I\'m not sure if someone can answer this because it\'s from a lab so I\'ll provi
ID: 900790 • Letter: I
Question
I'm not sure if someone can answer this because it's from a lab so I'll provide as much info as I can. It'll probably be confusing but I very much need help with this.
It's Gas Chromatography. The equation that's given is (conc in unknown) = (conc in std)(peak intensity of unknown) over (peak intensity of STD)
An example, unknown sample is analyzed and found to have peak intensity of 150 counts. An 18 ppm standard of the same compound shows a peak intensity of 125 counts. the concentration of the unknown can therefore be determined using equation 1 to be - conc of unknown = 18 ppm (150/125) = 22 ppm. In this case, suppose that when the same standard is then diluted by adding 1.0 mL of the sample with 10.0 mL of solvent, the GC does not detect the compound. (18 ppm) (1.0 mL) / (11.0 mL) = 1.6 ppm
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These following sets of numbers are important in this question- They are retention time and peak intensity
1 - 5.0 sec, 1556 counts
2 - 12.5 sec, 2073 counts
3- 18.3 sec , 1279 counts
4- 29.7 sec, 705 counts
5- 35.0 sec, 707 counts
6- 52.5 sec, 396 counts
For this experiment, I had to dilute the sample by 50% by adding 1.0 mL of solvent solution. After all of those steps were done, it would show several peaks on the monitor, and several of them were reduced by 50% in the diluted sample. However, other peaks werent much different and that's because of the compounds present in the solvent solution, rather than in the drug standards.
Ok, now for the follwing question, the peak intensity and retention times of the peaks above need to be looked at. For this question, the standard solution was diluted in half, so all of the peak intensities should be reduced by about half. Compare peak intensities observed in the numbers above, to the intensities of the peaks from running the diluted standards in the following question.. The peaks which were not reduced by half, are the solvent peaks and you should input the retention times for these solvent peaks for the answer to this question.
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What are the retention time for the two peaks, that can be assigned by this method to the solvent solution, rather than to te drug standards?
First solvent peak - ? s
Second solvent peak - ? s
Explanation / Answer
I try to solve the questions. I understand that there a 2 different questions.
For the first question I see that the calculation is made .I suppose you need only an explanation.
For the second question, can you say wich are the peaks that are not reduced by half when the sample was diluted?
Please comment.
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For the first part.
It is supposed that the peak intensity vary linearly with concentration for the same compound.
Equation explained:
18 ppm produce a peak intensity of………………125 counts
X ppm (sample)…………………………………150 counts
X = 18 ppm x 150 counts / 125 counts = 21.6 ppm (rounded to 22)
Or in the general form:
conc in unknown = (conc in std) x(peak intensity of unknown) / (peak intensity of STD)
If the sample is diluted 10 times and its new concentration (1.6 ppm) is not detected,
it means that the corresponding intensity (150/10 = 15 counts) is too low and covered or comparable with the noise of the recording baseline.
For the second part
You say:
“For this experiment, I had to dilute the sample by 50% by adding 1.0 mL of solvent solution [to 1 mL sample]. After all of those steps were done, it would show several peaks on the monitor, and several of them were reduced by 50% in the diluted sample. “
The injected sample has always the same volume (some few microL). The drug is in a very low proportion (ppm). So you have the same main volume of solvent in the injected samples and constant peak intensity if peak is due to solvent.
It is a common case that the first peak is from solvent (it is not retained in column).
You have to compare two records (I don’t see any image here). You give me data from only one record. If you have the second one, look to find 2 peaks that have the same intensity in both records.
I see that you have read the retention times (e.g. peak 1 have 5.0 sec). The only thing to do is to identify the 2 peaks and report their retention times.