Atomic Mass Spectrometry
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Nearly all elements in the periodic table can
be determined by mass spectrometry
More selective and sensitive than optical
instruments
Simple spectra
Isotope ratios
Much more expensive instrumentation
Still spectroscopy?
What is a mass spectrometer?
Illustration of the basic components of a mass spectrometry system.
Ionization
Source
Inlet
Mass
Analzyer
all ions
Detector
selected
ions
Data
System
Lets talk about mass!
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Atomic mass of Carbon
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Atomic mass of Chlorine
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Atomic mass of Hydrogen
Lets talk about mass!
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Atomic mass of Carbon
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Atomic mass of Chlorine
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12.000000000000000000000000000 amu
35.4527 amu
Atomic mass of Hydrogen
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1.00794 amu
1amu = 1 dalton (Da)
What about isotopes?
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Atomic mass of Carbon
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Atomic mass of Chlorine
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12.000 amu for 12C but 13.3355 for 13C
34.9688 amu for 35Cl and 36.9659 for 37Cl
Atomic mass of Hydrogen
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1.00794 amu for H and 2.0141 for D!
Just for clarification
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Atomic mass
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amu, atomic mass units (uma??)
“Da” or Dalton.
kD (kiloDalton for macromolecules)
1 amu = 1.66056*10-27 kg.
proton, mp = 1.67265*10-27 kg,
neutron, mn = 1.67495*10-27 kg.
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Ways to define and calculate the mass of an
atom, molecule or ion
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Average mass: calculated using the atomic weight, which
is the weighted average of the atomic masses of the
different isotopes of each element in the molecule.
Often used in stoichiometric calculations.
Nominal mass: calculated using the mass of the
predominant isotopes of each element rounded to the
nearest integer value that corresponds to the mass number.
Monoisotopic mass: calculated using the extract mass of
the most abundance isotope for each constituent element.
Use monoisotopic mass if possible in MS
Differences between Masses
C20H42
Nominal:
Monoisotopic:
Average:
1404.7039
C100H202
(20 x 12) + (42 x1) = 282 u
(100x12) + (202x1) = 1402u
(20 x12) + (42 x 1.007825) = 282.33 (100x12) + (202x1.007825) = 1403.5807
(20 x 12.011) + (42 x 1.00794) = 282.5535
(100x12.011)+(202x1.00794) =
Exact Masses of Some Common Elements and Their Isotopes:
Element
Symbol
Exact Mass (u)
Rel. Abundance %
Hydrogen
1H
1.007825037
100.0
Deuterium
2H or D
2.014101787
0.015
Carbon 12
12C
12.00000
100.0
Carbon 13
13C
13.003354
1.11223
Nitrogen 14
14N
14.003074
100.0
Nitrogen 15
15N
15.00011
0.36734
Oxygen 16
16O
15.99491464
100.0
Oxygen 17
17O
16.9991306
0.03809
Oxygen 18
18O
17.99915939
0.20048
Fluorine
19F
18.998405
100.0
Sodium
23Na
22.9897697
100.0
Silicon 28
28Si
27.9769284
92.23
Silicon 29
29Si
28.9764964
5.0634
Silicon 30
30Si
29.9737717
3.3612
Phosphorus
31P
30.9737634
100.0
Sulfur 32
32S
31.972074
100.0
Sulfur 33
33S
32.9707
0.78931
Sulfur 34
34S
33.96938
4.43065
Sulfur 36
36S
35.96676
0.02105
Chlorine 35
35Cl
34.968854
100.0
Chlorine 37
37Cl
36.965896
31.97836
(a) only one chlorine atom
(b) only one bromine atom
1:1
3:1
35Cl:
75.77
37Cl: 24.23
79Br:
50.69
81Br: 49.31
3:4:1
c) one chlorine and one bromine atom
Types of Atomic mass spectrometers
Atomic mass spectrometer
AMS
MMS
YO, Y(I), Y(II) EMISSION ZONES
COURTESY VARIAN
Dr. Houk Presentation, 2002
Sampler
Skimmer
Photo by A. L. Gray
Dr. Houk Presentation, 2002
AGILENT 7500
OMEGA LENS
Ionization Source
8,000 to 10,000 oC
Plasma Torches
Mass Analyzer (Quadrupole)
Two pairs of rods:
Attach + and - sides of a
variable dc source
Apply variable radiofrequency ac potentials to
each pair of rods.
Ions are accelerated into
the space between the rods by
a small potential (5-10V)
Ions having a limited range
of m/z value reach the
transducer.
Skoog et al., 1999, Instrumental Analysis
Ion trajectories in a Quadrupole
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A pair of positive rods (as lying in the xz plane).
In the absence of a dc potential:
Positive half of the ac cycle: Converge (ion in the channel will
tend to converge in the center of the channel during the
positive half of the ac cycle).
Negative half of the ac cycle: Diverge (ions will tend to
diverge during the negative half).
