I keep getting it wrong don\'t where I am doing my error A researcher was attemp
ID: 505460 • Letter: I
Question
I keep getting it wrong don't where I am doing my error
A researcher was attempting to quantify the amount of DDT (dichlorodiphenyltrichloroethane) in spinach with gas chromatography using a chloroform internal standard. To begin, the researcher examined a sample containing 7.44 m DDT standard and 2.10 mg/L chloroform as the internal standard, producing peak areas of 34 and 9933, respectively. Then, the researcher collected 5.81 g of spinach, homogenized the sample, and extracted the DDT using an established method (assume 100% extraction) producing a 2.88 mL volume of unknown sample. The researcher then prepared a sample that contained 0.750 mL of the unknown sample and 2.00 mL of 12.61 mg/L chloroform, which was diluted to a final volume of 25.000 mL. The sample was analyzed using GCMS, producing peak areas of 7291 and 12149 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
DDT Standard: 7.44 mg/L...........has a peak area of 3479 (measured standard volume is not given and so the total area mentioned is taken as for total material given).
Let 5.81 g of spinach gave a total of unknown concentration of the total sample volume 2.88 mL.
From the stock solution (2.88 ml) 0.75 ml is taken and mixed with 2.00 ml of 12.61 mg/L chloroform. This gives a total volume of 2.75 ml which is diluted to 25 mL by adding 22.25 mL water.
So in the final volume we have 0.75 mL solution of unknown quantity of DDT which is diluted to 25 ml.
From standard we know 7.44 mg/L DDT solution gives a peak area of 3479.
Unknown sample gave a peak area of 7291 and this is for 26.04% (0.75 ml) of total solution (2.88 ml) which is diluted for 25 ml.
So for total solution (2.88 ml) the peak area will be = 100 x 7291 / 26.04 = 27,999.23
From standard we know 3479 peak area is from 7.44 mg/L
So 27,999.23 peak area will be from 59.877 mg/L................that will be...........0.059877 mg/ml
So 2.88 mL will have a .............2.88 x 0.059877 = 0.172 mg
So 5.81 g of spinach will have 0.172 mg DDT
So 1 g of spinach will have 0.0296 mg
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Following are considered while calculating and arriving at the solution.......
i) It's assumed that the total area given is for the total volume of sample in the absence of injection volume into machine. For precise calculations one needs to know injection volume into the machine. Hence, in the absence of injection volume it's assumed that the area produced is for total volume of standard or unknown sample.
ii) The areas produced for CHCl3 and unknown sample are considered as non influencing and independent of each other. It's assumed that the CHCl3 area has no influence on the area of sample used.