains everything you (or your students) need to know about mass spectrometry for your A Level chemistry exam.

For more practice, look here for a Mass Spectrometry A Level extended writing question (with mark scheme), or here for practice questions (with answers).

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## Mass Spectrometry A Level Study Guide

A time of flight (TOF) mass spectrometer is an analytical instrument that is widely used to
determine:

• the abundance of each isotope of an element;
• the relative atomic mass of an element;
• the relative molecular mass of a compound.

The stages which a sample passes through within a TOF mass spectrometer are:

1. Ionisation. Samples are ionised to make positively charged ions. Positive ions can be accelerated using an electric field and can attract electrons from the detector in the final stage of mass spectrometry. There are two methods of ionisation:

• Electron Impact Ionisation. A sample of the chemical to be analysed (X) is vaporised and then bombarded with high energy electrons from an electron gun. These high energy electrons knock one electron off each particle of the sample creating gaseous, positively charged (1+) ions.

This process can be shown by the equation:
X(g) + e ➔ X+(g) + 2e

This type of ionisation is used when analysing elements or compounds with a small molecular mass. When the electron impact method is used to ionise larger molecules, it can lead to fragmentation (splitting) of the molecule.

• Electrospray Ionisation. A sample of the chemical to be analysed (X) is dissolved in a volatile solvent and passed through a thin, hollow needle connected to the positive terminal of a power supply. As the sample leaves the needle it gains a proton (H+ ion) from the solvent and is released into the spectrometer as a fine mist. The solvent evaporates leaving gaseous, positively charged (1+) ions.

This process can be shown by the equation:
X(g) + H+ ➔ XH+(g)

This method of ionisation rarely causes fragmentation of the sample, therefore can be used to analyse large molecules such as proteins. However, the mass of the proton in the XH+ ion must be taken into account when interpreting a spectrum where this method of ionisation has been used. For example, when analysing C6H12O6 using electrospray ionisation, the ion C6H12O6H+ would be formed. The mass spectrum would show a peak at m/z = 181, but the relative molecular mass of C6H12O6 is 180.

2. Acceleration. Negatively charged plates create an electric field which causes the positive ions in the mass spectrometer to accelerate. The energy transferred from the electric field to the ions is constant, so all ions have the same kinetic energy.

$KE = \frac{1}{2}mv^2$ therefore $v = \sqrt{\frac{2KE}{m}}$

Where KE is the kinetic energy in kJ, m is the mass of one ion in kg and v is the velocity of an ion in ms-1.

If all ions have the same kinetic energy, then their velocity depends on their mass. Lighter ions will have a greater velocity.

3. Ion Drift. Once accelerated, the ions move through a flight tube. Here, the ions become separated according to their mass. Lighter ions have a greater velocity and so will reach the end of the flight tube in a shorter time. Figure 1 represents a time of flight mass spectrometer.
1. Detection. At the end of the flight tube is a detector that consists of a negatively charged plate. When the positive ions reach the detector, they become discharged by drawing an electron from the detector, creating a current. The detector is connected to a computer which will record the time it has taken for a particular ion to reach it. This is known as the ion’s time of flight. The time of flight can be used to calculate the mass of the ion. All ions of the same type will reach the detector at the same time. If there are many of a particular ion, a large number of electrons will be drawn from the detector at the same time and therefore a large current will be recorded. The size of the current produced indicates the relative abundance of each ion in a sample.

A mass spectrum shows the mass to charge ratio, m/z (x-axis) and the relative abundance (y-axis) of each ion in a sample. Since the charge of most ions in the mass spectrometer is 1+, the m/z is equal to the mass of the ion.

### Mass Spectra of Elements

Many elements exist as a mixture of isotopes. Analysis of an element in a TOF mass spectrometer can determine the relative abundance of each isotope. Figure 2 shows the mass spectrum of neon, Ne.

Figure 2 shows that neon has three isotopes, 20Ne, 21Ne and 22Ne. Each isotope has the same
number of protons and electrons but a different number of neutrons and therefore a different
mass.

