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 with gas chromotography using a chloroform internal standard. To begin, the researcher examined a sample containing 7.16 mg/L DDT standard and 1.70 mg/L chloroform as the internal standard, producing peak areas of 4075 and 10369, respectively. Then, the researcher collected 5.73 g of spinach, homogenized the sample, and extracted the DDT using an established method (assume 100% extraction), producing a 3.62 mL volume of unknown sample. The researcher then prepared a sample that contained 1.00 mL of the unknown sample and 1.50 mL of 10.33 mg/L chloroform, which was diluted to a final volume of 25.00 mL. The sample was analzed using GCMS, producing peak areas of 7725 and 14743 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. Number mg DDT/g spinach

Explanation / Answer

Area of analyte / concentraiton of analyte = F *(Area of std / Concentration of std)

Calculate F initially:

4075/(7.16) = F*(10369)/(1.7)

F= 4075/(7.16) *1.7/10369 = 0.093309

so..

[S] = 10.33 mg/L --> 10.33*1.50/25 = 0.6198 mg/L

now..

for unkown mis

Area Analyte / Concentratio analyte = F * Area std / Concentration of std.

7725/X = 0.093309 * 14743 /0.6198

X = 7725*0.6198 /(0.093309*14743 ) = 3.4804 g/mL

since this is

V = 1 mL -->

X = 3.4804 * 25/1

X = 87.01 mg/L

so

X = 0.08701 mg/mL

for V = 3.62 mL

mass = X*V = 0.08701*3.62 = 0.31497 mg

DDT = mg of DDT / mass sample = 0.31497 / 5.73 = 0.054968 mg DDT/g spinach

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