Question

A known mixture of Compound X (1.52 mg/mL) and Compound Y (1.73 mg/mL) gave peak areas of 16.02 and 9.41 cm², respectively, after separation by isocratic HPLC. A solution was prepared by mixing 8.76 mg of Compound Y with 5.00 mL of unknown containing just Compound X, and diluting to 10.00 mL. Peak areas of 5.96 cm- (Compound X) and 4.86 cm (Compound Y) were measured. Find the concentration of Compound X in the unknown.

Answer 1

Calculation of response factor (F) .

We have equation, A _x / [X] = F ( A _s / [S] )

In this equation, A _x is area of analyte, [X] is concentration of analyte, F is response factor, A _s is area of standard,[S] is concentration of standard.

Given : Analyte is X and standard is Y

Peak Area of X = 16.02 cm ²

Concentration of X = 1.52 mg/ml

Peak Area of Y =9.41 cm ²

Concentration of Y = 1.73 mg/ml

$\therefore$ 16.02 cm ² / 1.52 mg/ml = F ( 9.41 cm ² / 1.73 mg/ml )

10.539 cm ² / (mg/ml) = F 5.439 cm ² / (mg/ml)

F = 10.539 cm ² / (mg/ml) / 5.439 cm ² / (mg/ml)

F = 1.94

Now, calculate concentration of standard (Y) in the mixture.

8.76 mg compound Y is present in 10.00 ml solution.

[Y] = 8.76 mg / 10.00 ml = 0.876 mg / ml

Put the values of response factor , [S] and peak area into equation A _x / [X] = F ( A _s / [S] ) we get

5.96 cm ² / [X] = 1.94 ( 4.86 cm ² / 0.876 mg/ml)

5.96 / [X] = 10.76

[X] = 5.96 / 10.76 = 0.5539 mg /ml

10.00 ml sample prepared have concentration of X = 0.5539 mg / ml.

Sample solution is prepared by diluting 5.00 ml of X to 10.00 ml.

Hence, original concentration of X in the unknown is calculated as shown below.

C stock x V stock = C dilute x V dilute

C stock = C dilute x V dilute / V stock =0.5539 mg /ml x 10.00 ml / 5.00 ml = 1.107 mg/ml = 1.11 mg / ml

ANSWER : concentration of X in the unknown = 1.11 mg /ml