a Sapling Learning macmillan learning The following questions apply to separatio
ID: 1057785 • Letter: A
Question
a Sapling Learning macmillan learning 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 215 mg of pentanol and 263 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 0.893:1.00. 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 For pentanol, the peak height was 25.3 mm and the width at half height was 4.0 mm. For 2,3-dimethyl-2- butanol, the peak height was 69.6 mm and the width at half-height was 2.1 mm. Assume each peak to be a Gaussian Number Pentanol mnm Number 2,3-Dimethyl-2-butanol: mnm (Scroll down for more questions.)Explanation / Answer
WE know that
Response Factor = Peak Area / Concentration
The concentration of pentanol is 215 mg dissolved in a volume of 10 ml. This is equal to [P] = 21.5mg / mL
The concentration of 2,3dimethyl butanol is 263 mg dissolvedin a volume of 10 ml. This is equal to [B] = 26.3 mg/mL
The peak area of pentanol = 0.893 [Ap]
The peak area of 2,3dimethyl butanol= 1 [Ab]
The correspondent response factors of 2,3dimethyl butanol
RF(2,3dimethyl butanol) = Peak Area / Concentration = 1 / 26.3 = 0.038
Respons factor for internal standard = 0.893 / 21.5 = 0.0415
b) peak area = peak height X 1/2 peak width
For pentanol
Area = 25.3 X 4mm^2 = 101.2 mm^2
For 2,3dimethyl butanol)
area = 69.6 X 2.1 = 146.16 mm^2
c) We know that
Conc. a reagent in a mixture
= Peak Area reagent x (Conc. internal Standard / Peak Area Internal standard) x (1 / Relative response factor)
Relative response factor = response factor of the reagent / Response factor of internal standard
RRF = 0.038 / 0.0415 = 0.916
Concentration of 2,3dimethyl butanol = 146.16 (61.1 mM / 101.2 ) (1/ 0.916)= 96.34 mM
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