Question

This question applies to the separation of a solution containing pentanol (MM 88.15 g/mol) and 2,3-dimethyl-2-butanol (MM 102.17 g/mol). Pentanol is the internal standard. Separation of a standard solution containing 211 mg of pentanol and 255 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 0.889:1.00. Calculate the response factor, F. for 2,3-dimethyl-2-butanol. F- Calculate the areas for pentanol and 2.3-dimethyl-2-butanol gas chromatogram peaks in an unknown solution. For pentanol, the peak height is 38.3 mm and the width at half-height is 4.72 mm. For 2,3-dimethyl-2-butanol, the peak height is 45.3 mm and the width at half-height is 4.78 mm. Assume each peak to be a Gaussian. pentanol: ma 2,3-dimethyl-2-butanol: If the concentration of pentanol in the unknown solution of is 18.5 mM, what is the 2,3-dimethyl-2-butanol concentration? concentration: mM

Answer 1

Liven data Mass = 285 mg Volume = tome 9 0:0100L Mut of 2,3-dimethyl 2-butanol = 102:1781moi. Area off aneliytei sjpad (Ax) concontration og analyte w Flitrea i standard signes C concentration of signal o (62 = MQSstrie moliwst XV. 2557103 102,17 x 010100 255X102 1.0217 : 16 0.249 molesll) (s) = mass mollitx voj = 2114163 88.15 80 000 - 2118103 0.8815 115) 20-239 molesl.) F - @)ýcs). Cx945 IX D'239 0.249x0,889 $.081

The area of gaussian Peak 51,064 x peale height xuidth at half height = 1.064 x 38.3.x 4.72 . > 192.34.mm? The area of pentanot and peak, 31,064 x 45,3 x 4178 = 230.39.mm. c) we know that, . [Ar] =f(ASO xD (S) F21081Axa 230-39 , C5 ) = 18.5 mm. Ag = 192.34 mm? 230139 = 1.081 (192:34) [x] 1815 230.39 - 11.238 x] (= 20.50 MM