Question

A standard solution containing 5.6 10-8 M iodoacetone and 2.5 10-7 M p-dichlorobenzene (an internal standard) gave peak areas of 306 and 877, respectively, in a gas chromatogram. A 3.00 mL unknown solution of iodoacetone was treated with 0.100 mL of 1.0 10-5 M p-dichlorobenzene and the mixture was diluted to 10.00 mL. Gas chromatography gave peak areas of 685 and 329 for iodoacetone and p-dichlorobenzene, respectively. Find the concentration of iodoacetone in the 3.00 mL of original unknown.

Answer 1

Suppose response factor of iodoacetone = Ria

Response factor of dichlorobenzene = Rdc

[iodoacetone] = 5.6*10-8 M

[dichlorobenzene] = 2.5*10-7 M

Peak area of iodoacetone = 306.

Peak area of dichlorobenzene = 877.

Ria = 306/(5.6 × 10-8) = 5.4 × 10⁹

Rdc = 877/ (2.5×10-7) = 3.508 × 10 ⁹

The appropriate formula for internal standard calculations is:

Ria = F x Rdc, F = 1.55

For unknown

[Iodoacetone] = x M and after dilution = 3/10 x M

[Dichlorobenzene] = 10-5 M, after dilution = 10-7 M

Now Ria = 685/(3x/10 ) and Rdc = 329/ 10-7 = 3.29 × 10 ⁹

Again

Ria = F Rdc

685/(3x/10) = 1.55 × 3.29 × 10⁹

^{x = 4.455 × 10-7 M = Unknow Original concentration of iodoacetone in 3 ml.}