Q 88CP - Chemistry: Molecular Nature Of Matter And Change

Question

Question: Silver has a face-cantered cubic structure with a unit cell edge length of $\mathbf{408}\mathbf{.}\mathbf{6}\mathbf{ }\mathbf{pm}$. Sterling silver is a substitutional alloy that contains $\mathbf{7}\mathbf{.}\mathbf{5}\mathbf{\%}$copper atoms. Assuming the unit cell remains the same, find the density of silver and of sterling silver.

Step-by-Step Solution

Answer

Silver has a density $10.50620 \mathrm{g}/{\mathrm{cm}}^3$ and sterling silver has a density of $10.1822 \mathrm{g}/{\mathrm{cm}}^3$.

Step 1: Explanation of the concept

Given the silver $\left(\mathrm{Ag}\right)$possesses a face-cantered cubic structure $\left(\mathrm{fcc}\right)$denotes that the same $\mathbf{\left(}\mathbf{Ag}\mathbf{\right)}$atoms are found at each corner and in the centre of each face, but not in the centre of the cube.

The $\mathbf{\left(}\mathbf{fcc}\mathbf{\right)}$ unit cell is composed of four atoms, as follows:

$\frac{\mathbf{Atoms}}{\mathbf{Unit}\mathbf{ }\mathbf{cell}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{8}}\mathbf{\times }\mathbf{8}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\times }\mathbf{6}\mathbf{=}\mathbf{4}$

Step 2: Calculating the density of Silver

To calculate the density of silver, the mass of$1 \mathrm{mol}$ including $6.023x{10}^{23}$. First, the number of atoms assembled in unit cells is calculated:

$$\begin{gathered} \mathrm{m}\left(\mathrm{Silver}\right)=\frac{4 \mathrm{Ag} \mathrm{atoms}}{\mathrm{unit} \mathrm{cell}}\times \frac{1 \mathrm{mol}}{6.022\times {10}^{23} \mathrm{Ag} \mathrm{atoms}}\times {\mathrm{M}}_{\mathrm{R}}\left(\mathrm{Ag}\right) \\ \mathrm{m}\left(\mathrm{Silver}\right)=\frac{4 \mathrm{Ag} \mathrm{atoms}}{\mathrm{unit} \mathrm{cell}}\times \frac{1 \mathrm{mol}}{6.022\times {10}^{23} \mathrm{Ag} \mathrm{atoms}}\times 107.90 \mathrm{g}/\mathrm{mol} \\ \mathrm{m}\left(\mathrm{Silver}\right)=7.176\times {10}^{-22} \mathrm{g} \end{gathered}$$ 

The volume of the $fcc$ unit cell - a cube, knowing that the edge length is  $408.60 \mathrm{pm}$ is calculated.

$$\begin{gathered} \mathrm{V}=\frac{{\left(408.60 \mathrm{pm}\right)}^3}{\mathrm{unit} \mathrm{cell}}\times {\left(\frac{{10}^{-10} \mathrm{cm}}{1 \mathrm{pm}}\right)}^3 \\ \mathrm{V}=6.8217\times {10}^{-23 } {\mathrm{cm}}^3 \end{gathered}$$

Finally, the silver density is:

$$\begin{gathered} \mathrm{d}\left(\mathrm{silver}\right)=\frac{\mathrm{m}}{\mathrm{V}}=\frac{7.167\times {10}^{-22} \mathrm{g}}{6.8217\times {10}^{-23} {\mathrm{cm}}^3} \\ \mathrm{d}\left(\mathrm{silver}\right)=10.5062 \mathrm{g}/{\mathrm{cm}}^3 \end{gathered}$$

 

Step 3: Calculating the density of sterling silver

For Sterling Silver, the molecular mass is calculated first as a weighted average of the atom masses present:

$$\begin{gathered} {\mathrm{M}}_{\mathrm{R}}\left(\mathrm{sterling} \mathrm{silver}\right)={\mathrm{M}}_{\mathrm{R}}\left(\mathrm{Ag}\right)\times \frac{\mathrm{m}\%\left(\mathrm{Cu}\right)}{100}+{\mathrm{M}}_{\mathrm{R}}\left(\mathrm{Cu}\right)\times \frac{\mathrm{m}\%\left(\mathrm{Cu}\right)}{100} \\ {\mathrm{M}}_{\mathrm{R}}\left(\mathrm{sterling} \mathrm{silver}\right)=107.9 \mathrm{g}/\mathrm{mol}\times \frac{\left(100-7.5\right)\%}{100}+63.55 \mathrm{g}/\mathrm{mol}\times \frac{7.5\%}{100} \\ {\mathrm{M}}_{\mathrm{R}}\left(\mathrm{sterling} \mathrm{silver}\right)=104.57375 \mathrm{g}/\mathrm{mol} \end{gathered}$$The mass of the object is used to calculate its density $1 \mathrm{mol}$ containing $6.023\times {10}^{23} \mathrm{atoms}$ First, the number of atoms assembled in unit cells is calculated:

$$\begin{gathered} \mathrm{m}\left(\mathrm{Sterling} \mathrm{Silver}\right)=\frac{4 \mathrm{Ag} \mathrm{atoms}}{\mathrm{unit} \mathrm{cell}}\times \frac{1 \mathrm{mol}}{6.022\times {10}^{23} \mathrm{Ag} \mathrm{atoms}}\times {\mathrm{M}}_{\mathrm{R}}\left(\mathrm{Sterling} \mathrm{silver}\right) \\ \mathrm{m}\left(\mathrm{Sterling} \mathrm{Silver}\right)=\frac{4 \mathrm{Ag} \mathrm{atoms}}{\mathrm{unit} \mathrm{cell}}\times \frac{1 \mathrm{mol}}{6.022\times {10}^{23} \mathrm{Ag} \mathrm{atoms}}\times 104.57 \mathrm{g}/\mathrm{mol} \\ \mathrm{m}\left(\mathrm{Sterling} \mathrm{Silver}\right)=6.946\times {10}^{-22} \mathrm{g} \end{gathered}$$Assuming that the unit cell does not change, the volume does not change.

 $\mathrm{V}=6.8217 \times {10}^{-23 }{\mathrm{cm}}^3$

Finally, the Sterling Silver density is

$$\begin{gathered} \mathrm{d}\left(\mathrm{silver}\right)=\frac{\mathrm{m}}{\mathrm{V}}=\frac{6.946\times {10}^{-22} \mathrm{g}}{6.8217\times {10}^{-23} {\mathrm{cm}}^3} \\ \mathrm{d}\left(\mathrm{silver}\right)=10.1822 \mathrm{g}/{\mathrm{cm}}^3 \end{gathered}$$ 

As a result, silver has a density $10.5062 \mathrm{g}/{\mathrm{cm}}^3$and sterling silver has a density of $10.1822 \mathrm{g}/{\mathrm{cm}}^3$.