Chemistry 421/821 – Third Exam Spring 2010 page 1
Name
Answer Key
1) (16 points)
a. (10 points) Please describe how a Flame Ionization Detector (FID) works for Gas
Chromatography.
An
FID
measures
an
eluting
analyte/compound/solute by the production of
ions when an analyte/compound/solute is
burned in a flame. The ions are collected at an
electrode to create a current. Thus, the
presence of a compound is identified by a spike
in current.
b. (3 points) Please identify the primary operational difference between the Flame Ionization
Detector (FID) and the Nitrogen-Phosphorus Detector (NPD).
FID works with any organic molecule, NPD is optimized for nitrogen and phosphorus
containing compounds. This is accomplished by including a small amount of alkali
metal in the flame.
c. (3 points) Please Identify one disadvantage of the FID detector.
FID is a destructive detector, the compound is burned in a flame and cannot be
collected.
Chemistry 421/821 – Third Exam Spring 2010 page 2
Name
Answer Key
2) (9 points) Please describe the important chemical characteristics (polar, non-polar, etc) of the
strong mobile phase for the following (Do not simply list a solvent):
a. (3 points) Adsorption liquid chromatography that uses silica (slightly acidic) stationary
phase.
Polar solvent
b. (3 points) Affinity chromatography where the stationary phase is an organic compound
that specifically binds an enzyme that will be purified from a cell extract.
Any buffer change (pH, salt, etc) that diminishes the affinity between the compound
and the enzyme or including the organic compound in the buffer to compete with the
stationary phase and displace the enzyme.
c. (3 points) Size exclusion chromatography
SEC does not have a weak or strong mobile phase, the solvent just transports the
compounds through the column.
3) (9 points) Please describe the important chemical characteristics (polar, non-polar, etc) of the
weak mobile phase for the following (Do not simply list a solvent):
a. (3 points) Adsorption gas chromatography that uses charcoal (non-polar) stationary phase.
GC does not have a weak or strong mobile phase, the carrier gas just transports the
compounds through the column.
b. (3 points) Reverse phase liquid chromatography using C18 stationary phase.
Polar solvent
c. (3 points) ion-exchange chromatography
Low concentration or absence of competing ions
4) (5 points) Please identify a detector for liquid chromatography and list one advantage and one
disadvantage of using the detector.
Multiple answers are possible, only one of the following is required:
1) Refractive Index Detector – universal, but low sensitivity
2) UV/Vis Absorbance Detector – higher sensitivity, but limited to UV/vis active
compounds. Can simplify chromatogram by selecting for compounds that absorb at
specific wavelength, but can give false impressions of purity
3) Fluorescence Detector – even higher sensitivity that UV/Vis, but significantly limited by
number of fluorescence compounds. Same selectivity advantage/disadvantage as UV/vis.
4) Conductivity Detector – high selectivity for ionic compounds, but low sensitivity.
5) Electrochemical Detector – extremely high selectivity and sensitivity, but requires redox
compounds and separate electrode for each analyte.
Chemistry 421/821 – Third Exam Spring 2010 page 3
Name
Answer Key
5) (10 points) You are purifying multiple samples with essentially identical composition using gas
chromatography. Specifically, you are using gas-liquid chromatography with a column composed
of a polysiloxane liquid adsorbed to a silica support. You collect two sequential chromatograms,
first (A) then (B). Please note the two boxed regions are essentially identical and is the expected
outcome of the purification.
(A)
(B)
a. (5 points) Please provide the most likely explanation for the deterioration in the
appearance of the (B) chromatogram.
The polysiloxane liquid adsorbed to the silica support is bleeding off the column,
actually a catastrophic deterioration of the column has occurred.
b. (5 points) How can you modify the column to prevent this problem from occurring again?
Chemical bond the polysiloxane liquid to the silica support, a reaction that forms a
Si-O-Si bond.
Chemistry 421/821 – Third Exam Spring 2010 page 4
Name
Answer Key
6) (10 points) The following chromatogram resulted from the analysis of an oil sample using a
reverse phase C18 column and a mobile phase composed of a 50:50 mixture of water:acetonitrile
running at an optimal flow rate.
a. (5 points) Without changing the column or instrumentation, what experimental technique
can be applied in an attempt to improve the resolution?
Instead of using an isocratic elution, apply gradient elution. Gradually change the
water:acetonitrile mixture from 100% water to 100% acetonitrile or some similar
variation.
b. (5 points) Supercritical fluid chromatography (SFC) may be an alternative solution to
question (a) above.
i. (4 points) What changes to the instrumentation are required?
