Midterm - Solutions

1. An HPLC experiment has the following parameters:
Length             15.0 cm
Flow rate         0.5 mL/min
VM                  1.37 mL
VS                   0.164 mL
  retention time (min) width of peak (W) (min)
solvent
unk-A
unk-B
unk-C
 3.1
8.4
11.1
13.8
 -
1.01
1.19
1.54
 
1. (a) Calculate the resolution between unknown peaks A and B, and between peaks B and C. (2 points)

1a
similarly, RS(B&C) = 1.98
 
1. (b) If RS α (N)1/2 and N α L, then RS α (L)1/2

1b

 
2. Answer 2 of the following 3 (a) to (c) (2 points each, 4 total):
 
2. (a) Explain electroosmotic flow.

- at "normal" pH, some SiOH groups on capillary walls are deprotonated to Si-O- (0.5)
- therefore, positive ions in the buffer are attracted to the capillary walls (0.5)
- and to the cathode at the end of the column (0.5)
- this drags the solvent and all other species (negatively charged and neutral) along with it (0.5)
 
2. (b) Explain how an electron-capture detector works

- 63Ni radioactive source decays by β-emission (emits a high-energy electron) (0.5)
- these particles hit the carrier gas (N2), creates a bunch more, lower energy electrons (0.5)
- electronegative groups on analyte molecules easily attract these electrons (0.5)
- loss of electrons to absorption is detected as current loss (0.5)
 
2. (c) Describe how a light-scattering HPLC detector works.

- solution comes off column into a nebuliser, forms a fine spray (0.5)
- solvent evaporated by the temperature and low pressure of the evaporation chamber (0.5)
- leaves behind small particles of clumped analyte molecules (0.5)
- light scatters off these particles (like dust in a sunbeam), drop in light intensity measured (0.5)
 
3. Answer the following (3 points each, 6 total):
 
3. (a) Samples for infrared spectroscopy are commonly prepared in one of these three forms: (i) nujol mull, (ii) solution, (iii) KBr pressed pellet.  Compare the relative merits (advantages and disadvantages) of 2 of these 3 methods.

(i) advantage - easy to prepare and use (0.5)
     disadvantages - not reproducible/quantitative (0.5) and several interfering peaks (0.5)

(ii) advantage - no interfering peaks (0.5)
     disadvantages - not reproducible/quantitative (0.5), requires a hydraulic press (0.5)

(iii) advantage - quantitative in concentration
     disadvantages - requires special cell, sample needs to be soluble, solvent may have interfering peaks (1)
 
3. (b) Describe the basis for size-exclusion chromatography, including a description of why size-exclusion gels have an upper and lower mass limit.

- gel contains little pores into which molecules can fit (0.5)
- small molecules fit into holes and so go in and out of pores, increasing distance travelled and therefore retention time (1)
- the larger the molecule, the fewer pores they will fit into, so pathlength and retention time decrease as size of molecule increases (0.5)
- therefore, any molecules larger than the upper limit (size of the largest pore) of the gel will come straight off with no separation (0.5)
- likewise, any molecule smaller than the smallest pore size will enter every pore in its path and will therefore have the same pathlength (and retention time) (0.5)

 
4.
 Answer the following (2 points each, 6 total)

 
4. (a) A phenol has a distribution constant between hexane and water of 6.25.  A contaminated water sample (50.0 mL) is 0.0600 M in the phenol.  What is the concentration remaining in the water after 5 extractions with hexane if each extraction uses a volume of 15.00 mL? (2 points)

4a

 
4. (b)
 Describe the injection system of EITHER a GC OR an HPLC instrument.

GC:
- must have a rubber septum to reseal after injection to avoid sucking in air (0.5)
- must be hot to volatilise entire sample simultaneously and quickly (therefore requires small sample size) (1)
- must have a connection to allow addition of carrier gas (0.5)

HPLC:
- need a sample loading loop where sample is drawn into a reservoir (1)
- there is a toggle between the reservoir and the column that is switched over to expose sample to the high pressure of the instrument (1)

 
4. (c)
 Explain why a molecule's emission (fluorescence) spectrum is always at longer wavelength than its absorption spectrum

- light is absorbed from the ground state to an excited vibrational and electronic state; there are many vibrational states possible (0.67)
- the excited vibrational states always relax very fast, therefore, the excited electron quickly loses a little energy in vibrational relaxation (0.66)
- the molecule then re-emits a photon to relax to the ground state which is of lower energy because some energy has been lost to vibrational relaxation (0.66)

 
5. Answer 2 of 3 (a)-(c).  For this question, you must supply specific details (instrument, sampling technique, detector, etc.) (5 points each, 10 total)

There are several options for each situation, I've supplied what I think is the simplest (but not necessarily the best) method.  The approximate breakdown in marks for (b) and (c) is as follows:

- type of instrument used (1)
- separation type (stationary phase and mobile phase) (1.5)
- detector type (1)
- sampling technique (is the sample injected "as is" or dissolved in solution, and how much?) (0.5)
- identifying (and quantifying for (b)) the peaks (1)
 
5. (a) Infrared spectroscopy is often used to analyse emissions from internal combustion engines (e.g., car engines).  Without resorting to chromatography, explain how you would use IR to determine the amount of CO, CO2, and SO2 that might be present in the exhaust.  Do any of these peaks overlap?  Why or why not?

- CO has a triple bond, CO2 has C=O double bonds, therefore they have different k so will not overlap (1)
- S has a different mass from C, therefore SO2 will not overlap with either CO2 (1)
- CO and SO2 have both different k and different masses, so a little harder to say definitively if they will overlap, but there's no reason to think they would everlap (1)
- therefore, the peaks for the three species will occur in different regions of the spectrum so simultaneous monitering should be possible (1)
- any general comment on difficulties in measuring or how an instrument should be set up will get you the final point (1)

   
5. (b) The "high test" gasoline at a local station claims to have 92 octane rating (meaning it is 92% "isooctane").  How would you determine the components (amount and identity) of this gasoline?  For this question, let's assume (it's a big assumption!) the following components could be in gasoline, all of which have an octane rating of 0 except isooctane.  [Followed be 5 compounds with their boiling points]

- instrument: gas chromatography
- separation method: probably should use capillary coated with stationary phase (capillary GLC)
            - inert gas as mobile phase (pick one of He, Ar, or N2)
- detector: for quantitation, probably need FID
            - could run standard solutions to determine location of peaks (identification) and detector response (quantitation) for each component
- some comment on sampling technique - e.g., can be injected neat (without dilution)
   
5. (c) You are a customs office that is trying to determine if the suspect's pills are percocet or a legal over-the-counter acetominophen pain-killer (without the active percocet component, 14-hydroxydihydrocodeinone). [structures for caffeine, acetominophen, and 14-hydroxydihydrocodeinone were included]

- sample must be made into a solution
- instrument: HPLC (1)
- separation method: non polar stationary phase (reverse phase) probably best
          - some comments on solvent required (say, a mixed water/acetonitrile or water/methanol mobile phase)
- detector: any of the following will work - IR, UV-vis absorption, light scattering, mass spec (but must pick one!)
          - some comment on identifying the peaks required (e.g., running standards)


 

Last updated, March 11, 2005
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