Problem 2

Heating a mixture of element A and fluorine (molar ratio 1 : 9, pressure ~1 MPa) to 900 °C gives three crystalline products B, C, D (all melting below 150 °C). The fluorine content of C is 36.7 % and of D is 46.5 % (by weight). Reaction of B with anhydrous

\ce{HOSO2F}

at -75 °C gives E + HF; X-ray diffraction maps of E show electron density maxima with values 52, 58, 104 and 350, roughly proportional to atomic numbers. 2.1 Assign these maxima to the elements in E (F, O, S, A) and determine the atomic number and identity of A. 2.2 Treating 450.0 mg of C with excess mercury liberates 53.25 cm

^3

of gas A at 101.0 kPa and 25 °C. Calculate the relative atomic mass of A and identify B, C, D and E. 2.3 Hydrolysis in water proceeds as follows: B hydrolyses to A,

\ce{O2}

and HF; C hydrolyses to A,

\ce{O2}

(in 4 : 3 molar ratio), aqueous

\ce{AO3}

and HF; D hydrolyses to aqueous

\ce{AO3}

and HF. Write the three balanced hydrolysis equations. 2.4 Quantitative hydrolysis of a mixture of B, C, D yields 60.2 cm

^3

of gas at 290 K and 100 kPa containing 40.0 %

\ce{O2}

by volume. The dissolved

\ce{AO3}

is titrated with 0.100 M

\ce{FeSO4}

, requiring 36.0 cm

^3

(

\ce{Fe^{2+}}

is oxidised to

\ce{Fe^{3+}}

and

\ce{AO3}

is reduced to A). Calculate the molar composition of the original mixture of B, C, D.

Step 5 of 5: Solve for XeF6 and mole percentages

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n(\ce{XeO3}) = \tfrac{1}{6}n(\ce{Fe^{2+}}) = 6.00\times 10^{-4}\ \text{mol};\quad c = 6.00\times 10^{-4} - \tfrac{1}{3}b = 1.00\times 10^{-4}\ \text{mol}

Analysis

Reduction $\ce{6Fe^{2+} + XeO3 + 3H2O -> 6Fe^{3+} + Xe + 6OH^-}$ requires 6 $\ce{Fe^{2+}}$ per $\ce{XeO3}$ , so $n(\ce{XeO3}) = (0.100 \times 36.0\times 10^{-3})/6 = 6.00\times 10^{-4}$ mol. Since $\ce{XeO3}$ comes from $(1/3)b + c$ , $c = n(\ce{XeF6}) = 6.00\times 10^{-4} - 5.00\times 10^{-4} = 1.00\times 10^{-4}$ mol. The mixture contains $0.50$ mmol $\ce{XeF2}$ (23.8 %), $1.50$ mmol $\ce{XeF4}$ (71.4 %), $0.10$ mmol $\ce{XeF6}$ (4.8 %).