1000 mL/2 mL = 500 fold concentration.

1000 mL/2 mL = 500 fold concentration.

Mrs/Miss Ekhosuehi,

The quantification of any kind of pollutants via only the show in your posting analytical scheme does not provide highly reliable and precise anyltical information due to a set of reasons, even using sample pretreatment steps such as SPE or any another of the available techniques. In particular, the major drawbacks, amongst other ones, of the coupling analytical scheme mentioned in your posting is when the pollutants are structuraly very similar organics, some heavy metals or some radionuclides, respectively. The same is true for metal organics and organometallics. Because of, (i) very frequently the HPLC peaks are strongly overlapped which results in a significant error contribution to the analytical information from the mathematical data-processing of the HPLC results; in addition to, (ii) a lack of standards.

Currently, only the mass spectrometric based quantification provides reliable, meaningful, accurate, precise, selective and sensitive analytical information about pollutants despite the type of the sample. Howerver, when the quantification is carried out within the framework of our more recently developed mass spectrometric quantitative stochastic dynamic method; and model formulas, which allow to achieve an absolute quantification, meaning coefficients of correlation between the theory (our model equations) and the experiment up to /r/ = 1. Please, consider discussion [A] and references [2,3], therein, with, respectively, the reference sections to references [2,3].

The method is tested at concentration levels of pg.(mL)-1 and (ng.)L-1 on organics in mixture.

[A]

[https://www.researchgate.net/post/How_can_we_detect_VOCs_in_exhaled_breath_sample_using_ISQ_7000_single_quadrupole_GC-MS_system]

Owing to the fact that the method is our authored development and represents a completely innovative strategy for quantification of mass spectrometric experimental variables, if you have any questions concerning its direct application to the analytical practice, in order to, determin pollutants in wastewater, then please do not hesitate to ask them, herein, and we shall provide to you a comprehensive guideline step-by-step.

Some detail on the theoretical background of our method compared with other available stochastic concepts, based on classical ideal systems we have provided to reference [1*] (below;) again within the framework of a quantitative analyses and a chemometric treatment. As can be seen, our mass spectrometric method is validated independently via chromatographic data, as well.

[1*] B. Ivanova, M. Spiteller, Stochastic dynamic mass spectrometric approach to quantify reserpine in solution, (2020) in press.

You can plot your curve as the original concentration of analyte that you spiked into the 2L for your standards and QCs. If you spike a stable isotope labeled internal standard into your original 2L, then you will also correct for extraction efficiency and other method vatiations. This will not work as well with an analog IS.

You can't know the exact concentration in the 100ul, the signal strength is dependent of instrument and sensibility. You need to plot a calibration curve for an expected domain of linearity, or you can use a quadratic curve for an extended domain. After you read the result, you multiply with 500. C1V1 = C2V2, in your case C2 = C1V1 / V2. C2 = C1 * 1000 / 2 so C2 = C1 * 500.

You correct the instrument and signal respose automaticaly with the internal stadard. If it is a stable isotope labelled IS, chemically identical to the analyte, it will also follow the analyte perfectly through the extraction procedure. The concetration is then relatede to the analyte/IS peak area ratio. In the Pharma industry we have been using stable isotiope labeled internal standards (ususally D, 15N or 13C, or combiation) for the last 30 years.

thank you for your answer Dan Vlad Popovici, please I need to verify, should I multiply or divide by 500?

Amen Ekhosuehi Think about the first reply you received. Your sample injected into your analytical system is 500x more concentrated. The opposite of multiplication is division, which suggests to get to the original concentration, you do a division.

Another way of thinking about it is you started with a large volume and collected the compound of interest into a small volume. Assuming your compound isn't gained or lost, the original sample has to be less concentrated, and you pick the operator that leads to a lower concentration.

Jack Silver, thank you very much. exactly my thought