Composition of the Oxides following the Law of Variable Proportions

The Law of Variable Proportions, also known as the Law of Multiple Proportions, states that when two elements combine to form different compounds, the mass ratio of one element to the fixed mass of the other element will always be a small whole number. In this case, we are given two different oxides of a metal, and we need to determine their composition.

1. Given Information:
- 1 gram of Oxide A in hydrogen gas gave 0.799 grams of the metal.
- 1 gram of Oxide B in hydrogen gas gave 0.888 grams of the metal.

2. Let's assume:
- The metal in both oxides is represented by the symbol M.
- The mass of oxygen in Oxide A is represented by x grams.
- The mass of oxygen in Oxide B is represented by y grams.

3. Determining the Composition:
- According to the Law of Variable Proportions, the mass ratio of M to oxygen in Oxide A should be a small whole number.
- Similarly, the mass ratio of M to oxygen in Oxide B should also be a small whole number.

4. Calculation:
- In Oxide A, the mass of M is 0.799 grams.
- In Oxide A, the mass of oxygen is 1 gram - 0.799 grams = 0.201 grams.
- Therefore, the mass ratio of M to oxygen in Oxide A is 0.799 grams / 0.201 grams = 3.98 ≈ 4.

- In Oxide B, the mass of M is 0.888 grams.
- In Oxide B, the mass of oxygen is 1 gram - 0.888 grams = 0.112 grams.
- Therefore, the mass ratio of M to oxygen in Oxide B is 0.888 grams / 0.112 grams = 7.9 ≈ 8.

5. Conclusion:
- The mass ratio of M to oxygen in Oxide A is approximately 4.
- The mass ratio of M to oxygen in Oxide B is approximately 8.

From this information, we can infer that the composition of Oxide A is M4O and the composition of Oxide B is M8O.

By following the Law of Variable Proportions and using the given data, we were able to determine the composition of the two oxides. The mass ratios of the metal to oxygen in both oxides were found to be whole numbers, confirming the validity of the law.