1 Chemistry 215 – Fundamentals of Analytical Chemistry Final Exam December 9, 2003 100 POINTS Name KEY SEE EQUATIONS AND DATA ON LAST PAGE COMPREHENSIVE PORTION (50 POINTS) PART I. DEFINITIONS (9 Points) Define the following terms as they relate to the topics covered in this course. 1. ppm (Write the full name and typical units) (3 Points) parts per million, mg/L or μg/mL 2. Kb (Write the general chemical reaction and the related mathematical expression for Kb) (3 Points) B + H2O ⇌ BH + + OH - [B] ] ][BH [OH K b + − = 3. Transmittance (3 Points) The fraction of incident light that passes through a sample.
Transcript
1
Chemistry 215 – Fundamentals of Analytical Chemistry Final Exam December 9, 2003
100 POINTS Name KEY
SEE EQUATIONS AND DATA ON LAST PAGE
COMPREHENSIVE PORTION (50 POINTS)
PART I. DEFINITIONS (9 Points) Define the following terms as they relate to the topics covered in this course.
1. ppm (Write the full name and typical units) (3 Points)
parts per million, mg/L or µg/mL
2. Kb (Write the general chemical reaction and the related mathematical expression for Kb) (3 Points)
B + H2O ⇌⇌⇌⇌ BH+ + OH- [B]
]][BH[OHKb
+−
=
3. Transmittance (3 Points)
The fraction of incident light that passes through a sample.
CHE 215 Final Exam December 9, 2003
2
PART II. CALCULATIONS (30 Points) Show all calculations in order to earn partial credit. For question 4, use the correct number of significant figures.
4. A sample of deicing salt was to be analyzed for chloride content using Fajan’s method. Recall that for this method, chloride in a sample is titrated with a standard silver solution to form a silver chloride precipitate:
Ag+ + Cl- → AgCl(s)
A 1.1034 g sample of the deicing salt was weighed into an Erlenmeyer flask and 50 mL of water, 0.1 g of dextrin, and 5 drops of dichlorofluorescein indicator was added to the flask. It took 39.44 mL of 0.5042 M AgNO3 to reach the endpoint of the titration.
a. Calculate the mass of chloride in the sample. (8 Points)
Since the reaction is 1:1 Ag:Cl, if we know the number of moles of Ag we then know the number of moles of Cl:
Moles Ag = Moles Cl = (0.03944 L)(0.5042 M AgNO3) = 0.01989 mol mass Cl = (0.01989 mol)(35.45 g/mol Cl) = 0.7051 g
b. Calculate the wt% chloride in the sample. (2 Points)
wt % = (0.7051 g Cl) / (1.1034 g sample) ×××× 100 = 63.90%
c. BONUS Based on this %Cl, what do you think the chemical formula of the deicing salt is? You must show calculations to earn full credit. (3 Points)
It’s likely to be CaCl2. Since there are 2 mol of Cl per mole of CaCl2:
63.89%100g/mol 110.98g/mol) (35.45 2 =×
CHE 215 Final Exam December 9, 2003
3
5. Calculate the pH if 10.00 mL of 0.200 M HCl is added to 50.00 mL of 0.0500 M
diethylamine (CH3CH2)2NH. (10 Points) (CH3CH2)2NH2+ ⇌ (CH3CH2)2NH + H+ Ka = 1.17 × 10-11 First, we need to know the number of moles of each species initially presnt: (CH3CH2)2NH: (0.05000 L)(0.0500 M) = 0.0025 mol H+: (0.01000 L)(0.200 M) = 0.0020 mol
We also need to know pKa: pKa = -log(1.17 ××××10-11) = 10.93
Now we can use the Henderson Hasselbalch Equation:
10.320.002000.00050log10.93pH =
+=
CHE 215 Final Exam December 9, 2003
4
6. A new water-soluble anti-diabetic drug with a molecular weight of 208.4 g/mol has an
absorption maximum at 281 nm and a molar absorptivity at this wavelength of 5.12 × 103 M-1cm-1. One tablet of the drug was crushed and quantitatively transferred to a 100 mL volumetric flask and the flask was filled to the mark with deionized water. This solution had an absorbance of 0.691 at 281 nm. A blank was prepared by treating a tablet with identical composition but without the anti-diabetic drug in an identical manner. The absorbance of this blank solution was 0.048 at 281 nm in a 1.00-cm cuvet. Calculate the mg of the anti-diabetic drug in the tablet. (10 Points)
Calculate the concentration of the final solution from Beer’s Law:
0.000126M)(1.00cm)cmM10(5.12
0.048)(0.691εbAC 113 =
×−== −−
To find the mass, multiply concentration by volume and molecular weight:
mg 2.63g
mg 1000mol
208.4g0.10000LL
mol 0.000126(mg) mass =
=
CHE 215 Final Exam December 9, 2003
5
PART III. MULTIPLE CHOICE (12 Points—2 Points Each) Choose the BEST answer for the following questions.
