IChO 2004 · Theory problem 4

Determining atomic masses

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The reaction of the element X with hydrogen leads to a class of compounds that is 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, respectively.

4.1 Determine the molar mass of X from this information. Give the chemical symbol of X, and the 3D-structure of the two products.

The following more complex case is of great historical interest.

The mineral Argyrodite is a stoichiometric compound that contains silver (oxidation state +1), sulphur (oxidation state -2) and an unknown element Y (oxidation state +4). The ratio between the masses of silver and Y in Argyrodite is \(m(\ce{Ag}) : m(\ce{Y}) = 11.88 : 1\). Y forms a reddish brown lower sulfide (oxidation state of Y is +2) and a higher white sulfide (oxidation state of Y is +4). The coloured lower sulfide is the sublimate obtained by heating Argyrodite in a flow of hydrogen. The residues are \(\ce{Ag2S}\) and \(\ce{H2S}\). To convert 10.0 g of Argyrodite completely, 0.295 \(\mathrm{dm^3}\) of hydrogen are needed at 400 K and 100 kPa.

4.2 Determine the molar mass of Y from this information. Give the chemical symbol of Y, and the empirical formula of Argyrodite.

The atomic masses are correlated with spectroscopic properties. To determine the vibrational frequency \(\tilde{\nu}\) expressed in wave numbers of chemical bonds in IR spectra chemists use Hooke's law which focuses on the frequency of the vibration (attention to units!):

\[ \tilde{\nu}=\frac{1}{2\pi c}\cdot\sqrt{\frac{k}{\mu}} \]

\(\tilde{\nu}\) - vibrational frequency of the bond, in wavenumbers (\(\mathrm{cm^{-1}}\))

\(c\) - speed of light

\(k\) - force constant, indicating the strength of the bond (\(\mathrm{N\,m^{-1}=kg\,s^{-2}}\))

\(\mu\) reduced mass in \(\ce{AB4}\), which is given by \[ \mu=\frac{3\,m(A)\,m(B)}{3\,m(A)+4\,m(B)} \]

\(m(A),m(B)\) - the masses of the two bond atoms

The vibrational frequency of the \(\ce{C-H}\) bond of methane is known to be 3030.00 \(\mathrm{cm^{-1}}\).

The vibrational frequency of the Z-analogue of methane is known to be 2938.45 \(\mathrm{cm^{-1}}\). The bond enthalpy of a \(\ce{C-H}\) bond in methane is 438.4 \(\mathrm{kJ\,mol^{-1}}\). The bond enthalpy of a \(\ce{Z-H}\) bond in the Z-analogue of methane is known to be 450.2 \(\mathrm{kJ\,mol^{-1}}\).

4.3 Determine the force constant \(k\) of a \(\ce{C-H}\) bond using Hooke's law.

Estimate the force constant \(k\) of a \(\ce{Z-H}\) bond, assuming that there is a linear proportionality between force constant and bond enthalpy.

Determine the atomic mass of Z from this information.

Give the chemical symbol of Z.

Official English text of the problem.

Answers

  • 4.1: \(M(X)\approx28.1\ \mathrm{g\,mol^{-1}}\); \(X=\ce{Si}\), silicon. The products are silane, \(\ce{SiH4}\), SMILES [SiH4], with tetrahedral geometry, and disilane, \(\ce{Si2H6}\), SMILES [SiH3][SiH3], with approximately tetrahedral coordination at each silicon and a preferred staggered conformation.
  • 4.2: \(M(Y)\approx72.6\ \mathrm{g\,mol^{-1}}\); \(Y=\ce{Ge}\), germanium. Argyrodite has empirical formula \(\boxed{\ce{Ag8GeS6}}\).
  • 4.3: \(k_{\ce{C-H}}=490.4\ \mathrm{N\,m^{-1}}\); \(k_{\ce{Z-H}}\approx503.6\ \mathrm{N\,m^{-1}}\). The estimated atomic mass is \(m(Z)\approx73.4\ \mathrm{u}\), identifying \(Z=\ce{Ge}\). Its methane analogue is germane, \(\ce{GeH4}\), SMILES [GeH4].

