Chapter 15

Liquid Chromatography

Problem 15.1: Albuterol (a drug used to fight asthma) has a lipid:water KOW (see “Profile—Other

Applications of Partition Coefficients”) value of 0.019. How many grams of albuterol would

remain in the aqueous phase if 0.001 grams of albuterol initially in a 100 mL aqueous solution

were allowed to come into equilibrium with 100 mL of octanol?

C (aq) ⇌ C (org)

Initial (g) 0.001 g 0 Change -y +y Equil 0.001-y y

𝐾𝑂𝑊 = 0.019 = 𝑦

100 𝑚𝐿(0.001−𝑦)

100𝑚𝐿⁄

1.9 x 10-5 – 0.019y = y 1.019y = 1.9 x 10-5

y = 1.864x10-5 g albuterol would be in the octanol layer Problem 15.2: (a) Repeat Problem 15.1, but instead of extracting the albuterol with 100 mL of octanol, extract the albuterol solution five successive times with only 20 mL of octanol per extraction. After each step, use the albuterol remaining in the aqueous phase from the previous step as the starting quantity for the next step. (Note: the total volume of octanol is the same in both extractions). (b) Compare the final concentration of aqueous albuterol to that obtained in Problem 15.1.

Discuss how the extraction results changed by conducting five smaller extractions instead of one

large one. What is the downside to the multiple smaller volume method in this exercise compared

to the single larger volume method in Problem 15.1?

(a) Each step will have this general form of calculation: 𝐾𝑂𝑊 =𝑦

20 𝑚𝐿(𝑔𝑖𝑛𝑖𝑡−𝑦)

100𝑚𝐿⁄

so we can calculate the g transferred to octanol from 𝑦 = (𝐾𝑂𝑊)(𝑔𝑖𝑛𝑖𝑡)

5+𝐾𝑂𝑊

We can use a spreadsheet to calculate the successive steps:

(b) In Prob 15.1, the final concentration of albuterol in the water layer would be (1-1.864 x 10-2)mg/0.1L = 9.8136 mg/L In part (a) above, the final concentration of albuterol in the aqueous layer is (1-1.878 x 10-2)mg/0.1L = 9.8122 mg/L In this particular case, the successive extractions only yielded slightly higher extraction of the albuterol, probably because the KOW is so low. In general, we would expect a higher net transfer, but the risk of doing multiple extractions is that errors associated with each step are propagated multiple times. Problem 15.3: Albuterol has a lipid:water KOW value of 0.019. If you were to conduct a chromatographic separation using a polar solvent (i.e., methanol) and a nonpolar stationary phase, would the albuterol’s distribution favor the mobile phase or the stationary phase? Explain your answer.

We know that 𝐾𝐷 = 𝐾𝑂𝑊 = [𝐶]𝑜𝑟𝑔

[𝐶]𝑎𝑞. A large KOW value would indicate a preferential affinity for

an organic or nonpolar phase vs. an aqueous or polar phase. The very low value of KOW indicates that the albuterol would have a higher affinity for the polar solvent than the nonpolar stationary phase. Problem 15.4: Caffeine has a lipid:water KOW value of 0.79. If a mixture of albuterol and caffeine

were to be separated using a polar solvent (i.e., methanol) and a nonpolar stationary phase,

which component would elute first (having the shorter retention time)? Would the two

components elute close to each other—or far apart? Explain.

First extraction: KOW= 0.019

Init (g) 0.001

g in Octanol: 3.78561E-06

g left in aq 0.000996214 Total albuteral transferred:

1.87853E-05

Second Extraction:

Init (g) 0.000996214

g in Octanol: 3.77128E-06

g left in aq 0.000992443

Third Extraction:

Init (g) 0.000992443

g in Octanol: 3.75701E-06

g left in aq 0.000988686

Fourth Extraction:

Init (g) 0.000988686

g in Octanol: 3.74278E-06

g left in aq 0.000984943

Fifth Extraction:

Init (g) 0.000984943

g in Octanol: 3.72862E-06

g left in aq 0.000981215

Caffeine (KOW=0.79) has a roughly 40 times greater affinity for the nonpolar stationary phase than does albuterol (KOW = 0.019). The albuterol will have a much shorter retention time than caffeine – they would not be expected to elute close to each other at all. Problem 15.5: Sometimes it is desirable to determine the retention (R) of an analyte as the

difference between the observed retention time (tr) and the void time (tm). Using the

chromatogram in Figure 15.4, determine the retention (R) of compounds A and B.

