Chem 454/554
Exam 3 Discussion
April 8, 2002
Answer 8 of the 10 questions-- but you must answer #7, 8 and 9.
Notice that 7-8-9 are worth 15 points, the others are 10 points
Yes, it is a 95 point exam.
Remark: Question 1-4 ask specific questions about the advantages of one method over another. It is important that you address the question, not just supply information about the topic.
1. (10) Give an example or application where Raman spectroscopy is more appropriate than IR absorption. After all, both provide more or less the same information -- identifying species by detecting the vibrational energy levels of the sample. Explain the reasons behind the advantage.
- Any of the following advantages could be cited.
- Probably the best advantage is the shift from IR to visible light. This lets us use laser light sources, ordinary windows and cells, solvents like water and PMT or diode array detectors.
- In the same light, you might address ability to work in aqueous environment (important with biological systems, for example.)
- New developments make compact, rugged, sealed instruments possible. This lets Raman be used in the field, in factory environment-- where conventional IR would be hopeless.
- There are a few vibrational bands that are IR inactive and Raman can provide additional information. (However, this isn't often an advantage.)
- in many other respects Raman is less desirable-- typically weaker signals.
2. (10) What is the basic advantage of FT-IR over conventional IR (with a monochromator.) Explain the reasons for the advantage.
- The major advantage is the multiplex feature-- all wavelengths are present and active throughout the measurement. Conventional IR (grating instruments) see only one wavelength at a time. This makes FT-IR faster.
- It is probably also an advantage that the interferometer is a simpler device and it can lead to lower cost instrumentation. (This is not a major factor, in practice.)
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It's tempting to assign all good properties as an edge to FT-IR instruments. Many papers said higher resolution, better sensitivity, wider spectral range. Those aren't really true advantages, but a consequence of the multiplex. For spectral scans of 5-8 minutes, FT-IR shows these features. In most cases, a grating instrument can be made with higher resolution and better sensitivity-- but it may be too slow for routine use.
3.(10) Why do fluorescence methods often have detection limits that are much lower than spectrophotometry. Explain the reason for the advantage.
- Absorption spectroscopy of dilute samples needs a precise measurement of the change in intensity due to absorption. This requires two measurements and a subtraction. The practical limit is when the difference approaches the noise level in the signal.
- Fluorescence light is proportional to the sample concentration at low concentration. The practical limit is that the measured intensity be significantly higher than the dark current of the light detector. Even very weak fluorescence (weak solutions) are detectable.
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An occasional advantage-- since fewer molecules fluoresce, there are usually fewer interfering species.
4. (10) When would ATR (Attenuated Total Reflection) be more appropriate than a conventional IR cell for measuring the IR Absorption spectrum of a sample. (Include a brief description of ATR.)
- ATR lets you measure spectrum directly from a wide range of samples-- just place sample into contact with the prism. (No need to dissolve, dilute, make KBr pellets, slice sample...)
- The savings occurs in sample preparation.
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again, it's tempting the say a better method must have better sensitivity (use with dilute solutions), ok with aqueous solutions, better resolution, etc. As a rule ATR does not have those advantages.
5. (10) Two important methods of producing ions for mass spectrometry are (a) electron impact and (b) chemical ionization. Briefly describe each method and indicate differences in the resulting mass spectra.
6. (10) What is meant by a normal mode vibration in a molecule?
- This answer needs to describe vibration in the molecule. (The normal modes are observed with spectroscopy but they are not the spectral lines or the spectroscopic process itself.)
- A normal mode involves the simultaneous movement of all (or many) atoms in the molecules; all atoms move at the same frequency and stay in phase.
7.(15 required) Describe two different types of Mass Analyzers used in mass spectrometry. Very briefly tell how each functions and one advantage of each method.
8. (15 required) Draw an energy level diagram for electronic states of a molecule. Show and label
a. excitation
b. fluorescence
c. phosphorescence
d. vibrational relaxation
e. an inter system crossing
d. singlet and triplet states
9. (15 required) Draw a diagram and label components for
one of the following:
a. A Fourier Transform IR Spectrometer
b. A Photoacoustic Spectrometer
c. A Spectrofluorimeter
10. (10 pts) The Raman spectrum of CCl4 is shown in figure 18.1 in the textbook (copy on the exam)
a. what does this tell you about the vibrational modes of CCl4
b. explain why the peaks on the right side are smaller than those on the left.
In particular, why is the -459 peak so small when the +459 is so large. The -218 and +218 peaks show much less difference.
(The large central peak is the Raleigh line; those on the right are Antistokes lines; those on the left are Stokes lines.)
Three different molecular vibrations are seen-- at 218, 314 and 459 cm-1.
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since CCl4 has nine modes there probably are others at higher frequency that don't show up on the figure; some modes won't be Raman active.
- We can rule out overtones and combination bands (e.g., overtones would be near 426, 614 and combinations near 532 cm-1.
- merely seeing the vibration tells us little about the vibration (besides energy and that it is one of the allowed symmetry modes.)
The anti-Stokes lines must originate from molecules already in a vibrationally excited state. Since there are few of these, the Stokes lines are much more intense.
We also expect the population of the excited states to be reduced as the energy of the vibration increases (Boltzmann exp{-E/kT} term.) Thus the 459 anti-Stokes is greatly reduced relative to the matching Stokes line.