A sample of copper sulphate pentahydrate, `CuSO_(4). 5H_2O` contains `3.782g` of `Cu`. How many grams of oxygen are in this sample?

Options

A. 0.952g
B. 3.809g
C. 4.761g
D. 8.570g

Answer

To find the grams of oxygen in a sample of copper sulfate pentahydrate (CuSO₄·5H₂O) that contains 3.782 g of copper (Cu), we can follow these steps:

### Step 1: Determine the molar mass of copper (Cu).
The molar mass of copper (Cu) is approximately 63.55 g/mol.

### Step 2: Calculate the number of moles of copper in the sample.
Using the formula:
\[ \text{Number of moles of Cu} = \frac{\text{mass of Cu}}{\text{molar mass of Cu}} \]
Substituting the values:
\[ \text{Number of moles of Cu} = \frac{3.782 \, \text{g}}{63.55 \, \text{g/mol}} \approx 0.0595 \, \text{mol} \]

### Step 3: Determine the composition of copper sulfate pentahydrate.
In one mole of CuSO₄·5H₂O, there are 4 moles of oxygen from the sulfate (SO₄²⁻) and 5 moles of oxygen from the water (H₂O). Therefore, the total number of moles of oxygen in one mole of CuSO₄·5H₂O is:
\[ \text{Total moles of oxygen} = 4 + 5 = 9 \, \text{mol} \]

### Step 4: Calculate the mass of oxygen corresponding to the moles of copper.
We know that 1 mole of Cu corresponds to 9 moles of oxygen. Therefore, the number of moles of oxygen corresponding to the moles of copper calculated in Step 2 is:
\[ \text{Number of moles of O} = 0.0595 \, \text{mol Cu} \times \frac{9 \, \text{mol O}}{1 \, \text{mol Cu}} \approx 0.5355 \, \text{mol O} \]

### Step 5: Calculate the mass of oxygen.
Using the molar mass of oxygen (O), which is approximately 16 g/mol:
\[ \text{Mass of O} = \text{Number of moles of O} \times \text{molar mass of O} \]
Substituting the values:
\[ \text{Mass of O} = 0.5355 \, \text{mol} \times 16 \, \text{g/mol} \approx 8.568 \, \text{g} \]

### Final Answer:
The mass of oxygen in the sample is approximately **8.57 g**.

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