Chemical Analysis - Francis Rouessac

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and the resolution R between these two peaks. Assume that the efficiency N is the same for these two solutes, i.e. the value given by the manufacturer.Knowing that the phase ratio for such a column may be expressed as: where SA is the specific area of the material contained in the column and SV is the specific volume of the pores, calculate the phase ratio in the case where SA = 330 m2 g‐1 and SV = 1 ml g‐1 (units adapted to the direct application of the relationship giving β).Comment. The factor 0.6 corresponds to the column’s packing rate with silica and 0.45 corresponds to the silica volume in ml for 1 gram.Knowing the phase ratio, calculate, for solutes 4 and 5, the Nernst distribution factor. Do these two values correspond to the elution order obtained?

      7 Using the internal normalization method, we want to determine the mass composition of a sample made of four butyric esters. A reference solution of these four esters (with known mass concentrations) leads to the following relative response factors for methyl ester (ME), ethyl ester (EE), and propyl ester (PE) as compared to butyl ester (BE): From the information provided by the chromatogram of the sample being assayed, find the mass composition of this mixture.Peak No.tR CompoundArea (arbitrary units)12.54Methyl ester (ME)2,340.123.47Ethyl ester (EE)2,359.035.57Propyl ester (PE)4,077.347.34Butyl ester (BE)4,320.7

      8 Assay of serotonin S (5‐hydroxytryptamine) using the internal standard method.We sample 1 ml of the solution to assay, to which we add 1 ml of a solution containing 3 ng of N‐methylserotonin (NMS). We treat this mixture to remove all other intrusive compounds from it. We perform solid‐phase extraction to isolate serotonin and its methyl derivative, diluted in a suitable phase.Why do we add the internal standard compound before the extraction stage?Calculate the response factor of serotonin with respect to N‐methylserotonin, knowing that the calibration chromatogram leads to the following results:‐ serotonin area30,885,982 μV squantity injected: 5 ng‐ N‐methylserotonin area30,956,727 μV squantity injected: 5 ngBased on the chromatogram of the sample solution, find what the serotonin concentration is in the starting sample, knowing that the serotonin peak area is equal to 2,573,832 μV s and the N‐methylserotonin peak area is 1,719,818 μV s.

      1 We have tR = tM(1+k). And tM = L/ū and k = KVS/VM, hence tR = (L/ū)(1+KVS/VM).

      2 Knowing that VR = tR∙D, we get: α = (VR(2) – VM)/(VR(1) ‐ VM), i.e. α = 1.2.

      3 To transform Eq. (P1.1) into Eq. (P1.2), we add k, from tM = tR/(1+k).

      4 When peaks are neighbours, base widths are generally comparable. By posing ω1 = ω2 and by expressing ω2 with its value as a function of N2, we obtain R = ¼ √N2 (tR(2) – tR(1))/tR(2). Then, we add k, by using tR = tM(1 + k); finally, we introduce α = k2/k1;The relationship given in the problem is obtained by rearranging the relationship for R (as a function of k and α) and substituting the relationship for Neff from Eq. (P1.2) of problem 1.3.

      5 From the base relationship R = 2(tR(2) – tR(1))/(ω1 + ω2), we can replace ω with its value depending on N, ω = 4tR /√N and since tR = tM(1 + k), we obtain Eq. (P1.5);We multiply the second member of (1), numerator and denominator by (k1 + k2) and we obtain α (α = k2/k1).

      6 ω0.1 = a + b = 2A ∙ a; A = (a + b)/2a => b = a(2A − 1) => a/b = 1/(2A − 1) => where N = 53,000 plates, tR = 1.58 min and A = 1.07, and thus we get:a = 0.0142 min.b = 0.0162 min.ω0.1 = 0.0304 min.k = (tR − tM)/ tMtherefore, k4 = 4.43 and k5 = 6.62; α4‐5 = k5 /k4 = 1.47; (this assumes that the peak is Gaussian)β = 136.6;K = k β; K4 = 605 and K5 = 884; K5 > K4 means that solute 5 has more affinity for the stationary phase than solute 4; therefore, it is more retained and exits later.

      7 %(ME) = 16.6; %(EE) = 16.6; %(PE) = 33.4; %(BE) = 33.4.

      8 By adding N‐methylserotonin before extraction, we do not have to take into consideration any potential loss of product due to the various manipulations. We suppose that the extraction yield is the same for these two compounds, which are very similar;kS/NMS = 1.002;serotonin mass: 45 ng/ml.

      Note

      1 1 The symbols used follow IUPAC recommendations – Pure and Applied Chemistry, 65(4), 819 (1993).

      Objectives

      Represent a GC device

      Choose the carrier gas

      Choose the column

      Compare the stationary phases

      Know the injection methods

      List the main detectors

      Optimize a separation

      Address new orientations of micro GC and fast GC

      Explain retention indexes and constants of stationary phases

      Analysis starts when a very small quantity of sample is introduced in either liquid or gas form into the injector, which has the dual function of vaporizing the sample and mixing it with the gas flow at the head of the column. The column is usually a narrow‐bore tube that coils around itself with a length that can vary from 1 m to over 100 m, depending upon the type and the contents of the stationary phase. The column, which can serve for thousands of successive injections, is housed in a temperature‐controlled oven. At the end of the column, the mobile phase (carrier gas) passes through a detector before it exits to the atmosphere. Some gas chromatograph models of reduced size have their own electrical supply, enabling them to operate in the field (Figure 2.1).

      In GC, there are four operational parameters for a given stationary phase: L, length of the column; u, velocity of the mobile phase (which affects the theoretical efficiency N); T, temperature of the column; and β, phase ratio, which affects the retention factor k. The settings of the chromatograph enable modifications in terms of T and u, and therefore both the efficiency of the column and the retention factors can be adjusted as well.

      (Source:

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