1. 4. Mass Spectrometry Objectives:
Know what information they can provide
Interpret simple mass spectra graphs
Know some uses of mass spectrometry

2. A mass spectrometer…. Draw the mass spectrometer
Describe the 5 step process AS A CARTOON STORY BOARD!
Extension:
Use your own words to describe the stages

3. Mass Spectrometry
The first mass spectrometer was built in 1918 by Francis W Aston, a student of J J Thomson, the man who discovered the electron. Aston used the instrument to show that there were different forms of the same element. We now call these isotopes.
In a mass spectrometer, particles are turned into positive ions, accelerated and then deflected by an electric or magnetic field. The resulting path of ions depends on their ‘mass to charge’ ratio (m/z).
Particles with a large m/z value are deflected least
those with a low m/z value are deflected most.
The results produce a mass spectrum which portrays the different ions in order of their m/z value.
Francis Aston
USES
Mass spectrometry was initially used to show the identity of isotopes.
It is now used to calculate molecular masses and characterise new compounds

4. Mass Spectrometer
DETECTOR
ION SOURCE
ANALYSER
A mass spectrometer consists of ... an ion source, an analyser and a detector.
PARTICLES MUST BE IONISED SO THEY CAN BE ACCELERATED AND DEFLECTED

5. How does it work? IONISATION
DETECTOR
ION SOURCE
ANALYSER
IONISATION
gaseous atoms are bombarded by electrons from an electron gun and are IONISED
sufficient energy is given to form ions of 1+ charge
ACCELERATION
ions are charged so can be ACCELERATED by an electric field
DEFLECTION
charged particles will be DEFLECTED by a magnetic or electric field
DETECTION
by electric or photographic methods

6. How does it work - Deflection
20Ne
m/z values
ABUNDANCE
1+ ions
2+ ions
20Ne
22Ne
Doubling the charge, halves the m/z value
Abundance stays the same
21Ne
22Ne
HEAVIER ISOTOPES ARE DEFLECTED LESS
the radius of the path depends on the value of the mass/charge ratio (m/z)
ions of heavier isotopes have larger m/z values so follow a larger radius curve
as most ions are 1+charged, the amount of separation depends on their mass
if an ion acquires a 2+ charge it will be deflected more; its m/z value is halved

7. What is a Mass Spectrum? 20Ne 90.92% MASS SPECTRUM OF NEON 21Ne 0.26%
20Ne %
21Ne 0.26%
22Ne 8.82%
MASS SPECTRUM OF NEON
In early research with a mass spectrograph, Aston (Nobel Prize, 1922) demonstrated that naturally occurring neon consisted of three isotopes Ne, 21Ne and 22Ne.
positions of the peaks gives atomic mass
peak intensity gives the relative abundance
highest abundance is scaled to 100% and other values are adjusted accordingly

8. Example calculation 1
Calculate the average relative atomic mass of neon using data on the previous page.
Out of every 100 atoms are 20Ne , are 21Ne and are 22Ne
Average = (90.92 x 20) + (0.26 x 21) + (8.82 x 22) = Ans. =
100
TIP In calculations of this type... multiply each relative mass by its abundance
add up the total of these values
divide the result by the sum of the abundances
* if the question is based on percentage abundance, divide by 100 but if
it is based on heights of lines in a mass spectrum, add up the heights
of the lines and then divide by that number (see later).

9. Test Question
A mass spectrum shows the presence of two isotopes of m/z values 38 and 40. Both have been formed as unipositive ions.
Redraw the diagram with the most abundant isotope scaled up to 100%.
Calculate the average relative atomic mass.
What would be a) the m/z values and b) the abundance if 2+ ions had been formed?
100%
60%
40%
ABUNDANCE
m/z values
ANSWERS ON NEXT PAGE

10. Test Question
A mass spectrum shows the presence of two isotopes of m/z values 38 and 40. Both have been formed as unipositive ions.
Redraw the diagram with the most abundant isotope scaled up to 100%.
Calculate the average relative atomic mass.
What would be a) the m/z values and b) the abundance if 2+ ions had been formed?
100%
60%
40%
ABUNDANCE
m/z values
The new values are 100 and 66.7 (see diagram)
New scale atoms of mass 38; abundance = 100
atoms of mass 40; abundance = 66.7
Average = (100 x 38) + (66.7 x 40) =
166.7
By doubling the charge to 2+, m/z value is halved; new peaks at 19 and 20. The abundance is the same.
100%
100%
66.7%
ABUNDANCE
m/z values

11. Now try these….
Now give the questions in the textbook a go on your own (page 14)