7. Six samples of canned tuna were measured for mercury content. The average of the six measurements was 11.4 ng/g and the standard deviation was 2.1 ng/g. What is the correct expression for the confidence interval (95% confidence level) of these measurements?
a. 5
)ng/g 1.2)(571.2(ng/g 4.11CI ±=
b. 6
ng/g) 1(2.571)(2.ng/g 11.4CI ±=
c. 6
)ng/g 1.2)(776.2(ng/g 4.11CI ±=
d. 5
)ng/g 1.2)(447.2(ng/g 4.11CI ±=
e. 6
)ng/g 1.2)(447.2(ng/g 4.11CI ±=
8. In laboratory experiment number 3, water hardness is determined by titrating Ca2+ and
Mg2+ with EDTA:
Ca2+ and Mg2+ + EDTA4- → CaEDTA2- and MgEDTA2-
This is an example of a. an acid-base titration. b. a precipitation titration. c. a redox titration. d. a complexation tritration e. a back titration.
9. Given that the H+ concentration of a solution is 2.50 × 10-6 M, what is the solution pH expressed to the appropriate number of significant figures?
a. 5.6 b. 5.60 c. 5.602 d. 5.6021 e. 5.60206
10. Which of the solutions should have the highest buffer capacity? a. 0.00010 M H2PO4- + 0.0002 M HPO42- b. 0.1 M H2PO4- c. 0.10 M H2PO4- + 0.2 M HPO42- d. 6 M HNO3 e. 6 M NaOH
CHE 215 Final Exam December 9, 2003
6
11. The purpose of the flame in atomic emission spectrometry is to provide
a. ground state atoms. b. ground state molecules. c. excited state molecules. d. exited state atoms.
12. Which statement about fluorescence is false? a. In general, fluorescence is a more sensitive technique than absorbance. b. Many types of molecules can absorb, but few types can fluoresce. c. Fluorescence intensity is proportional to concentration d. Fluorescence intensity is proportional to P0. e. All of these statements are true.
CHE 215 Final Exam December 9, 2003
7
COMPREHENSIVE PORTION (50 POINTS) PART IV. DEFINITIONS (9 Points) Define the following terms as they relate to the topics covered in this course.
13. Theoretical Plate (3 Points)
A hypothetical discreet section on a chromatography in which a solute molecule equilibrates between the mobile and stationary phase.
14. Stationary Phase (3 Points)
The substance that stays fixed inside a chromatographic column.
15. Isocratic Elution (3 Points)
Elution (in HPLC) with a single solvent or constant solvent mixture. PART V. CALCULATIONS (6 Points) Show all calculations in order to earn partial credit.
16. To the right is a portion of a chromatogram. The following questions pertain to this chromatogram.
a. Calculate the number of theoretical plates (N)
for the peak at 13.81 min. (3 Points)
plates 19,000min) (0.4
min) 16(13.81w
16tN 2
2
2
2r ===
b. Estimate the resolution between the peaks at 12.98 and 13.20 min. (3 Points)
0.55min) (0.4
min) 12.98-(13.20w
tresolutionave
r ==∆=
CHE 215 Final Exam December 9, 2003
8
PART VI. SHORT ANSWER (10 Points) **ANSWER ONLY TWO OF THE FOLLOWING THREE QUESTIONS**
17. Describe the difference between split and splitless injections for GC. Explain under what circumstances you would use each. (5 Points)
Splitless – all of the carrier gas (mobile phase) and therefore all of the sample is swept into the column. This is used for trace analyses and for packed columns.
