To demonstrate how the results you provided illustrate the law of constant proportions, we need to analyze the chemical reactions and the mass relationships involved. The law of constant proportions states that a chemical compound always contains its component elements in fixed ratio by mass, regardless of the source or how it was prepared. Let's break down the data step by step.
Step 1: Analyzing the Reduction of Cupric Oxide
Initially, we have 5.06 g of cupric oxide (CuO) that, when reduced by hydrogen, yields 4.04 g of metallic copper (Cu). This reaction can be represented as:
From this reaction, we can determine the mass of oxygen that was released during the reduction. The mass of oxygen in the cupric oxide can be calculated as follows:
- Mass of CuO = 5.06 g
- Mass of Cu = 4.04 g
- Mass of O = Mass of CuO - Mass of Cu = 5.06 g - 4.04 g = 1.02 g
Step 2: Establishing the Mass Ratio
Now, we can find the mass ratio of copper to oxygen in cupric oxide:
- Mass ratio of Cu to O = Mass of Cu / Mass of O = 4.04 g / 1.02 g ≈ 3.96
This ratio indicates that for every part of oxygen, there are approximately 3.96 parts of copper in cupric oxide, which is consistent with the formula of CuO, where the ratio of copper to oxygen is approximately 4:1 by mass.
Step 3: Analyzing the Reaction with Nitric Acid
Next, we look at the dissolution of 1.3 g of metallic copper in nitric acid, which produces cupric oxide. The reaction can be represented as:
- Cu + 2HNO₃ → CuO + 2NO₂ + H₂O
After the reaction, we obtain 1.63 g of cupric oxide. This means that the mass of copper that was converted into cupric oxide can be calculated as follows:
- Mass of CuO produced = 1.63 g
Step 4: Confirming the Mass Relationships
To confirm the mass of copper that corresponds to the mass of cupric oxide produced, we can again analyze the mass ratio. The molar mass of copper (Cu) is about 63.55 g/mol, and that of oxygen (O) is about 16.00 g/mol. The molar mass of cupric oxide (CuO) is approximately 79.55 g/mol.
Using stoichiometry, we can find the mass of copper that would produce 1.63 g of cupric oxide:
- 1 mole of CuO (79.55 g) contains 1 mole of Cu (63.55 g).
- Mass of Cu for 1.63 g of CuO = (63.55 g / 79.55 g) * 1.63 g ≈ 1.31 g.
This calculation shows that the mass of copper produced from the reaction is consistent with the mass of copper that was initially dissolved, supporting the law of constant proportions.
Final Thoughts
In summary, both reactions demonstrate that the ratios of the masses of the elements involved remain constant. The mass ratios of copper to oxygen in cupric oxide and the mass of copper that produces cupric oxide from nitric acid are consistent with the law of constant proportions. This illustrates that regardless of the process or source, the composition of a chemical compound remains fixed, reinforcing the fundamental principles of chemistry.
