Chemical Analysis. Francis Rouessac

      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 , , and , the areas of this new chromatogram obtained under the same operating conditions as previously. If , and represent the quantities of 1, 2, and IS introduced into the column, then:

      From the relative response factors calculated in the first experiment as well as from the known concentration of the internal standard in the sample, this leads to:

If we expand to n components, it is easy to calculate the mass concentration of the solute i using Eq. (1.44):

(1.44)

In addition, the percentage concentration of i can be expressed using Eq. (1.45):

(1.45)

      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.

1.17 INTERNAL NORMALIZATION METHOD

      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.

Let us suppose that we are trying to find the mass concentrations of three compounds 1, 2, 3 in a mixture (Figure 1.14). The analysis is again carried out in two steps.

      1.17.1 Calculation of the Relative Response Factors

A standard solution containing the three compounds 1, 2, and 3 at known concentrations C₁, C₂, and C₃ is prepared. The chromatogram corresponding to the injection of a volume V of this standard solution shows three peaks of area A₁, A₂, and A₃. These areas will be related to the masses m₁, m₂, and m₃ of the compounds in volume V, by three expressions of the Eq. (1.42) type.

Figure 1.14 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 K_1/3 and K_2/3 for compounds 1 and 2 with respect to 3 . As previously deduced:

      Given that m_i = C_i.V, then the following equations for K_1/3 and K_2/3 are obtained:

      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 as