Question

The following questions apply to separation of a solution containing pentanol (FM 88.15) and 2,3-dimethyl-2-butanol (FM 102.17). Pentanol is the internal standard.

Separation of a standard solution containing 231 mg of pentanol and 249 mg of 2,3-dimethyl-2-butanol in 10.0 mL of solution led to a pentanol:2,3-dimethyl-2-butanol relative peak area ratio of 0.927:1.00. Calculate the response factor, F, for 2,3-dimethyl-2-butanol.

(b) Calculate the areas for pentanol and 2,3-dimethyl-2-butanol gas chromatogram peaks in an unknown. For pentanol, the peak height was 69.9 mm and the width at half-height was 2.3 mm. For 2,3-dimethyl-2-butanol, the peak height was 47.6 mm and the width at half-height was 4.8 mm. Assume each peak to be a Gaussian.

       1) _______mm^2

       2)________mm^2

(c) If the concentration of pentanol in the unknown solution of part (b) was 61.5 mM, what was the 2,3-dimethyl-2-butanol concentration?

           _______________mM

Answer 1

solution ,

given

realative peak area = 0.927:1.00.

1] response factor is calculated by

F = peak area /Concentration in mg

F = (1/0.927) / (24.9/102.17)/(23.1/88.15)

F = 1.078/0.2440.262

F = 1.158

B)

areas for pentanol = (1.064)(69.9)(2.3)

= 171.05

areas for 2,3-dimethyl-2-butanol = (1.064)(47.6)(4.8)

= 243.10

C)

Here we have to find Cx concentration that is concentration of 2,3-dimethyl-2-butanol

using formula

Cx = (Ax/Ais)Cis/F

Ax = area for 2,3-dimethyl-2-butanol

Ais = areas for pentanol

Cis = concentration for unknown

Cx = (Ax/Ais)Cis/F

= (243.10/171.05)(61.5/1.158)

= ( 1.421)(53.10)

= 75.46mm