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 2 of 5: Confirm atomic mass and identify fluorides

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n(\ce{Xe}) = \dfrac{pV}{RT} = 2.17\times 10^{-3}\ \text{mol};\quad M = \dfrac{0.4500}{2.17\times 10^{-3}} = 207.4\ \text{g mol}^{-1} \;\Rightarrow\; \ce{XeF4}

Analysis

The liberated gas volume gives $n(\ce{Xe}) = 2.17\times 10^{-3}$ mol, hence $M(\text{C}) = 0.450 / 2.17\times 10^{-3} = 207.4$ g mol $^{-1}$ . With 36.7 % F, $m(\ce F) = 76.1$ g mol $^{-1}$ corresponds to 4 fluorine atoms, giving $M(\ce{Xe}) = 207.4 - 76.1 = 131.3$ g mol $^{-1}$ and C = $\ce{XeF4}$ . Then 46.5 % F in D gives $\ce{XeF6}$ , B is $\ce{XeF2}$ , and E is $\ce{XeF(OSO2F)}$ .