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I hope everyone is doing well and safe! Yesterday, I was fortunate enough to be taught about how to read a chromatography calibration curve through this stackexchange! Special thanks to those who explained to me patiently and meticulously. Anyways, I've tried to solve some questions regarding this field but ended up no-where with this one. Could someone run me through the procedure of how to solve this question? Huge thanks in advance!!! <3 <3

Info: A $\pu{1.0mL}$ aliquot of a drink was diluted to $\pu{50.0mL}$ with deionized water. A sample of the drink was run through the HPLC to produce the calibration curve below. In this experiment, the mass of caffeine in the drink was determined by HPLC. The peak area obtained for this diluted sample was $2800$ arbitrary units.

Question: What is the mass of caffeine in grams, in a $\pu{330mL}$ can of this drink?

Answer = $\pu{825mg}$

P.S.: Could someone explain to me the significance of the peak area ($2800$ arbitrary units) that is obtained? And thank you for the constant support and help. Much love to this community!!!

Calibration Curve of Caffeine Solutions

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    $\begingroup$ The arbitrary peak area allows you to use the calibration curve. Since the calibration curve doesn't go up to 825 mg you must take the dilution into account. $\endgroup$ – MaxW Sep 15 '20 at 17:40
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    $\begingroup$ Read my answer once again where I talk about running a sample in your previous question. $\endgroup$ – M. Farooq Sep 15 '20 at 17:51
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    $\begingroup$ To calculate the peak area, you must be in front of the curve defining the peak. Usually it is a curve $y(x)$ starting at $y$ = 0 on the left, then suddenly growing up like up the hill in the middle of the page, where $y$ > 0, and going down to $y$ = 0 at the right part of the page. Now you look for the two inflexion points, one before and one after the maximum value of $y(x)$. You draw the tangent to the curve through these points. These two curves define a triangle with the base line $y$ = 0. You calculate the surface of this triangle, in $cm^2$ or in $mm^2$. That is the peak area. $\endgroup$ – Maurice Sep 15 '20 at 19:06
  • $\begingroup$ Thanks you for the explanations. I really appreciate the help!! $\endgroup$ – Munchies Sep 16 '20 at 14:01
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Warning: This is also a homework question. However, this is more like mathematics (geometry) than chemistry, I decided to help you. Nonetheless, please keep in mind that we won't help you in all future perspectives, if you do not show your effort to solve the problem, regardless more mathematics involved.

As M. Farooq pointed out that this problem can be solved using the answers to your previous question, specially my old-fashioned approach. Here is how you could do it. Analyze your graph thoroughly to find two different points of graph where it perfectly crossed the cross-lines:

HPLC Calibration Curve

For me, these two points are: $(35,2000)$ and $(60,3500)$. Therefore the tangent of the slope ($\alpha$) of line is:

$$\tan \alpha = \frac{3500-2000}{60-35} = \frac{1500}{25}=60$$

Now, take another arbitrary point in the graph. Say it is $(x,y)$. Then:

$$\tan \alpha = \frac{y-2000}{x-35} = 60 \ \Rightarrow \ y = 60x - 100 \tag1$$

The equation $(1)$ is the straight-line equation of the graph.

When $y = 2800$, $x= \frac{2800+100}{60} = \pu{48.33 mg/L}$. This is for the diluted sample. I think it is fair to leave you some work to do. So you can back calculate to get your answer. When I did, it came close to $\approx \pu{798 mg/can}$.

Note: The error ($3.27\%$) is due to the finding of the slope. Also there are few flaws in the question. The calibration curve is made with HPLC traces of known standard caffeine solutions (not using diluted samples of the drink). The peak area of $2800$ arbitrary units was obtained by the injection of diluted sample of the drink ($1$ to $\pu{50 mL}$ dilution).

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  • $\begingroup$ Hello, thanks so much for taking your time to help me out. I really appreciate the help and it has definitely solidified my knowledge in chemistry. Again, thank you! Stay safe! $\endgroup$ – Munchies Sep 16 '20 at 14:00
  • $\begingroup$ I was trying to connect the ties between the 48.33mg/L of the diluted sample to the 798 mg/can but I'm not ending up with your answers and i'm getting 1292.5 mg/can. :( Could you point out the steps that you would take. Sorry about the inconvenience. $\endgroup$ – Munchies Sep 16 '20 at 14:03
  • $\begingroup$ I redid my calculations and it was very close to the actual answer. Could you tell me if my steps are correct or if there is another way of solving this? The dilution factor is 1:50 since a 1mL aliquot was diluted to 50mL with water. So I multiplied the 48.33mg/L by 50 = 2416.6 mg/L. Since the can contains 330mL, 2416.6 x 0.33 which leaves me with 797.445 mg/can? $\endgroup$ – Munchies Sep 16 '20 at 14:06
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    $\begingroup$ The value $\pu{48.33 mg}$ is in $\pu{1000 mL}$ of diluted sample. Thus, find how much in $\pu{50.0 mL}$. That value is in $\pu{1.0 mL}$ of original solution. Then find how much in $\pu{330 mL}$. What you have got is correct, but calculation operation is wrong. $\endgroup$ – Mathew Mahindaratne Sep 16 '20 at 14:49
  • $\begingroup$ Ahhh i see. Thanks so much for the help!! Really appreciate it. :) $\endgroup$ – Munchies Sep 16 '20 at 14:58

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