The following questions apply to separation of a solution containing pentanol (F
ID: 865260 • Letter: T
Question
The following questions apply to separation of a solution containing pentanol (FM 88.15) and 2,3- butanol (Fm 102.17). Pentanol is the internal standard. Separation of a standard solution containing 231 mg of pentanol and 233 mg of 2,3-dimethyl 2-butandol in 10.0 mL of solution led to a pentanol:2,3-dimethyl-2-butanol relative peak area ratio of 0.884:100. Calculate the response factor, F, for 2,3-dimethyl-2-butanol. Calculate the areas for pentanol and 2,3-dimenthyl-2-butanol gas chromatogram peaks in an unknown. For pentanol, For pentanol, the peak height was 28.4 mm and the width at half-height was 2.4 mm. For 2,3-dimethyl-2-butanol, the peak height was 29.4 mm and the width at half-height was 4.9mm. Assume each peak to be a Gaussian. If the concentration of pentanol in the unknown solution of part (b) was 64.4mM, what was the 2.3-dimethyl-2-butanol concentration?Explanation / Answer
(a) Separation of a standard solution containing 231 mg of pentanol and 233 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 0.884:1.00
Fi = (Ai/Ap) / (Ci/Cp)
Fi = (1.000/0.884) / [(233 mg/88150 mg.moL-1)/(231 mg/120170 mg.moL-1)]
Fi = (1.1312) / (1.3750)
Fi = 0.8227
(b) Calculate the areas for pentanol and 2,3-dimethyl-2-butanol gas chromatogram peaks in an unknown. For pentanol, the peak height was 28.4 mm and the width at half-height was 2.4 mm. For 2,3-dimethyl-2-butanol, the peak height was 29.4 mm and the width at half-height was 4.9 mm. Assume each peak to be a Gaussian.
Apentanol=Bx(H/2) =28.4mm x 2.4mm =68.16 mm2/2 =34.08 mm2
A2,3-dimethyl-2-butanol=Bx(H/2) =29.4mm x 4.9mm =144.06 mm2/2=72.03 mm2
(c) If the concentration of pentanol in the unknown solution of part (b) was 64.4 mM, what was the 2,3-dimethyl-2-butanol concentration?
F2,3-dimethyl-2-butanol = (A2,3-dimethyl-2-butanol/Apentaol) / (C2,3-dimethyl-2-butanol/Cpentanol)
0.8227= (72.03 mm2/34.08mm2) / (C2,3-dimethyl-2-butanol/64.4mM)
C2,3-dimethyl-2-butanol = [(72.03 mm2/34.08mm2) / 0.8227] x 64.4mM
C2,3-dimethyl-2-butanol = 165.47 mM
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.