# What is the molar mass of a compound with 0.5 atoms in its molecular formula?

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?

• 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? – Karl Jan 31 '17 at 10:03
• Thank you for your answer! I would like to calculate the theoretical specific capacity of LiFePO4 when the cathode is fully discharged (Li0.5FePO4). – Physther Jan 31 '17 at 10:06
• So? 0.5 mol of Li are 3.5 g. – Karl Jan 31 '17 at 10:23
• Well, that was the question. So I need to consider half of the mass, right? – Physther Jan 31 '17 at 10:26
• 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. – Karl Jan 31 '17 at 10:53

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.