Whether or not a positive ion strikes
the rod will depend upon the rate of
movement of ion along the z axis, its
m/z, and the frequency and magnitude
of the ac signal.
Skoog et al., 1999, Instrumental Analysis
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A pair of positive rods (Cont’d)
With dc potential:
Heavier ions: less affected by ac (largely by dc).
Lighter ions: deflected during negative cycle of ac.
The pair of positive rods: a high-pass mass filter for
positive ions traveling in the xz plane.
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The pair of negative rods
In the absence of the ac potential:
All positive ions will tend to strike the rods.
With ac potential:
For the lighter ions, however, this movement may be
offset by the positive half cycle of ac potential.
Thus, the pair of negative rods operates as a low-pass
mass filter.
The mass that can be analyzed can be varied by
adjusting the ac and dc potential.
How does it work?
-U - Vcoswt
.
m/z=900
.
m/z=1000
. m/z=990
U + Vcoswt
..
m/z=1010
m/z=1100
Typical mass spectrum
How good this is?
Linear calibrations over 4 orders of magnitude
Multi-elemental analysis of a standard
ICP-MS: a handy tool!
Spectral Interferences?
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Isobaric overlap
Due to two elements that have isotopes having
substantially the same mass
40Ar+
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and 40Ca+
Polyatomic
Due to interactions between species in the plasma
and species in matrix or atmosphere
56Fe
and 40Ar16O
44Ca and 12C16O16O.
Isobaric interferences?
Spectral Interferences?
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Refractory oxide
As a result of incomplete dissociation of the sample
matrix or from recombination in the plasma tail
MO+, MO2+, MO3+
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Doubly charged ions
Matrix Effects?
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Space charge effects
Advanced Analytical Chemistry – CHM 6157
Updated on 9/13/2006
® Y. CAI
Chapter 3
Florida International University
ICPMS
Isotope Dilution
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Isotope dilution is a super internal standard
addition method on the basis of isotope
ratios.
Add a known amount (spike) of a stable
enriched isotope of the element considered,
which has at least two stable isotopes 1 and
2, to the sample
Measure the isotope ratio of isotopes 1 and
2 in the Spike, the unspiked sample and
finally the spiked sample.
The concentration of the element of interest
can then be deducted from these isotopic
ratios and from the amount of spike added.
Advanced Analytical Chemistry – CHM 6157
Updated on 9/13/2006

® Y. CAI
Chapter 3
Florida International University
ICPMS
Advantages:
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Simplified chemical and physical separation
procedures
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Elimination (reduction) of matrix effects
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Elimination of the effect of instrumental drift
Advanced Analytical Chemistry – CHM 6157
Updated on 9/13/2006

® Y. CAI
Chapter 3
Florida International University
ICPMS
Theory
In principle, any element with at least two
isotopes that can be measured is suitable for
determination by isotope dilution. The two
selected are designed 1 and 2.
Three solutions will be used:
Sample (s)
sample (m)
Standard (t)
Spiked
Advanced Analytical Chemistry – CHM 6157
Updated on 9/13/2006
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1n
® Y. CAI
Chapter 3
Florida International University
ICPMS
is the number of moles of isotope 1 in the
sample.
2n is the number of moles of isotope 2 in the
s
sample.
1n is the number of moles of isotope 1 in the
t
standard.
2n is the number of moles of isotope 2 in the
t
standard.
Rs is the ratio of isotope 1 to isotope 2 in the
sample solution.
Rt is the ratio of isotope 1 to isotope 2 in the
standard.
Rm is the ratio of isotope 1 to isotope 2 in the
spiked sample.
s
Advanced Analytical Chemistry – CHM 6157
Updated on 9/13/2006
® Y. CAI
Chapter 3
Florida International University
ICPMS
Assuming the molecular sensitivity 1S/2S
of the MS for isotope 1 and 2 are the
same, then
For the sample solution:
Rs = 1ns/2ns
[1]
For the standard solution:
Rt = 1nt/2nt
[2]
Advanced Analytical Chemistry – CHM 6157
Updated on 9/13/2006
® Y. CAI
Chapter 3
Florida International University
ICPMS
For the spiked sample solution:
Rm = (1ns + 1nt)/(2ns + 2nt)
[3]
Substitution of equations 1 and 2 into equation 3:
Rm = (Rs2ns+ Rt2nt)/(2ns + 2nt)
[4]
Rearranged to:
2n = 2n (R -R )/(R -R )
[5]
s
t
m t
s m
Convert the number of moles of isotope 2 in the
sample to the total number of moles of the
elements in the sample.
ns =(2nt/θ2)(Rm-Rt)/(Rs-Rm)
[6]
θ2 is the isotopic abundance of isotope 2 in the
sample.
Advanced Analytical Chemistry – CHM 6157
Updated on 9/13/2006
® Y. CAI
Chapter 3
Florida International University
ICPMS
The mass of the element in the sample is
then given by:
Ms = M(2nt/θ2)(Rm-Rt)/(Rs-Rm)
[7]
M is the molecular weight of the element.