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 4 of 4: Force constants from IR frequencies

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k(\ce{C-H}) = (2\pi c\tilde{\nu})^2\mu = (2\pi\times3\times10^{10}\times3030)^2\times\frac{3(12.01)(1.008)}{3(12.01)+4(1.008)N_A}\cdot\frac{1}{1000} = 491.9\ \text{N m}^{-1};\quad k(\ce{Z-H}) = 491.9\times\frac{450.2}{438.4} = 505.2\ \text{N m}^{-1}

Analysis

Hooke's law with the $\ce{AB4}$ reduced mass $\mu = \frac{3m_A m_B}{3m_A + 4m_B}$ per molecule ( $m$ in kg, via $M/N_A$ ) gives $k(\ce{C-H}) = 491.9$ N m $^{-1}$ . Scaling by the bond-enthalpy ratio yields $k(\ce{Z-H}) = 505.2$ N m $^{-1}$ ; solving $\tilde{\nu} = \frac{1}{2\pi c}\sqrt{k/\mu}$ for the unknown mass gives $M(Z) = 72.7$ g mol $^{-1}$ — again germanium ( $\ce{GeH4}$ ).