Gas Chromatograghy

Question

If a sample is made using 0.145g of methanol and 0.554g of ethanol, what percent of the total peak area is expected for the methanol peak (that is, what is 100*(area of methanol peak/(area of methanol peak + area of ethanol peak)=?) 

Answer 1

The ratio of the areas is expected to be the same as the ratio of the mole percentages of the substances.

 

0.145 g methanol *  1 mole/32 g =  4.53 millimoles

0.554 g ethanol * 1mole/46g = 12.0 millimoles

 

Replacing moles for areas in the given equation yields    100*4.53/(12.0+4.53) = 27.4%