Gas Chromatography A standard solution containing 6.3×108 M iodoacetone and 2.0×

Question

Gas Chromatography

A standard solution containing 6.3×108 M iodoacetone and 2.0×107 M p-dichlrorobenzene (an internal standard) gave peak areas of 395 and 787, respectively, in a gas chromatogram. A 3.00-mL unknown solution of iodoacetone was treated with 0.100 mL of 1.6×105 M p-dicholorobenzene and the mixture was diluted to 10.00 mL. Gas chromatography gave peak areas of 633 and 520 for iodoacetone and p-dichlorobenzene, respectively. Find the concentration of iodoacetone in the 3.00 mL of original unknown.

Explanation / Answer

In the unknown preparation the concentration of internal standard iscalculated :
(0.1ml)*(1.0x10^-5M) = (3.1ml)(xM) and x = 3.22x10^-7
and then diluted to 10 mls, so concentration of IS is

: (3.1ml)(3.22x10^-7M) = (10 ml)(xM); x = 0.998x10^-7M
This would be the final conc. of IS in the 10 ml of diluted unknown.

Here's the problem: The IS in the first run was 2.3x10^-7M and gave peak area of 718. In the unknown run, it is 1.0x10^-7M and gives peak area of 260. Thus it is not linear. The 1.0x10^-7 should give peak area of 630 (1.6 is 80% of 2). So either the recovery was different in the unknown, or the response is not linear.

I could go on and calculate a "presumptive" value for the iodoacetone in the unknown, but it would be based on assumptions I'm not sure you are willing to make. In theory, the IS should correct for recovery, but since you didn't really "treat" the sample, I don't see where losses could occur. If, on the other hand, you want to continue, just assume 80% recovery of the unknown iodoacetone (based on 80% of the expected value of the IS), and proceed.

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.