CHM3122 F2010 MS.ppt

CHM3122 F2010 MS.ppt - Chemistry 3122: Applied Spectroscopy...

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Unformatted text preview: Chemistry 3122: Applied Spectroscopy Mass Spectrometry Module Prof. D.E. Fogg Fall 2010 2 Mass Spectrometry form beam of ions; separate into different mass-to-charge ratios (m/z) by applying electrostatic or magnetic Felds or both. nominal mass = integer mass (e.g. C 2 H 4 + , CH 2 N + both 28) exact mass = weighted sum of exact isotope masses (28.0313, 28.0187, respectively) Resolution: capability of MS to differentiate between masses R = m/ m (m is nominal mass of Frst peak; m = mass difference between adjacent peaks that are just resolved) e.g. to resolve C 2 H 4 + , CH 2 N + : R = 28 / (28.0313-28.0187) 2200 Less expensive low-resolution mass spectrometers (R 500- 1000) readily differentiate simple ions of different nominal masses. 3 Introduction and Reminders mass spectrometry tells us about molecular composition. Information about connectivity and functional groups is obtained more easily from NMR and IR. IR reports selectively on functional groups present NMR reports only non-spin active elements in the molecule. information from mass spec (and elemental analysis, see next) is fundamental to structural elucidation. Structural elucidation is much easier if you know the empirical formula Frst. MS and elemental analysis are complementary. The latter is not a spectroscopic method: it is discussed here for completeness. 4 Reminder: Molecular Composition by elemental analysis Procedure: Accurately weigh a few mg. of pure unknown, then completely decompose it into known species, each containing one of the elements to be analyzed. Then, accurately measure amounts of the various species evolved. commercial analysis usually limited to % C, H, N * complete combust sample in O 2 , then measure CO 2 , H 2 O, N 2 evolved by gas chromatography (N 2 by reduction of N oxides) This permits determination of empirical formulae, or relative proportions of C, H, N NB . cannot determine % O, as analysis carried out in O 2 : may determine by difference, if only CHNO present * Classical chemical analyses used for non-hydrocarbons ( Cl, Br, I, S, etc) . Flame emission or atomic absorption spectrometry for heavier elements 5 % Composition: Empirical Formula from EA data Calculate empirical formula: C 6.25 H 25 6.25 6.25 or CH 4 275 mg CO 2 x 1 g x 1 mol CO 2 x 1 mol C 1000 mg 44.01 g CO 2 1 mol CO 2 = 6.25 mmol C in original sample 225 mg H 2 O x 1 g x 1 mol H 2 O x 2 mol H 1000 mg 18.02 g H 2 O 1 mol H 2 O 25 mmol H in original sample = Check: (6.25 mmol C)(12.01mg/mmol C) = 75 mg C (25 mmol H)(1.008 mg/mmol H) = 25 mg H Sums to 100 mg: thus no other elements were present 100 mg 275 mg 225 mg 2 y excess O 2 C x H y x C O 2 + H 2 O 6 2 y x C O 2 + H 2 O C x H y O z 9.83 mg 23.26 mg 9.52 mg xs O 2 Percentage Composition (2): Oxygenated Hydrocarbons 23.26 mg CO 2 x 1 g x 1 mol CO 2 x 1 mol C 1000 mg 44.01 g CO 2 1 mol CO 2 = 0.5285 mmol C in original sample 9.52 mg H 2 O x 1 g x 1 mol H 2 O x 2 mol H...
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CHM3122 F2010 MS.ppt - Chemistry 3122: Applied Spectroscopy...

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