Chromatography Forum

Am I being stupid, only the results I'm getting seem a bit lower than expected?

I am extracting an analyte from plastic and quantifying with GC-MS using a IS. My calibration and chromatography is very good.

I tend to quantify the analyte at 0.5mg/l after extracting from a 5g sample. Therefore a simple calculation of :

0.5/(5 x 1000) = 0.0001mg/l = 0.1ug/l

Does that seem right, I would expect a higher result...

What are you calculating? The units in your calculations don't add up: (mg/L)/mg = L⁻¹. Which is not µg/L.

Quite, I just seem to be having a bit of brain freeze!

Bascially I have extracted 0.5mg/l of analyte from a 5g. I'm trying to calculate how much in mg/l or ppm there is in the sample.

Bascially I have extracted 0.5mg/l of analyte from a 5g.

Hm, but 0.5mg/L is concentration, it doesn't say how much you extracted. You need mg to compare it with the original (5g) sample mass. For that you need to multiply concentration by the volume of the solution:

xL * 0.5mg/L = ymg
(ymg / 5g) * 1000 = zppm

Cheers I think I have it now,

As I used an IS, the same amount in the 100ml extraction solvent as in the final standard solutions of 25ml. I would use the volume in the std solution.

So in my example:

0.5mg/l x (25ml/1000) =0.0025mg

(0.0025/5)*1000 = 2.5ppm - about the level I was expecting.

Thanks for that, weird how sometimes simple basics just throws you!

Cheers I think I have it now,

As I used an IS, the same amount in the 100ml extraction solvent as in the final standard solutions of 25ml. I would use the volume in the std solution.

So in my example:

0.5mg/l x (25ml/1000) =0.0025mg (0.5 * 0.025 = 0.0125)

(0.0025/5)*1000 = 2.5ppm - about the level I was expecting.

Thanks for that, weird how sometimes simple basics just throws you!

Are you saying the same concentration in the 100ml and 25ml or the same mass?

The internal standard should be a constant concentration and you more or less can ignore its concentration since they will cancel out in the calculations.

If you are getting a value of 0.5mg/L in the sample, and you have a final volume of 100ml then you have 0.5mg/L * 0.1L = 0.05mg per sample so that 0.05mg/5g=0.01mg/g of sample.

The instrument reading should give you the mg in what was injected, if you set up your calibration curve in units of standard concentration then you can bypass the need to calibrate from mass on column to mass in sample extract which makes it easier. The calibration curve will be in mg/L where the target analyte concentration varies but the internal standard concentration is held constant at the same concentration as used in the sample extract(either spiked into the sample and extracted or spiked into the final extract after extraction).

So the final calculation is something like:
(mg/L{instrument reading} * Volume of extract {in L})/Sample Weight {in grams} = Final Result mg/g

Cheers I think I have it now,

As I used an IS, the same amount in the 100ml extraction solvent as in the final standard solutions of 25ml. I would use the volume in the std solution.

So in my example:

0.5mg/l x (25ml/1000) =0.0025mg (0.5 * 0.025 = 0.0125)

(0.0025/5)*1000 = 2.5ppm - about the level I was expecting.

Thanks for that, weird how sometimes simple basics just throws you!

Are you saying the same concentration in the 100ml and 25ml or the same mass?

The internal standard should be a constant concentration and you more or less can ignore its concentration since they will cancel out in the calculations.

If you are getting a value of 0.5mg/L in the sample, and you have a final volume of 100ml then you have 0.5mg/L * 0.1L = 0.05mg per sample so that 0.05mg/5g=0.01mg/g of sample.

The instrument reading should give you the mg in what was injected, if you set up your calibration curve in units of standard concentration then you can bypass the need to calibrate from mass on column to mass in sample extract which makes it easier. The calibration curve will be in mg/L where the target analyte concentration varies but the internal standard concentration is held constant at the same concentration as used in the sample extract(either spiked into the sample and extracted or spiked into the final extract after extraction).

So the final calculation is something like:
(mg/L{instrument reading} * Volume of extract {in L})/Sample Weight {in grams} = Final Result mg/g

Thanks, I add the IS in the same mass as the standards with the extraction solvent.

Totally got my head round it now.

Thanks, I add the IS in the same mass as the standards with the extraction solvent.

Totally got my head round it now.

It should be the same mass per Liter in each, so that the concentration is equal.

If it is same total mass and different volumes then you have to include both the extract volume and the standard volume.

The main way to validate a calculation is to write in only the units with no numbers and then cancel each unit (example mg/ml x ml/l = mg/l) where a unit above and a unit below the division line cancel each other out. If you finish and the final units are not what you are looking for then there is an error in the calculation.