Compound A: R = tr – tm = 6.6-1.6 = 5.0 min Compound B: R = tr – tm = 10.4 – 1.6 = 8.8 min Problem 15.6: Sometimes you will see chromatographic peaks identified by the volume of mobile

phase that was required to elute the sample instead of the time it took for the sample to elute.

Using the retention times seen in Figure 15.4, calculate the retention volumes of compounds A

and B assuming a flow rate of 1 mL/min. Repeat this exercise assuming a flow rate of 0.7 mL/min.

Flow rate of 1 mL/min: Compound A: Vr = tr x rate = (6.6 min)x(1 mL/min) = 6.6 mL

Compound B: Vr = 10.4 min Flow rate of 0.7 mL/min: Compound A = (6.6 min)(0.7 mL/min) = 4.6 mL Compound B = 7.3 mL Problem 15.7: Determine the retention factors (kr values) for analytes A and B seen in Figure

15.4.

Compound A: kr = tr−tm

tm=

6.6 min − 1.6 min

1.6 min = 3.125 = 3.1

Compound B: kr = tr−tm

tm=

10.4 min − 1.6 min

1.6 min = 5.5

Problem 15.8: Calculate the retention factor for each peak in the chromatograph seen in Example

15.3 and create a results table that might be used in a report.

**Note: Met-Enk is labeled as having a tr=10.1, but it

clearly comes before the 10-minute mark, estimated at

around 8.7 min.

tr

Substance (min) kr

tm 2.1 ---

Gly-Tyr 3.3 0.57

Val-Tyr-Val 6.7 2.19

Met-Enk 10.1 3.81

Met-Enk (actual) 8.7 3.14

Problem 15.9: Use the chromatogram from Example 15.3 to determine the retention times and

retention volumes of each of the peaks in this sample and create a results table that might be

included in a report.

**Note: Met-Enk is labeled as having a tr=10.1, but it

clearly comes before the 10-minute mark, estimated at

around 8.7 min.

Problem 15.10: Use the chromatograms in panels (B) and (C) from Example 15.4 and estimate the resolution of the two-component system under each of these conditions. What challenges do you face for estimating σ for panel (B)? In terms of the value of the separation, discuss the significance of the resolution in panels (A), (B), and (C) with respect to the speed of the analysis (throughput), quantitative analysis, and qualitative analysis. *** Hint – scan or photo the image and then bring it up in Paint – you can use the pixel index (lower left) for each position and use interpolation to calculate the time of each point. Panel B: Wh,1 = 1.33 min Wh,1 = 2.355σ1 σ1 = 0.5656 Wb1 = 4σ = 2.262 Wh,2 = 1.35 min Wh,2 = 2.355σ1 σ2 = 0.5732 Wb2 = 4σ = 2.293

Rs = tr2−tr1

(Wb2+Wb1)/2=

3.5−2

(2.293+2.262)/2 = 0.66

The challenging thing about Panel B, and the thing that introduces error, is the need to mentally extend the peaks beyond the dip that lies between the two peaks Our measurement of width at half-height is based on an estimate of where the sides of the peaks are. Panel C: Wh,1 = 1.24 min Wh,1 = 2.355σ1 σ1 = 0.5265 Wb1 = 4σ = 2.106 Wh,2 = 1.25 min Wh,2 = 2.355σ1 σ2 = 0.5307 Wb2 = 4σ = 2.123

Rs = tr2−tr1

(Wb2+Wb1)/2=

4.75−2.0

(2.123+2.106)/2 = 1.3

The resolution increases from B to A to C. However, in order to improve resolution, we need to extend the run time, which will decrease throughput. Usually this will increase peak width. Depending upon your analytical needs, this might be an acceptable trade-off. Problem 15.11: In a chromatographic experiment, two components elute with retention times

of 11.52 and 12.76 minutes, respectively. The half-height widths of the peaks were determined

tr Vr

Substance (min) (mL)

tm 2.1 ---

Gly-Tyr 3.3 3.30

Val-Tyr-Val 6.7 6.70

Met-Enk 10.1 10.10

Met-Enk (actual) 8.7 8.70

to be 0.37 and 0.42 minutes, respectively. Assuming Gaussian peak shapes, estimate the

resolution between the two peaks.

