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 situation step by step.

### Step-by-Step Solution:

1. **Identify the Compounds**:
- Copper forms two oxides: cuprous oxide (Cu2O) and cupric oxide (CuO).
- In cuprous oxide (Cu2O), there are 2 atoms of copper for every 1 atom of oxygen.
- In cupric oxide (CuO), there is 1 atom of copper for every 1 atom of oxygen.

2. **Determine the Atomic Weights**:
- The atomic weight of oxygen (O) is approximately 16 grams.
- The atomic weight of copper (Cu) is approximately 63.5 grams.

3. **Calculate the Weight Ratios**:
- For cuprous oxide (Cu2O):
- Weight of Cu = 2 × 63.5 g = 127 g
- Weight of O = 16 g
- For cupric oxide (CuO):
- Weight of Cu = 1 × 63.5 g = 63.5 g
- Weight of O = 16 g

4. **Establish the Ratios**:
- In cuprous oxide (Cu2O), the ratio of weight of Cu to weight of O is 127 g : 16 g.
- In cupric oxide (CuO), the ratio of weight of Cu to weight of O is 63.5 g : 16 g.

5. **Simplify the Ratios**:
- The ratio of weights of Cu in Cu2O to Cu in CuO can be expressed as:
- 127 g (Cu2O) : 63.5 g (CuO) = 2 : 1 (after simplifying the ratio by dividing both by 63.5).

6. **Conclusion**:
- The weights of copper in the two oxides (Cu2O and CuO) demonstrate the law of multiple proportions. This law states that when two elements combine to form more than one compound, the weights of one element that combine with a fixed weight of the other element are in a ratio of small whole numbers.