Aresearcher used hplc to examine a bicomponent mixture containing 1.22 mg/l of compound a and 1.31 mg/l of compound b, which was added as an internal standard. this mixture produced peak areas for compounds a and b of 10919 and 5379 , respectively. using the above information, determine the response factor (f).

after establishing f, the researcher prepared a solution by combining 8.18 mg of b with 10.00 ml of an unknown solution containing only a and then diluted it to a final volume of 50.00 ml. the sample was examined using hplc and peak areas of 6065 and 9111 were observed for a and b, respectively.

determine the concentration of a (mg/ml) in the unknown solution.

Response factor A: 8950 L/mg

Response factor B: 4106 L/mg

A = 02497 mg/mL

Explanation:

The response factor (F) is defined as the ratio between the signal produced by an analyte and its concentration.

For A:

F(A) = 10919 / 1,22mg/L = 8950 L/mg

F(B) = 5379 / 1,31mg/L = 4106 L/mg

It is possible to obtain relative response factor (RRF) that is the ratio between F(A) and F(B), thus:

RRF = 8950 L/mg / 4106 L/mg = 2,180

RRF could be:

That is:

The concentration of B in mg/mL is:

8,18mg / (10,00mL + 50,00mL) = 0,1363 mg/mL

Replacing:

A = 0,04162 mg/mL -In the diluted solution-

The concentration in the unknown solution is:

0,04162 mg/mL × = 02497 mg/mL

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