Rs = tr2−tr1

(Wb2+Wb1)/2=

12.76−11.52

(0.37+0.42)/2 = 3.1

Problem 15.12: Use Equation 15.15 to determine Nt for each peak in the chromatogram found in Example 15.4, panel (C). Using values found in Problem 15.10:

Nt = 16 (tr

Wb)

2

= 16 (2.0

2.106)

2

= 14.4 Nt = 16 (tr

Wb)

2

= 16 (4.75

2.123)

2

= 80.1

Problem 15.13: Use Equation 15.16 to determine Nt for each peak in the chromatogram found in

Example 15.4, panel (A).

Nt = 5.54 (tr

Wh)

2

= 5.54 (2.0

1.3)

2

= 13.1 Nt = 5.54 (tr

Wh)

2

= 5.54 (4.14

1.2)

2

= 65.9

Problem 15.14: Relate the multistep extraction process described in Problem 15.2 to the concept

of theoretical plates.

In a chromatographic experiment, extraction is happening many times over as the solute band

moves through the stationary phase. Even the small (0.014%) increase in efficiency seen from

using 5 extractions over 1 extraction becomes important if it is repeated hundreds or thousands

of times throughout the experiment.

Problem 15.15: List at least three undesirable effects of band broadening.

Broader peaks are more likely to overlap with nearby peaks (decreasing resolution). Broader peaks yield less accurate quantitative results – the “tails” of the peaks spread out and blend with the baseline, so we lose accuracy when integrating for the area of the peaks. Broader peaks take more time to elute, decreasing throughput. Broader peaks have a less distinct maximum, yielding lower accuracy in determining tr. With broad peaks, fewer components of a mixture can be separated in a given period of time. Problem 15.16: Make a summary table, perhaps the size of an index card, that identifies the equations that define KD, Rs, Nt, kr, and a. Be sure to define the underlying terms (like tm, Wb, H, and s) that are used in these equations.

To be included Problem 15.17: In the analysis of some analgesic preparations using a polar stationary phase and a mobile phase composed of 10% water and 90% methanol, four compounds were seen to elute in the following order: caffeine, acetaminophen, aspirin, ibuprofen. The last two compounds were widely separated by several minutes. How might you decrease the time required for the experiment while maintaining adequate resolution? Both aspirin and ibuprofen contain carboxylic acid groups, which are highly polar. They will exhibit a high affinity for the polar stationary phase, which is why they took a long time to elute. If we switched from isocratic to gradient elution, starting with 10% water / 90% methanol, but then gradually increasing the polar (water) content of the mobile phase as the experiment proceeded, we would be able to pull the polar solutes from the column more effectively. Problem 15.18: If a nonpolar stationary phase were used in the above experiment, what might be the elution order? We would expect the order to be, generally, reversed, although if we stuck with the 90% methanol solution, we might not be able to separate caffeine and acetaminophen. Problem 15.19: When using a nonpolar stationary phase with 10% water and 90% methanol, the first two components co-elute in a single peak. How might you change the conditions of the experiment to resolve those components?

Changing the solvent will have the greatest chance of success. The simplest change would be to

try isocratic with just a change of solvent, perhaps 50/50 or more polar such as 70/30

water/methanol. Or, we could switch to gradient elution, starting with a higher polarity ratio

and then gradually moving toward the 10/90 ratio that both solutes seem to have an affinity

for.

Problem 15.20: You have conducted an LC analysis of three different headache medicines. The

column was a C18, the mobile phase was 50:50 methanol:water, and the detector was a UV-vis

spectrometer (output measured in AU or absorbance units). The detector response for four

different standards and for the three samples is presented here. Construct a calibration curve,

perform a linear regression, and determine the concentration of acetaminophen in each sample.

Problem 15.21: Assuming a flow rate of 1 mL/min, determine the retention volume of peak #6 from the profile “Analysis of Wine: Qualitative and Quantitative”. Peak #6 elutes at about tr = 17.1 min; with a 1mL/min flow rate, Vr = 17.1 mL. Problem 15.22: Describe how a gradient method could be used to shorten the run time for the

chromatograms seen in the profile “Analysis of Wine: Qualitative and Quantitative”. Be specific!

