`A` mixture of `CuSO_(4).5H_(2)O` and `MgSO_(4).7H_(2)O` was heated until all the water was driven off If `5.0g` of mixture gave `3g` of anhydrous salts, what was the percentage by mass of `CuSO_(4).5H_(2)O` in the original mixture :

Explanation

To solve the problem step by step, we will follow the outlined process to find the percentage by mass of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) in the original mixture.

### Step 1: Determine the mass of the mixture and anhydrous salts
- Given:
- Mass of the mixture = \( 5.0 \, \text{g} \)
- Mass of anhydrous salts = \( 3.0 \, \text{g} \)

### Step 2: Calculate the mass of water lost
- The mass of water lost during heating can be calculated as:
\[ \text{Mass of water} = \text{Mass of mixture} - \text{Mass of anhydrous salts} = 5.0 \, \text{g} - 3.0 \, \text{g} = 2.0 \, \text{g} \]

### Step 3: Write the equations for the components
- Let:
- Mass of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) in the mixture = \( x \, \text{g} \)
- Mass of \( \text{MgSO}_4 \cdot 7\text{H}_2\text{O} \) in the mixture = \( 5.0 - x \, \text{g} \)

### Step 4: Calculate the moles of water from each salt
- Molar mass of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \):
\[ \text{Molar mass} = 63.5 + 32 + 4 \times 16 + 5 \times 18 = 250 \, \text{g/mol} \]
- Molar mass of \( \text{MgSO}_4 \cdot 7\text{H}_2\text{O} \):
\[ \text{Molar mass} = 24.3 + 32 + 4 \times 16 + 7 \times 18 = 246 \, \text{g/mol} \]
- Moles of water from \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \):
\[ \text{Moles of water} = \frac{5}{250} \times x \]
- Moles of water from \( \text{MgSO}_4 \cdot 7\text{H}_2\text{O} \):
\[ \text{Moles of water} = \frac{7}{246} \times (5 - x) \]

### Step 5: Set up the equation for total moles of water
- Total moles of water lost = \( \frac{2.0}{18} \)
- Therefore, we can set up the equation:
\[ \frac{5}{250}x + \frac{7}{246}(5 - x) = \frac{2}{18} \]

### Step 6: Solve the equation
1. Multiply through by the least common multiple to eliminate denominators.
2. Rearranging and simplifying will yield a linear equation in terms of \( x \).
3. Solve for \( x \) to find the mass of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \).

### Step 7: Calculate the percentage by mass
- Once \( x \) is found, the percentage by mass of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) in the original mixture is calculated as:
\[ \text{Percentage} = \left( \frac{x}{5.0} \right) \times 100 \]

### Final Calculation
- After performing the calculations, we find \( x \approx 3.68 \, \text{g} \).
- Thus, the percentage is:
\[ \text{Percentage} = \left( \frac{3.68}{5.0} \right) \times 100 \approx 73.6\% \]

### Conclusion
- The percentage by mass of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) in the original mixture is approximately \( 74\% \).