converting ug/mL to mg/kg help needed

[joe_stevens]

This is probably a really silly question but I'm struggling to find out how to convert my results from ug/mL to mg/kg. I have internal standard in my calibration curve and samples. My sample size is 10 g. Any comments would be greatly appreciated.

Thanks
Joe

[Multidimensional]

Here you go. One of the best sources for info on the web!

"Conversion Factors microgram, nanogram, ppm, ppb and percent"
https://hplctips.blogspot.com/2011/08/c ... ogram.html

[dblux_]

This is probably a really silly question but I'm struggling to find out how to convert my results from ug/mL to mg/kg. I have internal standard in my calibration curve and samples. My sample size is 10 g. Any comments would be greatly appreciated.

Thanks
Joe

Such conversion is impossible unless you provide density of final solution.
If it is water solution, and concentration is low you may assume density 1 g/mL.

[Peter Apps]

This is probably a really silly question but I'm struggling to find out how to convert my results from ug/mL to mg/kg. I have internal standard in my calibration curve and samples. My sample size is 10 g. Any comments would be greatly appreciated.

Thanks
Joe

Such conversion is impossible unless you provide density of final solution.
If it is water solution, and concentration is low you may assume density 1 g/mL.

Whether you have an internal standard, and the size of the sample, are not relevant to a straightforward calculation as long as you know the density of the sample. Which makes me think that there is more to the problem than you are telling us.

Peter

[joe_stevens]

Here you go. One of the best sources for info on the web!

"Conversion Factors microgram, nanogram, ppm, ppb and percent"
https://hplctips.blogspot.com/2011/08/c ... ogram.html

Thanks for the link in was much appreciated.

[joe_stevens]

This is probably a really silly question but I'm struggling to find out how to convert my results from ug/mL to mg/kg. I have internal standard in my calibration curve and samples. My sample size is 10 g. Any comments would be greatly appreciated.

Thanks
Joe

Such conversion is impossible unless you provide density of final solution.
If it is water solution, and concentration is low you may assume density 1 g/mL.

Whether you have an internal standard, and the size of the sample, are not relevant to a straightforward calculation as long as you know the density of the sample. Which makes me think that there is more to the problem than you are telling us.

Peter

Thanks for your message I'm happy to provide any information that might assist. How do I calculate the density of my sample? I have found a publication on pesticides which describes the conversion as

Concentration (ug/mL) x volume of extraction solvent (mL) / sample size (g).

Is this what you are describing?

Thanks
Joe

[joe_stevens]

This is probably a really silly question but I'm struggling to find out how to convert my results from ug/mL to mg/kg. I have internal standard in my calibration curve and samples. My sample size is 10 g. Any comments would be greatly appreciated.

Thanks
Joe

Such conversion is impossible unless you provide density of final solution.
If it is water solution, and concentration is low you may assume density 1 g/mL.

Thanks for your message. I reconstituted my samples in 100 uL of MeOH:H20 60:40.

Is it possible to calculate density from this?

Thanks
Joe

[dblux_]

Thanks for your message. I reconstituted my samples in 100 uL of MeOH:H20 60:40.

Is it possible to calculate density from this?

Thanks
Joe

Nope, calculation of density of MeOH:H20 60:40 would be much more complicated than finding the density of such solution in density tables !

So lets assume that you found density I talk about. That will be density of a solvent.
When you dissolve something in it final density will change. If concentration of solution is low (and probably it is) then you may assume that density of the solution is roughly equal to the density of solvent.
Use no equations but your brain and proportions and having such data you will convert units as you need.

--
If density of solution is d [kg/L] then concentration x [µg/mL] equals x/d [mg/kg] - but remember our assumption about infinitely low concentration

[joe_stevens]

Thanks for your message. I reconstituted my samples in 100 uL of MeOH:H20 60:40.

Is it possible to calculate density from this?

Thanks
Joe

Nope, calculation of density of MeOH:H20 60:40 would be much more complicated than finding the density of such solution in density tables !

So lets assume that you found density I talk about. That will be density of a solvent.
When you dissolve something in it final density will change. If concentration of solution is low (and probably it is) then you may assume that density of the solution is roughly equal to the density of solvent.
Use no equations but your brain and proportions and having such data you will convert units as you need.

--
If density of solution is d [kg/L] then concentration x [µg/mL] equals x/d [mg/kg] - but remember our assumption about infinitely low concentration

Thank you so much for your help. If I was going to do this the correct way I should have calculated the density of my initial standard when it was prepared. This would give me a density value in g/mL. Then I can divide the points in my calibration curve by the density value to get a calibration curve in ug/g.

I was also thinking that I should recalculate the density for every dilutions of the standard I make until I produce the working solution for each standard. I don't think calculating the density of the internal standard is necessary. The standard and internal standard will be evaporated off and reconstituted in a different solvent. The density of the solvent that I reconstitute my standards and samples in shouldn't matter as I can calculate how much has been added from the density of the working solution.

Thanks again for all your comments. If I'm understanding this correctly please let me know.

Thanks
Joe

[Hollow]

For me it's not so clear what/where your concentration (mg/L) is?
Is it the concentration of your analyte in your specimen (your 10g) or just in the solution in the vial?

If it's case one, then you need to know just the density of the specimen and convert it by dividing the conc / density.

