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?

Answer 1

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