The peak at m/z = 20 is caused by 20Ne+ ions.
The peak at m/z = 21 is caused by 21Ne+ ions.
The peak at m/z = 22 is caused by 22Ne+ ions.

Relative atomic mass (Ar) is the weighted average mass of an atom of an element, on a scale where the mass of a 12C atom is exactly 12.00.

Data from the mass spectrum of an element can be used to calculate the Ar of that element.

$A_r = \frac{\sum{(\text{isotope mass}\;\times\;\text{isotope abundance)}}}{\text{total abundance}}$

From Figure 2, we can see that the abundance of the 20Ne isotope is 90.5, the abundance of the 21Ne isotope is 0.3 and the abundance of the 22Ne isotope is 9.2.

The total abundance is 100 (be sure to check the total abundance, it will not always be 100).

So, the relative atomic mass of neon, Ar is:

$A_r = \frac{(20\;\times\;90.5)\;+\;(21\;\times\;0.3)\;+\;(22\;\times\;9.2)}{(90.5\;+\;0.3\;+\;9.2)} = 20.2\;\text{(1.d.p.)}$

The calculated Ar should be comparable with the given Ar for that element in the periodic table.

When electron impact ionisation is used, it is possible that the high energy electrons knock
more than one electron off some of the sample particles, creating ions with a 2+ charge. When this occurs, you may see a very small peak representing the m/z of the 2+ ion.

Figure 3 shows the mass spectrum of argon, Ar. For example, the Ar of argon is 40. During
ionisation most argon atoms will lose just one electron and form 40Ar+ ions with m/z = 40. However, a very small number of atoms may lose two electrons to form 40Ar2+ with m/z = 20, hence a small peak is seen in this region of the mass spectrum of argon.

### Mass Spectra of Diatomic Elements

The mass spectra of diatomic elements show that isotopes can combine in different ways to form molecules with different masses. Figure 4 shows the mass spectrum of bromine, Br2.

Bromine has two isotopes, 79Br and 81Br, with an approximately equal abundance. In the molecule Br2, the two bromine atoms could be the same isotope or different isotopes. This leads to four possible combinations of isotopes in a Br2 molecule.

There is an equal chance of each 79Br atom bonding with another 79Br atom, to form a
Br2 molecule with a molecular mass of 158, or bonding with a 81Br atom to form a Br2
molecule with a molecular mass of 160.

There is also an equal chance of each 81Br atom bonding with a 79Br atom, to form a Br2
molecule with a molecular mass of 160, or bonding with another 81Br atom to form a Br2
molecule with a molecular mass of 162.

A probability table can be used to show the molecular masses of all the possible combinations of isotopes. For bromine, we see three peaks at m/z = 158, 160 and 162 in a ratio of 1:2:1 (as there are two possible combinations of the isotopes which have a molecular mass of 160).

### Mass Spectra of Compounds

Mass spectra of compounds can look quite different depending on the method of ionisation used. Figure 5 and Figure 6 show two spectra of pentane, C5H12.

If electron impact ionisation is used, fragmentation of the molecule can occur, giving rise to many peaks, seen in Figure 5.

The most prominent peak with the highest m/z is known as the molecular ion peak and indicates the relative molecular mass (Mr) of the compound.

There is sometimes a very small peak at m/z = Mr +1 due to the presence of 13C or 2H isotopes in the molecular ion.

The molecular ion peak in this spectrum is caused by C5H12+ ions which have a molecular mass of 72. The small peak at m/z = 73 is caused by C5H12 ions which contain a 13C or 2H isotope.

If electrospray ionisation is used, no fragmentation occurs and so a different spectrum will be seen, Figure 6.

In this spectrum the peak at m/z = 73 is caused by C5H12H+ ions (as the molecules gain a H+ ion during electrospray ionisation) and so the mass of a pentane molecule, C5H12, is 72.

The small peak at m/z = 74 is caused by C5H12H+ ions which contain a 13C or 2H isotope.

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