1 g mixture of cuprous oxide and cupric oxide was quantitatively reduced to `0.839` g of metallic copper. What was the weight of cupric oxide in the original sample ?
`(Cu = 63. 5, O = 16)`.

Answer

To solve the problem, we need to determine the weight of cupric oxide (CuO) in a 1 g mixture of cuprous oxide (Cu2O) and cupric oxide (CuO) that was reduced to 0.839 g of metallic copper (Cu).

### Step-by-Step Solution:

1. **Define Variables**:
Let \( x \) be the weight of cupric oxide (CuO) in grams. Therefore, the weight of cuprous oxide (Cu2O) will be \( 1 - x \) grams.

2. **Molar Mass Calculation**:
- The molar mass of cupric oxide (CuO) is:
\[ \text{Molar mass of CuO} = 63.5 \, \text{(Cu)} + 16 \, \text{(O)} = 79.5 \, \text{g/mol} \]
- The molar mass of cuprous oxide (Cu2O) is:
\[ \text{Molar mass of Cu2O} = 2 \times 63.5 \, \text{(Cu)} + 16 \, \text{(O)} = 143 \, \text{g/mol} \]

3. **Calculate Moles of Each Compound**:
- The moles of cupric oxide (CuO) in the mixture is:
\[ \text{Moles of CuO} = \frac{x}{79.5} \]
- The moles of cuprous oxide (Cu2O) in the mixture is:
\[ \text{Moles of Cu2O} = \frac{1 - x}{143} \]

4. **Copper Atom Conservation**:
- Each mole of cupric oxide produces 1 mole of copper, and each mole of cuprous oxide produces 2 moles of copper. Therefore, the total moles of copper produced from the mixture is:
\[ \text{Total moles of Cu} = \frac{x}{79.5} + 2 \times \frac{1 - x}{143} \]

5. **Set Up the Equation**:
- The total moles of copper produced is equal to the moles of copper obtained from the reduction, which is:
\[ \frac{0.839}{63.5} \]
- Thus, we can set up the equation:
\[ \frac{x}{79.5} + 2 \times \frac{1 - x}{143} = \frac{0.839}{63.5} \]

6. **Solve the Equation**:
- First, calculate the right side:
\[ \frac{0.839}{63.5} \approx 0.0132 \]
- Now substituting and simplifying the left side:
\[ \frac{x}{79.5} + \frac{2(1 - x)}{143} = 0.0132 \]
- Finding a common denominator (which is \( 79.5 \times 143 \)):
\[ \frac{143x + 2(79.5)(1 - x)}{79.5 \times 143} = 0.0132 \]
- Cross-multiplying gives:
\[ 143x + 159 - 159x = 0.0132 \times (79.5 \times 143) \]
- Simplifying leads to:
\[ -16x + 159 = 0.0132 \times 11317.5 \]
- Calculate the right side:
\[ 0.0132 \times 11317.5 \approx 149.5 \]
- Thus, we have:
\[ -16x + 159 = 149.5 \]
- Rearranging gives:
\[ -16x = 149.5 - 159 \]
\[ -16x = -9.5 \]
\[ x = \frac{9.5}{16} \approx 0.59375 \, \text{g} \]

7. **Calculate Weight of Cupric Oxide**:
- Since \( x \) represents the weight of cupric oxide (CuO), we find:
\[ \text{Weight of CuO} \approx 0.59375 \, \text{g} \]

### Final Answer:
The weight of cupric oxide in the original sample is approximately **0.59375 g**.