Starting from a perfect $\ce{MgO}$ crystal a solid solution of $\ce{Al2O3}$ and $\ce{MgO}$ is prepared with an atomic proportion of $15:85$ of $\ce{Al}:\ce{Mg}$. Knowing the number of vacancies per $\ce{Mg}$ atom is $0.088$, calculate the % variation of density.

My reasoning is as follows:

For every $3~\ce{Mg^{2+}}$ atoms, we have $2~\ce{Al^3+}$ to maintain electric neutrality and 1 vacancy to keep the number of lattice points. Then, in one cell of crystal, which is FCC, there are initially 4 $\ce{Mg^{2+}}$ atoms which will then be redistributed as 2 $\ce{Mg^{2+}}$, $\frac43$ $\ce{Al^{3+}}$ and $\frac23$ vacancies to fulfill stoichiometry.

Then, I calculate the initial density as: $$\rho_\mathrm o=\frac{(4 \ce{Mg^{2+}}\times\frac{\pu{24.31g}}{N_A \pu{at/mol}})+ (4 \ce{O^{2-}}\times\frac{\pu{16g}}{N_A \pu{at/mol}})}{a^3}$$ And the final density as: $$\rho_\mathrm f=\frac{(2 \ce{Mg^{2+}}\times\frac{\pu{24.31g}}{N_A \pu{at/mol}})+ (\frac43 \ce{Al^{3+}}\times\frac{\pu{26.98g}}{N_A \pu{at/mol}})+(4 \ce{O^{2-}}\times\frac{\pu{16g}}{N_A \pu{at/mol}})}{a^3}$$

When taking the difference and dividing by the initial density and multiplying by 100%, I get 31.9%, when the solution is 4%. What am I doing wrong?

  • $\begingroup$ Well, you forgot several things, but in particular where does the 15:85 ration of Al:Mg figure into your calculation? $\endgroup$ – Jon Custer Mar 1 '18 at 14:43
  • $\begingroup$ Update: I have successfully reached the right solution. Thanks again to @JonCuster $\endgroup$ – Bee Mar 1 '18 at 21:46
  • $\begingroup$ @Bee Could you please post a self-answer? $\endgroup$ – Gaurang Tandon Jul 14 '18 at 9:05

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