The information below is provided to assist LIME users who need to convert data. Please consult this page prior to contacting analysts. If you have questions about this page or the content herein, please contact msf17 at psu dot edu (Matthew Fantle).
Convert wt. % oxide to parts per million element (ppm)
NB: wt. % is g oxide per 100 g of solid (e.g., CaO, Na2O, Al2O3).
Divide the wt. % by 100 to obtain g oxide per g solid then:
NB: the mole ratio of element per mole of oxide is related to the stoichiometry of the oxide. For example, the mole ratio in CaO, MgO, SiO2, TiO2 is 1:1; in Na2O and K2O is 2:1; in Al2O3 and Fe2O3 is 2:1, etc.
Convert wt. % oxide to wt. % element
To convert from wt. % oxide to wt. % element, you follow the same procedure but cut off the calculation before you multiply by 1e6:
Perform dilution calculation
Some users will hand LIME analysts solutions that have been diluted from either an original sample solution (e.g., a river or groundwater) or a solid that was digested. The concentration that is handed back to the user by LIME is NOT the concentration relevant to the natural sample if a dilution was done. It is common to perform dilutions to bring analytes into the appropriate range for the instrument (NB: if you want accurate data for major and trace elements, multiple dilutions are often required.).
If your sample WAS diluted, you need to perform a dilution correction. LIME analysts do not do this as part of their normal data processing routine, since they do not have the data necessary to perform the calculation UNLESS they performed the dilutions.
In LIME, we perform all dilutions gravimetrically (by mass) — not by volume — which is more precise. If you perform dilutions, you must track the mass of the solution at all points at which an aliquot is taken, the sample is dried down and resuspended, and/or the sample is filtered (which is, in essence, an aliquot since 100% of the sample is never recovered in the filtrate).
One-step dilution of an aqueous sample
The calculation pathway for a simple one-step dilution is as follows; start with elemental concentrations in units of ppm:
The second term in the calculation above is often called the dilution factor. You can correct for multiple dilutions by performing a similar calculation in sequence. I give an example of a two-step calculation below. Note that the term aliquot refers to the mass of solution that you removed from the parent solution in order to make a dilution. So, if I have a 10 g solution and I pipette and weigh 0.952 g from that solution, the aliquot mass is 0.952 g. Say you hand the analyst 5 g of solution to analyze; this is NOT the total grams of analyzed solution. That would be 10 g, the total mass of the solution from which the aliquot was taken. In this calculation, you are back-calculating to the concentration in that solution; if you assumed 5 g was the total mass of solution, you would change the calculated concentration by a factor of two.
Dilution of a dissolved solid
If the user digested a solid, such as a rock, mineral, or soil, then it is common to calculate the concentration of a given element with respect to the mass of solid. This is a little bit trickier, as it involves back calculation to the original digestion. The idea here is that you are calculating the TOTAL MASS of each element that was in the original digestion, and then you will normalize that by the total mass dissolved. Thus, you need to weigh out the total mass of solid dissolved and you need to weigh the solution and aliquots taken from solution. The calculation below illustrates which values you must have weights for; start with the same calculation above to determine the concentration in the original digestion:
then multiply by the mass of the digestion solution; this gives you the total mass of the element in that solution. Then divide by the mass of the digested solid; you obtain the concentration of the element in units of 
It is critical to remember that you need a mass of solution and/or aliquot at any point that mass is lost. Mass may be lost for one of a number of reasons; for instance:
-
- Selective leaches: You centrifuged the leached solution and aliquoted only a portion of the supernatant, leaving the rest behind. There can be really good reasons for doing this, like you don’t want to get fine, difficult to pelletize solid particles in your leachate (in that case, however, I would really recommend filtration) and/or you want to carry over all of the solid to the next leaching step.
- Aliquots: Every time you take an aliquot of a solution (such as you typically do when you dilute), you leave mass behind. If you do not take that into account when you perform the calculation above, you are missing mass that came from the solid. Thus, your solid concentration is inaccurate.
- Filtration: When you filter a solution, there is always a hold up volume. Normally, the hold up volume is a small percentage of the total solution. However, if you are filtering small volumes or are filtering solutions that have high concentrations of some elements, then the amount of elemental mass lost can matter. In these cases, you need to weigh the total solution PRIOR to filtration, and the mass of the filtered solution. That allows you, in essence, to perform an additional ‘dilution’ calculation.
Note that one cannot account for loss by sorption to container walls or to solid particles in this manner. Only loss of solution, intentional or otherwise.
Two-step dilution of an aqueous sample
Finally, I will provide an example of a two-step dilution calculation, as this is also typical in situations in which users run a less diluted aliquot by ICP-OES for major elemental composition and a more diluted aliquot by ICP-MS for trace elements.
Start with the MOST diluted solution and calculate from there in the same manner as above:
In this case, we assume that the ICP-MS dilution was a serial dilution from the ICP-OES solution. If instead you diluted using the digested solution in a single step, you can perform a one-step dilution calculation.