Map The following questions apply to butanol (FM 102.17). Pentanol is the intern

Question

Map The following questions apply to butanol (FM 102.17). Pentanol is the internal standard. Thaholcang gzestions ntay to srho rntiaasolon containing pentand FM 8.15) and 2.3-dimethyez a) Separation of a standard solution containing 211 mg of pentanol and 265 mg of 2,3-dimethyl-2-butanol in 0 mL of solution led to a pentanol2.3-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 91,6 mm and the width at half-height was 1.4 mm. For 2,3-dimethyl-2- butanol, the peak height was 37.7 mm and the width at half-height was 3.1 mm. Assume each peak to be a Gaussian Number Pentanol: Number 2,3-Dimethyl-2-butanol: (Scroll down for more questions.) (c) If the concentration of pentanol in the unknown solution of part (b) was 55.0 mM, what was the 2,3- dimethyl-2-butanol concentration? O Previous Give Up & View Saiulion Previous Give Up & View Sclution Check Answer Next Exit Check Answer 0 Next Exit Hint

Explanation / Answer

(a) Response factor = area/concentration

concentration of 2,3-dimethyl-2-butanol = 265/10 = 26.5 mg/ml

Response factor becomes = 1/26.5 = 0.038

(b) Area = peak height x width at half height

So, area for pentanol = 91.6 x 1.4 = 128.24 mm^2

area for 2,3-dimehtyl-2-butanol = 37.7 x 3.1 = 116.87 mm^2

(c) If the concentration for pentanol was 55 mM, then the concentration for 2,3-dimethyl-2butanol would be,

50.14 mM

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