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BIOCHEMICAL ENGINEERING INSTRUMENTATION - EXAMPLE 29.5 : The following formula is used as a reference : (analyte signal) / (internal standard signal) = (f-factor) x (concentration of analyte) / (concentration of internal standard). A solution containing 3 mM of analyte and 4 mM of internal standard gave peak signals of 2 and 3 mamps respectively. Another similar solution containing 2 mM of analyte and 1 mM of internal standard gave peak signals of 1 and 4 mamps respectively. Find the average f-factor.
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Answer Posted / kangchuentat
BIOCHEMICAL ENGINEERING INSTRUMENTATION - ANSWER 29.5 : Let (f-factor) = (analyte signal) (concentration of internal standard) / [ (internal standard signal) (concentration of analyte) ]. For first solution, (f-factor) = (2 mamps) (4 mM) / [ (3 mamps) (3 mM) ] = 8 / 9. For second solution, (f-factor) = (1 mamps) (1 mM) / [ (4 mamps) (2 mM) ] = 1 / 8. Average : Average f-factor = (8 / 9 + 1 / 8) / 2 = (64 + 9) / (72 x 2) = 73 / 144. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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