Q22.84CP - Chemistry: Molecular Nature Of Matter And Change

Question

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?

Step-by-Step Solution

Answer

a) The amount of ore needed to build another statue of liberty is obtained as:$3.62\times {10}^{7}kg$  .

b) The mass % of chalcopyrite in the ore is obtained as:$\text{0.72\% }$

Step 1: Define Elements

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.

Step 2: Computation of how many kilograms of such ore is needed?

1. To begin, we must convert the mass of the copper from pounds to kilograms:

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

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:

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

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

Step 3: Evaluation of the mass % of chalcopyrite in the ore

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: $0.72\%$ .