Where in the chromatogram would you change the mobile phase and how would you change it?

Since it is a reverse phase column, nonpolar solutes are preferentially retained, and we would

be starting with a relatively polar mobile phase with gradient shift to less polar. The first

labelled (and presumably first important) peak comes off after 4 minutes, and then there is a 6-

minute gap before the next labelled peak, and another 6 minutes before the next. Coming off

relatively early in the chromatogram, these 3 compounds would be (generally speaking) more

polar than compounds 4-19. So we might start with a moderately polar mobile phase, hold it

isocratic for about 5 minutes, then begin a gradient toward a less polar composition, slowly at

first to allow for the separation of peaks 4 and 5, then more rapidly after peak 12 elutes in

order to close the gap between 13,14 and 15.

Problem 15.23: An aqueous sample contained traces of the compounds listed in Table 15.1. The

octanol–water distribution coefficients are tabulated as pKD values. Assume you analyzed an

aliquot of the aqueous sample using high-performance LC with a mobile phase that was 75%

methanol and 25% water on a cellulose column. Predict the elution order and explain your

answer.

Std (ppm) AU Linest Block

1.00 0.098 m= 0.090824 0

2.00 0.211 sm= 0.003527 #N/A

4.00 0.402 R2= 0.995496 0.032518

8.00 0.699

Samples Conc (ppm)

A 0.375 4.13

B 0.296 3.26

C 0.801 8.82 y = 0.0908xR² = 0.9847

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.00 2.00 4.00 6.00 8.00

Ab

sorb

ance

Concentration (ppm)

Cellulose is a normal phase stationary phase – that is, it is a relatively polar stationary phase. Polar substances will be retained longer, while less polar substances will move along with the mostly nonpolar (Methanol) mobile phase. The pKD values in the table are identified as KOW constants, so a higher pKD value will have a lower affinity for nonpolar phases and a higher affinity for polar substances. That is, those substances with higher pKD values will elute later. Order of elution: 2,2-DMP : n-Pentane : 3-methylpropane : 2,2-DMB : 2,3-DMB Problem 15.24: Based solely on molecular polarity, predict the most likely order of elution for

each set of three compounds in a normal phase chromatographic experiment.

(A) The benzoic acid is the most polar so will be most greatly retained; the ethyl benzene is the

least polar, so will be least retained.

Order of elution: ethyl benzene : 2-phenyl-ethanol : benzoic acid

(B) p-dichlorobenzene : m-dichlorobenzene : o-dichlorobenzene

Problem 15.25: Assume that an aqueous sample contains equal trace amounts of the compounds listed in Table 15.1. If you extract the aqueous sample with an aliquot of octanol:

(a) List the compounds in order of increasing concentration in the octanol phase at equilibrium. (b) Assume that you analyzed an aliquot of the aqueous sample using high-performance LC with a mobile phase that was 75% methanol and 25% water on a C8 column. Make a sketch of the predicted chromatogram showing the expected elution order. Label each peak. Compare and contrast your answer with the one you gave in Problem 15.21.

(a) lowest to highest: 2,3-DMB < 2,2-DMB < 3-methylpropane < n-pentane < 2,2-DMP

(b) This should refer to Prob. 15.23, not Prob. 15.21 In this case, we have a nonpolar column with a partially polar stationary phase. Based solely on the KD values given, we would expect the elution order to be reversed, and basically identical to the order of concentrations given in (a) 2,3-DMB : 2,2-DMB : 3-methylpropane : n-pentane : 2,2-DMP Problem 15.26: Based solely on molecular polarity, predict the most likely order of elution for

each set of three compounds for a reversed phase chromatographic experiment. Compare your

answer with the one you gave in Problem 15.22.

This problem should refer to Problem 15.24, not 15.22 As in Problem 15.25, based only on molecular polarity the elution order should be reversed with respect to Problem 15.24 because we are now using a nonpolar column rather than a polar one: (A) Order of elution: benzoic acid : 2-phenyl-ethanol : ethyl benzene

(B) o-dichlorobenzene : m-dichlorobenzene : p-dichlorobenzene

Problem 15.27: Describe schematically how you might incorporate a detection system in the

column chromatography system depicted in Figure 15.25. (There is more than one correct

answer.)

