The following questions apply to separation of a solution containing pentanol (F
ID: 695684 • Letter: T
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 and 277 mg of 2,3-dimethyl-2-butanol in I relative peak area ratio of 0.898:1.00. Calculate (a) Separation of a standard solution containing 213 mg of 0 mL of solution led to a pentanol:2,3-dimethyl-2 the response factor, F,for 2,3-dimethyl-2-butanol Number b) Calculate the areas for pentanol and 2,3-dimethyl-2-butanol peaks in an unknown or pentanol, the peak height was 28.8 mm and the width at half-height was 2.2 mm. For 2,3-dimethyl-2 butanol, the peak height was 27.2 mm and the width at half-height was 2.8 mm. Assume each peak to be a Gaussian Number Pentanol Number 2,3-Dimethyl-2-butanol L Scroll down for more questions.)Explanation / Answer
Separation on column
(a) peak are for pentanol = 0.898
peak area for 2,3-dimethyl-2-butanol = 1.00
concentration of pentanol in solution = 213 mg/88.15 g/mol x 0.010 L = 241.63 mM
concentration of 2,3-dimethyl-2-butanol in solution = 277 mg/102.17 g/mol x 0.010 L = 271.12 mM
So,
response factor (F) = (area/concentration)2,3-dimethyl-2-butanol/(area/concentration)pentanol
= (1.00/271.12)/(0.898/241.63) = 0.992
(b) area = height x width at half-height
Pentanol area = 28.8 mm x 2.2 mm = 63.36 mm^2
2,3-dimethyl-2-butanol area = 27.2 mm x 2.8 mm = 76.16 mm^2
(c) concentration of pentanol in unknown solution = 15 mM
Using response factor equation,
0.992 = (76.16/concentration of 2,3-dimethyl-2-butanol)/(63.36/15 mM)
concentration of 2,3-dimethyl-2-butanol in unknown solution = 18.17 mM
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