Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2017 / Paper 1 / Section B / 12 / iii / a / Solution

Four moles of lithium react per mole of iron disulfide. Using molar masses $M_{\mathrm{Li}}=6.94\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, $M_{\mathrm{Fe}}=55.845\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ and $M_{\mathrm{S}}=32.06\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ givesThus the stoichiometric mass ratio is 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.^[ref]