Five standards of various butyl alcohols diluted with methanol to 25 mL. One con

Question

Five standards of various butyl alcohols diluted with methanol to 25 mL. One contained n-butyl alcohol. Another contained only sec-butyl alcohol and methanol. A third only contained iso-butyl alcohol. A fourth contained only tert-butyl alcohol. The final sample contained equal amounts of each alcohol. These samples were then run on the GC to obtain chromatograms of the data. The table below shows peak areas that correspond to each of the butyl peaks (you will need to determine which peak corresponds to which alcohol by comparing the retention times of each of the single alcohol standards to the standard with all of the alcohols). An Unknown sample that contained the butyl alcohols was also run on the GC to obtain its chromatogram. The peak areas that correspond to each alcohol in the unknown shown in the table below.

Using the data in the table above, calculate the percentage of each butyl alcohol in the unknown sample.

Alcohols Peak Areas of Unknowns Peak Areas of Approximate Retention Times of Standard with All the Butyl Alcohols Butyl Alcohols (min) N-butyl Sec-butyl Iso-butyl Tert-butyl 4.4 3.8 4.0 3.3 0.595 0.558 0.635 0.635 0.236 0.459 0.145 0.260

Explanation / Answer

to slove this question add all the are we will get one number the divide each area with the number multiply by 100 give you the percentage

sso 0.236+0.459+0.145+0.26=1.1

so % of n butyl =0.236/1.1*100=21.45%

sec butyl=0.459/1.1*100=41.72%

isobuty=0.145/1.1*100=13.18%

tert butyl=0.26/1.1*100=23.63%

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