4f01

So far, we have have discussed various manipulations of the sample (weighing, dilution, etc.) and related calculations. Often two or more of these operations are combined into a single assay procedure. Here we will look at the calculation of a final assay result in such cases.

We can summarize our approach to various problems as follows:

1. Determine the concentrations of the analyte in the calibrators;

2. Prepare the sample solution for injection into the LC; determine the dilution factor, DF.

3. Inject calibrators and sample(s);

4. Calculate the sensitivity factor SF for the calibrators

(divide calibrator peak size by calibrator concentration or determine the slope of the calibration line);

5. Determine the concentration of analyte in the final sample solution that was injected into the LC;

(divide analyte peak size by sensitivity factor)

6. Correct the concentration of analyte to the original (undiluted) basis, using the formula

C1 = C2 x DF.

In our first example, consider the analysis of a liquid formulation for phthalic acid using a single-point calibration. We begin by preparing our calibrator; i.e., weighing out phthalic acid into a volumetric flask and filling to mark with solvent. The procedure for this method calls for 0.5 grams of phthalic acid to be dissolved in 50 mL of deionized water:

Concentration = weight / volume

= 0.500 g / 50 mL

= 0.01 g / mL

= 10 mg / mL

Our analytical procedure calls for first diluting the 1 mL of the liquid sample to 25 mL, followed by injecting the diluted sample into the LC system. The dilution factor for the sample is calculated from:

DF = (flask volume) / (pipette volume)

= 25 mL / 1 mL

= 25

Next we run our calibrator and obtain a peak at 5.4 minutes with an area of 20,000 counts. Because we are using a single-point calibration, we must calculate the sensitivity factor:

SF = area / concentration

= (20,000)/(10 mg / mL)

= 2000 mL / mg

Chromatogram of phthalic acid calibrator solution (10 mg/mL). The area of the peak at 5.4 minutes was measured as 20,000 counts.

When we run the diluted sample, we obtain a peak at 5.4 minutes with an area of 30,000 counts. Now we can calculate the concentration of diluted sample:

Concentration = Area / SF

= 30000 / 2000 mL / mg

= 15 mg/mL

Chromatogram of sample containing phthalic acid. The area of the peak at 5.4 mnutes was measured at 30,000 counts.

What is the concentration of phthalic acid in the original (undiluted) sample? We obtain this from the concentration of the dilute sample and the dilution factor:

C1 = C2 X DF

= 15 mg / mL X 25

= 375 mg / mL

= 375 g / L

In a second example, consider the analysis of a solid sample for dimethylchickenwire (DMCW). We begin by weighing out a certain quantity of DMCW and dissolving it in a certain volume of solvent. Let's assume the following values for calibrator and samples 1 and 2:

	Calibrator	Sample #1	Sample #2
weight	0.100 g	1.050 g	0.995 g
volume	25 mL	25 mL	25 mL
concentration	4.00 mg / mL	42.00 mg / mL	39.80 mg / mL

Now, three chromatograms are run with the following results:

	Calibrator	Sample #1	Sample #2
Retention Time	4.68 min	4.68 min	4.68 min
Area	72,000	63,000	off-scale

We first calculate the sensitivity factor for the DMCW peak in the calibrator:

SF = Area / Concentration

= (72,000) / (4.00 mg / mL)

= 18,000 mL / mg

Next we calculate the concentration of DMCW in diluted sample #1:

concentration of DMCW = Area / SF

= 63,000 / 18,000 mL / mg

= 3.50 mg / mL.

The concentration of the total sample in the solution is 42.0 mg / mL, so the fraction of DMCS in the original solid sample #1 is:

3.5 / 42.0

= 0.0833 or 8.33% by weight.

Sample #2 was off-scale, so it must be diluted and rerun. Five mL of sample was diluted into a 25 mL flask. The dilution factor is:

DF = (flask volume)/(pipette volume)

= 25 mL / 5 mL

= 5.00

The diluted sample was reinjected. The resulting peak at 4.68 minutes had a peak area of 36,000.

The concentration of DMCW in this diluted sample of #2 was then:

Concentration of DMCW = Area / SF

= 36000 / 18,000 mL / mg

= 2.00 mg / mL.

The concentration of DMCW in the original (undiluted) sample is given by the concentration in the diluted sample and the dilution factor:

C1 = C2 X DF

= 2.00 mg / mL X 5

= 10.0 mg / mL

The concentration of the total sample in the solution is 39.80 mg / mL, so the fraction of DMCS in the original solid sample #1 is:

10.0 / 39.8

= 0.251 or 25.1% by weight.