Cu forms two oxides cuprous and cupric oxides, which law can be proved by the weights of Cu and O?

Options

A. Constant composition
B. Multiple proportions
C. Reciprocal proportions
D. Definite proportions

Answer

To determine which law can be proved by the weights of copper (Cu) and oxygen (O) in the formation of cuprous (Cu2O) and cupric (CuO) oxides, we can analyze the mass ratios of the elements involved in the compounds.

### Step-by-Step Solution:

1. **Identify the Oxides**:
- Cuprous oxide (Cu2O)
- Cupric oxide (CuO)

2. **Determine the Molar Masses**:
- Molar mass of Copper (Cu) = 63.5 g/mol
- Molar mass of Oxygen (O) = 16 g/mol

3. **Write the Chemical Equations**:
- For cuprous oxide:
\[ 2 \text{Cu} + \text{O}_2 \rightarrow 2 \text{Cu}_2\text{O} \]
- For cupric oxide:
\[ 2 \text{Cu} + \text{O}_2 \rightarrow 2 \text{CuO} \]

4. **Calculate the Mass Ratios**:
- For cuprous oxide (Cu2O):
- Mass of Cu in Cu2O = 2 × 63.5 g = 127 g
- Mass of O in Cu2O = 16 g
- Ratio of Cu to O in Cu2O = 127 g : 16 g

- For cupric oxide (CuO):
- Mass of Cu in CuO = 63.5 g
- Mass of O in CuO = 16 g
- Ratio of Cu to O in CuO = 63.5 g : 16 g

5. **Simplify the Ratios**:
- For cuprous oxide (Cu2O):
- Ratio = 127 g / 16 g = 7.9375 (approximately 8)
- For cupric oxide (CuO):
- Ratio = 63.5 g / 16 g = 3.96875 (approximately 4)

6. **Establish the Whole Number Ratio**:
- The ratio of the masses of copper in the two oxides can be expressed as:
- Cu2O : CuO = 127 : 63.5
- This simplifies to a ratio of 2:1 for the copper in both oxides when considering the fixed mass of oxygen.

7. **Conclusion**:
- The law that can be proved by the weights of Cu and O in these oxides is the **Law of Multiple Proportions**. This law states that when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in a ratio of small whole numbers.

### Final Answer:
The law that can be proved by the weights of Cu and O in cuprous and cupric oxides is the **Law of Multiple Proportions**.