29th March, 2023 Inject reference standard of known concentration and purity Perform 5 injection of reference standard. Reason is assessing precision of instrument Replicates are A, B and C, inject each three times Factors for each sample that has been replicated three times: Retention time, area, height, symmetry of peak Response Factor = peak area/reference standard amount Amount/Concentration of sample = peak area/response/reference factor Similar to AAS, but a single point calibration is used Multiple calibration points on GC Equation of the line = y=100.92x + 0.3562 Sample Area = 327 To find the concentration (x), insert sample area in place of y and solve for x Concentration (x) = 3.24 10ml in a 25ml flask with A = 327 Final concentration = Dilution Factor x concentration Final concentration = 2.5 x 3.24 = 8.075 If sample is 98% pure Final concentration = 1/0.98 x 8.075 Final concentration = 8.24 Three different types of calibration: external method, standard addition, internal standard Internal Standard Its natural for your instrument to drift and to deal with this an internal standard needed. It’s a compound that does not interfere with the analyte of interest and has a different retention time. Propanol can be used as an internal standard (screenshot in tablet) When ethanol standard is prepared, it can be spiked with propanol Sample must be spiked with propanol as well and then compare factors (retention time, peak area). How internal standard approach is used? Internal response factor = (area of internal standard x amount of specific compound of interest)/amount of internal standard x area of specific compound) Amount of specific compound = (amount of internal standard x area of specific compound x internal response factor)/area of internal standard Example: (Screenshot in tablet) Benzene and toluene as the internal standard Conc of Toluene = 2000 Conc of Benzene = 1000 Area of T within calibration/standards = 120,000 Area of B within calibration/standards = 67,000 Area of sample of T = 122000 Area of sample of B = 43000 Calculate Internal response factor = 0.896 Amount of Benzene in sample = (2000 x 43000 x 0.896)/122000 Different area values are due to uncertainties Internal standard should be chemically and physically similar to the analyte It should not be present in the sample, not trace of toluene in the benzene It should be highly pure and it is a good estimate of how much to spike your sample, above .3 or .5 of the maximum concentration of the analyte of interest