The column needs to be placed in an oven and a restrictor (back-pressure
device) needs to be added to control the temperature and pressure of the
mobile phase.
ii. (1 points) Your answer to (a) above has an analog (or similar method) in SFC,
what is it?
A pressure gradient
Chemistry 421/821 – Third Exam Spring 2010 page 5
Name
Answer Key
7) (10 points) Give the following slab electrophoresis gel separating various proteins from complex
mixtures:
a. (5 points) What factors determine the relative position of each band in the slab? (i.e., why
is protein OmpA near the bottom of the gel and protein YaeT is near the top of the gel?)
The charge-to-size ratio of an analyte (protein) is an important factor in its mobility.
A small, high charged protein migrate faster; large, low charged protein migrates
slower. Basically, this indicates OmpA has a significantly smaller molecular-weight
relative to YaeT
b. (5 points) What is a likely explanation for the bands corresponding to proteins OmpC and
OmpA being significantly broader than the bands for proteins YaeT, FepA and CirA
(ignore any contributions due to concentration differences)?
Joule heating – uneven heating in the slab caused by the electric field resulting in
temperature variations across the slab and band broadening.
Chemistry 421/821 – Third Exam Spring 2010 page 6
Name
Answer Key
8) (21 points) Given the following electrochemical cell for a potentiometric experiment:
SCE||H+(1.00x10-2M),V3+ (3.00x10-2M), VO2+(2.00x10-3M)|Pt
a. (10 points) Calculate the cell potential (ignore activity coefficients) given a 0.244V
potential for the SCE reference electrode.
Ecell = Eindicator – Ereference
Indicator ½-reaction: VO2+ + 2H+ + e-
 V3+ + H2O
Eo = 0.359
Ecell = (0.359 – 0.0592log [V3+]/[VO2+][H+]2) – 0.244
Ecell = (0.359 – 0.0592log [3.00x10-2]/[2.00x10-3][1.00x10-2]2) - 0.244
Ecell = (0.359 – 0.0592log[2.00x10-2]/(2.0x10-7)) – 0.244
Ecell = (0.359 – 0.0592log(1.00x105)) – 0.244
Ecell = (0.359 – (0.0592)(5.1761)) – 0.244
Ecell = (0.359 – 0.306) – 0.244
Ecell = 0.053 – 0.244 = -0.191
Chemistry 421/821 – Third Exam Spring 2010 page 7
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Answer Key
b. (5 points) If the Pt electrode is replaced with a liquid membrane electrode that binds V3+
(hint: similar to the dialkyl phosphate liquid membrane electrode example discussed in
class that bound Ca2+), how would the mechanism of detecting the indicator potential
change? (i.e., how does the liquid membrane electrode work relative to a Pt electrode?)
The Pt electrode is a metallic indicator electrode of the fourth kind and only provides
a source/sink for electrons. Conversely, a liquid membrane electrode works by
sensing the potential that develops at the two interfaces of the organic membrane
caused by the differences in the V3+ concentrations in the analyte solution and the
fixed reference solution. A potential arises du to a separation in charge, two different
V3+ concentrations results in a charge differential across the membrane.
c. (3 points) If you used activity coefficients to solve question (a) above (assume all the
activity coefficients are approximately the same value), how would the cell potential
change (decrease, increase, no change)?
Increase - Since all the activity coefficients are equal and less than 1, the effect would
be to increase the effective equilibrium constant (the concentrations are multiplied by
a number > 1, one small number divided by the product of two small numbers) which
increases the indicator electrode potential and thus the overall cell potential (less
negative).
d. (3 points) You experimentally measure the potential for the above electrochemical cell and
the measured potential still differs from your calculated cell potential, even using activity
coefficients. Suggest one possible source of error between the calculated and measured cell
potentials.
Liquid junction potentials, and/or any currents in the electrochemical cell, and/or a
polarization effects. i.e., E = IR
Chemistry 421/821 – Third Exam Spring 2010 page 8
Name
Answer Key
9) (10 points) Suppose you want to separate Cd from Zn using a coulometric titration at a constant
potential (potentiostatic). The mixture initially contains 0.0100 M Cd(NH3)42+, 0.0100 M
Zn(NH3)42+ and a relatively constant concentration of 1.00 M free NH3. Given the following two
1/2 – reactions:
a. (5 points) What cell potential would you need to maintain to reduce 99.9% of the
Cd(NH3)42+ (final concentration of 1x10-5M) without reducing Zn(NH3)42+?
b. (5 points) At what cell potential would you begin to reduce Zn(NH3)42+?