Ethanol and methanol are separated in a capillary GC column with retention times

Question

Ethanol and methanol are separated in a capillary GC column with retention times of 370 and 385 s respectively and base widths of 16.0 and 17.0 s. An unretained air peak occurs at 10.0 s. Calculate the retention factor and the resolution. (Use longest peak to calculate N) Ethanol and methanol are separated in a capillary GC column with retention times of 370 and 385 s respectively and base widths of 16.0 and 17.0 s. An unretained air peak occurs at 10.0 s. Calculate the retention factor and the resolution. (Use longest peak to calculate N)

Explanation / Answer

Answer:

Given retention time for ethanol tR1=370 s and for methanol tR2=385 s

and retention time for unretained air peak t0=10s

Retention factor (k)=(tR1-t0)/t0=(370-10)/10=36 for ethanol

Retention factor (k)=(tR2-t0)/t0=(385-10)/10=38.5 for methanol

Average retention factor=(36+38.5)/2=37.25.

Average retention factor=37.25.

Base widths for ethanol Wb1=16 s and for methanol Wb2=17 s.

Resolution (Rs)=(tR2-tR1)/0.5(Wb1+Wb2)

Rs=(385-370s)/0.5(16+17)

Rs=0.909~0.91.

Number of theoretical plates N=16(tR/Wb)2

For longest peak i.e for methanol tR=385s and Wb=17 s

Then N=16(285/17)2=8206 plates.

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