Response (arbitrary units) Hexane r.t. = 1.25 min Methane r.t. = 0.50 min T Solute 1 r.t. = 1.78 min Heptane Octane r.t. = 2.47 min r.t. = 5.66 min Solute 2 r.t. 4.83 min 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 Time (min)

Response (arbitrary units) Hexane r.t. = 1.25 min Methane r.t. = 0.50 min T Solute 1 r.t. = 1.78 min Heptane Octane r.t. = 2.47 min r.t. = 5.66 min Solute 2 r.t. 4.83 min 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 Time (min)

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

ChapterA: Scientific Notation And Experimental Error

Section: Chapter Questions

Problem 9P

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Question

calculate the kovats retention indices for solutes 1 and 2 from the data shown and assume tm=0.50 min (methane)

Response (arbitrary units)
Hexane
r.t. = 1.25 min
Methane
r.t. = 0.50 min
T
Solute 1
r.t. = 1.78 min
Heptane
Octane
r.t. = 2.47 min
r.t. = 5.66 min
Solute 2
r.t. 4.83 min
0 0.5
1
1.5
2
2.5
3
3.5
4 4.5
5
5.5
6
Time (min)

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