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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?

Short Answer

Expert verified
For the packed column, the optimum flow rate is \( u = \sqrt{\frac{B}{C}} \) and H is minimized. For the open-tubular column, without A, calculate \( u \) and H similarly using \\ \( H = \frac{B}{u} + Cu \). Compare theoretical plates by \( N = \frac{L}{H} \).

Step by step solution

01

Understanding the Van Deemter Equation

The Van Deemter equation is used to describe the relationship between the height equivalent to a theoretical plate (HETP or Hi) and mobile phase velocity (u). It is given by the equation: \[ H = A + \frac{B}{u} + Cu \] where A represents the eddy diffusion path, B/u is the longitudinal diffusion, and Cu is the mass transfer term.
02

Setting Up the Graph

To graph H versus u, we calculate H for different values of u between 5 and 120 mL/min using the Van Deemter equation: \[ H = 1.65 + \frac{25.8}{u} + 0.0236u \]. Calculate H for several values of u and then plot these calculated points on a graph.
03

Identifying Dominant Terms

For different flow rates (u), identify which term (A, B/u, or Cu) in the Van Deemter equation is dominant. Typically, the B/u term is more significant at low flow rates, the A term can be significant over a wide range, and the Cu term is usually more pronounced at high flow rates.
04

Finding Optimum Flow Rate for Packed Column

The optimum flow rate occurs when H is minimized. To find this flow rate, take the derivative of the Van Deemter equation with respect to u and set it to zero: \[ H'(u) = -\frac{B}{u^2} + C = 0 \]. Solving for u gives us \( u = \sqrt{\frac{B}{C}} \). Substitute B and C to find the optimum u, then calculate H at this u.
05

Calculating Optimum for Open-Tubular Column

Without the A term, the Van Deemter equation becomes \( H = \frac{B}{u} + Cu \). Repeat the optimization process: \[ H'(u) = -\frac{B}{u^2} + C = 0 \]. Solve for u with only B and C to find the new optimum u for this situation, and calculate the corresponding H.
06

Comparing Theoretical Plates

The number of theoretical plates \( N \) is related to the height of a theoretical plate \( H \) by \( N = \frac{L}{H} \), where L is the column length. Compare values of H from steps 4 and 5 to see how much more efficiency (greater N) is achieved with the open-tubular column compared to the packed column.

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Key Concepts

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

Theoretical Plates
The concept of theoretical plates is crucial in chromatography and helps us understand the efficiency of a separation process. A theoretical plate can be thought of as a hypothetical zone where solute equilibrium is achieved between two phases.
The more theoretical plates a column has, the more efficient it is in separating compounds.
  • The number of theoretical plates, often denoted by \( N \), indicates how well the column can separate different components.
  • An ideal column with an infinite number of plates would perfectly separate components.
  • Mathematically, \( N \) is related to the height of a theoretical plate \( H \), through the equation \( N = \frac{L}{H} \), where \( L \) is the length of the column.
When we reduce the value of \( H \), the column's efficiency increases, meaning you get better separation with fewer meters of column length. In the Van Deemter equation, the various terms influence \( H \), which in turn affects the number of theoretical plates.
Optimum Flow Rate
Achieving an optimum flow rate is vital for efficient separation in gas chromatography (GC). The Van Deemter equation helps us understand that different terms are affected by changes in the flow rate \( u \).
  • Low flow rates increase longitudinal diffusion (term \( \frac{B}{u} \)), resulting in wider peaks and less efficient separation.
  • High flow rates enhance the mass transfer effect (term \( Cu \)), causing band broadening and reduced separation quality.
To find the optimum flow rate, you must balance these effects. Mathematically, the optimum flow rate is where the derivative of the Van Deemter equation equals zero:
\[H'(u) = -\frac{B}{u^2} + C = 0\]This condition occurs when \( u = \sqrt{\frac{B}{C}} \). At this flow rate, \( H \) is minimized, meaning the separation is most efficient, providing the best compromise between different sources of dispersion and ensuring sharp, well-resolved peaks.
GC Separation
Gas chromatography (GC) is a powerful technique for separating volatile components in a sample. In GC, the sample is volatilized and carried by an inert gas (the mobile phase) through a column packed with a stationary phase.
Separation occurs due to differing affinities of the compounds for the mobile and stationary phases. Compounds with higher affinity for the stationary phase take longer to elute.
  • GC separation efficiency is described using the Van Deemter equation, which considers multiple factors that can cause peak broadening.
  • Understanding the contributions of eddy diffusion, longitudinal diffusion, and mass transfer is key to optimizing GC separations.
  • The goal is to achieve narrow peaks that are well-separated, something that is highly influenced by the conditions outlined in the Van Deemter equation.
Improving separation involves selecting the right column, adjusting the flow rate, and maintaining suitable temperature conditions to ensure the sample components elute in well-defined peaks, allowing for accurate identification and quantification.
Mass Transfer Term
The mass transfer term in the Van Deemter equation (\( Cu \)) is essential for understanding peak broadening in chromatography. This term represents the finite time required for solute molecules to move between the stationary and mobile phases.
High flow rates can worsen this effect because molecules have less time to equilibrate between phases as they pass through the column rapidly.
  • When the \( Cu \) term is dominant, it indicates that the separation process is limited by the speed at which solutes can transfer between phases.
  • Smooth, efficient mass transfer results in less band broadening and sharper peaks.
  • Reducing the mass transfer effect involves optimizing the flow rate close to the point where \( Cu \) is minimized.
An understanding of the mass transfer term allows chromatographers to fine-tune the process, altering the flow rate to ensure solutes spend an ideal amount of time in the column. This ensures you maximize separation efficiency and achieve clear, distinct peaks for each component in your mixture.

