Single Point Internal Standard in Gas Chromatography

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Single Point Internal Standard in Gas Chromatography

The internal standard method accounts for any variances in gas chromatograph
performance. The analyte chosen for the internal standard has a predictable
retention time and area, allowing it to be used to determine if abnormalities have
occurred.
The Single Point Internal Standard method requires at least two analyses. The first
analysis contains a known amount of internal standard and the compounds of
interest. Calculate the response factor using Equation 1.

Internal Response Factor = AreaIS * AmountSC / AreaSC * AmountIS (Equation 1)

IS = Internal standard
SC = Specific Compound of Interest

Then add a known amount of the internal standard to the sample containing analytes of unknown concentrations. Calculate the amount of the unknown analyte using Equation 2.

Amount of Specific Compound = AmountIS *AreaSC * IRFSC / AreaIS (Equation 2)

IS = Internal standard
SC = Specific Compound of Interest
IRF = Internal Response Factor.

Example:
Prepare a sample containing 19.433 ppm of isopropyl alcohol (the internal standard) and 9.5µg/mL ethanol (the analyte). Then inject the sample. The resulting peak areas are 111.83235 for isopropyl alcohol and 44.12193 for ethanol. Using Equation 1 the response factor for the ethanol is:
IRF = 111.83235 * 9.5 / 19.5 * 44.12193 = 1.2348

Inject the sample containing 19.433 of isopropyl alcohol and an unknown amount of ethanol using the same chromatography conditions. The resulting areas are 113.85451 for isopropyl alcohol and 57.25642 for ethanol. Calculate the amount of ethanol present using Equation 2.

Amount of ethanol = 57.25642 * 19.5 * 1.2348 / 113.85451 = 12.1 ppm.

Note: If the unknown concentration sample range will be high from the standard amount, we will use multiple point internal standard method.
Example: If an unknown concentration of ethanol will be 20 ppm or high, we must use multiple point internal standard method.