Ores with as little as $\mathbf{\text{0.25\%}}$by mass of copper are used as sources of the metal. (a) How many kilograms of such an ore would be needed for another Statue of Liberty, which contains $\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}$lb of copper? (b) If the mineral in the ore is chalcopyrite (${\mathbf{\text{FeCuS}}}_{\mathbf{\text{2}}}$), what is the mass % of chalcopyrite in the ore?

An element is a pure material made up entirely of atoms with the same number of protons in their nuclei, as defined by chemistry. Chemical elements, unlike chemical compounds, cannot be broken down into smaller substances by chemical reactions.

$$\begin{gathered} {\text{m(Cu)=2.0×10}}^5\text{lb} \\ 1lb=0.\text{4535924kg} \\ \text{m(Cu) =}{\mathrm{2.0\times 10}}^5\times 0.\text{4535924kg} \\ \text{m(Cu) =90718.480} \\ kg=9.07\times {10}^{4}kg \end{gathered}$$

We can compute the amount of ore necessary to extract this much copper if just$0.25$percent of the mass of the ore contains copper:

$$\begin{gathered} \text{m(Cu)=}9.07\times {10}^{4}kg \\ \text{w(Cu, ore ) =}\frac{{\displaystyle \text{m(Cu)}}}{{\displaystyle \text{m(ore}}}\text{×100\%} \\ \text{m(ore)=}\frac{9.07\times {10}^{4}kg}{0.25\%}\text{×100\%} \\ \text{m( ore ) =36287392kg} \\ =3.62\times {10}^{7}kg \end{gathered}$$

Therefore, the value is:$3.62\times {10}^{7}kg$ .

In chalcopyrite, the mass percent of copper is:

$$\begin{gathered} \text{w}\left({\text{Cu,FeCuS}}_{\text{2}}\right)\text{=}\frac{{\displaystyle \text{Ar(Cu)}}}{{\displaystyle \text{Mr}\left({\text{FeCuS}}_{\text{2}}\right)}}\text{×100} \\ \text{w}\left({\text{Cu,FeCuS}}_{\text{2}}\right)\text{=}\frac{{\displaystyle \text{63.546}}}{{\displaystyle \text{183.5210}}}\text{×100} \\ \text{w}\left({\text{Cu,FeCuS}}_{\text{2}}\right)\text{=34.6\%} \end{gathered}$$

We can determine the mass percent of chalcopyrite in the ore if we know the ore contains$0.25$percent Cu:

$$\begin{gathered} \text{w( Cu,chalcopyrite )}=\frac{{\displaystyle \text{m(Cu)}}}{{\displaystyle \text{m( chalcopyrite )}}} \\ \text{m( chalcopyrite )=}\frac{{\displaystyle \text{m(Cu)}}}{{\displaystyle \text{w( Cu,chalcopyrite )}}} \\ \text{w(Cu, ore ) =}\frac{{\displaystyle \text{m(Cu)}}}{{\displaystyle \text{m( ore )}}} \\ \text{m( ore )=}\frac{{\displaystyle \text{m(Cu)}}}{w(\mathrm{Cu},\mathrm{ore})} \\ \text{w( chalcopyrite, ore ) =}\frac{{\displaystyle \text{m( chalcopyrite )}}}{{\displaystyle \text{m( ore )}}}\text{×100} \\ \text{w( chalcopyrite, ore ) =}\frac{{\displaystyle \frac{\text{m( Cu )}}{\text{w( Cu,chalcopyrite )}}}}{{\displaystyle \frac{\text{m( Cu )}}{\text{w( Cu,ore )}}}}\text{×100} \\ \text{w( chalcopyrite, ore ) =}\frac{\text{0.25}}{\text{34.6}}\text{×100} \\ \text{w( chalcopyrite, ore ) =0.72\%} \end{gathered}$$

Therefore, the mass % is: $0.72\%$.