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Introduction to Mass Spectrometry: Analytical Chemistry Techniques

Learn about mass spectrographs, isotopes, mass abundance, and solving weighted average mass questions. Explore methods like chromatography, NMR, titration, and more.

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Introduction to Mass Spectrometry: Analytical Chemistry Techniques

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  1. Objectives In this session you will learn: • What is a mass spectrograph and how does it work? • What are isotopes, a review • What are the values of isotopes found on the mass spectrograph and how they can they be used to solve weighted average mass questions • Understand relate mass abundance and solve questions arising from them.

  2. Analytical Chemistry Chemical Methods Instrumentals Chromatography Infrared NMR • Titration • Gravimetric Analysis • Qualitative Identification • Reaction Rates Mass Spectrometry

  3. NASA and the Mass Spectrometer https://www.youtube.com/watch?v=_L4U6ImYSj0 Or Royal Chemistry Society Mass Spectrograph Film https://www.youtube.com/watch?v=J-wao0O0_qM

  4. Ionization Chamber The electrically heated metal coil gives off electrons which travel across to the electron trap. The particles in the vaporized sample are bombarded with electrons that then knock out one electrons from the sample making the sample 1+. M + e-  M+ + 2e- Note: (a 2 e- loss is unlikely as the sample is now positive and it is very hard to remove another e-) These positive ions are persuaded out into the rest of the machine by the ion repeller which is a metal plate carrying a slight positive charge

  5. Masses Symbol : AMU A dalton or the Atomic mass unit (amu) is defined as 1/12 the mass of a carbon-12. The atomic mass unit, abbreviated as either "amu" or "u". You can convert grams into “u” using this conversion factor: 1 u = 1.661 × 10-24 g Or 0.000000000000000000000001661 grams We are measuring the weigh of atoms which are so small that writing these numbers is too much effort, so we say 1u. How many grams of carbon are in 1 mole, calculate using the fact that 1/12 of Carbon is an amu or u , we know this 1 u = 1.661 × 10-24 g ? Answer: Well 1 amu is like 1 nucleon and electron, 1/12 of carbon, thus carbon has 12 amu, if 1.661 ×10-24 g is 1 amu; then 12 amu’s =12 x 1.993 x 10-23 , a mole is 6.022 x 1023 Thus 1.993 x 10-23 X 6.022x10-23 = 12.00g , why is this not 12.01? Answer: Molar mass is a WEIGHTED average of ALL the isotopes, not just 12C, but 13C too

  6. All Masses Herein Try this: How many amu’s are in 196.7 g of gold? Answer: 1.661 × 10-24 g :196.7g = 1.18 X 1026 1amu x amu Note if we divide 1.18x1026 / 196.7 = 6.02 x 1023 or 1 mole

  7. Mass Spectrometry • Ionization: A very tiny sample is vaporized and bombed with e- to give a molecular ion M+ • Acceleration: The ions are accelerated by an electrical field (to all have same kinetic energy. • Deflection: The ions are then deflected by a magnetic field according to their masses. The lighter they are, the more they are deflected. 4. Detection: The beam of ions passing through the machine is detected electrically e- gun

  8. Mass Spectrometry You change the magnetic field strength to guide the light and heavy particles around the bend to the detector.“Incorrect” mass particles, fragments without a charge to deflect would strike the magnet, NOT the detector The need for a vacuum is important so that the ions will not be destroyed by air molecules of N2 or O2.

  9. Mass Spectrometer Block Diagram Summary: acquiring a mass spectrum Ionization Mass Sorting (filtering) Detection Ion Source Ion Detector Mass Analyzer • Form ions • (charged molecules) Sort Ions by Mass (m/z) • Detect ions 100 75 Inlet • Solid • Liquid • Vapor 50 25 0 1330 1340 1350 Mass Spectrum

  10. MOLECULAR MASS DETERMINATION USING MASS SPECTROMETRY When a molecule is ionised it forms a MOLECULAR ION. Most of the positive ions formed will carry a charge of +1 because it is much more difficult to remove further electrons from an already positive ion, so positive +2 is so much harder & unlikely This molecular ion can also undergo FRAGMENTATION Only particles with a positive charge will be deflected and detected. The resulting spectrum has many peaks. The final peak (M+) shows the molecular ion (highest m/z value) and indicates the molecular mass. The rest of the spectrum provides information about the structure. IONISATION MOLECULAR ION FRAGMENTION RE-ARRANGEMENT FRAGMENTION

