Does aliquot matter for final concentration?

I am wondering if my calculation for getting the final concentration is right:


40 mL of solvent A was taken to extract a sample xy. From that 30 mL were taken and evaporated to dryness. Afterwards, 1 mL of a solvent was added for reconstitution. 100 uL were taken and were diluted to 1000 uL, analysis was performed with external calibration.
How to calculate the final concentration?


My thoughts were the following:
Take the obtained concentration from the calibration curve and multiply by the dilution factor 10 since 100 uL was diluted to 1000 uL. Now, I am struggling a bit. Is this already the final concentration or should I also consider the factor from taking 30 mL out of 40 mL?
Thank you very much for your ideas on this question!




$$c\_\mathrm{r} = \frac{V\_\mathrm{d} }{ V\_\mathrm{i}} \cdot c\_\ce{s}$$


$$c\_\mathrm{e} = \frac{V\_\mathrm{r} }{ V\_\mathrm{a}} \cdot c\_\mathrm{r}$$


* $c\_\mathrm{e}$ is analyte [molar concentration](https://en.wikipedia.org/wiki/Molar_concentration) in extract
* $c\_\mathrm{r}$ is analyte molar concentration in reconstituted extract
* $c\_\mathrm{s}$ is analyte molar concentration in the final sample,  

being compared to the calibration curve.
* $V\_\mathrm{a}$ is volume of the extract aliquote
* $V\_\mathrm{r}$ is volume of the reconstituted extract
* $V\_\mathrm{i}$ is volume of the extract injected for final dilution
* $V\_\mathrm{d}$ is volume of the final diluted solvent


$$n\_\mathrm{e} = V\_\mathrm{e}\cdot c\_\mathrm{e} = V\_\mathrm{e} \left(\frac{V\_\mathrm{r} }{ V\_\mathrm{a}} \cdot \frac{V\_\mathrm{d} }{ V\_\mathrm{i}} \cdot c\_\ce{s}\right)$$


* $n\_\mathrm{e}$ is the analyte molar amount in the whole extract
* $V\_\mathrm{e}$ is the total extract volume


All can be rearranged for mass concentrations and masses.


$$m\_\mathrm{e} = V\_\mathrm{e}\cdot \rho\_\mathrm{e} = V\_\mathrm{e} \left(\frac{V\_\mathrm{r} }{ V\_\mathrm{a}} \cdot \frac{V\_\mathrm{d} }{ V\_\mathrm{i}} \cdot \rho\_\ce{s}\right)$$


* $m\_\mathrm{e}$ is the analyte mass in extract
* $\rho\_\mathrm{e}$ is the analyte [mass concentration](https://en.wikipedia.org/wiki/Mass_concentration_(chemistry)) in extract
* $\rho\_\mathrm{s}$ is the analyte mass concentration  

in the final sample, being compared to the calibration curve.


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Is it also necessary to consider a third equation like:


$$c\_{fe} = \frac{c\_{e}\* V\_{e}}{V\_{fe}} 
$$


with $ c\_{fe}$ as finale analyte molar concentration in extract


with $ V\_{fe}$ as total original volume of the extract


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30 ml of extraction solution was dried and reconstituted to 1 ml, then diluted 100 ul to 1000ul so the *concentration* is now:


Extraction solution × 30 ml/1 ml × 100ul/1000 ul = Concentrated 3x



```
      ----> Sample run in analytical instrument ---->

```

What ever result you get from your analysis as grams analyte/ml must be *divided* by 3 to get the *concentration* of analyte in the extraction solution.


Now to find out *total grams* of analyte extracted in your 40 ml of extraction solution:



> 
> grams analyte/ ml × 40 ml = grams analyte extracted
> 
> 
> 


Now, in order to determine the *original* concentration of whatever solution you extracted *from* (assuming 100% extraction efficiency$^1$), you take grams analyte and divide by volume of original solution *before* the 40 ml extraction solution was mixed with it (also assuming a 2 phase extraction with a separatory funnel).


So you need to know your original volume of solution.


$^1$ good luck with that, HPLC may be a better choice


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It's good to keep in mind the distinction between ***amount*** and ***concentration***.


Take the step where you go from 40 mL to 30 mL. Since you're not diluting, you're not changing the concentration of the solution, but you are changing the amount. The way you figure out the reduced amount of compound you have in the new sample is by looking at the ratio of volumes.


But take the 30 mL to 1 mL step. You (hopefully) haven't changed the amount of substance you have when you evaporated and reconstituted things, but you have changed the concentration. To figure out the new concentration, this is where you pull out the old `c1 * v1 = c2 * v2` -- this isn't a magic invocation, it's just a representation of the fact that the amount of substance hasn't changed. (The amount of substance in a solution being the concentration times the volume - you do the calculation before and after and set them equal to each other, as the amount of substance hasn't changed.)


Your whole process can be thought of as a series of changes in the amount and concentration:


* 40 mL -> 30 mL -- Constant concentration; change of amount
* 30 mL -> 1 mL -- Constant amount; change of concentration
* 1 mL -> 100 uL -- Constant concentration; change of amount
* 100 uL -> 1000 uL -- Constant amount; change of concentration


So what you can do is take the value (the assay concentration) for the final solution and then work backwards through the steps, converting to concentration or total amount as necessary.


If you think about it stepwise, making sure you understand whether each step is a constant concentration step or a constant amount step, you can apply the relevant conversion factors backwards for each step until you reach the value you want. The benefit here is that this process is robust to (and forces you to think about) whether you want the amount of substance in the original sample or the concentration in the original sample (or if, for some reason, you need the amount in your 30 mL sample) - just add in an extra conversion at the appropriate location. It also generalizes to more complex sampling procedures.


(There's certainly potentials for shortcuts and rules of thumbs here to speed up calculations -- but if you're confused or uncertain, it's best to go back to first principles and work things out stepwise.)


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