91Ó°ÊÓ

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.

$$\begin{gathered} {\text{³¾(°ä³Ü)=2.0×10}}^5\text{lb} \\ 1lb=0.\text{4535924kg} \\ \text{m(Cu) =}{\mathrm{2.0×10}}^5×0.\text{4535924kg} \\ \text{m(Cu) =90718.480} \\ kg=9.07×{10}^{4}kg \end{gathered}$$

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:

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

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

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: $0.72\%$.