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Atomic Mass Spectrometry Nearly all elements in the periodic table can be determined by mass spectrometry Nearly all elements in the periodic table can.

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Presentation on theme: "Atomic Mass Spectrometry Nearly all elements in the periodic table can be determined by mass spectrometry Nearly all elements in the periodic table can."— Presentation transcript:

1 Atomic Mass Spectrometry Nearly all elements in the periodic table can be determined by mass spectrometry Nearly all elements in the periodic table can be determined by mass spectrometry More selective and sensitive than optical instruments More selective and sensitive than optical instruments Simple spectra Simple spectra Isotope ratios Isotope ratios Much more expensive instrumentation Much more expensive instrumentation

2 Still spectroscopy?

3 What is a mass spectrometer? Illustration of the basic components of a mass spectrometry system. Ionization Source Mass Analzyer Detector Inlet all ions selected ions Data System

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7 Lets talk about mass! Atomic mass of Carbon Atomic mass of Carbon Atomic mass of Chlorine Atomic mass of Chlorine Atomic mass of Hydrogen Atomic mass of Hydrogen

8 Lets talk about mass! Atomic mass of Carbon Atomic mass of Carbon 12.000000000000000000000000000 amu 12.000000000000000000000000000 amu Atomic mass of Chlorine Atomic mass of Chlorine 35.4527 amu 35.4527 amu Atomic mass of Hydrogen Atomic mass of Hydrogen 1.00794 amu 1.00794 amu 1amu = 1 dalton (Da)

9 What about isotopes? Atomic mass of Carbon Atomic mass of Carbon 12.000 amu for 12 C but 13.3355 for 13 C 12.000 amu for 12 C but 13.3355 for 13 C Atomic mass of Chlorine Atomic mass of Chlorine 34.9688 amu for 35 Cl and 36.9659 for 37 Cl 34.9688 amu for 35 Cl and 36.9659 for 37 Cl Atomic mass of Hydrogen Atomic mass of Hydrogen 1.00794 amu for H and 2.0141 for D! 1.00794 amu for H and 2.0141 for D!

10 Just for clarification Atomic mass Atomic mass amu, atomic mass units (uma??) amu, atomic mass units (uma??) “Da” or Dalton. “Da” or Dalton. kD (kiloDalton for macromolecules) kD (kiloDalton for macromolecules) 1 amu = 1.66056*10 -27 kg. 1 amu = 1.66056*10 -27 kg. proton, mp = 1.67265*10 -27 kg, proton, mp = 1.67265*10 -27 kg, neutron, mn = 1.67495*10 -27 kg. neutron, mn = 1.67495*10 -27 kg.

11 Ways to define and calculate the mass of an atom, molecule or ion 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. 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. 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. Monoisotopic mass: calculated using the extract mass of the most abundance isotope for each constituent element. Use monoisotopic mass if possible in MS

12 Differences between Masses C 20 H 42 C 100 H 202 Nominal: (20 x 12) + (42 x1) = 282 u(100x12) + (202x1) = 1402u Monoisotopic: (20 x12) + (42 x 1.007825) = 282.33(100x12) + (202x1.007825) = 1403.5807 Average: (20 x 12.011) + (42 x 1.00794) = 282.5535(100x12.011)+(202x1.00794) = 1404.7039

13 Exact Masses of Some Common Elements and Their Isotopes: ElementSymbol Exact Mass (u) Rel. Abundance % Hydrogen1H1.007825037100.0 Deuterium 2H or D 2.0141017870.015 Carbon 12 12C12.00000100.0 Carbon 13 13C13.0033541.11223 Nitrogen 14 14N14.003074100.0 Nitrogen 15 15N15.000110.36734 Oxygen 16 16O15.99491464100.0 Oxygen 17 17O16.99913060.03809 Oxygen 18 18O17.999159390.20048 Fluorine19F18.998405100.0 Sodium23Na22.9897697100.0 Silicon 28 28Si27.976928492.23 Silicon 29 29Si28.97649645.0634 Silicon 30 30Si29.97377173.3612 Phosphorus31P30.9737634100.0 Sulfur 32 32S31.972074100.0 Sulfur 33 33S32.97070.78931 Sulfur 34 34S33.969384.43065 Sulfur 36 36S35.966760.02105 Chlorine 35 35Cl34.968854100.0 Chlorine 37 37Cl36.96589631.97836

14 (a) only one chlorine atom(b) only one bromine atom c) one chlorine and one bromine atom 3:1 1:1 3:4:1 35 Cl: 75.77 37 Cl: 24.23 79 Br: 50.69 81 Br: 49.31

15 Types of Atomic mass spectrometers

16 Atomic mass spectrometer AMS MMS

17 YO, Y(I), Y(II) EMISSION ZONES COURTESY VARIAN Dr. Houk Presentation, 2002

18 Sampler Skimmer Photo by A. L. Gray Dr. Houk Presentation, 2002

19 AGILENT 7500 OMEGA LENS

20 Plasma Torches 8,000 to 10,000 o C Ionization Source

21 Mass Analyzer (Quadrupole) Skoog et al., 1999, Instrumental Analysis  Two pairs of rods: Attach + and - sides of a variable dc source Apply variable radio- frequency 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.

