4. Mass Spectrometry

Objectives:1. Know what information they can provide2. Interpret simple mass spectra graphs3. Know some uses of mass spectrometry

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

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.

Mass Spectrometry

USES

Mass spectrometry was initially used to show the identity of isotopes.

It is now used to calculate molecular masses and characterise new compounds

Francis Aston

A mass spectrometer consists of ... an ion source, an analyser and a detector.

ION SOURCE

ANALYSER

DETECTOR

Mass Spectrometer

PARTICLES MUST BE IONISED SO THEY CAN BE ACCELERATED AND DEFLECTEDPARTICLES MUST BE IONISED SO THEY CAN BE ACCELERATED AND DEFLECTED

How does it work?

ION SOURCE

ANALYSER

DETECTOR

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

How does it work - Deflection

• 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

20Ne21Ne

22Ne

HEAVIER ISOTOPES ARE DEFLECTED LESS

0 4 8 12 16 20 m/z values

AB

UN

DA

NC

E

1+ ions2+ ions

20Ne

22Ne

Doubling the charge, halves the m/z valueAbundance stays the same

In early research with a mass spectrograph, Aston (Nobel Prize, 1922) demonstrated that naturally occurring neon consisted of three isotopes ... 20Ne, 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

MASS SPECTRUM OF NEON

19 20 21 22 23

20Ne 90.92%

21Ne 0.26%

22Ne 8.82%

What is a Mass Spectrum?

Calculate the average relative atomic mass of neon using data on the previous page.

Out of every 100 atoms... 90.92 are 20Ne , 0.26 are 21Ne and 8.82 are 22Ne

Average = (90.92 x 20) + (0.26 x 21) + (8.82 x 22) = 20.179 Ans. = 20.18

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 ifit is based on heights of lines in a mass spectrum, add up the heightsof the lines and then divide by that number (see later).

Example calculation 1

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?

Test Question

0 10 20 30 40 m/z values

AB

UN

DA

NC

E 60%

40%

100%

ANSWERS ON NEXT PAGE

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?

Test Question

0 10 20 30 40 m/z values

AB

UN

DA

NC

E 60%

40%

100%

0 10 20 30 40 m/z values

AB

UN

DA

NC

E

100%

66.7%

100%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) = 38.80 166.7

By doubling the charge to 2+, m/z value is halved; new peaks at 19 and 20. The abundance is the same.

Now try these….

• Now give the questions in the textbook a go on your own (page 14)