1. Suppose you do a GC analysis on a sample consisting of 15% A, 60% B, and 25%

Question

1. Suppose you do a GC analysis on a sample consisting of 15% A, 60% B, and 25% C by weight. A and B are volatile compounds that elute from the GC column while C is nonvolatile and does not come off the column and is therefore not detected by the detector. Assuming you do not know there is nonvolatile material in the sample, what will you conclude from the GC analysis? What percentages of A and B would you calculate?

2. To get around the problem presented in question 2, the internal standard technique is used. An internal standard is a volatile compound that does not interfere with the GC analysis.

-10.0 mg of internal standard compound (Std) added to 50.0 mg of the mixture of A, B,

and C (problem 2) to give a mixture with a total mass of 60.0 mg.

-The sample is analyzed by GC.

-Relative Peak Areas: Std (1.00), A (0.75), B (3.00). Remember C does not elute.

-Based off of the 10.0 mg of internal standard added to the sample, what are the weight

percentages of A and B in the original sample?

Explanation / Answer

1) If C does not get detected then the normalized peaks will be such that A = 20% and B = 80%

2) wt % of compund = (peak area of compound/total area)*100

Total peak area = Std + A + B = 1 +0.75+3 = 4.75

wt % A = (0.75/4.75)*100 = 15.78%

wt % B = (3.00/4.75)*100 = 63.16%

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