I would like to ask how to correctly draw a calibration curve in Excel.
I have 6 standard solutions at different concentrations (ng/µL): 0.02 ; 0.2 ; 0.5 ; 1 ; 1.5 ; 2. They are spiked with two internal standards, each is fixed at the same concentration (1ng/µL). I have injected 6 times those 6 spiked solutions into the GC-MS and got the peak areas of the standard analyte and two internal standards (so 36 values of peak area for each of those 3 compounds). And then, I calculated the ratio Area (analyte) / Area of each internal standard, I also got 36 values for each ratio.
I would like to draw 2 calibration curves:
Area(analyte)/A(internal std 1) = f ([analyte] / [internal std 1])
Area(analyte)/A(internal std 2) = f ([analyte] / [internal std 2])
To do this, should I directly take all those 36 values of a ratio (for example, A(analyte) / A(internal std 1) ) for Y axis, and 6 values of ratio of concentrations for X axis ? Or should I first calculate the mean value of peak area ratio for each concentration, and then take those 6 mean values for Y axis and 6 values of ratio of concentrations for X axis ?
I would not use mean values (I have only used them in setting specifications). Eventually, you will want to set you LOQ (LOD too) which is based on the standard deviation (3 replicates). Since y = mx + b or (detector response) = slope x (analyte concentration) + y intercept, I question why you want 2 calibration curves?
I want to add the first internal standard before sample treatment and the second after treatment. The calibration curve with the 1st internal standard will provide the amount of analyte in the initial sample before treatment, and the curve with the 2nd internal standard its recovered amount. So the ratio between those two amounts is the % recovery.
I injected 6 times the same series of standard solutions. 3 replicates mean 3 different series of standard solutions ?
The answer depends on your industry! If it is pharmaceuticals you are mandated by USP section <621> to inject 5 replicates and have a %RSD not exceeding 2.0% If your %RSD is higher than 2.0% (5.0% is typical for low concentrations of related substances) then you have to inject 6 replicates.
What is the purpose of your %Recovery. I presume you’re going to use it as a recovery factor.
I work in environmental chemistry. So 3 replicates would be enough.
I don’t know what recovery factor means… %Recovery is the percent of compound of interest that can be obtained after treatment (extraction, etc.).
For example, with the 1st internal standard, the amount of compound of interest detected by GC-MS is 10 ng, meaning there is 10 ng of this compound in the initial sample before treatment.
With the 2nd internal standard, the amount of compound of interest detected by GC-MS is 9 ng, meaning there is 9 ng of this compound in the initial sample after treatment.
So the % of recovery is 9/10*100% = 90%
That’s appropriate. I have used a %recovery factor in SPE. In the validation spiked samples were analysed before SPE and analyzed afterwards. Thus in real samples, API/% recovery.
In your case, there may be a peak underneath your target molecule that interferes with quantitation and the only way to eliminate it is to treat your samples. Be sure to put a ‘spike’ in your GC-MS run to ensure your treatment works (for example = heating block doesn’t work).