1. Atomic structure
L.O. To be able to recall definitions of relative atomic, isotopic and molecular mass.
To demonstrate an understanding of the basic principles of mass spectrometry.
To be able to interpret mass spectrometry data.
51.5
11.2
17.1
17.4
2.8

2. Why? What do we mean by ‘relative’?
‘In proportion’ to or ‘in relation’ to something else
In this case masses are measured relative to 1/12th the mass of a carbon-12 atom
Why?

3. Some definitions you need………..
Relative atomic mass (Ar): The average mass (considering the abundance of each isotope) of the atoms of that element relative to 1/12th the mass of a C-12 atom.
Relative isotopic mass: The mass of 1 atom of an isotope relative to 1/12th the mass of a C-12 atom.
Relative molecular mass (Mr): The sum of all the relative atomic masses of the molecule’s constituents.

4. How do we know the relative masses of atoms and molecules?
Mass Spectrometry

5. What does a mass spectrometer do?
2. Ionisation:
Neutral atoms (or molecules) converted into positively charged ions by removing an electron
3. Acceleration:
The ions are accelerated so they have the same kinetic energy.
4. Deflection:
The ions are separated according to their relative mass (m) to relative charge (z) ratio, (m/z). Lighter ions are deflected more than heavier ions.
Higher charged ions (+2) deflected more than lower (1+).

6. What does a mass spectrometer do?
5. Detection:
"m/z" and the % abundance for each ion is measured.
A MASS SPECTRUM = graph of “m/z” versus % abundance

7. to remove air molecules which would also be measured
to form SEPARATE atoms / molecules
A Mass Spectrometer

8. Ionisation Why must there be a vacuum?
High energy electrons
Positive ions repelled
Electron gun – electrically heated coil
Electron trap (+)
Vapourised sample
Ion repeller (+)
Ionisation
High-energy electrons fired at sample
particle collisions happen and an electron is knocked off atom/molecule
A positive ion is produced
e.g. M(g) e-  M+(g) + 2e-
Why must there be a vacuum?

9. The result is a fine focussed beam
High speed beam of ionised sample
Ionisation chamber at volts
Final plate at 0 volts
Intermediate plate
Acceleration
+ve ions repelled by very positive chamber
They are accelerated through 3 slits
The result is a fine focussed beam
Why is ionisation and acceleration needed?

10. Electromagnet
Ion stream C
Ion stream A
Ion stream B 
Mixed ion stream from accelerating unit
Deflection
High speed beam of +ve ions deflected by strong, variable magnetic field (increased steadily)
The result is: ions of INCREASING m/z value deflected in turn onto the detector.
Does ion stream A or ion stream C have lowest m/z ratio?

11. Detection and measurement
Each ion reaching the detector is neutralised by an e- jumping from metal box
Ion stream B
Metal box
Wire to amplifier
 tiny current produced
 current measures abundance of that ion.
Output is called a ‘mass spectrum’
Relative
abundance
m/z
showing abundance (or detector current) against m/z ratio for each isotope.

12. Using mass spec data
L.O. To be able to deduce relative atomic and molecular mass from mass spectrometry data.
51.5
11.2
17.1
17.4
2.8

13. Using mass spec data to find Ar: RUBIDIUM
% Isotopic
Abundance
72.2%
27.8%
Relative (mass/charge) ratio
Using mass spec data to find Ar: RUBIDIUM
Calculate Ar(Rb)
= average mass of a Rb atom relative to the mass of a 12C atom
= total relative mass of 100 atoms
100
= 85.6 (3sf)
= (72.2 x 85) + (27.8 x 87)
100

14. = total relative mass of sample number of atoms in sample
Example 2 : ZIRCONIUM
51.5
11.2
17.1
17.4
2.8
Ar(Zr)
= total relative mass of sample
number of atoms in sample
= (51.5 x 90) + (11.2 x 91) + (17.1 x 92) + (17.4 x 94) + ( 2.8 x 96)
= 91.3 (3sf)

15. Example 3 : Lead Ar(Pb) = = 207.2
(1.5 x 204) + (23.6 x 206) + (22.6 x 207) + (52.3 x 208) =
( )
Ar(Pb) =

16. Mass spec gives us so much more than just relative Ar ………..
Deducing relative molecular mass
Molecular ion formed: M → [M]+ + e–
m/z value for [M]+ gives the Mr.

17. propyl methanoate

18. butane
CH3CH2CH2CH3
C4H10  [C4H10]+ + e–
m/z 58
[C4H10]+

19. FRAGMENTATION
Molecular ion fragments due to covalent bonds breaking: [M]+ → X+ + Y
Lots of information from fragmentation patterns – can often work out structure
Loss of methyl (CH3), water (H20) and ammonia (NH3)
The more stable the ion, the greater the peak intensity.

20. butane CH3CH2CH2CH3 C4H10  [C4H10]+ + e–
m/z 58
[C4H10]+  [CH3CH2CH2]+ + CH3
m/z 43
[CH3CH2CH2]+
[C4H10]+  [CH3CH2]+ + CH2CH3
m/z 29
[CH3CH2]+
[C4H10]+