Dear Jie,

please have a look at the links.

a) Definition according to the wikipedia link:

"The equivalent weight of a chemical element is the gram atomic weight divided by its valence"

b) Definition according to the second link(A Dictionary of Units of Measurement):

"equivalent or equivalent weight (Eq): a unit of relative amount of substance used in chemistry. One equivalent weight of an element, compound, or ion is the weight in grams of that substance which would react with or replace one gram of hydrogen. Since one gram of hydrogen is very nearly equal to one mole and since hydrogen has one electron free to react with other substances, 1 Eq of a substance is effectively equal to one mole divided by the valence of the substance (the number of electrons the substance would engage in participating in the reaction). In practice, this is a large unit and measurements are more likely to be in milliequivalents (mEq or meq)."

For dissiociated molecules it is quite clear: take the molecular mass of the cation or anion divided by the valence.

But for your CaCO3, which probably does not dissociate in your resin; is such a definition meaningful? (dividing molecuar mass of CaCO3 by 2 (=100/2=50).

???  So 1 eq  (CaCO3) =  50g ; and 1meq (CaCO3)=0,05g ???

I think let's wait for comments from other colleagues.

https://en.wikipedia.org/wiki/Equivalent_(chemistry)

http://www.unc.edu/~rowlett/units/dictE.html

Dear Gerhard Martens,

Thanks for your answers.

     I probably understand the calculation of the meq of CaCO3 after reading your detailed answers and the wikipedia.

• First, the eq of compound is generally calculated by that of the representive-ion in the compound.  CaCO3 is in the form of Ca2+(in solution) when I do separation method. That is to say, the calculation of eq of  Ca2+ can solve the problem.
• Then, meq  of ion=  mmol of ion* valence. As an example, 100 mg CaCO3. meq of Ca2+ =the mmol of  Ca* valence=100*0.4mg/40g.mol-1*2=2meq.
• Last, I think there may be a confusion between the " equivalent weight" and "equivalent".

             equivalent weight=element weight/valence

             equivalent=mol* valence

      There may be someting wrong with my understanding. Let us wait for comments from other colleagues.

Dear Jie Lin,

You can find a lot in our contribution published in JAAS (Vanhoecke et al., 2015, Single-step chromatographic isolation of lithium from whole-rock carbonate and clay for isotopic analysis with multi-collector ICP-mass spectrometry J. Anal. At. Spectrom., 2015,30, 2533-2540). Here, we also developped a protocol to measure Li isotopes in carbonates.

The resin that we used in our JAAS paper (see Vanhoecke et al., 2015) had a capacity of 1.7meq/mL. So suppose you are using a resin volume of 2mL. Your resin capacity is then 3.4meq. If it is recommended to only use 10% of the resin capacity, as is in your case, your cation load cannot exceed 0.1*3.4meq which equals 0.34meq. Are you using pure carbonate (CaCO3?). Then, for a rough estimation (not taking into account traces of Sr, Mg), you only need to take into account the Ca2+ cation. Note that eq is the amount of substance it takes to combine with 1 mole of hydrogen ions. We know that calcium weighs approximately 40g. Since 1 equivalent of calcium is half a mole, we can see that 1 equivalent weighs approximately 20 g, and since 1meq (milliequivalents) is 10-3 eq, one meq of Ca weighs approximately 20mg. Since in this example, your cation load (Ca2+ in case of pure calcium carbonate) cannot exceed 0.34meq, you cannot have more than 20mg*0.34 = 6.8 mg of Ca2+ in your solution.

So you will have to calculate how much CaCO3 you are dissolving in what amount of acid and then calculate the volume you need to load. There is also useful data in our publication.

Hope this helps!

Best,

Joke Belza

Dear Joke Belza,

Thanks for your answers.

 I have already read your paper (Vanhoecke et al., 2015, Single-step chromatographic isolation of lithium from whole-rock carbonate and clay for isotopic analysis with multi-collector ICP-mass spectrometry J. Anal. At. Spectrom., 2015,30, 2533-2540) as soon as it was published online, and it gave me much more knowledgement abot the Li separation in carbonate.  And we also want to try 8 ml resin to separate Li from other natrix elements. And I also want to ask a question: have you ever thought about  separating Ca firstly and then Li?

thank you again!

Dear Jie Lin,

Yes, I have thought about that :). Back then, I was thinking about the modified procedure from De Muynck (2009 JAAS, 24(11). p.1498-1510) because we had used it before in one of our articles to measure Sr isotopes, in which we separated Sr from Ca. If you look at the acid dependency of the k factors (see graphs when scrolling down at http://www.eichrom.com/products/extraction.aspx) for this resin, Ca easily comes off the resin when using 7M HNO3, so indeed you would be able to get rid off the heavy matrix. But here as well you will have to mind the load, which is usually a fraction of what you are dissolving, and you basically need all of your Li in order to achieve a sufficiently high concentration to measure. For this resin, there is no information for Li, but it could be tested off course, and you can maybe have an idea by comparing with Na. However, this means you have to dissolve your sample in 7M HNO3. Such severe acid attack might be used on carbonates if they are extremely pure, without any clay or organic fraction, but otherwise, there will be Li leaching from the non-carbonate fractions and this will severely compromise your Li isotopic composition from the pure carbonate fraction. This is also discussed in our Li paper. I am using the Sr resin example now, but I have not considered other resins, so it might work if you can manage to use another resin/eluens/acid strength. Another option is dissolve the carbonate in low concentrations, centrifuge and pipet off the solutions from the clay, evaporate, and then dissolve in higher acid concentrations. But even then the risk is high that you will leach some Li from non-carbonate fraction, and you are also adding more steps and more possible contamination to your procedure (Li and Na in other resins??). If you start heating and using high acid concentrations, always keep the leaching from non-carbonate fractions in mind.

If you think I could be of any help, please do feel free to contact me by mail.

Good luck!!

Dear Joke Belza,

Thank you very very much to provide detailed answers for me. Let's keep in touch!

caco3 = 3*16+12+40=100 @