Using an Internal Standard with an HPLC,

A Simple Calculation Method

 

There are many reasons to use an internal standard with any chromatographic system.  The most common is to eliminate variations with the injection volume.  With new autosampler technologies this reason has largely disappeared but there remain a number of applications where it is a good idea to subject the standard and the analyte to the same sample introduction methods.

 

Because an internal standard method uses the ratio of responses for standard and analyte, quantitation does not depend on the amount of material injected.

 

In this example the internal standard is caffeine and the analyte is aspirin. 

 

First we make up a standard solution containing a known concentration of analyte and a known concentration of internal standard.

 

For the known standards of aspirin, Solve for F

 

Area internal standard peak                                     Concentration of internal standard

---------------------------------------    =    F   ------------------------------------------------------

Area of aspirin standard peak                                   Concentration of aspirin standard

 

Note that in the equation above we know everything but F, which is a dimensionless constant.


 

 

 

Plugging in some rounded off numbers from a recent HPLC run we get the following:

 

13          4

---   = F ------

8            5

 

Note that even though we have more aspirin (5 mg/ml) than caffeine (4 mg/ml) the caffeine peak was bigger. 

 

Solving for F we get 2.03

 

Now we run the solutions containing a known amount of internal standard (caffeine at 4 mg/ml) and an unknown amount of analyte (aspirin at ?)

 

Area internal standard peak                                      Concentration of internal standard

---------------------------------------    =    F          --------------------------------------------------

Area of aspirin unknown peak                                 Concentration of aspirin unknown

 

 

Now we solve for the concentration of aspirin in our unknown sample (again these are rounded off numbers.)

 

13               4

---   = 2.03 ------

5                 X

 

X = 3.12 mg / ml of aspirin.


Internal standards can be confusing but by taking the calculations one step at a time and working carefully, they can be a valuable addition to your chromatography skills.


This information posted by:

Kevin Olsen
Instrumentation Specialist
Chemistry and Biochemistry Support Staff
Montclair State University
Montclair, NJ, 07043
973-655-4076
http://blake.montclair.edu/~olsenk/