Atomic Mass Spectrometry 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 Mass Detector Source Analzyer selected Data Inlet all ions ions System Lets talk about mass! Atomic mass of Carbon Atomic mass of Chlorine Atomic mass of Hydrogen Lets talk about mass! Atomic mass of Carbon 12.000000000000000000000000000 amu Atomic mass of Chlorine 35.4527 amu Atomic mass of Hydrogen 1.00794 amu 1amu = 1 dalton (Da) What about isotopes? Atomic mass of Carbon 12.000 amu for 12C but 13.3355 for 13C Atomic mass of Chlorine 34.9688 amu for 35Cl and 36.9659 for 37Cl Atomic mass of Hydrogen 1.00794 amu for H and 2.0141 for D! Just for clarification Atomic mass 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. 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. 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 C100H202 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 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 3:1 1:1 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 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. Skoog et al., 1999, Instrumental Analysis 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). 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 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. 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=990 . U + Vcoswt . . m/z=1000 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? Isobaric overlap Due to two elements that have isotopes having substantially the same mass 40Ar+ 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? Refractory oxide As a result of incomplete dissociation of the sample matrix or from recombination in the plasma tail MO+, MO2+, MO3+ Doubly charged ions Matrix Effects? Space charge effects Advanced Analytical Chemistry – CHM 6157 ® Y. CAI Florida International University Updated on 9/13/2006 Chapter 3 ICPMS Isotope Dilution 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 ® Y. CAI Florida International University Updated on 9/13/2006 Chapter 3 ICPMS Advantages: Simplified chemical and physical separation procedures Elimination (reduction) of matrix effects Elimination of the effect of instrumental drift Advanced Analytical Chemistry – CHM 6157 ® Y. CAI Florida International University Updated on 9/13/2006 Chapter 3 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) Standard (t) Spiked sample (m) Advanced Analytical Chemistry – CHM 6157 ® Y. CAI Florida International University Updated on 9/13/2006 Chapter 3 ICPMS 1n is the number of moles of isotope 1 in the s 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. Advanced Analytical Chemistry – CHM 6157 ® Y. CAI Florida International University Updated on 9/13/2006 Chapter 3 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  For the standard solution: Rt = 1nt/2nt  Advanced Analytical Chemistry – CHM 6157 ® Y. CAI Florida International University Updated on 9/13/2006 Chapter 3 ICPMS For the spiked sample solution: Rm = (1ns + 1nt)/(2ns + 2nt)  Substitution of equations 1 and 2 into equation 3: Rm = (Rs2ns+ Rt2nt)/(2ns + 2nt)  Rearranged to: 2n = 2n (R -R )/(R -R )  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)  θ2 is the isotopic abundance of isotope 2 in the sample. Advanced Analytical Chemistry – CHM 6157 ® Y. CAI Florida International University Updated on 9/13/2006 Chapter 3 ICPMS The mass of the element in the sample is then given by: Ms = M(2nt/θ2)(Rm-Rt)/(Rs-Rm)  M is the molecular weight of the element.