[Halogen876]

Hello,

I've been struggling with the following question:

An electronic integrator provides the following data for a set of standards and an unknown analyzed by GLC:

Sample                                                          Reading (C6H12)                Reading (C6H6)
0.00100mol C6H12 + 0.000100mol C6H6                937                                  83
0.00100mol C6H12 + 0.000250mol C6H6                1042                                231
0.00100mol C6H12 + 0.00050mol C6H6                  868                                  384
0.00100mol C6H12 + 0.001000mol C6H6                945                                  836
0.00100mol C6H12 + 1.00mL of unknown               998                                  291

The cyclohexane is used as an internal standard. All samples are diluted to 10.0mL with a suitable solvent. Show whether the instrument response is linear in concentration. Compute the concentration of benzene in the original sample.

I divided all the moles of benzene by 0.01L to get the concentrations of benzene in mol/L and I graphed those concentarions vs. the benzene readings. It looks fairly linear with a possible deviation in the last data point. I got the equation of the line and plugged in 291 for the y value and solved for x and then multiplied by 10 to take into account the dilution factor. I tried using all the points and also leaving out certain ones but the correct answer is 0.329mol/L whereas no matter how I play around with the numbers I keep getting numbers that are closer to 0.35 than 0.33 so I'm close but it seems like I must be missing something...I didn't use the internal standard numbers for anything but I'm not sure how that could come into it...if anyone could point anything out to me, I'd really appreciate at!

[Arkcon]

Yes, I did notice right away you didn't seem to be using the internal standard for anything, to say it like you said.  Lets try to figure out what we can do with those results.  You have the same exact amount of cyclohexane in each sample, including the unknown.  Carried to 5 significant figures, if we are to believe that.  But you don't get the same reading for each.  Does that give you a hint as to what you are to do?

I'm going to start working with this data set in a spreadsheet program, lets see what you can come up with.

[Halogen876]

I did notice that there was exactly the same amount of standard in each solution and that the readings did vary quite a bit. The reason I neglected to do anything with the information though, was because I figured that since there was exactly the same amount of standard in each solution, that it shouldn't affect anything. I did find it a little odd that the standard readings varied so much but, relatively speaking, they're all in the same relative range so I guessed that perhaps the instrument just wasn't very precise  but I guess there's probably more to it than that... I've been thinking about what you said, but I'm still not really sure what a good explanantion of those readings being so different would be...

[Arkcon]

What's an internal standard for?  If you class notes and textbook can't help you, try Wikipedia.

[Halogen876]

I know that the purpose of an internal standard is to add a known amount of a substance (distinct from the analyte) to the mixture for the purpose of calibration.

I know that the equation is: R_x/R_IS=F[X]/[IS].

I could use some of the data from the first 4 trials to solve for F and then use that value of F to plug back into the equation and slove for [X]. If I'm on the right track with this, my question now is would you find all the Fs between each solution and average them, since I worked out a couple of them and they're similar but not identical?

[Arkcon]

I know that the purpose of an internal standard is to add a known amount of a substance (distinct from the analyte) to the mixture for the purpose of calibration.

Wow.  That's a lot of words, but it says almost nothing.  The "purpose of calibration" means almost anything, and everything.  No, internal standards are meant to do one thing.

I know that the equation is: R_x/R_IS=F[X]/[IS].

I could use some of the data from the first 4 trials to solve for F and then use that value of F to plug back into the equation and slove for [X]. If I'm on the right track with this, my question now is would you find all the Fs between each solution and average them, since I worked out a couple of them and they're similar but not identical?

Not following you here.  Maybe you should try plugging in one example, and see if we can all follow along.

[Halogen876]

I've got it figured out. I see now what you were getting at. Here is what I did:

I used the equation R_x/R_IS=F[X]/[IS]

I used the first solution and filled in 83 for R_x, 937 for R_IS, 0.0100mol/L for [X] and 0.100mol/L for [IS]. I solved for F and got 0.8858.

Then I used the same equation and filled in that F value, 291 and 998 for the 2 readings and 0.100mol/L for the concentration of C₆H₁₂. I solved for the concentration of benzene and I got 0.0329mol/L. Taking into account the dilution factor, that means the original concentration was 0.329mol/L, which is the correct answer.

Thank you very much for all your help - I really appreciate it! If you or anybody else who's been following this doesn't follow anything I've done, I'd be happy to attempt to clarify it.

[Arkcon]

That's essentially what I did, but without the benefit of your calculation.  Question is, do you know why you did this?

[Halogen876]

Yes, I think now that I do understand the theory behind the internal standard technique and the equation. I understand that the ratio of the 2 signals is directly proportional to the ratio of the concentrations and the key is to find the proportionality constant. I know that the internal standard is a substance that behaves similarly to (but does not interfere with) the analyte but which is different enough to provide a distinct signal. The theory is that anything that affects the strength of one signal will affect the strength of the other signal in the same way so that the ratio will (or at least should) stay the same.