To convert grams of a substance into moles, you'll use the following formula:

\[
\text{Moles} = \frac{\text{Mass in grams}}{\text{Molar mass}}
\]

First, we need to find the molar mass of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \).

### Step 1: Calculate the molar mass of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \)
- Copper (Cu): \( 63.55 \, \text{g/mol} \)
- Sulfur (S): \( 32.07 \, \text{g/mol} \)
- Oxygen (O): \( 16.00 \, \text{g/mol} \times 4 = 64.00 \, \text{g/mol} \)
- Water (H₂O): \( (2 \times 1.01 + 16.00) \, \text{g/mol} = 18.02 \, \text{g/mol} \times 5 = 90.10 \, \text{g/mol} \)

Adding these together:

\[
\text{Molar mass of } \text{CuSO}_4 \cdot 5\text{H}_2\text{O} = 63.55 + 32.07 + 64.00 + 90.10 = 249.72 \, \text{g/mol}
\]

### Step 2: Calculate the number of moles
Given:
- Mass = 499 g
- Molar mass = 249.72 g/mol

\[
\text{Moles of } \text{CuSO}_4 \cdot 5\text{H}_2\text{O} = \frac{499 \, \text{g}}{249.72 \, \text{g/mol}} \approx 2.00 \, \text{moles}
\]

So, 499 g of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) is approximately **2.00 moles**.

Would you like more details or have any questions? Here are five related questions you might find interesting:

1. How would you calculate the number of molecules in 2.00 moles of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \)?
2. How do you determine the empirical formula of a hydrated compound?
3. What is the role of water in hydrated salts like \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \)?
4. How would you find the percentage composition of water in \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \)?
5. Can you convert moles back to grams? How would you do that for 1.5 moles of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \)?

**Tip:** Understanding molar mass is crucial for solving stoichiometry problems, which involve the calculation of reactants and products in chemical reactions.

Learn how to convert 499 grams of CuSO4 · 5H2O into moles with a detailed step-by-step solution. This problem covers chemical stoichiometry and molar mass calculations, essential for understanding chemical compositions and reactions.

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