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Most popular questions from this chapter

Ohta and Tanaka reported on an ion-exchange chromatographic method for the simultaneous analysis of several inorganic anions and the cations \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) in water. \({ }^{28}\) The mobile phase includes the ligand 1,2,4 -benzenetricarboxylate, which absorbs strongly at \(270 \mathrm{nm}\). Indirect detection of the analytes is possible because its absorbance decreases when complexed with an anion. (a) The procedure also calls for adding the ligand EDTA to the mobile phase. What role does the EDTA play in this analysis? (b) A standard solution of \(1.0 \mathrm{mM} \mathrm{NaHCO}_{3}, 0.20 \mathrm{mM} \mathrm{NaNO}_{2}, 0.20\) \(\mathrm{mM} \mathrm{MgSO}_{4}, 0.10 \mathrm{mM} \mathrm{CaCl}_{2},\) and \(0.10 \mathrm{mM} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) gives the following peak areas (arbitrary units). \(\begin{array}{lcccc}\text { ion } & \mathrm{HCO}_{3}^{-} & \mathrm{Cl}^{-} & \mathrm{NO}_{2}^{-} & \mathrm{NO}_{3}^{-} \\ \text {peak area } & 373.5 & 322.5 & 264.8 & 262.7 \\\ \text { ion } & \mathrm{Ca}^{2+} & \mathrm{Mg}^{2+} & \mathrm{SO}_{4}^{2-} & \\\ \text { peak area } & 458.9 & 352.0 & 341.3 & \end{array}\) Analysis of a river water sample (pH of 7.49 ) gives the following results. \(\begin{array}{lcccc}\text { ion } & \mathrm{HCO}_{3}^{-} & \mathrm{Cl}^{-} & \mathrm{NO}_{2}^{-} & \mathrm{NO}_{3}^{-} \\ \text {peak area } & 310.0 & 403.1 & 3.97 & 157.6 \\ \text { ion } & \mathrm{Ca}^{2+} & \mathrm{Mg}^{2+} & \mathrm{SO}_{4}^{2-} & \\ \text { peak area } & 734.3 & 193.6 & 324.3 & \end{array}\) Determine the concentration of each ion in the sample. (c) The detection of \(\mathrm{HCO}_{3}^{-}\) actually gives the total concentration of carbonate in solution \(\left(\left[\mathrm{CO}_{3}^{2-}\right]+\left[\mathrm{HCO}_{3}^{-}\right]+\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]\right) .\) Given that the \(\mathrm{pH}\) of the water is \(7.49,\) what is the actual concentration of \(\mathrm{HCO}_{3}^{-}\) ? (d) An independent analysis gives the following additional concentrations for ions in the sample: \(\left[\mathrm{Na}^{+}\right]=0.60 \mathrm{mM} ;\left[\mathrm{NH}_{4}^{+}\right]=0.014\) \(\mathrm{mM}\); and \(\left[\mathrm{K}^{+}\right]=0.046 \mathrm{mM}\). A solution's ion balance is defined as the ratio of the total cation charge to the total anion charge. Determine the charge balance for this sample of water and comment on whether the result is reasonable.

The amount of caffeine in an analgesic tablet was determined by HPLC using a normal calibration curve. Standard solutions of caffeine were prepared and analyzed using a \(10-\mu L\) fixed-volume injection loop. Results for the standards are summarized in the following table. \begin{tabular}{cc} concentration \((\mathrm{ppm})\) & signal (arb. units) \\ \hline 50.0 & 8354 \\ 100.0 & 16925 \\ 150.0 & 25218 \\ 200.0 & 33584 \\ 250.0 & 42002 \end{tabular} The sample is prepared by placing a single analgesic tablet in a small beaker and adding \(10 \mathrm{~mL}\) of methanol. After allowing the sample to dissolve, the contents of the beaker, including the insoluble binder, are quantitatively transferred to a \(25-\mathrm{mL}\) volumetric flask and diluted to volume with methanol. The sample is then filtered, and a \(1.00-\mathrm{mL}\) aliquot transferred to a \(10-\mathrm{mL}\) volumetric flask and diluted to volume with methanol. When analyzed by HPLC, the signal for caffeine is found to be \(21469 .\) Report the milligrams of caffeine in the analgesic tablet.

A series of polyvinylpyridine standards of different molecular weight was analyzed by size-exclusion chromatography, yielding the following results. \begin{tabular}{cc} formula weight & retention volume (mL) \\ \hline 600000 & 6.42 \\ 100000 & 7.98 \\ 20000 & 9.30 \\ 3000 & 10.94 \end{tabular} When a preparation of polyvinylpyridine of unknown formula weight is analyzed, the retention volume is \(8.45 \mathrm{~mL}\). Report the average formula weight for the preparation.

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.

The concentrations of \(\mathrm{Cl}^{-}, \mathrm{NO}_{3}^{-},\) and \(\mathrm{SO}_{4}^{2-}\) are determined by ion chromatography. A \(50-\mu \mathrm{L}\) standard sample of \(10.0 \mathrm{ppm} \mathrm{Cl}^{-}, 2.00 \mathrm{ppm}\) \(\mathrm{NO}_{3}^{-},\) and \(5.00 \mathrm{ppm} \mathrm{SO}_{4}^{2-}\) gave signals (in arbitrary units) of \(59.3,\) \(16.1,\) and 6.08 respectively. A sample of effluent from a wastewater treatment plant is diluted tenfold and a \(50-\mu \mathrm{L}\) portion gives signals of 44.2 for \(\mathrm{Cl}^{-}, 2.73\) for \(\mathrm{NO}_{3}^{-},\) and 5.04 for \(\mathrm{SO}_{4}^{2-} .\) Report the parts per million for each anion in the effluent sample.

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