The following questions apply to separation of a solution containing pentanol (F
ID: 996120 • Letter: T
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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.
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. (a) Separation of a standard solution containing 231 mg of pentanol and 269 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 0.916:1.00. Calculate the response factor, F, for 2,3-dimethyl-2-butanol. Number b) Calculate the areas for pentanol and 2,3-dimethyl-2-butanol gas chromatogram peaks in an unknown. or pentanol, the peak height was 46.7 mm and the width at half-height was 3.3 mm. For 2,3-dimethyl-2- butanol, the peak height was 24.7 mm and the width at half-height was 2.1 mm. Assume each peak to be a Gaussian Number mm2 Pentanol: Number 2,3-Dimethyl-2-butanol: L mim (Scroll don for more questions.,) (c) If the concentration of pentanol in the unknown solution of part (b) was 94.9 mM, what was the 2,3- dimethyl-2-butanol concentration?Explanation / Answer
We need to calculate the molar concentration of pentanol
M(pentanol) = [231x10-3 g /88.15 g /mol]/10x10-3 L = 0.2621M
M (butanol ) = [269x10-3 g /102.17 g /mol]/10x10-3 L = 0.2633 M
The Response factor is = area pentanol /concentration pentanol = 0.916/0.2621M = 3.4948
b) areas: Use the formulas provided
peak area pentanol = 1.604 x peak height x w1/2 = 1.604 x 46.7 x 3.3 = 247.192 mm2
peak area 2,3-dimethyl-2-butanol = 1.604 x peak height x w1/2 = 1.604 x 24.7 x 2.1 = 83.199 mm2
C ) 247.192 /94.9 mM x 1.00/0.916 = 2.6048 mM
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