Internal standards are conmonly used in chromatography to calculate the any part

Question

Internal standards are conmonly used in chromatography to calculate the any particular detector, the relative response factor for the analyte compared to the internal standard mu be determined first. Calibrating the linearity of the response factor for an analyte compared to the internal standard requires making a series of solutions with the same concentration of standard, and a varying concentration of analyte. Plotting the response of the analyte relative to the standard (peak area o analyte/peak area of standard) versus the concentration of the analyte relative to the standard (lanalyteNstandard]) should produce a straight-line graph whose slope is the response factor, F area of analyte signal eoncentr are a of analyte signal concentration of analyte area of standard signa concentration of standard The table at the right gives some data for a chromatographic analysis of an analyte, using an internal standard (Peak areas of each are measured by a mass spectrometer). The volume injected for each sample is similar, but not exactly the same. Analyte Standard Analyte Standard Peak Area 1.0 5.0 10.0 10.0 2975 3653 6189 10.0 | 10.0 | 2999|2803 (a) Calculate the peak area ratio (analyte/standard) and the concentration ratio [analyteV[standard] for each of the samples, and calculate F for each sample (peak area ratio/concentration ratio Sample Concentration ratio Peak area ratio (analyt e/standard) (analyte/standard) Number Number Previous Give Up & View Solution O check Answer Next Exit

Explanation / Answer

sample   

concerntration ratio(1) = 1/ 10 = 0.1    

concerntration ratio (2) = 5/10 = 0.2    

concerntration ration (3)     = 10/10 = 1

peak area ratio (1) = 305 / 2975 = 0.1

peak area ratio (2) = 3653 / 6189 = 0.6

peak area ratio (3) = 2999 / 2803 = 1.06

F= ratio of peak area / ratio of concerntration

F1= 1

F2= 3

F3= 1.06

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