Split – A large portion of the carrier gas is vented to waste. Therefore, only a small fraction (~1:10 to 1:250) is injected into the chromatography column. This is used primarily for capillary columns.
18. Name and describe two advantages of open tubular columns over packed columns in gas chromatography. (5 Points)
1. Better separations – there is essentially no multiple path band broadening, and the equilibration time between stationary and mobile phases is fast, giving a smaller C term. 2. Lower operating pressures – do not need to force flow through packed column, so much longer columns can be used. This results in more theoretical plates.
19. Sketch a block diagram showing the components of a typical gas chromatograph. Label and briefly describe each component. (5 Points)
Injector: allows introduction of sample Column Oven: controls column T Column: Where the separation takes place Detector: Produces signal as solutes elute from the column Computer/Recorder: Displays data and often controls instrument.
CHE 215 Final Exam December 9, 2003
9
PART VII. MATCHING! (10 Points) Select the ONE best answer.
O 20. A relatively simple, universal gas chromatography detector
G 21. A gas chromatography detector that has no response to water.
K 22. The “A” term of the van Deemter Equation
I 23. The “B” term of the van Deemter Equation
F 24. The “C” term of the van Deemter Equation
N 25. A polar gas chromatography stationary phase
D 26. The solution or gas (mobile phase + solutes) entering the chromatographic column.
C 27. The solution or gas exiting the column.
B 28. The most selective kind of chromatography; it employs specific interactions between an immobilized molecule and a solute
H 29. A type of chromatography in which charged solutes are attracted to the stationary phase by electrostatic forces.
A. Adsorption Chromatography
B. Affinity Chromatography
C. Eluate
D. Eluent
E. Elution
F. Equilibration between Stationary and
Mobile Phases Term
G. Flame Ionization Detector
H. Ion Exchange Chromatography
I. Longitudinal Diffusion Term
J. Molecular Exclusion Chromatography
K. Multiple Path Term
L. Partition Chromatography
M. Polydimethylsiloxane
N. Polyethylene Glycol
O. Thermal Conductivity Detector
CHE 215 Final Exam December 9, 2003
10
PART VIII. MULTIPLE CHOICE (16 Points—2 Points Each) Choose the BEST answer for the following questions.
30. Which of the following most accurately depicts the variation of H vs. flow rate (i.e. the
van Deemter equation)? a.
b.
c.
d.
31. Of the following compounds, which would you expect to elute last from a gas chromatography column?
a. Methanol (CH3OH) b. Ethanol (CH3CH2OH) c. n-Propanol (CH3CH2CH2OH) d. n-Butanol (CH3CH2CH2CH2OH) e. n-Pentanol (CH3CH2CH2CH2CH2OH)
32. Of the following compounds, which would you expect to give the largest peak on a flame ionization column? Assume equal concentrations of each compound.
a. Methanol (CH3OH) b. Ethanol (CH3CH2OH) c. n-Propanol (CH3CH2CH2OH) d. n-Butanol (CH3CH2CH2CH2OH) e. n-Pentanol (CH3CH2CH2CH2CH2OH)
Flow Rate
H
Flow Rate
H
Flow Rate
H
Flow Rate
H
CHE 215 Final Exam December 9, 2003
11
33. In HPLC, when the mobile phase composition is changed during the separation, this is called
a. isothermal. b. isocratic elution. c. temperature programming. d. gradient elution.
34. Smaller plate heights mean a. better separation. b. poorer separation. c. lower numbers of theoretical plates. d. Both (a) and (c) are correct. e. Both (b) and (c) are correct.
35. In gas chromatography, if a solute is retained too long it can be made to elute more quickly by
a. using a longer column. b. using a higher column temperature. c. using a lower column temperature. d. using a more polar solvent.
36. In HPLC, smaller stationary phase particles result in a. higher operating pressures for the same flow rate. b. smaller plate heights. c. better separations. d. higher column costs. e. All of the above are correct.
37. In HPLC, the “HP” stands for a. Hewlett-Packard. b. high pressure. c. high performance.
