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Haddad and associates report the following retention factors for the reversed- phase separation of salicylamide and caffeine. \({ }^{25}\) \(\begin{array}{ccccccc}\% \text { methanol } & 30 \% & 35 \% & 40 \% & 45 \% & 50 \% & 55 \% \\ k_{\text {sal }} & 2.4 & 1.6 & 1.6 & 1.0 & 0.7 & 0.7 \\\ k_{\text {caff }} & 4.3 & 2.8 & 2.3 & 1.4 & 1.1 & 0.9\end{array}\) (a) Explain the trends in the retention factors for these compounds. (b) What is the advantage of using a mobile phase with a smaller \(\% \mathrm{v} / \mathrm{v}\) methanol? Are there any disadvantages?

Short Answer

Expert verified
As methanol percentage increases, retention factors decrease, improving elution speed. Lower methanol enhances separation but lengthens run time.

Step by step solution

01

Understanding Retention Factors

The retention factor, denoted as \( k \), is a measure of how long a compound is retained in the chromatography column. A higher \( k \) value indicates that the compound interacts more with the stationary phase and thus spends more time in the column.
02

Analyzing Retention Factor Trends

As the percentage of methanol increases from 30% to 55%, the retention factors \( k_{\text{sal}} \) (for salicylamide) and \( k_{\text{caff}} \) (for caffeine) both decrease. This trend indicates that as the polarity of the mobile phase increases (more methanol), the compounds are retained less by the stationary phase and elute faster.
03

Explaining the Trend

Methanol is a polar solvent, and increasing its proportion in the mobile phase increases the overall polarity of the mobile phase. Salicylamide and caffeine are less retained because the polar mobile phase tends to dissolve and carry these polar compounds more effectively, reducing their retention times.
04

Discussing Advantages of Lower Methanol Percentage

A lower percentage of methanol (such as 30%) leads to higher retention factors (\( k = 2.4 \) for salicylamide and \( k = 4.3 \) for caffeine), which implies better separation between the compounds. It can be advantageous as it might enhance resolution between closely eluting compounds.
05

Discussing Disadvantages of Lower Methanol Percentage

The disadvantage of using a lower percentage of methanol is longer analysis time since both compounds elute more slowly. This can lead to peak broadening, reducing separation efficiency over longer runs, and may not be suitable for analyzing columns that may degrade over extended periods.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reversed-Phase Chromatography
Reversed-phase chromatography is a common method used to separate and analyze compounds in a mixture. In this technique, the stationary phase is usually non-polar, while the mobile phase is polar.
This setup contrasts with normal-phase chromatography, where the stationary phase is polar. Reversed-phase chromatography is particularly popular because it can separate molecules that have varying degrees of hydrophobicity.
Compounds are separated based on their interactions with the stationary and mobile phases. Molecules that are less polar will tend to interact more with the non-polar stationary phase, while polar molecules will more readily move with the polar mobile phase.
Thus, reversed-phase chromatography is ideal for separating species based on differences in polarity and is widely used in chemical and biological analyses.
Stationary Phase
In reversed-phase chromatography, the stationary phase is the non-moving material inside the column. It is non-polar, typically made up of hydrophobic materials like C18 chains bonded to silica.
The stationary phase holds back less polar molecules longer than polar ones due to hydrophobic interactions, thus aiding in their separation.
The effectiveness and selectivity of the stationary phase are crucial for achieving good separation because they dictate how different compounds will be retained and eluted.
  • Retention Science: Retention in the stationary phase depends on the interaction strength between a compound and the non-polar material.
  • Tuning Separation: By choosing different types or modifications of stationary phases, separations can be tuned for specific compounds.
Mobile Phase
The mobile phase in reversed-phase chromatography is usually a mixture of water with a polar solvent, such as methanol or acetonitrile. The mobile phase flows through the column, carrying the sample mixture along.
The polar nature of the mobile phase contrasts with the non-polar stationary phase, creating a dynamic in which compounds are carried and separated depending on their polarity.
Adjusting the composition of the mobile phase is a common way to optimize the separation of different compounds:
  • Composition Effects: Increasing the proportion of methanol in the mobile phase increases overall polarity, which can decrease retention factors as shown in the given exercise.
  • Elution Impact: A more polar mobile phase can carry polar compounds more swiftly, leading to faster elution times.
Polarity
Polarity plays a central role in determining how substances behave in chromatographic separations. It refers to the distribution of electrical charge over atoms joined by a bond.
Polar molecules, like water or methanol, have asymmetric charge distribution resulting in positive and negative sides. Non-polar molecules, meanwhile, have more balanced charge distribution.
Understanding polarity is crucial in chromatography, particularly in reversed-phase chromatography, because the interaction of compounds with the stationary and mobile phases depends significantly on their polarity.
  • Intermolecular Interactions: Polar compounds tend to dissolve better in polar solvents, impacting how quickly they move through the column with the mobile phase.
  • Separation Optimization: By carefully selecting mobile and stationary phases of different polarities, specific separation goals can be achieved.
Chromatographic Separation
Chromatographic separation refers to the process of separating individual components of a mixture in chromatography. In reversed-phase chromatography, this separation is based primarily on differences in polarity and interaction strength with the stationary and mobile phases.
The goal is to elute distinct compounds at separate times to allow for their analysis. Various factors like the composition of mobile phase, the type of stationary phase, and the compound's polarity influence how well separation occurs.
For effective chromatographic separation, several adjustments might be necessary:
  • Gradient Elution: Gradually changing the mobile phase composition can lead to better separation of compounds, particularly if they elute closely together.
  • Resolution Balance: Aiming for good resolution without unnecessarily long elution times is key to efficient chromatographic work.

