A procedure to determine nicotine in urine utilizes an internal standard method
ID: 820676 • Letter: A
Question
A procedure to determine nicotine in urine utilizes an internal standard method for quantitation. A 10.0 mL urine sample containing no nicotine was spiked with 10 microL of a solution containing 100 microg/mL of the internal standard 5-aminoquinoline (5-AQ) and 10 microL of a solution containing 50 microg/mL of nicotine. A 5.0 mL volume of dichloromethane (DCM) was then added and the sample shaken for 10 minutes. Extraction efficiencies for both compounds were assumed to be 100%. The DCM fraction was then removed and a 1.0 microL aliquot injected for GC analysis. The areas obtained for the 5-AQ and nicotine from the resulting chromatogram were 345 and 470 relative area units respectively. Extraction of a urine sample known to contain nicotine was performed by spiking a 10.0 mL sample only with 10 microL of the 5-AQ internal standard prior to extraction with 5.0 mL of DCM. The DCM was then evaporated down to 1.0 mL prior to using a 1.0 microL injection for the analysis. The peak areas for this chromatogram for the 5-AQ and nicotine were 1740 and 680 units respectively. Calculate the concentration of nicotine in the urine sample in microg/mL.
Explanation / Answer
Initial volume = 10 mL
5- AQ added= 10 micro L= 0.01 mL
concentration of 5 AQ = 100 micro g/ mL
Amount of 5 AQ= 0.01 *100 = 1 micro g
Nicotine added= 10 micro L= 0.01 mL
Concentration of nicotine = 50 micro g /mL
Amount of nicotine= 0.5 micro g
DCM added = 5 mL
Total volume = 15.02 mL
DCM removed . So, 5ml reduced
Aliquot added = 1 micro liter= 0.001 mL
Total volume now= 15.02-5+0.001= 10.021 mL
Concentration of 5 AQ= 1/10.021= 0.09979 micro g /mL
PEAK area= 345
concentration= Peak area / total area (A1)
So A1= 345/ 0.09979=3457.26
Concentration of nicotine= 0.5/ 10.021= 0.04989 mico g/mL
PEAK area= 470
Total area:
A2=470 / 0.04989 = 9420.72
ANALYSIS:
Sample= 10mL
5 AQ added= 10 micro L= 0.01 mL
DCM left= 1 mL + 0.001 = 1.001 mL
Total volume= 11.011 mL
Concentration of 5-AQ= 1740 / A1= 1740 /3457.26
= 0.5032 micro g/ mL
Concentration of nicotine = 680 / A2= 680/ 9420.72
= 0.07218 micro g /mL
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