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I'm trying to calculate the theoretical specific capacity of the $\ce{LiFePO4}$ cathode for a fully discharged state, let's say when the cathode has the following notation $\ce{Li_{0.5}FePO4}$. And I'm using the following formula:

$Q=\frac{nF}{3.6M}$ where F is the Faraday constant

The question is slightly related to this thread. The question is, what is the molar mass (M) of $\ce{Li_{0.5}FePO4}$.

Which one is the correct one for finding the specific capacity:

  • M = 157.751 g/mol, as if Lithium would be taken as a whole? or
  • M = 154.285 g/mol, as if only half of Lithium would be considered?

Can anyone help me with this?

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    $\begingroup$ Counting the Li in full is of course totally wrong. Point is, as I already told you with your earlier question, this is a nonstochiomtric compound (the 0.5 is an approximate). It makes little sense to calculate its molar mass. What do you want it for? $\endgroup$
    – Karl
    Jan 31, 2017 at 10:03
  • $\begingroup$ Thank you for your answer! I would like to calculate the theoretical specific capacity of LiFePO4 when the cathode is fully discharged (Li0.5FePO4). $\endgroup$
    – Physther
    Jan 31, 2017 at 10:06
  • $\begingroup$ So? 0.5 mol of Li are 3.5 g. $\endgroup$
    – Karl
    Jan 31, 2017 at 10:23
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    $\begingroup$ Well, that was the question. So I need to consider half of the mass, right? $\endgroup$
    – Physther
    Jan 31, 2017 at 10:26
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    $\begingroup$ Please change the question so it actually describes a problem, including the math you intend to do. At the moment, you seem to be just guessing, and Im guessing at your guesses. Otherwise this question should get closed. $\endgroup$
    – Karl
    Jan 31, 2017 at 10:53

1 Answer 1

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You should use 0.5 lithium atoms in the molecular mass. Alternatively you can multiply the formula indices by two to obtain $\ce{LiFe_2(PO_4)_2}$. This formula describes exactly the same compound but it's much more readable, for instance is now clear that the substance contains one $\ce{Li(I)}$, one $\ce{Fe(II)}$, one $\ce{Fe(III)}$ and two phosphate anions per $2n$ electrons transfered.

Here the value of $n$ is multiplied by 2 because you also have your stoichometry multiplied by 2. If you use $\ce{Li_{0.5}Fe(PO_4)}$ to calculate $M$ then the value of $n$ stays unchanged.

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  • $\begingroup$ Thank you for clarification. Multiplying it by 2 is a nice way to understand it. $\endgroup$
    – Physther
    Jan 31, 2017 at 19:37

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