Four Compounds Methanol, butyl acetate, isopropanol, and 2-butanone. Trial #3 Ru
ID: 929068 • Letter: F
Question
Four Compounds Methanol, butyl acetate, isopropanol, and 2-butanone.
Trial #3
Run the following mixture of the four compounds increasing Temp slightly with a short hold time
Start Temp
30 degree’s C
Hold Time
1 min
Ramp Time
10 degree’s C/min
Final Temp
50 degree’s C
Hold Time
2 min
Total Time
3 min
Pressure
7kPa
Which of the four peaks represents isopropyl alcohol?
How does does the boiling point of isopropyl alcohol compare with the boiling point of the other three compounds?
Explain why the isopropyl alcohol peak appears where it does in the gas chromatogram?
Please explain, thank you!
Trial #3
Run the following mixture of the four compounds increasing Temp slightly with a short hold time
Start Temp
30 degree’s C
Hold Time
1 min
Ramp Time
10 degree’s C/min
Final Temp
50 degree’s C
Hold Time
2 min
Total Time
3 min
Pressure
7kPa
Explanation / Answer
The peak centered aroung 1.3 is for i-PrOH.
The boiling point of i-PrOH is below 2-butanone and butyl acetate but higher than methanol.
In gas chromatography the compounds are separated on the basis of its boiling oint among other properties. So as the boiling point of i-PrOH is higher than MeOH it appears right after that.
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