Question

Drug money extract in the 5.0 mL methanol containing cocaine gave a signal of 4.27 mV in a GC-FID experiment. Then 500 L of 20.0 ppm cocaine were added to the 5.0 mL methanolic extract. This spiked solution sample gave a signal of 7.98 mV. Find the concentration of cocaine in the original extract, in ppm, in mol/L, in g.

Answer 1

molar mass of cocaine = 303.353 g/mol

molarity of 20 ppm solution of cocaine = 20 mg/L = 20/303.353 = 0.066 mmol/L

moles of cocaine added = molarity x volume

                                       = 0.066 mmol/L x 0.500 L

                                       = 0.033 mmol

Molarity of cocaine in 505 mL (0.505 L) solution = moles/L

                                       = 0.033/0.505

                                       = 0.065 mmol/L

We have 0.065 mmol/L solution gave a signal of 7.98 mV in GC-FID

So, 4.27 mV would be for a cocaine concentration = 4.27 x 0.065/7.98

                                        = 0.035 mmol/L in 5 mL of sample solution

Concentration in mol/L = 3.5 x 10^-5 mol/L

Concentration in ppm = 0.035 x 303.353

                                        = 10.62 mg/L

                                        = 10.62 ppm

Concentration in g = molarity x volume x 303.353

                                         = 0.035 mmol/L x 0.005 L x 303.353 g/mol x 1/1000

                                         = 5.31 x 10^-5 g

                                        = 0.01062 g