  11. Guess how Propane would fragment The simplest way: [CH3CH2CH3].+  [CH3CH2]+ + .CH3 (radical will NOT hit detector) 29 15 or [CH3CH2].+ [CH3]+ ion WILL hit detector 29 15

  12. Label this please, do you remember the last slide? Try on your own ANSWERS ON THE NEXT PAGE

  13. [CH3CH2CH3]+ and some fragments BASE PEAK = 29 [CH3CH2]+ 28 [CH2CH2]+ 15 [CH3] M+ = 44 [CH3CH2CH3]+ M-1 = 43. (1 less H) [CH3CH2CH2]+

  14. A B Answer: Please click The only difference between these spectra is A has a larger peak at 57. A little Math 72-57=15. A methyl weighs 15 amu, so which one is more likely to cleave off a methyl, draw them BOTH out to see. A=2-methylbutane (secondary carbocations are more stable than the primary CH 3 that would form in pentane cleaves a methyl)

  15. The Mass Spectrum 1. The mass spectrum is presented as ion abundance vs. m/z ratio (mass) 2. The most abundant ion formed in ionization gives the tallest peak– this is the base peak the most common fragment. It is ALWAYS the highest peak. All other peak intensities are relative to the base peak as a percentage 3. If a molecule loses only one electron when ionized, a molecular ion is observed that gives its molecular weight – this is designated as M+ on the spectrum base peak, m/e 43 M+= 114

  16. Isotopes Most elements have more than one stable isotope. For example, most carbon atoms have a mass of 12 , but in nature, 1.1% of C atoms have an extra neutron, making their mass 13. Why do we care? We can use the M+1 peak to determine the number of carbons in a sample #C = (M+1+)% / (M+ % x 0.011) isotope peak molecular ion peak This is because 13C occurs 1.1% in nature for every normal 12C atom. If our molecule is ethane with 2 carbons then we will have a greater number of 13C as well. If we have 2 carbons = then 13C = 2.2% and with 3 C’s = 13C 3.3% etc, so the M+1 peak gets larger the more 13C carbons when we increase our normal 12C carbons. See the next slide…

  17. THE MASS SPECTRUM - THE MOLECULAR ION The small peak (M+1) at 115 due to the natural abundance (about 1%) of carbon-13. The height of this peak relative to that for the molecular ion depends on the number of carbon atoms in the molecule. The more carbons present, the larger the M+1 peak. 20 40 60 80 100 Abundance % 114 . m/z 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140

  18. M+1 Peak created by Looking at that graph it appears, or let us assume that the M+1 peak is at 2.2% and the M+ is at 25% (let us assume). So if we assume no functional groups, let us use this information to get the formula #C = (M+1) % / (M+ x 0.011) = 2.2 / (25 x 0.011) = 8 carbons This is IMPORTANT for now we can estimate the mass 8 carbons carry = 8 x12 =96 We then subtract this from the total mass (M+) or molecular ion mass, which was 114 – 96 = 18 left over for 18 hydrogen's, so it is octane C8H 18

  19. Carbon 12 Explained

  20. # of Carbons Say for example, this last slide had a mass spectrograph that recorded a relative abundance set of peaks at M+1 ion peak at 4% and a M+ peak at 72%, and also happens to have a mass of 72 as well A) Calculate the number of carbon atoms in the sample? B) If this was a straight chain aliphatic, what might be the possible formula? Answer: a) # of C = [M+1] / 0.011 [M+] = 4 / (.011) 72 = 5.05 or 5 carbons b) With 5 carbons x 12 = 60 and the TOTAL ion at M+ = 72 then 72 – 60 = 12 left for hydrogen's Thus C5H12

  21. Weighted Average Atomic Mass from Mass Spectrograph 6 min Film Clip on Mass Spectrometer and Isotopes and Calculating Molar mass https://www.youtube.com/watch?v=5NnYwlDp6vQ

  22. QUESTIONS ALL SLIDES TO FOLLOW WILL TAKE YOU THROUGH CAMBRIDGE EXAM QUESTIONS BASED ON ISOTPOIC MASS SPECTROETER DATA. • We will find the weighted average atomic mass • We will use mass atomic spectra driven date for atomic weights to tell the difference between to similarly massed molecules

