/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 A series of polyvinylpyridine st... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
The average formula weight is approximately 50,000 using interpolation.

Step by step solution

01

Understand the Relationship

In size-exclusion chromatography, larger molecules elute faster and have lower retention volumes than smaller molecules. Hence, a smaller retention volume corresponds to a larger molecular weight. We need to interpolate the formula weights for unknown retention volumes.
02

Create a Calibration Curve

Plot the given formula weights against the retention volumes. The molecular weight (y-axis) decreases as the retention volume (x-axis) increases, indicating a trend that can be approximated by a linear or a polynomial relationship.
03

Analyze Data

Look at the given data: - For 6.42 mL, the formula weight is 600,000. - For 7.98 mL, the formula weight is 100,000. - For 9.30 mL, the formula weight is 20,000. - For 10.94 mL, the formula weight is 3,000. Calculate the empirical relationship between retention volume and formula weight using an appropriate statistical method like linear regression.
04

Perform Interpolation

Estimate the molecular weight for the unknown retention volume of 8.45 mL. Use interpolation with the generated trendline to find the corresponding molecular weight value. This might involve calculating a line or curve equation from the calibration data you plotted before.
05

Calculate Approximate Formula Weight

From interpolation, the formula weight corresponding to a retention volume of 8.45 mL is determined using the linear or polynomial equation derived in Step 3.

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Key Concepts

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

Molecular Weight Analysis
Molecular weight analysis in the context of size-exclusion chromatography is an essential process for determining the size of molecules in a sample. In size-exclusion chromatography, substances are separated based on their hydrodynamic volume. Larger molecules pass through the column more quickly because they cannot enter the smaller pores of the stationary phase and thus have lower retention volumes. Smaller molecules, on the other hand, enter the pores and take longer to elute, resulting in higher retention volumes.
The molecular weight of a sample is an important property that influences its physical characteristics and behavior in solution. Knowing the molecular weight allows scientists to determine the size of polymers and proteins and compare them to known standards for accurate analysis.
In our original exercise, we performed a molecular weight analysis using polyvinylpyridine standards with known molecular weights. By comparing these standards to the sample of unknown molecular weight, we can estimate its size using the principles of size-exclusion chromatography.
Retention Volume
Retention volume in chromatography refers to the volume of the mobile phase required to elute a solute from the column. It is a crucial indicator of molecular size in size-exclusion chromatography.
In the original exercise, various polyvinylpyridine standards with different molecular weights provided different retention volumes. For instance, a formula weight of 600,000 had a retention volume of 6.42 mL, whereas a weight of 3,000 had a retention volume of 10.94 mL. This demonstrates that larger molecules have lower retention volumes as they navigate the column faster.
By understanding these retention times, we can determine the molecular weight of unknown samples by comparing the retention volume of the unknown to those of the standards.
Calibration Curve
A calibration curve is created by plotting the known molecular weights against their respective retention volumes in size-exclusion chromatography. This graph allows us to visualize the relationship between molecular weight and retention volume, and most importantly, to predict unknown values.
In our exercise, the calibration curve needed to be drawn using data points from the polyvinylpyridine standards. As retention volumes increase, the molecular weights decrease, indicating a certain inverse relationship. Each point on the curve corresponds to a known standard's molecular weight and respective retention volume.
This curve helps when predicting an unknown's molecular weight, as you can interpolate the retention volume obtained from the analysis and determine its molecular size based on the established trend.
Linear Regression
Linear regression is a statistical method used to model the relationship between a dependent variable and one or more independent variables. It has a significant application in creating a calibration curve for size-exclusion chromatography.
In the original exercise, linear regression can help us approximate the curve that best fits the relationship between formula weights and retention volumes. By plotting the data points and applying linear regression, we obtain a mathematical equation that describes the line of best fit. This line can be used to estimate unknown molecular weights when given their retention volumes.
For instance, given the retention volume of 8.45 mL for our unknown sample, we can use the linear regression equation derived from the calibration curve to determine its approximate molecular weight. This technique ensures that we use the most accurate line possible to make predictions.

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Most popular questions from this chapter

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.

Bohman and colleagues described a reversed-phase HPLC method for the quantitative analysis of vitamin \(\mathrm{A}\) in food using the method of standard additions. \({ }^{27}\) In a typical example, a \(10.067-\mathrm{g}\) sample of cereal is placed in a 250 -mL Erlenmeyer flask along with \(1 \mathrm{~g}\) of sodium ascorbate, \(40 \mathrm{~mL}\) of ethanol, and \(10 \mathrm{~mL}\) of \(50 \% \mathrm{w} / \mathrm{v} \mathrm{KOH}\). After refluxing for \(30 \mathrm{~min}, 60 \mathrm{~mL}\) of ethanol is added and the solution cooled to room temperature. Vitamin \(\mathrm{A}\) is extracted using three \(100-\mathrm{mL}\) portions of hexane. The combined portions of hexane are evaporated and the residue containing vitamin A transferred to a \(5-\mathrm{mL}\) volumetric flask and diluted to volume with methanol. A standard addition is prepared in a similar manner using a \(10.093-\mathrm{g}\) sample of the cereal and spiking with \(0.0200 \mathrm{mg}\) of vitamin \(\mathrm{A}\). Injecting the sample and standard addition into the HPLC gives peak areas of, respectively, \(6.77 \times 10^{3}\)and \(1.32 \times 10^{4}\). Report the vitamin \(\mathrm{A}\) content of the sample in milligrams/100 g cereal.

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.

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.

Loconto and co-workers describe a method for determining trace levels of water in soil. \(^{19}\) The method takes advantage of the reaction of water with calcium carbide, \(\mathrm{CaC}_{2}\), to produce acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}\). By carrying out the reaction in a sealed vial, the amount of acetylene produced is determined by sampling the headspace. In a typical analysis a sample of soil is placed in a sealed vial with \(\mathrm{CaC}_{2}\). Analysis of the headspace gives a blank corrected signal of \(2.70 \times 10^{5} .\) A second sample is prepared in the same manner except that a standard addition of \(5.0 \mathrm{mg} \mathrm{H}_{2} \mathrm{O} / \mathrm{g}\) soil is added, giving a blank-corrected signal of \(1.06 \times 10^{6} .\) Determine the milligrams \(\mathrm{H}_{2} \mathrm{O} / \mathrm{g}\) soil in the soil sample.

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