2. In order to check to see that the percent recovery for a new sample preparati

Question

2. In order to check to see that the percent recovery for a new sample preparation method was constant over the range expected for pesticide residues in corn stalks (normally 1.0 5.0 ppb, depending on the time since application), a series of 20 g samples of untreated corn stalks were spiked with aliquots of a highly diluted solution of the pure pesticide. The spiked cornstalks together with 25 mL of acetonitrile were placed in a powerful blender and agitated for 45 s. The acetonitrile extract was filtered and then analyzed directly by gas chromatography. Comparison of the residue samples with standard samples produced the results below: Sample Instrument Response a. Calculate the concentration (in ppb) that would (arb. unit x 10 s) be expected in each of the solutions analyzed for Standards the samples. 1.00 ppb b. Plot the response versus concentration curve for 2.00 ppb 4.25 the standards. c. Based on the graph in part b, what was the 3.00 ppb 6.52 concentration in each of the solutions from the 4.00 ppb 8.57 residue samples? 10.76 5.00 ppb d. Calculate the percent recovery for each sample. Is Samples 200 g corn stalks x mL of standard the recovery constant over the range tested? containing 2.5 100 mL M LaL 2.10 1.00 mL spike 4.15 2.00 mL spike 6.28 3.00 mL spike 8.43 4.00 mL spike 10.28 5.000 mL spike

Explanation / Answer

a)the concentration of the standard used to spike=25 ppb

that is because the standard used is 2.5 micrograms per 100 ml when it is converted to ppb it gives 25 ppb

ppb means parts per billion that means mg per 106 ml

2.5 micrograms=2.5* 10-3 mg

100 ml contains 2.5* 10-3 mg therefore 1ml contains 2.5* 10-3 mg/100=2.5* 10-5 mg

106 ml contain=2.5* 10-5 mg*106 ml=25 ppb

to spike the samples 1,2,3,4 and 5 ml of this standard is used.

therefore for 1.0 ml= 25*10-6 mg

this was extracted in 25 ml of acetonitrile,therefore 25*25*10-6 mg/25 ml =1*10-6 mg/ml = 1 ppb

therefore for 2,3,4 and 5 the expected values will be 2, 3, 4 and 5 ppb respectively

b)

c) using the graph

for 2.10 =0.9805

for 4.15=1.9322

for 6.28=2.921

for 8.43=3.919

for 10.28=4.778

d)percentage recovery=amount recovered/amount spiked*100

1.0 ml

0.9805/1*100=98.05%

similary,

for 2.0 ml =96.61%

for 3.0ml=97.37%

for 4.0 ml=97.975%

for 5.0 ml=95.56%

Recovery is not constant as % recovery for each sample varies.

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