Chapter 11.13, Problem 80AAP

To determine

The density of strontium tin oxide $\left({\text{SrSnO}}_3\right)$ in grams per cubic centimeter.

Answer to Problem 80AAP

The density of strontium tin oxide $\left({\text{SrSnO}}_3\right)$ in grams per cubic centimeter is $\text{6}\text{.03}\text{g}/{\text{cm}}^3$.

Explanation of Solution

Write the expression to calculate lattice constant of strontium tin oxide structure $\left(a\right)$.

 $a=2\left(r_{{\text{Sn}}^{\text{4+}}}+r_{{\text{O}}^{-}}\right)$                                                                                                      ...... (I)

Here, ionic radius of ${\text{Sn}}^{\text{4+}}$ ion is $r_{{\text{Sn}}^{\text{4+}}}$ and ionic radius of ${\text{O}}^{-}$ ion is $r_{{\text{O}}^{-}}$.

Write the expression to calculate mass of unit cell of strontium tin oxide $\left(m\right)$.

 $m=\frac{n_{{\text{Sr}}^{\text{2+}}}{M}_{{\text{Sr}}^{\text{2+}}}+n_{{\text{Sn}}^{\text{4+}}}{M}_{{\text{Sn}}^{\text{4+}}}+n_{{\text{O}}^{-}}{M}_{{\text{O}}^{-}}}{N_A}$                                                                      ...... (II)

Here, number of atoms and molar mass of ${\text{Sr}}^{\text{2+}}$ ion are $n_{{\text{Sr}}^{\text{2+}}}$ and $M_{{\text{Sr}}^{\text{2+}}}$, number of atoms and molar mass of ${\text{Sn}}^{\text{4+}}$ ion are $n_{{\text{Sn}}^{\text{4+}}}$ and $M_{{\text{Sn}}^{\text{4+}}}$, number of atoms and molar mass of ${\text{O}}^{-}$ ion are $n_{{\text{O}}^{-}}$ and $M_{{\text{O}}^{-}}$ and Avogadro's number is $N_A$.

Write the expression to calculate density of strontium tin oxide in grams per cubic centimeter $\left(\rho \right)$.

 $\begin{array}{c}\rho =\frac{m}{V}\\ =\frac{m}{a^3}\end{array}$                                                                                                                  ...... (III)

Here, volume of unit cell is $V$.

Conclusion:

Substitute $0.074\text{nm}$ for $r_{{\text{Sn}}^{\text{4+}}}$ and $0.132\text{nm}$ for $r_{{\text{O}}^{-}}$ in Equation (I).

 $\begin{array}{c}a=2\left(0.074\text{nm}+0.132\text{nm}\right)\\ =0.412\text{nm}\\ =4.12\times {10}^{-8}\text{cm}\end{array}$

Substitute $1$ for $n_{{\text{Sr}}^{\text{2+}}}$, $87.62\text{g}/\text{mol}$ for $M_{{\text{Sr}}^{\text{2+}}}$, $1$ for $n_{{\text{Sn}}^{\text{4+}}}$, $118.7\text{g}/\text{mol}$ for $M_{{\text{Sn}}^{\text{4+}}}$, $3$ for $n_{{\text{O}}^{-}}$, $16\text{g}/\text{mol}$ for $M_{{\text{O}}^{-}}$ and $6.023\times {10}^{23}\text{ions}/\text{mol}$ for $N_A$ in Equation (II).

 $\begin{array}{c}m=\frac{\left(1\right)\left(87.62\text{g}/\text{mol}\right)+\left(1\right)\left(118.7\text{g}/\text{mol}\right)+\left(3\right)\left(16\text{g}/\text{mol}\right)}{6.023\times {10}^{23}\text{ions}/\text{mol}}\\ =4.22\times {10}^{-22}\text{g}\end{array}$

Substitute $4.22\times {10}^{-22}\text{g}$ for $m$ and $4.12\times {10}^{-8}\text{cm}$ for $a$ in Equation (III).

 $\begin{array}{c}\rho =\frac{4.22\times {10}^{-22}\text{g}}{{\left(4.12\times {10}^{-8}\text{cm}\right)}^3}\\ =6.03\text{g}/{\text{cm}}^3\end{array}$

Thus, the density of strontium tin oxide $\left({\text{SrSnO}}_3\right)$ in grams per cubic centimeter is $\text{6}\text{.03}\text{g}/{\text{cm}}^3$.