Concentration in mobile phase??

[olufadi1]

This may be a silly question, but it has been confusing me.

I am making some injections of known concentrations for my experiments. For some of the calculations I want to make, I need the concentration of my analytes in the mobile phase to produce the correct graph.

Do I need to solve some differential equation (since conc. is a function of time and position in the column)? Or could I just use the void volume of the column as a factor that is diluting my known injection concentration?

Thanks for any assistance!

[Gerhard Kratz]

The time when your Peak starts and Ends will give you the volume in which your compound is solved. If flow is 1ml per min and this time is half a Minute your analyt is disolved in half a ml. Maybe that Explanation is too easy and some colleagues will give you the right equation. We will see.

[olufadi1]

The time when your Peak starts and Ends will give you the volume in which your compound is solved. If flow is 1ml per min and this time is half a Minute your analyt is disolved in half a ml. Maybe that Explanation is too easy and some colleagues will give you the right equation. We will see.

That seems to makes sense. I will try that out with some of my calculations and see if the results make sense. Thank you.

[tom jupille]

The time when your Peak starts and Ends will give you the volume in which your compound is solved. If flow is 1ml per min and this time is half a Minute your analyte is dissolved in half a ml.

That assumes that your starting point was the peak area (i.e., that you have already integrated the peak). If you are using peak height, and you have a Gaussian peak, you need to divide by 2 (the maximum amplitude is approximately twice the average).