Chemistry 421/821 – Second Exam Spring 2009 page 1Name ANSWER KEY

1)(20 points)Match the Structures with the IR Spectra.

2) (16 points) Identify and explain one difference in the components (light source, detectors, monochromator, etc) between the following pairs of instruments:

  1. (8 points) IR spectrometer and Raman spectrometer.

Raman requires an intense light source because Raman scattering is only 0.001% of light source, but doesn’t have to be in IR region since observe frequency changes allowing Raman to use UV/vis detectors (photomultiplier tubes). Conversely, IR needs an IR light source (Nernst glower, globar,etc) and a thermal-type detector, change in resistance, conductance, or polarization as a result of a change in temperature.

  1. (8 points) Atomic Absorption spectrometer and Atomic Emission spectrometer.

The major difference between an AA and AE spectrometer is the light source. AA requires a separate hollow cathode lamp for each element detected, the emission spectrum of an element is used to detect an element. This supplies the extremely narrow bandwidth of light required to be able to detect an absorbance. With a typical continuous source of UV/vis light, only a very small decrease in the total signal area would be observed. Conversely, AE does not require a separate light source, the flame or other device that is used to atomize the sample is also used to increase the population of excited atoms. AE detects the light emitted from these excited atoms as they relax back to the ground state.

3)(25 points) For the molecule alanine (C3H7NO2):

  1. (5 points)How many vibrational modes are expected?

3N-6 = (3)(13)-6=33

  1. (5 points) Identify one IR active and oneRaman active vibration for alanine.

IR active: any asymmetric bond vibrations

Raman active: any symmetric bond vibration

  1. (5 points)Thealanine IR spectra exhibits a C=O absorbance at 1694 cm-1 and a C-O absorbance at 1364 cm-1, calculate a ratio for the force constants.

, since the atoms involved in the two bonds are identical, the reduced mass () is also identical. So,

Also, since the conversion from wavenumber to frequency involves multiplication by c, which cancels. Wavenumbers can be used to calculate the proportionality.

  1. (10 points) Both IR and Raman spectroscopy measure vibrational modes in a molecule through different processes. Please explain the differences using energy diagrams.

IR spectroscopy is based on the direct absorption of a photon of IR radiation where a molecule undergoes a transition to an excited vibrational state. An active transition requires a change in dipole. Conversely, Raman spectroscopy results from a momentary distortion of electrons distributed around a bond, an active transition requires a change in polarizability.

IRRAMAN

IR – transition from ground to excited Raman – transition to virtual state followed

vibrational state through an absorption by relaxation to ground state,

of a photon. where a change in the vibrational

state has occurred before

relaxation takes place.

A Raman spectrum is also dependent on transitions between ground and excited vibrational states, but unlike IR this does not result from a direct absorption between these vibrational states. Instead, Raman spectroscopy measures a frequency shift that is identical to the energy gap between the vibrational states. First, an absorption occurs to a virtual state that is below the lowest excited electronic state but significantly larger than typical vibrational excited states. Like UV/vis, there are a number of excited vibrational states associated with each virtual state. The vast majority of these excited virtual states relax back to the ground state without any change in energy (Rayleigh scattering). But some relax back to the ground state with a change in energy by relaxing back to either a higher vibrational state (Stokes, lower energy) or lower vibrational energy state (Anti-stokes, higher energy). The Stokes and Anti-stokes transitions have a low probability requiring an intense light source (laser) to observe. Also, since Raman spectroscopy is measuring a frequency shift, the light source does not have to be in the IR frequency range.

4)(10 points) You are interested in using atomic absorption spectroscopy to determine if a sample contains potassium (ionization potential: 4.339 eV). You have a choice between a flame (T=2000K) or a DC spark (T=60000K) light source, which should you use?Please explain and show all work to support your choice!

The flame is the best choice because the low ionization potential of K means that at higher temperatures a majority of K will be converted to ions:

At the high temperature of the DC spark, 0.40 of K is converted to ions!

5)( 19 points) Given the following chromatogram, baseline peak width of ~ 2 min. and a column length of 10 cm:

Calculate:

(a)(4 points)capacity factor for solutes A and B

capacity factor k’ = (tR –tM)/tM

k’(A) = (9.75 min.)/3 min = 3.25

k’(B) = (17.33 min)/3 min. = 5.78

(b)(3 points) separation factor between solutes A and B

 = k’B/k’A

 = 5.78/3.25 = 1.78

(c)(3 points) resolutionbetween solutes A and B

(d)(3 points)how would you classify the performance of the column based on the separation factor and resolution?

A value of  > 1.1 is indicative of a good separation and a Rs value > 1.5 indicates baseline separation, thus the column is performing well in separating solutes A & B. But, the relatively large resolution factor suggests the column is not operating at optimal efficiency.

(e)(3 points)number of theoretical plates for solute B

N = 16 (tR/Wb)2= 16(20.33/2)2 = 1653

(f)(3 points) plate height for solutes B

H= L/N = 10cm/1653 = 0.0060 cm

6)(10 points) Select the best/correct answer:

a)(2 points) Factors involved in peak broadening for Atomic Absorption Spectroscopy and Liquid Chromatography include:

  1. Heisenberg's UncertaintyPrincipal, Spin Diffusion, Nuclear Relaxation and Diffusion
  2. Heisenberg's UncertaintyPrincipal, Longitudinal diffusion, Doppler Effect and Eddy diffusion
  3. Diffusion, Chemical anisotropy, Chemical degradation and the Doppler Effect
  4. Doppler Effect and Diffusion and Chemical degradation and Longitudinal diffusion

b)(2 points) The relative ranking of energy gaps between the electronic, vibration and rotation ground and excited states is:

  1. rotation > vibration > electronic
  2. vibration > rotation > electronic
  3. electronic > vibration > rotation
  4. electronic > rotation > vibration

c)(2 points) The retention time of an analyte in chromatography is related to:

  1. column length
  2. flow rate
  3. interaction between mobile and stationary phases
  4. all of the above

d) (2points) An advantage of FTIR is:

  1. increase in speed, S/N and resolution
  2. multiple compounds can be analyzed simultaneously
  3. temperature independent
  4. all of the above

e) (2points) The Two-line method, Source Modulation method and the Zeeman effect are important in atomic absorption spectroscopy because they:

  1. allow for multiple compounds to be analyzed simultaneously
  2. increase the atomization rate
  3. correct for absorbance from molecular species and the background created by the flame
  4. decrease the formation of metal oxides