A standard solution containing 6.8 multiply.gif 10-8 M iodoacetone and 2.3 multi

Question

A standard solution containing 6.8 multiply.gif 10-8 M iodoacetone and 2.3 multiply.gif 10-7 M p-dichlorobenzene (an internal standard) gave peak areas of 333 and 718, respectively, in a gas chromatogram. A 3.00 mL unknown solution of iodoacetone was treated with 0.100 mL of 1.0 multiply.gif 10-5 M p-dichlorobenzene and the mixture was diluted to 10.00 mL. Gas chromatography gave peak areas of 363 and 260 for iodoacetone and p-dichlorobenzene, respectively. Find the concentration of iodoacetone in the 3.00 mL of original unknown.

Explanation / Answer

In the unknown prep, the concentration of internal standard (IS) is calculated as follows:
(0.1ml)(1.0x10^-5M) = (3.1ml)(xM) and x = 3.22x10^-7
This is then diluted to 10 mls, so conc. 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.

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