August 12, 2024, 12:10:00 PM
Forum Rules: Read This Before Posting

### Topic: Help with Discrepancy about right concentration calculation in lab (HPLC)??  (Read 838 times)

0 Members and 1 Guest are viewing this topic.

#### Fun8420!

• Very New Member
• Posts: 1
• Mole Snacks: +0/-0
##### Help with Discrepancy about right concentration calculation in lab (HPLC)??
« on: June 26, 2024, 07:58:30 AM »
Hello, we have a lab discussion about how to properly calculate a specific concentration, basically getting two different results. If i filter 5000 ml of a solution ("solution S") containing one specific compound ("compound A") onto one filter. Then I extract compound A from this one filter and re-dissolve the dry extract into 0.1 ml solvent. The final HPLC analysis including a prepared standard row now tells me I have 12.9 nM compound A in my 0.1 ml extract. How much is the concentration of "compound A" in the 5000 ml original solution S and how much compound A would i find in general per liter of solution S? The results we get would either be 0.000258 nM or 2.58 nM. Can anybody help with the correct results and calculation?

#### DrCMS

• Chemist
• Sr. Member
• Posts: 1299
• Mole Snacks: +211/-81
• Gender:
##### Re: Help with Discrepancy about right concentration calculation in lab (HPLC)??
« Reply #1 on: June 26, 2024, 09:53:19 AM »
If compound A is dissolved to make solution S then you should not removed anything when filtering that solution.

If you do not have a solution but rather a suspension of A; then yes A can be filtered out.
However if A has some solubility in S then only some of compound A will be filtered out.

With only the information you have given here there is no way anyone can answer your question.

#### Hunter2

• Sr. Member
• Posts: 2259
• Mole Snacks: +185/-49
• Gender:
• Vena Lausa moris pax drux bis totis
##### Re: Help with Discrepancy about right concentration calculation in lab (HPLC)??
« Reply #2 on: June 26, 2024, 10:21:20 AM »
Beside the specific problems with HPLC, the calculation is a simple dilution

The correct answer is 0.000258 nM.
First nM is Nanomole per liter

You got an result of 12.9 nM means 12.9 * 10^(-9) mol/l = 12.9 * 10^(-9) mmol/ml.
The volume of the sample is 0.1 ml
So your extract is then 12.9 * 10^(-9) mmol/ml* 0.1 ml = 12.9 *10^(-10) mmol.
This extract was obtained from 5000 ml means the concentration was 12.9 *10^(-10) mmol / 5000 ml = 2,58 *10^(-13) mmol/ml = 2,58 *10^(-13) mol/l
Convert in nM means divided by 10(-9) gives you 0.000258 nM