# What mass in kg of "CuFeS"_2" ore is required to obtain "325 g" of pure copper?

Jun 8, 2017

$\text{0.938 kg CuFeS"_2}$ is needed to obtain $325$ g of pure $\text{Cu}$.

#### Explanation:

The chemical formula for chalcopyrite, ${\text{CuFeS}}_{2}$ indicates that each mole of the compound contains one mole of copper.

The following represents the method that will be used to determine how many kg of ${\text{CuFeS}}_{2}$ must be mined to obtain $\text{325g Cu}$.

color(red)("mass Cu"$\rightarrow$color(blue)("mole Cu"$\rightarrow$color(green)("mol CuFeS"_2"$\rightarrow$color(magenta)("mass CuFeS"_2"

color(purple)("Molar Masses"

The first thing you need to do is determine the molar masses for copper and chalcopyrite.

The molar mass of ${\text{CuFeS}}_{2}$ is $\text{183.511 g/mol}$
https://www.ncbi.nlm.nih.gov/pccompound?term=CuFeS2

The molar mass of $\text{Cu}$ is $\text{63.546 g/mol}$ (periodic table)

color(red)("Mass Cu" to color(blue)("Moles Cu"

Divide the given mass of $\text{Cu}$ by its molar mass by multiplying by the inverse of the molar mass.

325color(red)cancel(color(black)("g Cu"))xx(1"mol Cu")/(63.546color(red)cancel(color(black)("g Cu")))="5.11 mol Cu"

color(blue)("Moles Cu" to color(green)("Moles CuFeS"_2"

Multiply moles $\text{Cu}$ by the mole ratio between $\text{Cu}$ and $\text{CuFeS"_2}$ with $\text{CuFeS"_2}$ in the numerator.

5.11color(red)cancel(color(black)("mol Cu"))xx(1"mol CuFeS"_2)/(1color(red)cancel(color(black)("mol Cu")))="5.11 mol CuFeS"_2"

color(green)("Moles CuFeS"_2" to color(magenta)("Mass CuFeS"_2"

5.11color(red)cancel(color(black)("mol CuFeS"_2))xx(183.511"g CuFeS"_2)/(1color(red)cancel(color(black)("mol CuFeS"_2)))="938 g CuFeS"_2" rounded to three sig figs

color(orange)("Convert mass in grams to kilograms".

938color(red)cancel(color(black)("g")) "CuFeS"_2xx(1"kg")/(1000color(red)cancel(color(black)("g")))="0.938 kg CuFeS"_2"