Solution

The problem gives no table of constants. The calculations use the standard reference molar masses \[ M(\ce{H})=1.008,\quad M(\ce{C})=12.01,\quad M(\ce{Ag})=107.87,\quad M(\ce{S})=32.06 \quad \mathrm{g\,mol^{-1}}, \] together with \[ R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}},\qquad c=2.99792458\times10^{10}\ \mathrm{cm\,s^{-1}},\qquad 1\ \mathrm{u}=1.66054\times10^{-27}\ \mathrm{kg}. \]

4.1

Methane, \(\ce{CH4}\), SMILES C, and ethane, \(\ce{C2H6}\), SMILES CC, have the stoichiometric X-analogues \(\ce{XH4}\) and \(\ce{X2H6}\).

Let their amounts be \(2t\) and \(t\), respectively. The mixture therefore contains \[ n(X)=2t+2t=4t,\qquad n(\ce{H}\text{ atoms})=4(2t)+6t=14t. \]

The mass gained is the mass of hydrogen: \[ m(\ce{H})=5.628-5.000=0.628\ \mathrm{g}. \] Consequently, \[ t=\frac{0.628\ \mathrm{g}} {14(1.008\ \mathrm{g\,mol^{-1}})} =0.04450\ \mathrm{mol}, \] and \[ M(X)=\frac{5.000\ \mathrm{g}}{4t} =28.09\ \mathrm{g\,mol^{-1}} \approx\boxed{28.1\ \mathrm{g\,mol^{-1}}}. \]

This identifies silicon, \(\boxed{\ce{Si}}\).

The balanced formation equations, with hydrogen \(\ce{H2}\), SMILES [H][H], are \[ \ce{Si + 2H2 -> SiH4}, \] \[ \ce{2Si + 3H2 -> Si2H6}. \] For the specified mixture, their combined stoichiometry is \[ \ce{4Si + 7H2 -> 2SiH4 + Si2H6}. \]

The requested three-dimensional structures are:

  • Silane, \(\ce{SiH4}\), SMILES [SiH4]: silicon lies at the centre of a regular tetrahedron, with the four hydrogen atoms at its vertices. The bond angles are approximately \(109.5^\circ\).
  • Disilane, \(\ce{Si2H6}\), SMILES [SiH3][SiH3]: a single silicon–silicon bond joins two groups, each containing silicon bonded to three hydrogen atoms. Each silicon has approximately tetrahedral coordination. In the preferred staggered conformation, looking along the silicon–silicon bond, the three rear hydrogen atoms are rotated by \(60^\circ\) relative to the three front hydrogen atoms.

The SMILES specify connectivity; the tetrahedral geometry and staggered conformation are supplied by these descriptions.

4.2

Y is reduced from oxidation state \(+4\) to \(+2\), requiring two electrons per Y atom. Oxidation of one hydrogen molecule supplies these two electrons. Thus, \[ n(Y)=n(\ce{H2}). \]

Using the ideal-gas equation, with \(1\ \mathrm{kPa\,dm^3}=1\ \mathrm{J}\), \[ n(\ce{H2}) =\frac{pV}{RT} =\frac{(100\ \mathrm{kPa})(0.295\ \mathrm{dm^3})} {(8.314\ \mathrm{J\,mol^{-1}\,K^{-1}})(400\ \mathrm{K})} =8.8706\times10^{-3}\ \mathrm{mol}. \]

Let \(n=n(Y)\) and \(M=M(Y)\). The mass ratio gives \[ m(Y)=nM,\qquad m(\ce{Ag})=11.88nM, \] so \[ n(\ce{Ag})=\frac{11.88nM}{M(\ce{Ag})}. \]

Charge neutrality, using the stated oxidation states, requires the amounts of atoms in the mineral to satisfy \[ n(\ce{Ag})+4n(Y)=2n(\ce{S}). \] Hence, \[ n(\ce{S}) =n\left(\frac{11.88M}{2M(\ce{Ag})}+2\right). \]

Adding the masses of all three elements gives \[ 10.0\ \mathrm{g} =nM+11.88nM+ n\left(\frac{11.88M}{2M(\ce{Ag})}+2\right)M(\ce{S}). \] Therefore, \[ M= \frac{\dfrac{10.0\ \mathrm{g}}{n}-2M(\ce{S})} {12.88+\dfrac{11.88M(\ce{S})}{2M(\ce{Ag})}} \approx\boxed{72.6\ \mathrm{g\,mol^{-1}}}. \]

This identifies germanium, \(\boxed{\ce{Ge}}\). The reddish brown lower sulfide is germanium(II) sulfide, \(\ce{GeS}\), and the white higher sulfide is germanium(IV) sulfide, \(\ce{GeS2}\).

