Question

A) A chromatographic analysis is being performed using internal standard calibration. The unknown "U" and the internal standard "S" give well-separated peaks. A calibration sample has the concentrations given below, resulting in the peak areas given below. Calculate the value of the response factor F.

Substance	concentration(mg/l)	peak area
U	19.8	2153
S	34.1	3645

What is F?

B) A test sample is prepared with the concentration of internal standard given below, resulting in the peak areas given below. Calculate the concentration of the unknown in the test sample.

Substance	Concentration (mg/L)	Peak area
U	?	3258
S	28.6	1994

? = ______ mg/L

Answer 1

Response factor, F is given by the ratio between signal produced by an analyte and the amount of analyte that produces the signal.

So, (area of analyte signal / area of standard signal) = F (concentration of analyte /concentration of standard)

Therefore .. A)

(2153/3645)=F (19.8/34.1)

0.5907 = F (0.5806)

So F = 0.5907/0.5806 = 1.0174

Therefore F is 1.0174.

B)

3258/1994 = F (U/28.6)

1.6339 = 1.0174 ( U/28.6)

So U is equal to (28.6 * 1.6339)/1.0174 =45.93 mg/L which is the concentration of test sample.