In case two, don't mess with densities of sample and ref solutions but calculate the mass-concentration (g/L) of raw specimen at the same point (eg. in the vial) where your mass-conc of analyte is (taking into account all dilution/concentration steps).
Then just simply divide (conc analyte) / (conc specimen) to get mass fraction g/kg (of course check the units accordingly)

[joe_stevens]

For me it's not so clear what/where your concentration (mg/L) is?
Is it the concentration of your analyte in your specimen (your 10g) or just in the solution in the vial?

If it's case one, then you need to know just the density of the specimen and convert it by dividing the conc / density.

In case two, don't mess with densities of sample and ref solutions but calculate the mass-concentration (g/L) of raw specimen at the same point (eg. in the vial) where your mass-conc of analyte is (taking into account all dilution/concentration steps).
Then just simply divide (conc analyte) / (conc specimen) to get mass fraction g/kg (of course check the units accordingly)

Thanks for your comments. It's the concentration of my standards (ref solutions) in my vials. I then do I calibration curve to get the values for my specimens. Internal standard is added at the start of the extraction so this should account for any dilution/concentration steps. I'm not sure if I've got this right. I start with 10g of sample and this is reconstituted in 100 uL of solvent into the vial before it is injected. This would produce a mass-conc of 10000g/L.

Thanks
Joe

[Hollow]

...this would produce a conc of 100000g/L...
(10g/0.1 ml) so it's 100 g/ml = kg/L (in your vial).

Now that you know the conc of your analyte in the vial you can just divide this two, µg/ml = mg/L so it will give you mg Analyt / kg specimen.

If it's that what you wanted? And in case your calibration curve is correct...

In fact it's this one you found, just refactored for a single division:
Concentration (ug/mL) x volume of extraction solvent (0.1 mL) / sample size (10 g).



in general:

(mass-conc of your analyte in vial) = (mass of specimen) x (mass ratio of analyte in specimen) x (dilution factor (if any)) / (volume of extraction solvent, used to prepare the solution in vial)

the left side you will obtain from calibration curve and raw data, so that you can isolate for the mass-ratio on the right side.

[James_Ball]

...this would produce a conc of 100000g/L...
(10g/0.1 ml) so it's 100 g/ml = kg/L (in your vial).

Now that you know the conc of your analyte in the vial you can just divide this two, µg/ml = mg/L so it will give you mg Analyt / kg specimen.

If it's that what you wanted? And in case your calibration curve is correct...

In fact it's this one you found, just refactored for a single division:
Concentration (ug/mL) x volume of extraction solvent (0.1 mL) / sample size (10 g).



in general:

(mass-conc of your analyte in vial) = (mass of specimen) x (mass ratio of analyte in specimen) x (dilution factor (if any)) / (volume of extraction solvent, used to prepare the solution in vial)

the left side you will obtain from calibration curve and raw data, so that you can isolate for the mass-ratio on the right side.

The conversion he is looking for is the cancellation of units. ug/ml x ml/g. ml in the first cancels ml in the second (one above and one below the divisor) which simply leaves ug on top and g on bottom to give ug/g, then you can convert by multiplying by 1000 to convert ug to mg or g to kg or whatever multiple you need to make the conversion.

[joe_stevens]

...this would produce a conc of 100000g/L...
(10g/0.1 ml) so it's 100 g/ml = kg/L (in your vial).

Now that you know the conc of your analyte in the vial you can just divide this two, µg/ml = mg/L so it will give you mg Analyt / kg specimen.

If it's that what you wanted? And in case your calibration curve is correct...

In fact it's this one you found, just refactored for a single division:
Concentration (ug/mL) x volume of extraction solvent (0.1 mL) / sample size (10 g).



in general:

(mass-conc of your analyte in vial) = (mass of specimen) x (mass ratio of analyte in specimen) x (dilution factor (if any)) / (volume of extraction solvent, used to prepare the solution in vial)

the left side you will obtain from calibration curve and raw data, so that you can isolate for the mass-ratio on the right side.

Thank you so much for your help it was very useful and you have made it very clear to me.

Thanks
Joe

[joe_stevens]

...this would produce a conc of 100000g/L...
(10g/0.1 ml) so it's 100 g/ml = kg/L (in your vial).

Now that you know the conc of your analyte in the vial you can just divide this two, µg/ml = mg/L so it will give you mg Analyt / kg specimen.

If it's that what you wanted? And in case your calibration curve is correct...

In fact it's this one you found, just refactored for a single division:
Concentration (ug/mL) x volume of extraction solvent (0.1 mL) / sample size (10 g).



in general:

(mass-conc of your analyte in vial) = (mass of specimen) x (mass ratio of analyte in specimen) x (dilution factor (if any)) / (volume of extraction solvent, used to prepare the solution in vial)

the left side you will obtain from calibration curve and raw data, so that you can isolate for the mass-ratio on the right side.

The conversion he is looking for is the cancellation of units. ug/ml x ml/g. ml in the first cancels ml in the second (one above and one below the divisor) which simply leaves ug on top and g on bottom to give ug/g, then you can convert by multiplying by 1000 to convert ug to mg or g to kg or whatever multiple you need to make the conversion.

Thank you for the information.

Thanks
Joe