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 1 of 4: Molar mass of X

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5.000 = n_1 M + n_2 M,\quad 5.628 = n_1(M+4.04) + n_2(2M+6.06),\quad n_1 = 2n_2 \Rightarrow M(\mathrm{X}) = 28.14\ \text{g mol}^{-1}\ (\ce{Si})

Analysis

Let $n_1, n_2$ be the moles of $\ce{XH4}$ and $\ce{X2H6}$ . Mass of X: $5.000 = M(n_1 + 2n_2)$ ; total mass adds the hydrogen: $5.628 = n_1(M+4.04)+n_2(2M+6.06)$ ; and $n_1 = 2n_2$ . Solving gives $M = 28.14$ g mol $^{-1}$ — silicon. The products $\ce{SiH4}$ and $\ce{Si2H6}$ are tetrahedral at each Si, exactly like methane and ethane.