= Solution
Four moles of <lithium> react per mole of <iron disulfide>. Using molar <masses> $M_{\rm Li}=6.94\,\mathrm{g\,mol^{-1}}$, $M_{\rm Fe}=55.845\,\mathrm{g\,mol^{-1}}$ and $M_{\rm S}=32.06\,\mathrm{g\,mol^{-1}}$ gives
$$
\boxed{\frac{m_{\rm Li}}{m_{\rm FeS_2}}=\frac{4(6.94)}{55.845+2(32.06)}=0.2314}.
$$
Thus the stoichiometric <mass> ratio is \b[Li:FeS₂ = 1:4.322], or approximately 18.79 wt% <lithium> and 81.21 wt% <iron disulfide> within the combined active reactants. This excludes <electrolyte>, current collectors, casing and any deliberate excess of one reactant. The underlying conversion reaction, rather than the nominal cell <voltage>, sets that ratio.<Cylindrical Primary Lithium Handbook and Application Manual: Lithium Iron Disulfide by Energizer>[printed p.4, the two-step discharge reaction sums to four Li per FeS₂]{ref}
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