To get around the problem presented in question 2, the internal standard techniq

Question

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

Solution :-

Area of the A= 0.750

Area of the internal standard = 1.00

Area of the B= 3.00

Therefore total area = 1.00 +0.750 + 3.00 = 4.750

now using the total area and the area of the each we can find their percentage

as follows

Weight percent of A = (area of A/ total area)*100%

                                = (0.750 / 4.750)*100%

                                = 15.79 %

weight percent of the B = (area of B/total area)*100%

                                    = (3.00/4.750)*100%

                                    = 84.21% B

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