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Measuring of concentration by HPLC

Discussions about HPLC, CE, TLC, SFC, and other "liquid phase" separation techniques.

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Hi all,

I have a guestion about calculation of concentration. of amaunt of X in sample as mg/g.

It was measured on analitical balance sample with m=0,5 000g
Extraction with 15 ml solvent
Then the extract was filtrated in buchner and washed with 1-2 ml H2O.
After that 20 ml solvent was added.
5 ml of solution was taken and evaporated 1 ml.
After that 7 ml of solvend was added and put in HPLC for analysis.
The results from HPLC is 1mg/ ml.
What is the concentration in sample as mg/ g?

What is the formula that I can use?

Thank you!
Hi all,

I have a guestion about calculation of concentration. of amaunt of X in sample as mg/g.

It was measured on analitical balance sample with m=0,5 000g
Extraction with 15 ml solvent
Then the extract was filtrated in buchner and washed with 1-2 ml H2O.
After that 20 ml solvent was added.
5 ml of solution was taken and evaporated 1 ml.
After that 7 ml of solvend was added and put in HPLC for analysis.
The results from HPLC is 1mg/ ml.
What is the concentration in sample as mg/ g?

What is the formula that I can use?

Thank you!
Between 115,2 and 118,4 mg/g ( ! 1-2 mL of water !)

Was the solvent miscible with water ? No information - I assumed it was.

Never use formulas for concentration calculation. Simply start thinking logically.
Thank you!

Yes the solvent is misible in wather.

How do calculate the concentration?
Thank you!

Yes the solvent is misible in wather.

How do calculate the concentration?
1. analyte was extracted with 15 mL of solvent and 1 mL of water was added to supernatant then 20 mL of solvent was added
total volume was 15+1+20 = 36 mL
---------------
2. 5 mL was taken, evaporated to 1 mL and diluted to 8 mL and analysed
---------------
3. concentration from HPLC was 1mg/mL so in 8 mL there was 8 mg of analyte
(in fact 8 mg in 5 mL of solution - see point 2.)
---------------
4. 36 mL/ 5 = 7,2

7,2 x 8 mg = 57,6 mg in 0,5 g of sample
so in 1 g there was 115,2 mg of analyte
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