Question

A researcher was attempting to quantify the amount of dichlorodiphenyltrichloroethane (DDT) in spinach with gas chromatography using a chloroform internal standard. To begin, the researcher examined a sample containing 6.37 mg / L DDT standard and 3.20 mg / L chloroform as the internal standard, producing peak areas of 5019 and 8179 , respectively.
Then, the researcher collected 5.29 g of spinach, homogenized the sample, and extracted the DDT using an established method (assume 100% extraction), producing a 2.40 mL volume of unknown sample. The researcher then prepared a sample that contained 0.750 mL of the unknown sample and 1.25 mL of 11.45 mg/L chloroform, which was diluted to a final volume of 25.00 mL. The sample was analzed using GCMS, producing peak areas of 6821 and 14061 for the unknown and chloroform, respectively.

Requried:
Calculate the DDT concentration in the spinach sample. Express the final answer as milligrams DDT per gram of spinach.

Answer 1

Answer:

0.0136mg DDT / g spinach

Explanation:

Quantification in chromatography by internal standard has as formula:

RF = Aanalyte×Cstd / Astd×Canalyte (1)

Where RF is response factor, A is area and C is concentration

Replacing with first experiment values:

RF = 5019×3.20mg/L / 8179×6.37mg/L

RF = 0.308

In the next experiment, final concentration of chloroform was:

11.45mg/L × (1.25mL / 25.00mL) = 0.5725mg/L

From (1), it is possible to write:

Aanalyte×Cstd / Astd×RF = Canalyte

Replacing:

6821×0.5725mg/L / 14061×0.308 = Canalyte

Canalyte = 0.9017mg/L

as the sample was made from 0.750mL of extract. Concentration of extract is:

0.9017mg/L × (25.00mL / 0.750mL) = 30.06mg/L. As the extract has a volume of 2.40mL:

30.06mg/L × 2.40x10⁻³L = 0.07213mg of DDT in the extract

As the extract was made from 5.29g of spinach:

0.07213mg of DDT in the extract / 5.29g spinach = 0.0136mg DDT / g spinach