Question

A solution containing 3.48 mM X (analyte) and 1.67 mM S (internal standard) gave peak areas of 3483 and 10069, 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.

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

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

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

Answer 1

a) the response factor for the analyte:

The response factor F can be expressed as

F =  [ Area of sample x Con.of standard ] / [Area of standard x Con.of sample]

   = [3483 x 1.67] / [ 10069 x 3.48]

   = 0.1660

b) The concentration (mM) of X in the 10.0 ml of mixed solution.:

From given data

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

Concentration of standard in 10 ml = [8.47] / [10]

                                                        = 0.847 mM

From the above formula F =  [ Area of sample x Con.of standard ] / [Area of standard x Con.of sample]

Con. of unknown = [ Area of sample x Con.of standard ] / [Area of standard x Unknown Response factor]

                             =    [5428 x 0.847] / [4431 x 0.166]

                             = 6.2505 mM

c) The concentration (mM) of X in the original unknown:

          If we use M1V1=M2V2

          M1=?, V1= 5 ml, M2= 6.2505, V2= 10.0 ml

the original unknown concentration M1 =       [6.2505 x 10] / [2]

                                                               =       12.501 mM