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.

[MichaelVW]

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?

You're right that if you prepare a calibration standard as described, the IS and Component A are both present in the total solution at 1%. But your colleague is right too. Another way to describe what you've created is "A 1.01% w/w solution of Analyte A in matrix, to which 1g of internal standard has been added per 99g of solution."

The real question, in terms of getting correct results, is how your samples are prepared. The important thing is to treat your samples and calibration solutions the same way, and to calculate dilution factors correctly.

If you take 99g of sample and add 1g of internal standard, you'll get the right concentration (with a dilution factor of 1). Pretty straightforward.

If you take 1g of sample and 98 grams of matrix, then add 1g of internal standard... should your dilution factor be 98, 99, or 100?

99! The internal standard's mass does not factor into the dilution factor math.

This post is long already but let me give you an example of another way to do this, which a lot of environmental labs use:

I create a 1ml solution of a 5 ppm standard, then add 10ul of a 40 ppm internal standard. I tell the instrument it's 5 ppm, and the internal standard is 40 ppm.

I prepare 1ml of sample. I add 10ul of internal standard. The multiplier I give the instrument completely ignores the 10ul of internal.

The instrument gives me the correct concentration of the parent sample. Here's part that may be a surprise. If I'm using a single point or average RF curve, and some monkey goes into my calibration table and replaces the "40" for the internal standard with "3.14159" or something - my concentration will still be calculated correctly.