Given data:
The following data are given in the question.
Fractional distillation (FD):
Distillate weight = 0.30; n-octane (by GC) = 23.74%
Residue weight = 0.33; n-octane (by GC)= 62.72%
Temperature @ first distillate collected = 66oC
Temperature @ distillation discontinued = 80oC
Simple distillation (SD):
Distillate weight = 1.10; n-octane (by GC)= 37.28%
Residue weight = 1.03; n-octane (by GC)= 76.25%
Temperature @ first distillate collected = 80oC
Temperature @ distillation discontinued = 92oC
Solution:
Key points:
Weight % of octane / isooctane in the distillate of FD:
Let us consider, RUN # 146 as the chromatogram for distillate. Based on the key point 3, isooctane eluted first, at RT 0.792 min (62.72%) and n-octane eluted second, at RT 1.281 min (37.29%). Let us consider the weight is in g
Weight of isooctane in distillate = Distillate weight x (% of isooctane / 100) = 0.30g x (62.72 / 100)
= 0.30g x 0.6272 = 0.18816 = 0.19g
Weight of octane in distillate = Distillate weight x (% of octane / 100) = 0.30g x (37.29 / 100) = 0.30g x 0.3729
= 0.11187 = 0.12g
Hence the distillate from fractional distillation contain 0.12g of octane and 0.19g of isooctane.
Weight % of octane / isooctane in the distillate of SD:
Let us consider, RUN # 147 as the chromatogram for distillate. Based on the key point 3, isooctane eluted first, at RT 0.781 min (23.75%) and n-octane eluted second, at RT 1.237 min (76.25%). Let us consider the weight is in g
Weight of isooctane in distillate = Distillate weight x (% of isooctane / 100) = 1.10g x (23.75 / 100)
= 1.10g x 0.2375 = 0.26125 = 0.26g
Weight of octane in distillate = Distillate weight x (% of octane / 100) = 1.10g x (76.25 / 100) = 1.10g x 0.7625
= 0.83875 = 0.84g
Hence the distillate from simple distillation contain 0.84g of octane and 0.26g of isooctane.
please show work based on the given data. the problem/question is given under the sub heading...