Question

A solution containing octanol was analyzed by GC using the method of internal standard. The internal standard was hexanol.

Control sample: A control solution was prepared that had 10.0 mM octanol and 2.00 mM hexanol. This sample produced a chromatograph with peak areas of 4500 for octanol and 5000 for hexanol.

Unknown sample: A spiked unknown was prepared by adding 2.00 mL of 10.00 mM hexanol to 5.00 mL of the unknown octanol solution and diluted to 10.00 mL with solvent. This solution gave peak areas of 3500 and 4000 for octanol and hexanol, respectively.

Calculate the concentration of octanol in the original (undiluted) unknown solution?

Answer 1

Concentration of octanol in the original (undiluted) unknown solution = 19.44 mM

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In GC, response factor F, is given by the relation, (Ax/Cx) = F (Ais/Cis)

Where, Ax and Ais are area under the curve of Analyte and Standard, Cx and Cis are concentration of Analyte and Standard.

The form response F can be calculated by the control experiment F = {(Ax)(Cis)}/{(Ais)(Cx)}

Cis (hexanol) = 2.00 mM, Cx(octanol) = 10.0 mM, Ais(hexanol) = 5000, and Ax(octanol) = 4500

F = (4500 x 2.00)/(5000 x 10) = 0.18

Unknown sample run,

Hexnanol: Cis = 2.00 mL of 10.00 mM, Ais = 4000

Octanol: Cx = 5.00 mL, Ax = 3500

Dilution = 10.0 mL (5 +2 = 7 mL Sample and 10.00-7.00 = 3 mL solvent)

Cis after dilution =   10.00 mM x 2 mL/ 10 mL = 2 mM

Cx in unknown sample; (Ax/Cx) = F (Ais/Cis); (3500/Cx) = 0.18 (4000/2)

Cx = 3500 / 360 = 9.7222 mM

Octanol (Cx) in original sample = 9.7222 mM x 10.00 mL / 5.00 mL = 19.4444 mM