  23. Stick Diagram or Mass Spectrograph That means that molybdenum consists of 7 different isotopes. Assuming that the ions all have a charge of 1+, that means that the masses of the 7 isotopes and they are: 92, 94, 95, 96, 97, 98 (highest peak, most abundant) and 100. Click here to check these values: https://www.webelements.com/molybdenum/isotopes.html

  24. Molybdenum Calculate the Weighted average atomic mass (molar mass) for Mo Mo = (91.906809 x 0.1484) + (93.9050853 + 0.0925) + (94.9058411 x 0.1592) + (95.9046785 x 0.1668) + (96.9060205 x0.0955) + (97.9054073 X 0.2413) + (99.907477 x 0.0963) = 95.9400

  25. Weighted average atomic mass from a spectrograph for Zr : 51.5 11.2 17.1 17.4 2.8 Calculate the weighted average atomic mass from the spectrograph for Zr =91.32

  26. Which Molecular Formula is Which???? QUESTION: Two organic compounds have a relative formula mass of 44 - propane, R = propane and S = ethanal. Using a high resolution mass spectrometer, distinguish between these two compounds. The following very accurate atomic masses are given in the table below C3H8 = 44.0624 Mass Propane C3H8 = (3 x 12.0000 + 8 X 1.0087) = CH3CHO = 44.0261 Mass Ethanal CH3CHO = (2 x 12.0000 + 4 X 1.0087+16) =

  27. Which Molecular Formula is Which???? Question 2: The molecular masses of 2 acids R =98 g/mol and S= 98 g/mol are known to 2 figures, using the accurate atomic masses from mass spectrometers (given in the table below. Find the difference between phosphoric and sulfuric acid? Sulfuric acid is H2SO4 = (1.0078 x 2 + 32.0221 + 15.9949 x 4) = 98.0173 Phosphoric acid is H3PO4 = (1.0078 x 3 + 30.9738 + 15.9949 x 4) = 97.9768

  28. Relative Mass The mass spectrum of Tellurium is set below, calculate the relative atomic mass of Te. Notice we do not have any RELATIVE numbers given, but as they are all relative we can make up any numbers and the other will be proportional to them by scale Relative Abundance M/Z ratio 124 126 128 130 M/Z ratio R.A.M = (mass isotope x RAM.Abundance 1) + (mass isotope x RAM or Abundance 2) …. Total Abundance (20+40+70+60) Adding the toatal abundance is 190 so ex: 124 is 20/190 or 10.5 % = (124 x 20/190) + (126 x 40/190) + (126x70/190) + (130 x 60/190) / 190 = 127.8

  29. Isotopes Magnesium has 3 isotopes 24, 25, and 26 , by mass number, if the atomic mass is 24.3 And we know that the relative abundance from a mass spectrometer reading is 10% for 25 Mg, determine the other 2 isotopes that would appear on a mass spectrograph. Why do these 3 isotopes have the same chemical properties? Answer : (90% remains unaccounted for) 24.3 = ( 25 x 0.10) + ( 24 * X) + ( 26 x (0.90 –x)) yes you can use X and Y And solve , but this way is easier. 24.3 = 2.50 + 24x + 23.40 – 26 x 24.3 = 25.9 – 2x so -1.60 = -2x thus 0.80 = x so x is 80% the other is 0.90 – x or 0.90 – 0.80 = 0.10 or 10% Thus we had 10% originally, then calculated another 80%, leaving 10% for the other These isotopes have the same chemical properties because chemical properties is related to electron configuration, and the all have the same electronic configuration and The same reactive valence electrons

  30. Empirical Formula V Molecular Formula A unknown compound X has the empirical formula of C2H3O2 and is thought to be a carboxylic acid. It has a M+ ion peak on a mass spectrograph at 118, what is the empirical formula? Answer: The empirical formula of C2H3O2 = 59 g/mol Multiple = Molar mass of MF / Molar mass of EF Multiple = 118 g/mol / 59 g/mol = 2 Thus the MF is C4H6O4

  31. Mass Spectrum of Ethanoic acid or chloroethane Is this the spectra of ethanoic acid or chloroethane? Answer Mr ethanoic = 60g/mol, the other is 65 g/mol, the spectra show the molecular ion as 60 So it is ethanoic acid

  32. Do you know • What is a mass spectrograph and how does it work? • What are isotopes? • What the values of isotopes found on the mass spectrograph are and how they can they be used to solve weighted average mass questions • Understand relate mass abundance and can you solve questions arising from them.

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