Chemical Analysis. Francis Rouessac

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Chemical Analysis - Francis Rouessac


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      The second step of the analysis is to obtain a chromatogram for a given volume of a solution containing the sample to quantify and to which has been added a known quantity of internal standard IS. This will yield upper A prime Subscript 1 , upper A prime Subscript 2 , and upper A prime Subscript upper I upper S , the areas of this new chromatogram obtained under the same operating conditions as previously. If m prime Subscript 1 Baseline comma m prime Subscript 2 Baseline , and m prime Subscript upper I upper S represent the quantities of 1, 2, and IS introduced into the column, then:

StartFraction m prime Subscript 1 Baseline Over m Subscript upper I upper S Baseline EndFraction equals upper K Subscript 1 slash upper I upper S Baseline dot StartFraction upper A prime Subscript 1 Baseline Over upper A prime Subscript upper I upper S Baseline EndFraction a n d normal StartFraction m prime Subscript 2 Baseline Over m prime Subscript upper I upper S Baseline EndFraction equals upper K Subscript 2 slash upper I upper S Baseline dot StartFraction upper A prime Subscript 2 Baseline Over upper A prime Subscript upper I upper S Baseline EndFraction

      From the relative response factors calculated in the first experiment as well as from the known concentration of the internal standard in the sample, upper C prime Subscript upper I upper S Baseline comma this leads to:

upper C prime Subscript 1 Baseline equals upper C prime Subscript upper I upper S Baseline upper K Subscript 1 slash upper I upper S Baseline dot StartFraction upper A prime Subscript 1 Baseline Over upper A prime Subscript upper I upper S Baseline EndFraction a n d upper C 2 equals upper C prime Subscript upper I upper S Baseline upper K Subscript 2 slash upper I upper S Baseline dot StartFraction upper A prime Subscript 2 Baseline Over upper A prime Subscript upper I upper S Baseline EndFraction

      This method becomes even more precise if several injections of the solution and of the sample are carried out. In conclusion, this general and reproducible method nevertheless demands a good choice of internal standard, which should have the following characteristics:

       it must be stable, pure, and not exist in the initial sample;

       it must be measurable, giving a well‐resolved elution peak on the chromatogram;

       its retention time must be close to that (or those) of the solute(s) to be quantified;

       its concentration must be close to or above that of the analytes in order to be in the detector’s linear response range;

       it must be inert with respect to the compounds in the sample.

      The advantage of this method is not needing perfect reproducibility of injections, which makes manual injections possible if the device is not equipped with an automatic injector. The disadvantage, however, is the choice of internal standard, which extends the development time of the analysis. However, with mass detectors that do not require perfect resolution, coelution of the internal standard and of the solute to quantify is possible, if we can recognize peaks caused by each of the two compounds on the mass spectrum of the mixture.

      This method, also called percentage normalization, is used for mixtures in which all components have been identified and have been assigned a well‐resolved peak on the chromatogram. This method uses relative response factors, as in the internal standard method. The biggest difference here is that the solute used as a reference to calculate the relative response factor is part of the mixture to quantify.

      1.17.1 Calculation of the Relative Response Factors

Schematic illustration of analysis by internal normalization method.

      One of the compounds, 3 for example, is chosen as the substance for internal normalization. This compound 3 will serve to calculate the relative response factors K1/3 and K2/3 for compounds 1 and 2 with respect to 3 . As previously deduced:

upper K Subscript 1 slash 3 Baseline equals StartFraction upper K 1 Over upper K 3 EndFraction equals StartFraction m 1 dot upper A 3 Over m 3 dot upper A 1 EndFraction a n d upper K Subscript 2 slash 3 Baseline equals StartFraction upper K 2 Over upper K 3 EndFraction equals StartFraction m 2 dot upper A 3 Over m 3 dot upper A 2 EndFraction

      Given that mi = Ci.V, then the following equations for K1/3 and K2/3 are obtained:

upper K Subscript 1 slash 3 Baseline equals StartFraction upper C 1 dot upper A 3 Over upper C 3 dot upper A 1 EndFraction a n d normal upper K Subscript 2 slash 3 Baseline equals StartFraction upper C 2 dot upper A 3 Over upper C 3 dot upper A 2 EndFraction

      1.17.2 Chromatogram of the Sample – Calculation of the Concentrations

      The next step consists in injecting a specimen of the mixture to be measured containing compounds 1, 2, and 3 . Labelling the elution peaks asСкачать книгу