Atomic Structure Topic 2

2.2  The mass spectrometer             2.2.1  Describe and explain the operation of a mass spectrometer.             2.2.2  Describe how the mass spectrometer may be used to determine the relative  atomic mass using the 12C scale. Note that the mass spectrometer can do a lot more than you will learn here. It is one of the most powerful analytical techniques available in chemistry. That’s why they keep mentioning it on CSI.

You need to be able to draw a labelled diagram of a mass spec. You also need to be able to explain what happens during each of the labelled stages.

Stages of Mass Spectrometry

Vaporisation A vaporised sample of the substance to be analysed is injected into the instrument at low pressure. This allows individual atoms or molecules to be analysed. The pressure is kept low by using a vacuum pump. This ensures that molecules from the air don’t interfere with our results.

Ionisation The vaporised atoms (or molecules) are bombarded with high speed electrons from an “electron gun”. These knock away valance electrons to leave a positive ion. THE IONS PRODUCED IN A MASS SPEC ARE ALWAYS +VE , EVEN FOR SPECIES WHICH NORMALLY FORM –VE IONS.

E.g. X(g) + e-  X+(g) + 2e-

Acceleration The +ve ions are accelerated by 2 electrically charged plates. What charge must be on the plates? The ions pass through a hole in the plates.

Deflection The fast moving ions are deflected by a strong magnetic field at right angles to their direction of travel. Take care not to confuse the stage with an electric field and the stage with the magnetic field!

We can use physics to explain why the magnetic field deflects the ions – but you don’t need to for IB chemistry! The amount that the ions are deflected by depends on (a) The mass of the ion. Heavy ions are deflected less (b) The charge on the ion. More about this later!

Very light ions are deflected too much and strike the walls at A. Very heavy ions are not deflected as much and strike the walls at C. Goldilocks ions (B) have just the right mass and are deflected just the right amount!

Detection By changing the strength of the electromagnet, ions of different mass can be detected. Using this method the mass of ions can be measured very accurately. Is this a good measure of the mass of an atom?

Because the mass of an atom is so low (e. g Because the mass of an atom is so low (e.g. a carbon atom has a mass of 1.99 x 10 -23 g), we compare everything to a convenient standard. The mass of individual atoms is expressed relative to an atom of carbon-12, which is given a mass of exactly 12. This is the Relative Atomic Mass.

The results obtained from a mass spectrometer look like this:

There are 2 lines on this graph, so . . . there are 2 isotopes in the sample being analysed. The X-axis is labelled m/z This is mass / charge on the ion The charge on ions produced in the mass spec is nearly always 1 So m/z gives a measure of relative atomic mass.

The Y-axis is labelled relative abundance. This may be expressed as relative numbers (e.g. 3:1) Or it may be expressed as a % (e.g. 75%:25%) For our graph, 3 out of every 4 atoms has a mass of 35 and one atom has a mass of 37. The average relative atomic mass for the sample analysed is: [(3x35) + (1x37)] / 4 =35.5

The mass spectrum we looked at was for Cl Deduce the mass spectrum for the gas Cl2

You need to be happy doing calculations with this data. Example: Rubidium has two isotopes Rubidium-85 and Rubidium 87 which have relative abundances of 72% and 28% respectively. In 100 atoms there are 72 Rb atoms with a mass of 85 and 28 Rb atoms with a mass of 87 Total mass of the rubidium atoms is:- (72 x 85) + (28 x 87) =8556 Therefore the average mass = 85.56 Rubidium has a relative atomic mass of 85.56

The only remaining complication is that sometimes an ion with a charge of 2+ is formed. Remember that the x-axis is m/z So an ion with a mass of 24 and a charge of 2+ would give a peak at 12. The IB board set lots of questions about mass spec. Learn it properly!

A harder example!

Boron( average mass 10.81) consists of 2 isotopes: 10B and 11B. Find the abundance of each isotope. Let x = % of 10B Therefore % of 11B = (100-x) Average mass = [10x + 11(100-x)] / 100 1081 = 10x + 1100 – 11x X = 1100 – 1081 X = 19 10B = 19% 11B = 81%