A researcher was attemping to quantify the amount of DDT (dichlorodiphenyltrichl

Question

A researcher was attemping to quantify the amount of DDT (dichlorodiphenyltrichloroethane) in spinach using gas chromotography mass spectrometry (GCMS). First attempts using external calibration proved difficult, as reproducibility and external standard agreement were poor. To overcome these problems, the researcher used the internal standard calibration method with chloroform as the internal standard. To begin, the researcher examined a sample containing 5.58 mg/L DDT standard and 3.22 mg/L chloroform as the internal standard, producing peak areas of 4833 and 10769, respectively. Using the above information, determine the response factor (F). After establishing the response factor of the instrument, the researcher collected 5.22 g of spinach, homogenized the sample, and extracted the DDT using an established method (assume 100% extraction), producing a 2.19 mL volume of unknown sample. The researcher then prepared a sample that contained 0.750 mL of the unknown sample and 1.00 mL of 12.82 mg/L chloroform, which was diluted to a final volume of 25.00 mL. The sample was analyzed using GCMS, producing peak areas of 7217 and 15027 for the unknown and chloroform, respectively. Calculate the DDT concentration in the spinach sample. Express the final answer as milligrams DDT per gram of spinach.

Explanation / Answer

(A) Given the mass of DDT standard, ms = 5.58 mg

peak area of the DDT standard, As = 4833

mass of Chloroform internal standard, mis = 3.22 mg

peak area of the Chloroform internal standard, Ais = 10769

Response factor can be calculated from the following formulae

Response factor, F = (As / ms) / (Ais / mis)

= (As *  mis) / (Ais * ms) = (4833 * 3.22 mg) / (10769 * 5.58 mg) = 0.259 (answer)

(B): Mass of chloroform in 25 mL of the sample = (12.82 mg / 1000 mL) * 1 mL = 0.01282 mg

Hence concentration( in mg/L) of Chloroform in 25 mL of the sample, mis = (0.01282 mg / 25 mL) * (1000 mL / 1L)

= 0.5128 mg/L

peak area of the Chloroform internal standard, Ais = 15027

Let the concentration of unknown DDT sample be 'ms'

Peak area of unknown DDT sample, As = 7217

Response factor, F = 0.259 = (As / ms) / (Ais / mis) = (As *  mis) / (Ais * ms)

=> 0.259 = (7217 * 0.5128 mg/L ) / (15027 * ms )

=> ms = 0.951 mg/L

Hence mass of DDT in 25 mL of the sample = (0.951 mg / 1000 mL) x 25 mL = 0.023775 mg

Hence 0.023775 mg of DDT is present in 0.750 mL ofof the unknown sample

5.22 g spinach produces the volume of the unknown sample = 2.19 mL

Hence 1 g of spinach prooduces the volume of unknown sample = 2.19 / 5.22 = 0.420 mL

0.023775 mg of DDT is present in 0.750 mL of the unknown sample

Hence mass of DDT that would be present in  0.420 mL(1 g spinach) of the unknown sample

= (0.023775 mg / 0.750mL) x 0.420 mL = 0.0133 mg DDT / g spinach (answer)

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