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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 , , 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.