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Chemistry 215 – Fundamentals of Analytical Chemistry
Exam 4
April 14, 2004 50 POINTS Name KEY
SEE EQUATIONS AND DATA ON LAST PAGE
PART I. DEFINITIONS (9 Points) Define the following terms as they relate to the topics covered in this course.
- Eluate (3 Points)
The gas or liquid exiting the column, containing the mobile phase and the solutes.
- Isothermal Elution (3 Points)
A chromatographic separation carried out at a single temperature (usually in gas chromatography).
- Gradient Elution (3 Points)
A chromatographic separation in which the mobile phase composition changes during the run. Usually used in liquid chromatography.
PART II. CALCULATIONS AND SHORT ANSWER (33 Points) Show all calculations in order to earn partial credit.
- (14 Points) This question refers to the two chromatograms shown below.
a. One of the chromatograms was recorded with capillary gas chromatography, the other with high performance liquid chromatography. Label each chromatogram with the technique that was used for the separation. Explain in the space below how you arrived at your conclusion. (4 Points)
Although the plate height in GC is larger than in HPLC, the fact that GC columns can be much longer means that the total number of theoretical plates for GC is much larger. Therefore, the separation efficiency for GC is better, resulting in significantly narrower peaks. The peaks on the right- hand chromatogram are much narrower, indicating that this is probably the capillary GC separation. There are two other acceptable answers: column size and solute volatility.
b. Circle the correct technique (2 Points):
The technique that normally has the lowest plate height ( H ):
Capillary GC HPLC
The technique that normally has the largest number of theoretical plates ( N ):
Capillary GC HPLC
HPLC
Capillary GC
- (9 Points) This question refers to the two chromatograms shown below.
These chromatograms were recorded using reverse phase HPLC.
a. Give an example of a mobile phase that might be used for this type of separation (i.e. reverse phase). (2 Points)
Water/methanol, water/acetonitrile, etc.
b. Give an example of a detector that might be used for these compounds. ( Points)
uv-vis absorption – they are all aromatic organic compounds
c. BONUS! Give an example of a stationary phase that might be used for these separations. (2 Points)
C18 (ODS), C8, C4, Phenyl are all acceptable
d. Chromatogram B was recorded on the same column and with the same sample as chromatogram A, but the separation is very different. This is because the mobile phase was changed between the two separations. Explain how the mobile phase was changed to affect the separation so dramatically. Be specific. Use the back of this sheet if necessary. (5 Points)
Chromatogram B was recorded using a mobile phase with a lower eluent strength than chromatogram B. For example, Chromatogram A may have been recorded in 90% Methanol and 10% water, whereas Chromatogram B may have been recorded in 50% Methanol and 50% water. You must mention that the % organic modifier in Chromatogram A is greater than that in Chromatogram B.
Peaks:
- Benzyl alcohol 5. Ethyl benzoate
- Phenol 6. Toluene
- 3,4-Dimethoxyacetophenone 7. 2,6-Dimethoxytoluene
- Benzoin 8. o -Methoxybiphenyl
Chromatogram A
Chromatogram B
- (10 Points) The van Deemter equation relates the column efficiency (measured as H ) to the mobile phase flow rate ( u ):
Cu
u
B
H = A + +
a. Sketch a typical van Deemter plot that shows how H depends on u. Clearly label the axes. (4 Points)
b. There are three terms in the van Deemter equation: the A term, the B term, and the C term, each describing a different type of contribution to the band broadening. Explain the type of band broadening described by 2 of the 3 terms. (6 Points)
A Term: Multiple path term—solute molecules can take different paths through the stationary phase, so some take longer to go through column than others. This is term is independent of flow rate.
B Term: Longitudinal diffusion term—the longer time a solute stays on the column the more time it has to diffuse, therefore spreading out its concentration. This term is inversely proportional to flow rate (i.e. longer time on column with slower flow rate and vice versa)
C Term: Mass transfer term—this arises from slow equilibrium for solutes partitioning between the mobile and stationary phases. If the equilibrium is slow, while a solute molecule is in the stationary phase other molecules will have traveled down the column a certain distance. The overall result is band spreading that is proportional to the flow rate.
H
u
PART III. MULTIPLE CHOICE (10 Points—2 Points Each) Choose the BEST answer for the following questions.
- Of the following compounds, which would you expect to elute first from a gas chromatography column? a. Methanol (CH 3 OH) (lowest boiling point) b. Ethanol (CH 3 CH 2 OH) c. n-Propanol (CH 3 CH 2 CH 2 OH) d. n-Butanol (CH 3 CH 2 CH 2 CH 2 OH) e. n-Pentanol (CH 3 CH 2 CH 2 CH 2 CH 2 OH)
- Of the following compounds, which would you expect to elute last from a reverse-phase liquid chromatography column? a. Methanol (CH 3 OH) b. Ethanol (CH 3 CH 2 OH) c. n-Propanol (CH 3 CH 2 CH 2 OH) d. n-Butanol (CH 3 CH 2 CH 2 CH 2 OH) e. n-Pentanol (CH 3 CH 2 CH 2 CH 2 CH 2 OH) (least polar)
- Which type of injector is most commonly used for capillary GC chromatography? a. Purge and trap b. Splitless injector c. Split injector (for injecting small volumes) d. Loop injector
- Which of the following is not a common detector for gas chromatography? a. Refractive index detector (an HPLC detector) b. Flame ionization detector c. Thermal conductivity detector d. Mass spectrometry e. All of these are common detectors for gas chromatography.
- In gas chromatography, if two solutes with short retention times co-elute (i.e. are not resolved), what is the simplest way to attempt to resolve the peaks? a. Use a longer column b. Use a higher column temperature c. Use a lower column temperature (at lower T elution times are longer) d. Use a more polar solvent
BONUS QUESTION! TWO ANSWERS ARE CORRECT; CIRCLE THE CORRECT ANSWERS AND CROSS OUT THE INCORRECT ANSWERS! (4 Points, 1 for each correct circle or cross-out)
- In normal phase HPLC, a. the mobile phase is polar and the stationary phase is nonpolar. b. the mobile phase is nonpolar and the stationary phase is polar. c. eluent strength is increased by adding a more polar solvent to the mobile phase. d. eluent strength is increased by adding a less polar solvent to the mobile phase.
Equations and Data
2
2 2 1 / 2
w
t w
t N = r^ = r N
L
H =
ave
r w
∆ t resolution=
Cu u
B
H = A + +
