Exam question 4:
4. (15 points) A sample of water is to be tested for the pesticide chlordane. Deuterated chlordane is used as an internal standard. The internal standard behaves the same in the separation as the analyte, but has a different mass, which distinguishes it from the analyte in the mass spectrometer. The internal standard (IS) and the chlordane give the same response per gram in the detector. 1000.0 ml of the water sample is spiked with 20.0 µg of deuterated chlordane and is passed through a SPE cartridge. The analyte and standard are eluted from the cartridge with 20.00 ml of solvent. 0.02 µl of this extract is injected into the GCMS. The following data are obtained:
| Solution | Amount injected | Peak areas |
| Standard solution containing 2.0 g/ml IS | 20 ul | d-chlordane 4859 |
Sample extract (20 ml of solution) | 20 ul | d-chlordane 2135 chlordane 5802 |
(a) In 20 ml of the extract we expect, if the extraction was 100% efficient, 20 ug of the IS (d-chlordane).
How much did we actually get?
Use the first injection of a known amount of IS which gives 4859 to compare with the second injection which gives 2135.
[(2 ug/ml)/4859] x 2135 = 0.87 ug/ml. We expected 1.0, and got 0.87 so we got 87% recovery
(b) Next we can use either the external standard or the internal standard to find the amount of Chlordane (call it C for now) in the extract.
External: [2 ug/ml / 4859] x 5802 = 2.4 ug/ml in the extract
or IS method: The sample extract we have already found contained 0.87 ug/ml.so:
[0.87 ug/ml / 2135] x 5802 = 2.4 ug/ml in the extract (Same answer as above).
And 20 ml extract x 2.4 ug/ml = 48 ug C
(c) 48 ug C recovered /.87 (since we only recovered 87%) = 55 ug in original sample.