22 Ion trajectories in a Quadrupole  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). 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). Negative half of the ac cycle: Diverge (ions will tend to diverge during the negative half). Skoog et al., 1999, Instrumental Analysis 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.

23  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.

24  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.

25 How does it work?

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27 Typical mass spectrum

28 How good this is? Linear calibrations over 4 orders of magnitude Multi-elemental analysis of a standard

29 ICP-MS: a handy tool!

30 Spectral Interferences? Isobaric overlap Isobaric overlap Due to two elements that have isotopes having substantially the same mass 40 Ar + and 40 Ca + Polyatomic Polyatomic Due to interactions between species in the plasma and species in matrix or atmosphere 56 Fe and 40 Ar 16 O 44 Ca and 12 C 16 O 16 O.

31 Isobaric interferences?

32 Spectral Interferences? Refractory oxide Refractory oxide A As a result of incomplete dissociation of the sample matrix or from recombination in the plasma tail MO +, MO 2+, MO 3+ Doubly charged ions Doubly charged ions

33 Matrix Effects? Space charge effects Space charge effects

34 Advanced Analytical Chemistry – CHM 6157® Y. CAIFlorida International University Updated on 9/13/2006Chapter 3ICPMS Isotope Dilution Isotope dilution is a super internal standard addition method on the basis of isotope ratios. 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 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. 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. The concentration of the element of interest can then be deducted from these isotopic ratios and from the amount of spike added.

35 Advanced Analytical Chemistry – CHM 6157® Y. CAIFlorida International University Updated on 9/13/2006Chapter 3ICPMS  Advantages: Simplified chemical and physical separation procedures Simplified chemical and physical separation procedures Elimination (reduction) of matrix effects Elimination (reduction) of matrix effects Elimination of the effect of instrumental drift Elimination of the effect of instrumental drift

36 Advanced Analytical Chemistry – CHM 6157® Y. CAIFlorida International University Updated on 9/13/2006Chapter 3ICPMS  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)Standard (t)Spiked sample (m)

37 Advanced Analytical Chemistry – CHM 6157® Y. CAIFlorida International University Updated on 9/13/2006Chapter 3ICPMS 1 n s is the number of moles of isotope 1 in the sample. 1 n s is the number of moles of isotope 1 in the sample. 2 n s is the number of moles of isotope 2 in the sample. 2 n s is the number of moles of isotope 2 in the sample. 1 n t is the number of moles of isotope 1 in the standard. 1 n t is the number of moles of isotope 1 in the standard. 2 n t is the number of moles of isotope 2 in the standard. 2 n t is the number of moles of isotope 2 in the standard. R s is the ratio of isotope 1 to isotope 2 in the sample solution. R s is the ratio of isotope 1 to isotope 2 in the sample solution. R t is the ratio of isotope 1 to isotope 2 in the standard. R t is the ratio of isotope 1 to isotope 2 in the standard. R m is the ratio of isotope 1 to isotope 2 in the spiked sample. R m is the ratio of isotope 1 to isotope 2 in the spiked sample.

38 Advanced Analytical Chemistry – CHM 6157® Y. CAIFlorida International University Updated on 9/13/2006Chapter 3ICPMS Assuming the molecular sensitivity 1 S/ 2 S of the MS for isotope 1 and 2 are the same, then For the sample solution: R s = 1 n s / 2 n s [1] For the standard solution: R t = 1 n t / 2 n t [2]

39 Advanced Analytical Chemistry – CHM 6157® Y. CAIFlorida International University Updated on 9/13/2006Chapter 3ICPMS For the spiked sample solution: R m = ( 1 n s + 1 n t )/( 2 n s + 2 n t )[3] Substitution of equations 1 and 2 into equation 3: R m = (R s 2 n s + R t 2 n t )/( 2 n s + 2 n t ) [4] Rearranged to: 2 n s = 2 n t (R m -R t )/(R s -R m )[5] Convert the number of moles of isotope 2 in the sample to the total number of moles of the elements in the sample. n s =( 2 n t /θ 2 )(R m -R t )/(R s -R m )[6] θ 2 is the isotopic abundance of isotope 2 in the sample.

40 Advanced Analytical Chemistry – CHM 6157® Y. CAIFlorida International University Updated on 9/13/2006Chapter 3ICPMS The mass of the element in the sample is then given by: M s = M( 2 n t /θ 2 )(R m -R t )/(R s -R m ) [7] M is the molecular weight of the element.


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