There are a variety of ways we could do this. We could use a fiber optic to bring light from a source to the exit tube below the stopcock with a return fiber on the opposite side returning the transmitted light to the detector. We could use a low-flow peristaltic pump to pump the eluent through a UV-Vis or FTIR flow-cell. Problem 15.28: Using Figure 15.29, make a plot of the instrument’s response as a function of the concentration of compound A. The concentrations used to obtain the chromatograms were 1.0, 2.0, and 3.0 mM respectively. This type of plot is called a sensitivity curve. To be included

Problem 15.29: Using Figure 15.29 and the plot you made in Problem 15.28: (a) Estimate the minimum detection quantity for compound A. (b) Estimate the instrument’s response to a 4 mM sample of compound A.

(c) Estimate the concentration of a sample that exhibited a retention time of 4.25 minutes and a

peak height of 0.366 AU.

To be include Exercise 15.1: Define the following terms:

a) Exclusion limit

b) Isocratic

c) Gradient

d) Mobile phase strength

e) Distribution constant

f) Stationary phase

g) Reverse phase

h) Normal phase

i) Ion exchange

j) Size exclusion

k) Eddy diffusion

l) Selectivity factor

m) Retention time

n) Elute

o) Resolution

p) Theoretical plates

q) Plate height

r) Void volume

s) van Deemter equation

t) Wb

u) Wh

v) Column efficiency

w) Hydrodynamic volume

x) Baseline resolution

Answers are in the text – to be included. Exercise 15.2: Assume that an aqueous sample contained traces of the following compounds listed in the table and that you extracted the aqueous sample with an aliquot of octanol. The distribution coefficients are given as pKD values.

(a) List the compounds in order of increasing concentration in the octanol phase at equilibrium. (b) Assume that you analyzed an aliquot of the aqueous sample using HPLC with a mobile phase

of 75% methanol and 25% water on a C8 column. Predict the elution order and make a sketch of the predicted chromatogram. Label each peak.

(a) The pKD values in the table are identified as KOW constants, so a higher pKD value will have a lower affinity for nonpolar phases and a higher affinity for polar substances. That is, those substances with higher pKD values be less soluble in octanol. Concentrations, lowest to highest: m-Xylene < p-Xylene < o-Xylene < Toluene < Benzene (b) In this case, we have a nonpolar column with a partially polar stationary phase. Based solely on the KD values given, we would expect the elution order to be basically identical to the order of concentrations given in (a) m-Xylene : p-Xylene : o-Xylene : Toluene : Benzene Exercise 15.3: Assume that an aqueous sample contained traces of the following compounds listed in the table and that you extracted the aqueous sample with an aliquot of octanol. The distribution coefficients are given as pKD values. (a) Assume that you analyzed an aliquot of the aqueous sample using HPLC with a mobile phase of 75% methanol and 25% water on a C8 column. Make a sketch of the predicted chromatogram and label each peak. (b) How would the chromatogram change if you repeated the experiment using a mobile phase of 50% methanol and 50% water?

To be included Exercise 15.4: Purification of a fragment of BRCA1 (amino acids 230–534) produced the chromatogram shown.

(a) Report retention times and retention volumes for each numbered peak assuming a flow rate of 1 mL/ min and a flow rate of 0.3 mL/min. (b) Assuming a tm value of 3 minutes, use the half-height method to estimate the resolution between peak 69 and 71. What inferences can you draw from this resolution value?

(a) Retention times are, presumably, as given: 49, 62, 69 and 71 minutes.

At a flow rate of 1 mL/minute, the Vrs would be the same: 49, 62, 69, 71 mL.

At a flow rate of 0.3 mL/minute, the Vrs would be 14.7, 18.6, 20.7, 21.3 mL

(b) Peak at 69 min: Wh = 69.1 – 68.7 = 0.4 min Wh = 2.355σ σ = 0.16985 Wb1 = 4σ = 0.679

Peak at 71: Wh = 71.2-70.6 = 0.6 min Wh = 2.355σ σ = 0.25478 Wb1 = 4σ = 1.019

𝑅𝑆 = (71−69)

(0.679+1.019)/2 = 2.36

Since Rs > 1.5 we can infer that the two peaks are fully resolved.

