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 4 of 5: Solve for XeF2 and XeF4

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\begin{cases} a + \tfrac{2}{3}b = 1.50\times 10^{-3} \\ \tfrac{1}{2}a + \tfrac{1}{2}b = 1.00\times 10^{-3} \end{cases} \;\Rightarrow\; a = 0.50\times 10^{-3},\ b = 1.50\times 10^{-3}\ \text{mol}

Analysis

Total gas is $n = (100\times 10^3 \times 60.2\times 10^{-6}) / (8.314 \times 290) = 2.50\times 10^{-3}$ mol, of which $n(\ce{O2}) = 1.00\times 10^{-3}$ mol and $n(\ce{Xe}) = 1.50\times 10^{-3}$ mol. With $a = n(\ce{XeF2})$ and $b = n(\ce{XeF4})$ : $a + b = 2.00\times 10^{-3}$ and $a + (2/3)b = 1.50\times 10^{-3}$ , which solves to $a = 0.50$ mmol and $b = 1.50$ mmol.