Problem 4

Determining atomic masses. (4.1) The reaction of element X with hydrogen gives compounds analogous to hydrocarbons. 5.000 g of X form 5.628 g of a molar 2:1 mixture of the stoichiometric X-analogues of methane and ethane. Determine the molar mass of X, give its symbol and the 3D structures of the two products. (4.2) The mineral argyrodite is a stoichiometric compound containing silver (oxidation state +1), sulfur (−2) and an unknown element Y (+4). The mass ratio is

m(\ce{Ag}):m(\ce{Y}) = 11.88:1

. Y forms a reddish-brown lower sulfide (Y in +2) and a white higher sulfide (Y in +4). When argyrodite is heated in a stream of hydrogen, the coloured lower sulfide sublimes; the residues are

\ce{Ag2S}

and

\ce{H2S}

. Complete conversion of 10.0 g of argyrodite needs 0.295 dm

^3

\ce{H2}

at 400 K and 100 kPa. Determine the molar mass of Y, its symbol, and the empirical formula of argyrodite. (4.3) IR frequencies (wavenumbers) follow Hooke's law

\tilde{\nu} = \frac{1}{2\pi c}\sqrt{k/\mu}

where, for a tetrahedral

\ce{AB4}

molecule,

\mu = \frac{3m(A)m(B)}{3m(A)+4m(B)}

. The C–H vibration of methane is 3030.00 cm

^{-1}

and that of the Z-analogue of methane is 2938.45 cm

^{-1}

; the bond enthalpies are 438.4 and 450.2 kJ mol

^{-1}

. Determine the force constant

k

of a C–H bond, estimate

k

of the Z–H bond assuming proportionality to bond enthalpy, and find the atomic mass and symbol of Z.

Step 3 of 4: Molar mass of Y and formula of argyrodite

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\frac{10.0\,b}{8.871\times10^{-3}} = a\times107.87 + b\,M + (a/2+2b)\times32.07,\quad a\times107.87 = 11.88\,b\,M \Rightarrow M(\mathrm{Y}) = 72.6\ \text{g mol}^{-1}\ (\ce{Ge}),\ \ce{Ag8GeS6}

Analysis

The mass balance per $b$ mol of $\ce{H2}$ is $1127\,b = 107.87a + bM + 32.07(a/2+2b)$ g mol $^{-1}$ and the mass ratio gives $107.87a = 11.88\,bM$ . Solving yields $M = 72.6$ g mol $^{-1}$ (germanium) and $a:b = 8:1$ , i.e. argyrodite is $\ce{Ag8GeS6}$ — the mineral in which Clemens Winkler discovered germanium in 1886.