Internal Standard Conc. Calculation

[AnaChem1983]

Hi all,

I have a question regarding the calculation of concentrations in calibrants for an internal standard method of calibration? I hadn’t really thought about how to go about this beyond what I’ve always done until someone challenged me about it yesterday as they were doing it differently. For context, the concentration units for the analysis are % (w/w).

I would normally do it as follows. If I added, say, 1g of component A to 98g of matrix solution along with 1g of internal standard, I’d say that the concentration of component A and the internal standard are both 1% (w/w). The total mass of the solution is 100g (1+1+98) and both component and ISTD are 1g each. Therefore, (1/100)*100 = 1%. So far so logical, to me anyway!

But a colleague put it to me that the concentration of the analyte in this calibration solution should actually be 1.01 %(w/w) on the basis that the internal standard isn’t actually in the samples being measured. It’s only added by the chemist. Therefore, the calculation should be (1/99)*100 = 1.01%.

This seems illogical to me. And, ultimately, you might argue that only a pedant would care about the difference between 1 and 1.01 (guilty as charged!!) but I can foresee issues further down the line unless I box this off in my head.

I was wondering what your thoughts were regarding this?

[Consumer Products Guy]

I was wondering what your thoughts were regarding this?

Once added, what is important is the amount of internal standard added compared to the amount of analyte.

The volume for the solution is not relevant.

[rb6banjo]

I don't follow your colleague's line of thinking. Final concentrations are always based on the total volume/mass of the sample to be analyzed.

To calibrate with an IS, you must add it to all samples. So, for all of your samples you will need 1.0 g in 100 mL of solution to be analyzed (let's assume it's a volume of sample). So, 1.0/100 x 100 = 1.0% (wt/vol) is correct for your IS. Your dilution factor is 100/99 = 1.01 for all samples you analyze if you prepare them this way. Multiply the answer you determine by 1.01 to get the best estimate of concentration in the sample.

If you are adding your analyte to a sample for calibration, and there's analyte in the sample, then you have a different problem. Let's say you have 0.2% analyte already in the sample, and you add your 1.0 g of analyte in addition to your 1.0 g of internal standard to make 100 mL of solution. Then for the analyte:

C = (98*0.002 + 1.0)/100 x 100 = 1.196% analyte in the solution. (not 1.2%, but close).

You still have 1.0% IS in the solution because it's 1.0 g IS to 100 mL total final volume.