A student used acetone to clean a sample vial, but forgot to allow it to evaporate...

A student used acetone to clean a sample vial, but forgot to allow it to evaporate completely before pouring the graduated cylinder of fraction 2 into the vial—the fraction was collected correctly from the distillation apparatus but then became contaminated with acetone prior to GC analysis. This experimental error was quite evident via GC: there was an additional peak on the graph. Would there be any effect on the calculated percent composition of the two esters for this fraction? Briefly explain why or why not

Expert Solution

The calculated percentages of both the esters in the mixture will be lower than the actual value. This is due to the fact that the percentage of a compound in a mixture is obtained from GC by analyzing the area under the curve (peak) for the compound. Assuming the original mixture had only two esters, the GC spectrum would have shown only two peaks corresponding to the two esters. The sum total of the areas under the peaks remain constant during the GC run since the sum represents the total percentage composition of the mixture (which must be 100%).

Due to the contamination, the GC spectrum will show an additional peak and now the total area under the peaks is split into three (one of the peaks is simply the contaminant acetone). Hence, the area under the peak for the two esters will be reduced and consequently, the percentages of each ester will be ester. Thus, the percentage composition of the mixture will be less than 100% and it will appear as if the mixture contained a GC-inert material.

queen_honey_blossom answered 2 years ago

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