The percentage of `H_(2)O` in `CuSO_(4)5H_(2)O` is

Options

A. zero
B. 28.7
C. 50
D. 36

Answer

To find the percentage of water (H₂O) in copper sulfate pentahydrate (CuSO₄·5H₂O), we can follow these steps:

### Step 1: Determine the molecular weight of water (H₂O)
- The molecular weight of water can be calculated as follows:
- Hydrogen (H) has an atomic mass of approximately 1 g/mol, and there are 2 hydrogen atoms in water.
- Oxygen (O) has an atomic mass of approximately 16 g/mol.

Therefore, the molecular weight of H₂O is:
\[ \text{Molecular weight of H₂O} = (2 \times 1) + 16 = 2 + 16 = 18 \text{ g/mol} \]

### Step 2: Calculate the total weight of water in CuSO₄·5H₂O
- Since there are 5 water molecules in copper sulfate pentahydrate, the total weight of water is:
\[ \text{Weight of H₂O in CuSO₄·5H₂O} = 5 \times 18 = 90 \text{ g/mol} \]

### Step 3: Determine the molecular weight of copper sulfate pentahydrate (CuSO₄·5H₂O)
- The molecular weight of CuSO₄·5H₂O can be calculated as follows:
- Copper (Cu) has an atomic mass of approximately 64 g/mol.
- Sulfur (S) has an atomic mass of approximately 32 g/mol.
- Oxygen (O) in sulfate (SO₄) has 4 oxygen atoms, contributing \(4 \times 16 = 64\) g/mol.
- The total molecular weight of CuSO₄ is:
\[ \text{Molecular weight of CuSO₄} = 64 + 32 + 64 = 160 \text{ g/mol} \]
- Now, adding the weight of the 5 water molecules:
\[ \text{Molecular weight of CuSO₄·5H₂O} = 160 + 90 = 250 \text{ g/mol} \]

### Step 4: Calculate the percentage of water in CuSO₄·5H₂O
- The percentage of water in the compound can be calculated using the formula:
\[ \text{Percentage of H₂O} = \left( \frac{\text{Weight of H₂O}}{\text{Molecular weight of CuSO₄·5H₂O}} \right) \times 100 \]
- Substituting the values we calculated:
\[ \text{Percentage of H₂O} = \left( \frac{90}{250} \right) \times 100 = 36\% \]

### Final Answer
The percentage of H₂O in CuSO₄·5H₂O is **36%**.

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