The silver-to-germanium atomic ratio is \[ \frac{n(\ce{Ag})}{n(\ce{Ge})} =\frac{11.88M(\ce{Ge})}{M(\ce{Ag})} \approx\frac{11.88(72.6)}{107.87} \approx8. \] Charge neutrality then gives \[ \frac{n(\ce{S})}{n(\ce{Ge})}=\frac{8+4}{2}=6. \] Thus the empirical formula is \[ \boxed{\ce{Ag8GeS6}}. \]

The balanced conversion is \[ \boxed{\ce{Ag8GeS6 + H2 -> 4Ag2S + GeS + H2S}}. \]

Here \(\ce{Ag2S}\) is silver(I) sulfide, and \(\ce{H2S}\) is hydrogen sulfide, SMILES S. The solid sulfides and argyrodite have extended structures, so a discrete-molecule SMILES would not describe their solid-state structures.

4.3

For methane, use the reduced-mass expression supplied for \(\ce{AB4}\): \[ \mu_{\ce{CH4}} =\frac{3m(\ce{C})m(\ce{H})} {3m(\ce{C})+4m(\ce{H})}. \] Substituting atomic masses, \[ \mu_{\ce{CH4}} =\frac{3(12.01)(1.008)} {3(12.01)+4(1.008)}\ \mathrm{u} =0.9065508\ \mathrm{u} =1.50536\times10^{-27}\ \mathrm{kg}. \]

Rearranging Hooke’s law, \[ k=4\pi^2c^2\tilde{\nu}^{\,2}\mu. \] Because the wavenumber is in \(\mathrm{cm^{-1}}\), use \(c\) in \(\mathrm{cm\,s^{-1}}\): \[ k_{\ce{C-H}} = 4\pi^2 (2.99792458\times10^{10}\ \mathrm{cm\,s^{-1}})^2 (3030.00\ \mathrm{cm^{-1}})^2 (1.50536\times10^{-27}\ \mathrm{kg}). \] Thus, \[ \boxed{k_{\ce{C-H}}=490.4\ \mathrm{N\,m^{-1}}}. \]

Under the assumed proportionality between force constant and bond enthalpy, \[ \frac{k_{\ce{Z-H}}}{k_{\ce{C-H}}} =\frac{450.2}{438.4}. \] Using the unrounded first result, \[ \boxed{k_{\ce{Z-H}}\approx503.6\ \mathrm{N\,m^{-1}}}. \]

The reduced mass of the Z-analogue follows from \[ \frac{\mu_{\ce{ZH4}}}{\mu_{\ce{CH4}}} = \frac{k_{\ce{Z-H}}}{k_{\ce{C-H}}} \left(\frac{\tilde{\nu}_{\ce{C-H}}} {\tilde{\nu}_{\ce{Z-H}}}\right)^2. \] Therefore, \[ \mu_{\ce{ZH4}} = 0.9065508 \left(\frac{450.2}{438.4}\right) \left(\frac{3030.00}{2938.45}\right)^2 \mathrm{u} =0.9898645\ \mathrm{u}. \]

Writing \(m_Z=m(Z)\) and \(m_H=m(\ce{H})\), \[ \mu_{\ce{ZH4}}=\frac{3m_Zm_H}{3m_Z+4m_H}. \] Solving for \(m_Z\), \[ m_Z=\frac{4\mu_{\ce{ZH4}}m_H} {3(m_H-\mu_{\ce{ZH4}})}. \] Retaining guard digits because the denominator contains a small difference, \[ m_Z= \frac{4(0.9898645)(1.008)} {3(1.008-0.9898645)}\ \mathrm{u} \approx\boxed{73.4\ \mathrm{u}}. \]

This estimated mass identifies \(\boxed{Z=\ce{Ge}}\), consistent with the germanium mass determined in 4.2. The methane analogue is germane, \(\ce{GeH4}\), SMILES [GeH4], with tetrahedral geometry.

The inferred \(73.4\ \mathrm{u}\) is an estimate: the assumed proportionality between bond enthalpy and force constant is approximate.

The solution was checked against the official solution.