﻿ Calculation - amount of analyte - Chromatography Forum

## Calculation - amount of analyte

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### Calculation - amount of analyte

Hi all,

Sorry I’m having complete and total brain freeze with this calculation.

I have a 15g sample of plastic that I’m testing for a leachable analyte. The plastic is put into 1L of water for 48 hours and then tested with HPLC.

HPLC result obtained is 5mg/kg. Is there a way to calculate the actual total mg of analyte in the 15g plastic I.e expressed as mg/sample?

Thanks.

### Re: Calculation - amount of analyte

If HPLC reports mg/kg, there must be some calculation involved.
Check what is implemented.

If you had to enter the sample amount somewhere it might be the result is already on sample basis.

The raw data of the chromatogram is a peak-area (or height).
by calibration this is normaly converted into a concentration (mass per volume) of your analyte in the vial.
then this analyte concentration has to be divided by the concentration of the matrix (sample per volume) in the vial. This gives you the mass fraction of analyte in the sample. Adjust the mass units as needed (%, mg/g, mg/kg...)

### Re: Calculation - amount of analyte

Hollow wrote:
If HPLC reports mg/kg, there must be some calculation involved.
Check what is implemented.

If you had to enter the sample amount somewhere it might be the result is already on sample basis.

The raw data of the chromatogram is a peak-area (or height).
by calibration this is normaly converted into a concentration (mass per volume) of your analyte in the vial.
then this analyte concentration has to be divided by the concentration of the matrix (sample per volume) in the vial. This gives you the mass fraction of analyte in the sample. Adjust the mass units as needed (%, mg/g, mg/kg...)

Hi Hollow,

Yep we have a calibration curve made with standards. Our concentration result is 5mg/kg in relation to the concentration of the plastic sample in 1L of water.

I’m hoping for a way to be able to say “the whole sample contains x grams of analyte,” rather than mass/volume.

### Re: Calculation - amount of analyte

but 5 mg/kg is a mass fraction, not a concentration...

general remark that I try to teach my students as well:
the more one uses the units and terms correctly, the less confusion is introduced...
concentration = something per volume
amount; mass fraction = something per mass

If your calibration standard are really in the unit of mg analyte/kg solute, then your result is also in this unit. So you must check, if this is the case or if there are additional calculations being made.

If the calibration is in mg/kg, then you would need to know the density of your solute, which in this case (water) can be assumed as 1 kg/L (or use the exact value from reference sources).
Then you have your matrix-concentration of 15 g sample in 1 kg water = 15 g/kg
from that, you measured an analyte-concentration of 5 mg/kg.
So 5 mg/kg / 15 g/kg = 5 mg analyte /15 g sample (= mass fraction = 0.33 mg/g sample)

But double check if this is correct and that the 15 g are not already contained in the calculation for the 5 mg/kg
(e.g. because you always use 15 g / 1L, so this factor is already implemented).

Just do the calculation by hand from the peak areas of your sample and calibration runs.

### Re: Calculation - amount of analyte

What does the kg in your units refer to?

If it's per kg of sample, 15 g = 0.015 kg. So 5 mg/kg * 0.015 kg = 0.075 mg total.

If it's per kg of extract, assuming 1 L = 1 kg as per water. So 5 mg/kg * 1 kg = 5 mg total.

But I don't know how you'd calibrate for either of those. Normally you'd calibrate for a concentration from your HPLC, e.g. mg/L. If you're getting 5 mg/L, then 5 mg/L * 1 L of extraction fluid = 5 mg in the total extract/per component tested.

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