Two minerals that contain Cu are `CuFeS_(2)` and `Cu_(2)S`. What mass of `Cu_(2)S` would contain the same mass of Cu as is contained in 125 lb of `CuFeS_(2)`?

Answer

To solve the problem, we need to find the mass of `Cu_(2)S` that contains the same mass of copper as is present in 125 lb of `CuFeS_(2)`. We will follow these steps:

### Step 1: Calculate the molar mass of `CuFeS_(2)`.

The molar mass of `CuFeS_(2)` can be calculated by adding the atomic masses of each element in the compound:

- Copper (Cu): 63.55 g/mol
- Iron (Fe): 55.85 g/mol
- Sulfur (S): 32.07 g/mol (there are 2 sulfur atoms)

\[ \text{Molar mass of } CuFeS_2 = 63.55 + 55.85 + (2 \times 32.07) = 63.55 + 55.85 + 64.14 = 183.54 \text{ g/mol} \]

### Step 2: Calculate the mass of copper in 125 lb of `CuFeS_(2)`.

First, convert 125 lb to grams (1 lb = 453.592 g):

\[ 125 \text{ lb} = 125 \times 453.592 \text{ g} = 56700 \text{ g} \]

Now, find the mass of copper in `CuFeS_(2)`:

The fraction of copper in `CuFeS_(2)` is given by:

\[ \text{Fraction of Cu} = \frac{\text{Molar mass of Cu}}{\text{Molar mass of } CuFeS_2} = \frac{63.55}{183.54} \]

Now, calculate the mass of copper in 56700 g of `CuFeS_(2)`:

\[ \text{Mass of Cu} = \text{Total mass} \times \text{Fraction of Cu} = 56700 \times \frac{63.55}{183.54} \]

Calculating this gives:

\[ \text{Mass of Cu} \approx 56700 \times 0.346 = 19667.2 \text{ g} \]

### Step 3: Calculate the molar mass of `Cu_(2)S`.

The molar mass of `Cu_(2)S` is calculated as follows:

- Copper (Cu): 63.55 g/mol (there are 2 copper atoms)
- Sulfur (S): 32.07 g/mol

\[ \text{Molar mass of } Cu_2S = (2 \times 63.55) + 32.07 = 127.10 + 32.07 = 159.17 \text{ g/mol} \]

### Step 4: Calculate the moles of copper in `Cu_(2)S`.

Since we want to find out how much `Cu_(2)S` would contain the same mass of copper (19667.2 g), we first need to calculate the moles of copper in `Cu_(2)S`:

The mass of copper in `Cu_(2)S` is:

\[ \text{Mass of Cu in } Cu_2S = 2 \times 63.55 = 127.10 \text{ g} \]

### Step 5: Set up the equation to find the mass of `Cu_(2)S`.

Let \( x \) be the mass of `Cu_(2)S` needed. The mass of copper in \( x \) grams of `Cu_(2)S` can be expressed as:

\[ \frac{127.10}{159.17} \times x = 19667.2 \]

### Step 6: Solve for \( x \).

Rearranging gives:

\[ x = \frac{19667.2 \times 159.17}{127.10} \]

Calculating this gives:

\[ x \approx \frac{3132971.54}{127.10} \approx 24609.4 \text{ g} \]

### Step 7: Convert grams to pounds.

Finally, convert grams back to pounds:

\[ x \text{ (in lb)} = \frac{24609.4}{453.592} \approx 54.2 \text{ lb} \]

### Final Answer:

The mass of `Cu_(2)S` that would contain the same mass of copper as is contained in 125 lb of `CuFeS_(2)` is approximately **54.2 lb**.

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