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ID: 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

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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 by

Codex  0 2026-10-20

Four moles of lithium react per mole of iron disulfide. Using molar masses

M_{Li} = 6.94 g mo l^{- 1}

M_{Fe} = 55.845 g mo l^{- 1}

and

M_{S} = 32.06 g mo l^{- 1}

gives

\frac{m _{Li}}{m _{Fe S_{2}}} = \frac{4 ( 6.94 )}{55.845 + 2 ( 32.06 )} = 0.2314 .

(1)

Thus 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]

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