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Most popular questions from this chapter

The analysis of \(\mathrm{NO}_{3}^{-}\) in aquarium water is carried out by CZE using \(\mathrm{IO}_{4}^{-}\) as an internal standard. A standard solution of \(15.0 \mathrm{ppm} \mathrm{NO}_{3}^{-}\) and 10.0 ppm \(\mathrm{IO}_{4}^{-}\) gives peak heights (arbitrary units) of 95.0 and 100.1 , respectively. A sample of water from an aquarium is diluted 1: 100 and sufficient internal standard added to make its concentration \(10.0 \mathrm{ppm}\) in \(\mathrm{IO}_{4}^{-}\). Analysis gives signals of 29.2 and 105.8 for \(\mathrm{NO}_{3}^{-}\) and \(\mathrm{IO}_{4}^{-},\) respectively. Report the \(\mathrm{ppm} \mathrm{NO}_{3}^{-}\) in the sample of aquarium water.

Moody studied the efficiency of a GC separation of 2 -butanone on a dinonyl phthalate packed column. \(^{16}\) Evaluating plate height as a function of flow rate gave a van Deemter equation for which \(A\) is \(1.65 \mathrm{~mm}\), \(B\) is \(25.8 \mathrm{~mm} \cdot \mathrm{mL} \min ^{-1},\) and \(C\) is \(0.0236 \mathrm{~mm} \cdot \min \mathrm{mL}^{-1}\) (a) Prepare a graph of \(H\) versus \(u\) for flow rates between \(5-120 \mathrm{~mL} / \mathrm{min}\). (b) For what range of flow rates does each term in the Van Deemter equation have the greatest effect? (c) What is the optimum flow rate and the corresponding height of a theoretical plate? (d) For open-tubular columns the \(A\) term no longer is needed. If the \(B\) and \(C\) terms remain unchanged, what is the optimum flow rate and the corresponding height of a theoretical plate? (e) Compared to the packed column, how many more theoretical plates are in the open-tubular column?

The composition of a multivitamin tablet is determined using an HPLC with a diode array UV/Vis detector. A \(5-\mu L\) standard sample that contains 170 ppm vitamin C, 130 ppm niacin, 120 ppm niacinamide, 150 ppm pyridoxine, 60 ppm thiamine, 15 ppm folic acid, and 10 ppm riboflavin is injected into the HPLC, giving signals (in arbitrary units) of, respectively, \(0.22,1.35,0.90,1.37,0.82,0.36,\) and \(0.29 .\) The multivitamin tablet is prepared for analysis by grinding into a powder and transferring to a \(125-\mathrm{mL}\) Erlenmeyer flask that contains \(10 \mathrm{~mL}\) of \(1 \%\) \(\mathrm{v} / \mathrm{v} \mathrm{N} \mathrm{H}_{3}\) in dimethyl sulfoxide. After sonicating in an ultrasonic bath for \(2 \mathrm{~min}, 90 \mathrm{~mL}\) of \(2 \%\) acetic acid is added and the mixture is stirred for \(1 \mathrm{~min}\) and sonicated at \(40^{\circ} \mathrm{C}\) for \(5 \mathrm{~min}\). The extract is then filtered through a \(0.45-\mu \mathrm{m}\) membrane filter. Injection of a \(5-\mu \mathrm{L}\) sample into the HPLC gives signals of 0.87 for vitamin C, 0.00 for niacin, 1.40 for niacinamide, 0.22 for pyridoxine, 0.19 for thiamine, 0.11 for folic acid, and 0.44 for riboflavin. Report the milligrams of each vitamin present in the tablet.

Suppose you need to separate a mixture of benzoic acid, aspartame, and caffeine in a diet soda. The following information is available. \begin{tabular}{lcccc} & \multicolumn{3}{c} {\(t_{\mathrm{r}}\) in aqueous mobile phase of \(\mathrm{pH}\)} \\ compound & 3.0 & 3.5 & 4.0 & 4.5 \\ \hline benzoic acid & 7.4 & 7.0 & 6.9 & 4.4 \\ aspartame & 5.9 & 6.0 & 7.1 & 8.1 \\ caffeine & 3.6 & 3.7 & 4.1 & 4.4 \end{tabular} (a) Explain the change in each compound's retention time. (b) Prepare a single graph that shows retention time versus \(\mathrm{pH}\) for each compound. Using your plot, identify a pH level that will yield an acceptable separation.

A mixture of \(n\) -heptane, tetrahydrofuran, 2 -butanone, and \(n\) -propanol elutes in this order when using a polar stationary phase such as Carbowax. The elution order is exactly the opposite when using a nonpolar stationary phase such as polydimethyl siloxane. Explain the order of elution in each case.

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