Chemical Analysis - Francis Rouessac

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alt="upper A prime Subscript 1 Baseline comma"/> upper A prime Subscript 2 Baseline comma and upper A prime Subscript 3 will give direct access to the percentage mass composition of the mixture represented by x1, x2, and x3 via three equations of the following form:

x Subscript i Baseline percent-sign equals StartFraction upper K Subscript i slash 3 Baseline dot upper A prime Subscript i Baseline Over upper K Subscript 1 slash 3 Baseline dot upper A prime Subscript 1 Baseline plus upper K Subscript 2 slash 3 Baseline dot upper A prime Subscript 2 Baseline plus upper A prime Subscript 3 Baseline EndFraction times 100 w i t h i equals 1 o r 2 o r 3 period

      The condition of normalization is that x1 + x2 + x3 = 100

      If the procedure is extrapolated to n components normalized to compound j (internal reference), a general expression for the response factor of a given compound i can be obtained:

      (1.46)upper K Subscript i slash j Baseline equals StartFraction upper C Subscript i Baseline dot upper A Subscript j Baseline Over upper C Subscript j Baseline dot upper A Subscript i Baseline EndFraction

      It is also possible to determine Ki/j by plotting a concentration–response curve for each of the solutes.

      In a mixture containing n solutes, if upper A prime Subscript normal i designates the area of the elution peak of compound i, and if the internal reference is j, then the content of compound i will obey the following equation:

      (1.47)x Subscript i Baseline percent-sign equals StartFraction upper K Subscript i slash j Baseline dot upper A prime Subscript i Baseline Over sigma-summation Underscript i equals 1 Overscript n Endscripts upper K Subscript i slash j Baseline dot upper A prime Subscript 1 Baseline EndFraction times 100

      This method is ideal to monitor rates of advancement of a chemical reaction and, at the same time, to follow the conversion kinetics.

      1 Chromatography is a separation method and also an analytical method. It is a very reliable method, with continued innovation in order to separate, purify, identify, and quantify chemical species.

      2 It is based on the adsorption or partition, depending on various parameters and interactions, of chemical species between two immiscible phases, one fixed or stationary (SP) and the other mobile (MP).

      3 There are many chromatographic techniques that differ in the process used, the nature of phases, physico‐chemical equilibria, the physical state, etc., and just as many classifications are possible depending on the criterion retained!

      4 To each compound, we assign a distribution factor K = CS /CM as a function of its concentration in the two phases. For each type of chromatography, thermodynamic relationships apply. From K (which is not a constant), we can calculate ΔH, ΔS, and ΔG for the separation in question.

      5 A document inseparable from any chromatographic analysis, the chromatogram, or elution curve, represents the graphic recording of concentration over time. It provides information on the analysed sample and on the quality of the separation in progress.

      6 One peak = one compound is the ideal situation. At column outlet, we treat the ideal peak as a Gaussian peak (hence the parameters σ, δ, ω), but the reality may be different. An ideal chromatogram has peaks that do not overlap and which are at least somewhat spaced out. A peak alone does not enable the absolute identification of a compound.

      7 A number of parameters can be used to better characterize each separation: efficiency, retention, separation, and resolution. The flow rate of the mobile phase has an influence on efficiency. Many empirical equations, not reported in this chapter, are offered to mathematically define the behaviour of analytes during elutions with a gradient.

      8 The speed of progression of the mobile phase with respect to the stationary phase has an essential influence on separation quality. A hyperbolic function, formed from the addition of three independent terms, known as the Van Deemter equation, demonstrates the existence of an optimal linear speed, i.e. leading to a minimum value of HETP.

      9 First established for GC, and later adapted to HPLC, the three terms of the previous equation refer to preferential paths (turbulent diffusion), to longitudinal diffusion, and to resistance to mass transfer.

      10 The quantification of analytes using chromatography is founded on the relationship between the mass and the area of the corresponding peak on the chromatogram. There are three methods, called the external standard method, internal standard method, and internal normalization method. The first is by far the most used because it is the quickest when it comes to processing a large number of analyses. Chromatograms equipped with a sample holder carousel and an automatic injection device make this method very reliable.

      1 For a given solute, show that the analysis time (which we can assume is equal to the retention time of the most retained compound) depends, amongst other things, on the column length L, on the average linear speed ū of the mobile phase, and on volumes VS and VM, which respectively designate the volume of the stationary phase and the volume of the mobile phase.

      2 Calculate the separation factor (or selectivity factor) between two compounds, 1 and 2, whose retention volumes are 6 ml and 7 ml, respectively. The dead volume of the column is 1 ml. Show that this factor is equal to the ratio of the distribution coefficients K2/K1 of these compounds (tR(1) < tR(2)).

      3 Equation (P1.2) is sometimes employed to calculate Neff. Show that this relation is equivalent to the more classical Eq. (P1.1):(P1.1) (P1.2)

      4 The resolution factor R for two solutes 1 and 2, whose elution peaks are adjacent, is sometimes expressed by Eq. (P1.3):(P1.3) (P1.4) If it is assumed that the two adjacent peaks have the same width at the baseline (ω1 = ω2), then show that Eq. (P1.4) is equivalent to Eq. (P1.3) for resolution.The effective plate number Neff may be calculated as a function of the separation factor α for a given value of the resolution, R. Derive this relationship.

      5 Show that if the number of theoretical plates N is the same for two neighbouring compounds 1 and 2, then the classical expression is thus:(P1.5) Then show that if Eqs. (P1.5) and (P1.6) are equivalent:(P1.6)

      6 The characteristics of a C‐18 column, offered by a manufacturer for ultra‐rapid HPLC, are:Length: 50 mm, diameter: 2 mm, particle diameter: 2 μm. The test chromatogram accompanying the column shows six well‐spaced peaks in less than 2 min.Solute No.012345tR (min)0.210.310.420.761.141.58On the chromatography report, the following information can also be found:For solute No. 5: N = 106,000 plates per metre; asymmetry at 10% of height A = 1.07. Remember that for an asymmetrical peak, the number of theoretical plates N is calculated from the following equation: and that For peak No. 5, calculate width ω0.1 measured at 10% of its height (from the baseline) and values a and b.Knowing that peak No. 0 corresponds to a nonretained

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