Question

An orebody contains pyrite and chalcopyrite in the same molar proportions. The percentage concentration of Cu in the ore will be ________. (Round off to the nearest integer) (Use atomic weight, in g/mol, of Cu = 63.55, Fe = 55.85, S = 32.06)

Answer 1

Correct Answer: 21

Calculating the Percentage Concentration of Cu in the Ore 

Step 1: Understanding the Problem

We are given that the orebody contains pyrite (FeS₂) and chalcopyrite (CuFeS₂) in the same molar proportions. We need to calculate the percentage concentration of Cu in the ore.

Step 2: Molar Masses of the Compounds

The molar masses of pyrite and chalcopyrite are calculated as follows:

• Pyrite (FeS₂): Molar mass = Fe (55.85 g/mol) + 2 × S (32.07 g/mol) = 55.85 + 64.14 = 119.99 g/mol
• Chalcopyrite (CuFeS₂): Molar mass = Cu (63.55 g/mol) + Fe (55.85 g/mol) + 2 × S (32.07 g/mol) = 63.55 + 55.85 + 64.14 = 183.54 g/mol

Step 3: Calculate the Molar Proportion of Cu in the Ore

Given that pyrite and chalcopyrite are present in the same molar proportions, let's assume we have 1 mole of each compound.

• In pyrite, there is no Cu, so the Cu content from pyrite is 0.
• In chalcopyrite, there is 1 mole of Cu, contributing 63.55 g of Cu.

Step 4: Calculate the Total Mass and Cu Mass in the Ore

The total mass of the ore from 1 mole of pyrite and 1 mole of chalcopyrite is:

\[ \text{Total mass of ore} = 119.99 \, \text{g (pyrite)} + 183.54 \, \text{g (chalcopyrite)} = 303.53 \, \text{g} \]

The mass of Cu from chalcopyrite is 63.55 g.

Step 5: Calculate the Percentage Concentration of Cu

The percentage concentration of Cu is calculated as:

\[ \text{Percentage concentration of Cu} = \frac{\text{Mass of Cu}}{\text{Total mass of ore}} \times 100 = \frac{63.55}{303.53} \times 100 \approx 20.95\% \]

Step 6: Round to the Nearest Integer

The percentage concentration of Cu is approximately 21% (rounded to the nearest integer).

Final Answer:

The percentage concentration of Cu in the ore is: 21%.