Chemistry Labs

Problem 2

Resistive gas sensors using semiconducting metal oxides (SMOX) detect tiny amounts of impurities. The mixed oxide X, which has the normal spinel structure A2+B23+O42−\mathrm{A^{2+}B^{3+}_2O^{2-}_4}, is obtained by decomposing in air the crystalline hydrated metal oxalate ZCX2OX4 ⋅ kHX2O\ce{ZC2O4*\mathit{k}H2O} of metal Z. When heated to 140 °C the hydrate loses 19.7 % of its mass; further heating in air at 500 °C yields 2.407 g of black X together with 3.8 dm3^3 of COX2\ce{CO2} (at 101325 Pa, 500 °C). (a) Determine the formula of X and the value of kk. (b) In the spinel structure the OX2−\ce{O^{2-}} ions form a face-centred cubic lattice; A2+\mathrm{A^{2+}} cations occupy part of the tetrahedral voids and B3+\mathrm{B^{3+}} cations part of the octahedral voids. Calculate the percentage of tetrahedral sites occupied in a normal spinel AB2O4\mathrm{AB_2O_4}.
Step 3 of 4: Hydration number
18.0 k146.9+18.0 k=0.197⇒k=2(CoCX2OX4 ⋅ 2 HX2O)\dfrac{18.0\,k}{146.9 + 18.0\,k} = 0.197 \Rightarrow k = 2\quad (\ce{CoC2O4*2H2O})
Analysis

The 19.7 % mass loss at 140 °C is the water: 18.0k58.9+88.0+18.0k=0.197\frac{18.0k}{58.9 + 88.0 + 18.0k} = 0.197 gives 18.0k=0.197(146.9+18.0k)18.0k = 0.197(146.9 + 18.0k), i.e. k≈2k \approx 2, so the precursor is CoCX2OX4 ⋅ 2 HX2O\ce{CoC2O4*2H2O}.