## Chemistry: Molecular Approach (4th Edition)

$\underline{49 percent}$
Since $1\text{ kg}=1000\text{ g}$, convert the mass into grams as follows: $\left( 4.36\text{ kg} \right)\left( \frac{1000\text{ g}}{1\text{ kg}} \right)=4360\text{ g}$ Now, assume that the volume of copper is $x\text{ c}{{\text{m}}^{3}}$. Thus, the volume of lead is calculated as follows: \begin{align} & {{V}_{\text{T}}}={{V}_{\text{Pb}}}+{{V}_{\text{Cu}}} \\ & 427\text{ c}{{\text{m}}^{3}}={{V}_{\text{Pb}}}+x\text{ c}{{\text{m}}^{3}} \\ & {{V}_{\text{Pb}}}=\left( 427-x \right)\text{ c}{{\text{m}}^{3}} \end{align} Now, the mass of copper is calculated as follows: ${{m}_{\text{Cu}}}=\left( {{d}_{\text{Cu}}} \right)\left( {{V}_{\text{Cu}}} \right)$ Here, ${{d}_{\text{Cu}}}$ is the density of copper and ${{V}_{\text{Cu}}}$ is the volume of copper. Substitute $8.96\text{ g/c}{{\text{m}}^{3}}$ for ${{d}_{\text{Cu}}}$ and $x\text{ c}{{\text{m}}^{3}}$ for ${{V}_{\text{Cu}}}$: \begin{align} & {{m}_{\text{Cu}}}=\left( 8.96\text{ g/c}{{\text{m}}^{3}} \right)\left( x\text{ c}{{\text{m}}^{3}} \right) \\ & =8.96x\text{ g} \end{align} Now, the density $\left( {{d}_{\text{Pb}}} \right)$ of lead is $\text{11}\text{.4 g/c}{{\text{m}}^{3}}$. Calculate the mass of the lead as follows: ${{m}_{\text{Pb}}}=\left( {{d}_{\text{Pb}}} \right)\left( {{V}_{\text{Pb}}} \right)$ Substitute $\text{11}\text{.4 g/c}{{\text{m}}^{\text{3}}}$ for ${{d}_{\text{Cu}}}$ and $\left( 427-x \right)\text{ c}{{\text{m}}^{3}}$ for ${{V}_{\text{Cu}}}$: \begin{align} & {{m}_{\text{Cu}}}=\left( \text{11}\text{.4 g/c}{{\text{m}}^{3}} \right)\left( 427-x \right)\text{c}{{\text{m}}^{3}} \\ & =11.4\left( 427-x \right)\text{ g} \end{align} The total mass is written as follows: \begin{align} & {{m}_{\text{T}}}={{m}_{\text{Pb}}}+{{m}_{\text{Cu}}} \\ & 4360\text{ g}=8.96x\text{ g}+11.4\left( 427-x \right)\text{ g} \\ & x=\frac{4867.8-4360}{2.44} \\ & =208.11 \end{align} Thus, the volume of copper is $208.11\text{ c}{{\text{m}}^{3}}$. The percentage of the total volume of the sphere that is copper is as follows: $\text{percentage}=\frac{{{V}_{\text{Cu}}}}{{{V}_{\text{T}}}}\times 100$ Substitute $208.11\text{ c}{{\text{m}}^{3}}$ for ${{V}_{\text{Cu}}}$ and $\text{427 c}{{\text{m}}^{3}}$ for ${{V}_{\text{T}}}$: \begin{align} & \text{percentage}=\frac{208.11\text{ c}{{\text{m}}^{3}}}{427\text{ c}{{\text{m}}^{3}}\text{ }}\times 100 \\ & =49\,\text{percent} \end{align} The percentage of the total volume of the sphere is $49\,percent$.