A copper ore contains 30 wt.% chalcopyrite (\(CuFeS_2\)) and the remaining gangue material. Assuming no copper is present in the gangue material, the amount of copper in the ore (rounded off to one decimal place) is ______________ wt.%.
Given: Atomic weights of Fe, Cu and S are 56, 63.5, and 32 g/mol, respectively.
Show Hint
Find the mass fraction of Cu inside CuFeS2 using the atomic weights, then scale it by the 30 wt.% chalcopyrite content of the ore.
Step 1: Understanding the Concept:
We are told what fraction of the ore is the copper mineral chalcopyrite, and we need the copper content of the whole ore. Working with an actual mass basis makes the arithmetic concrete.
Step 2: Key Formula or Approach:
Take $100$ kg of ore as the basis. Find how much chalcopyrite that contains, then use the atomic weights to split that chalcopyrite mass into its Cu, Fe and S parts.
Step 3: Detailed Explanation:
In $100$ kg of ore, the chalcopyrite present is
\[ 100 \times 0.30 = 30 \text{ kg} \]
The formula weight of $CuFeS_2$ is
\[ 63.5+56+2(32) = 183.5 \text{ g/mol} \]
Out of every $183.5$ kg of chalcopyrite, $63.5$ kg is copper. So the copper inside our $30$ kg of chalcopyrite is
\[ 30 \times \frac{63.5}{183.5} = 30 \times 0.3460 = 10.38 \text{ kg} \]
Since our basis was $100$ kg of ore, this mass of copper is directly the weight percent of copper in the ore.
Step 4: Final Answer:
The copper content of the ore is close to $10.4$ wt.%.
\[ \boxed{10.4\text{ wt.\%}} \]
