A solution containing 3.31 mM X (analyte) and 1.66 mM S (internal standard) gave

Question

A solution containing 3.31 mM X (analyte) and 1.66 mM S (internal standard) gave peak areas of 3423 and 10463, respectively, in a chromatographic analysis. Then 1.00 mL of 8.47 mM S was added to 5.00 mL of unknown X, and the mixture was diluted to 10.0 mL. This solution gave peak areas of 5428 and 4431 for X and S, respectively. Answer the following questions and enter your results with numerical values only.

Calculate the response factor for the analyte. (keep four significant figures)

Find the concentration (mM) of X in the 10.0 mL of mixed solution.

Find the concentration (mM) of X in the original unknown.

Explanation / Answer

1. Concentration of X, CX = 3.31 mM

Concentration of S, CS = 1.66 mM

Peak area of X, AX = 3423

Peak area of S, AS = 10463

Response factor of the analyte, F = (AX X CS) / (AS X CX) = (3423X 1.66 mM) / (10463 X 3.31 mM) = 0.1641

Response factor of the analyte, F = 0.1641

2. Peak Areas after dilution, AX = 5428 & AS = 4431

CS after dilution = (8.47 mM X 1.00 mL) / 10.00 mL = 0.847 mM

Response factor of the analyte, F = 0.1641

F = (AX X CS) / (AS X CX)

CX = (AX X CS) / (AS X F) = (5428X 0.847 mM) / (4431 X 0.1641) = 6.32 mM

Concentration (mM) of X in the 10.0 mL of mixed solution = 6.32 mM

3. CX in the original unknown = (6.32 mM X 10.00 mL) / 5.00 mL = 12.62 mM

Concentration (mM) of X in the original unknown = 12.62 mM

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