Exercise 15.5: One form of “gradient” chromatography one can use in an ion exchange separation is to alter the pH of the mobile phase over the course of the experiment. Likewise, you can select a pH that will render a weak acid either mostly ionized or mostly in its associated form. Use pKa values for the following weak acids to predict the percent ionization of each of the weak acids in a mobile phase with pH of 5. (Hint: The Henderson-Hasselbalch equation will prove very useful in solving this problem.)

We will select one acid, formic, to demonstrate how to solve this type of problem. The Henderson-Hasselbalch equation is

pH = pKa + log(𝑚𝑜𝑙𝑒𝑠 𝐶𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝐵𝑎𝑠𝑒

𝑚𝑜𝑙𝑒𝑠 𝐶𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝐴𝑐𝑖𝑑)

5.0 = 3.75 + log(𝑚𝑜𝑙𝑒𝑠 𝐶𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝐵𝑎𝑠𝑒

𝑚𝑜𝑙𝑒𝑠 𝐶𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝐴𝑐𝑖𝑑)

(𝑚𝑜𝑙𝑒𝑠 𝐶𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝐵𝑎𝑠𝑒

𝑚𝑜𝑙𝑒𝑠 𝐶𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝐴𝑐𝑖𝑑) = 17.78

Assuming we start with only formic acid, nCA = (nCA)init – nCB, or nCB = (nCA)init - nCA

((𝑛𝐶𝐴)𝑖𝑛𝑖𝑡−𝑛𝐶𝐴

𝑛𝐶𝐴) = 17.78

(nCA)init – nCA = 17.78nCA (nCA)init = 18.78 nCA nCA = 0.0532 (nCA)init The acid is 5.32% ionized Exercise 15.6: Determine Nt and H for each numbered peak in the chromatogram.

We will calculate the values for peak #2 to demonstrate the calculation. Because peak 2 is not fully resolved, we will use the half-height method to find Wb. Wh = 9.8 – 9.33 = 0.47

Nt = 5.54 (tr

Wh)

2

= 5.54 (9.58

0.47)

2

= 2301

A column length is not given, without which H cannot be calculated. We will assume a column length of 15 cm. Then, using Eq, 15.13

𝐻 = 𝐿

𝑁=

15 𝑐𝑚

2301 = 0.006519 cm = 65 μm

Exercise 15.7: Calculate the resolution and retention factor (kr) for each peak in the chromatogram found in Exercise 15.6. To be included Exercise 15.8: For each of the parameters given on the right-hand side of Equation 15.19, indicate what effect increasing the parameter has on the expected resolution between two closely spaced peaks. Increasing α and Nt will increase the resolution. Increasing kr will increase Rs, but as kr increases above 1, its effect diminishes. Exercise 15.9: Describe at least two advantages of a gradient chromatography method compared to an isocratic chromatography method.

In general, using gradient elution allows us to separate multiple peaks in less time, increasing throughput. Also, by careful programming of the solvent percentages, we can achieve separation between peaks that we could not normally do so. Exercise 15.10: Use one of these manufacturer’s websites (Agilent Technologies, Amersham Pharmacia Biotech, Beckman Coulter, Bioanalytical Systems, Gilson, Hitachi, PerkinElmer, Shimadzu Scientific, Varian, Waters Corporation) to find the chemical composition and structure of at least two normal phase stationary materials. To be included Exercise 15.11: Describe various ways one might alter the chromatographic experiment in order to improve the resolution of the two, circled peaks seen in the following chromatogram. Justify your suggestions.

To be included Exercise 15.12: In the HPLC analysis of caffeine in an energy drink by standard addition, the sample was prepared by pipetting 5.00 mL of the drink into a 50 mL volumetric flask, adding a fixed volume of a 1.205 mg/mL caffeine standard solution, and then diluting to volume using the HPLC mobile phase. The table summarizes the sample preparation protocol and the results obtained from the three injections:

Volume of energy drink added (mL)

Volume of caffeine standard added (mL)

Final dilution volume (mL)

Height of caffeine peak in chromatogram (arbitrary units)

Area of caffeine peak in chromatogram (arbitrary units)

5.00 0.00 50.00 24.79 408.3

5.00 0.50 50.00 46.77 831.7

5.00 2.00 50.00 80.40 1278.6

a. Use the peak height data to estimate the original concentration of caffeine in the energy drink. b. Use the peak area data to estimate the original concentration of caffeine in the energy drink.

c. Of the two values calculated in a and b above, which would you consider more reliable? Explain your answer. To be included