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(a) Make a graph showing retention times of peaks 6, 7, and 8 in Figure 25-12 as a function of %acetonitrile (%B). Predict the retention time of peak 8 at 45% B.

(b) Linear-solvent-strength model: In Figure 25-12, tm = 2.7 min. Compute k for peaks 6, 7, and 8 as a function of %B. Make a graph of log k versus Φ, where Φ= %B/100. Find the equation of a straight line through a suitable linear range for peak 8. The slope is -S and the intercept is log kw. From the line, predict tr for peak 8 at 45% B and compare your answer with (a).

(c) Gradient elution: A linear eluent gradient from 40 to 80% acetonitrile over 30 min is performed on the column in Figure 25-12. Assuming a dwell volume of 0 mL, use your data from (b) to plot the retention factor of peaks 6 and 8 during the gradient. What are the general characteristics of the plot?

(d) Why are the peaks in a gradient separation sharp?

Short Answer

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(a)

Fig1: graph showing retention times of peaks 6, 7, and 8

The retention time of peak 8 at 45% B is 38.6 min

(b)

Fig2: graph showing log k vs solvent composition of peaks 6, 7, and 8

From the trend line of peak 8 the linear equation obtained

Therefore, atΦ=0.45, k=14.65 and tr=42.255 min. The value obtained from graph in part (a) is 38.6 min which is slight less than that obtained in part (b).

(c)

For both the peaks retention factor is high at first and then decreases gradually following exponential behaviour until the compounds are almost unretained at the end of the gradient. The graph obtained with exponential trend line and R2 value is shown below.

(d) In gradient elution the zones are sharpened due to the compression caused by the different retention factors at the leading and trailing part of the peak. By the time a compound is eluted, it is almost unretained and has the peak width of a weakly retained compound.

Step by step solution

01

Explanation regarding part (a)

(a)

From figure 25-12 we observed the retention time of peak 6,7 and 8 and plotted the values as a function of %B(acrylonitrile). The table used to plot the graph is shown above.

The graph for peak 6, 7 and 8 obtained from the table is plotted below. It is asked topredict the retention time of peak 8 at 45% B. As per assumption at 45% B peak 8 may be eluted halfway between 40% B and 50% B which is around 45 min.But as per the curve fitting it suggests 38.6 min is a more accurate time. The value obtained is shown by dotted line here.

Fig1: graph showing retention times of peaks 6, 7, and 8

02

Explanation regarding part (b)

(b)

In Figure 25-12, tm = 2.7 min. We need to find k for peaks 6, 7, and 8 as a function of %B and need to plot a graph of log k versus Φ, where Φ= %B/100. The data regarding log k value is shown in the table below.

The retention factor (k) can be calculated using the following equation

k=tr-tmtm

tr = Retention time

The graph using the above table is plotted below:

Fig2: graph showing log k vs solvent composition of peaks 6, 7, and 8

Now we need to find the equation of a straight line through a suitable linear range for peak 8.

From the trend line of peak 8 the linear equation obtained logk=-3.2682ϕ+2.6364

For 45 % B lets plug inΦ=0.45 in the linear equation

localid="1660827850931" logk=-3.2682×0.45+2.6364logk=1.16571k=14.65k=tr-tmtmtr-tmtm=14.65tr-2.72.7=14.65tr=42.255

Therefore, atΦ=0.45, k=14.65 and tr=42.255 min. The value obtained from graph in part (a) is 38.6 min which is slight less than that obtained in part (b)

03

Explanation regarding part (c)

A linear eluent gradient from 40 to 80% acetonitrile over 30 min is performed on the column in Figure 25-12. Assuming a dwell volume of 0 mL, data from (b) was used to plot the retention factor of peaks 6 and 8 during the gradient. From the graph it was obtained that for both the peaks retention factor is high at first and then decreases gradually following exponential behavior until the compounds are almost unretained at the end of the gradient. The graph obtained with exponential trend line and R2 value is shown below.

Fig 3: graph showing retention times of peaks 6, 7, and 8 for gradient elution

04

Explanation regarding part (d)

(d) In gradient elution the zones are sharpened due to the compression caused by the different retention factors at the leading and trailing part of the peak. By the time a compound is eluted, it is almost unretained and has the peak width of a weakly retained compound.

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Most popular questions from this chapter

In monolithic columns60 the stationary phase is a single porous piece of silica or polymer filling the entire column and synthesized within the column from liquid precursors. Monolithic columns offer similar plate height to HPLC particles, but with less resistance to flow. Therefore, faster flow or longer columns can be used. The figure shows separation of isotopic molecules on a long monolithic column. Packed columns have too much resistance to flow to be made so long.

Separation of isotopic molecules on a 440-cm-long monolithic C18-silica column eluted withCH3CN/H2O(30: 70 vol/vol) at 308C. [Data from K. Miyamoto, T. Hara, H. Kobayashi, H. Morisaka, D. Tokuda, K. Horie, K. Koduki, S. Makino, O. Nuñez, C. Yang, T. Kawabe, T. Ikegami, H. Takubo, Y. Ishihama, and N. Tanaka, “High-Efficiency Liquid Chromatographic Separation Utilizing Long Monolithic Silica Capillary Columns,” Anal. Chem. 2008, 80, 8741.]

(a) Unretained thiourea is eluted in 41.7 min. Find the linear velocity ux (mm/s).

(b) Find the retention factor k forC6D6

(c) Find the plate number N and plate height forC6D6

(d) Assuming that the peak widths forC6H5Dand C6D6are the same as that of C6D6, find the resolution of C6H5Dand C6D6.

(f) If we just increased the column length to increase N, what value of N and what column length would be required for a resolution of 1.000?

(g) Without increasing the length of the column, and without changing the stationary phase, how might you improve the resolution?

(h) When the solvent was changed fromCH3CN/H2O(30:70 vol/vol) toCH3CN/CH3OH/H2O(10:5:85), the relative retention for C6H5D andC6D6increased to 1.0088 and the retention factor for C6H6 changed to 17.0. If the plate number were unchanged, what would be the resolution?

To which kinds of analytes do these liquid chromatography detectors respond?

(a) ultraviolet

(b) refractive index

(c) evaporative light scattering

(d) charged aerosol

(e) electrochemical

(f) fluorescence

(g) nitrogen chemiluminescence

(h) conductivity

Question: Explain how to use a gradient for the first run to decide whether isocratic or gradient elution would be more appropriate.

25-3 What length of column packed with2μmparticles is needed to yield the same plate number as in Fugure 25.12? How long woald the separation take?

The figure shows the separation of two enantiomers on a chiral stationary phase.


Sepation of enantiomers of Ritalin by HPLC with a chiral stationary phase.[Data from R.Bakhitar,L.Ramos,and F.L.S.Tse, “Quantification of methlylphenidate in plasma using chiral Liquid-chromatography/Tandem mass spectrometry: Application to Toxicokinetric studies,”Anal Chim Acta 2002,469,261.]

(a)From trandw1/2 find N for each peak.

(b) Fromtrandw1/2find the resolution.

(c)Giventm=1.62min, use Equation23-23with the average N to predict the resolution.

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