This law was proposed by Dalton in 1803 . According to this law , if two elements combune to form more than one compounds ,the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole number.
For exemple, hydrogen combines with oxygen to form two compounds ,namely, water and hydrogen peroxide.
Here, the masses of oxygen (i.e, 16g and 32g) which combine with a fixed mass of hydrogen (2g) bear a simple raio, i.e, 16 :32 or 1 :2 .
An element forms two oxides .In one oxide ,one gram of the oxide containd 0.5g of the element. In another oxide, 4g of the oxide contains 0.8g of the element . Show that these data confirm the law of multiple proportion.

Answer

To show that the data confirms Dalton's law of multiple proportions, we will analyze the given information step by step.

### Step 1: Understand the given data
We have two oxides of an element (let's call it M). The data provided is:
- In the first oxide, 1 gram of oxide contains 0.5 grams of element M.
- In the second oxide, 4 grams of oxide contains 0.8 grams of element M.

### Step 2: Calculate the mass of oxygen in each oxide
For the first oxide:
- Total mass of oxide = 1 gram
- Mass of element M = 0.5 grams
- Therefore, mass of oxygen = Total mass of oxide - Mass of element M
\[ \text{Mass of oxygen} = 1 \, \text{g} - 0.5 \, \text{g} = 0.5 \, \text{g} \]

For the second oxide:
- Total mass of oxide = 4 grams
- Mass of element M = 0.8 grams
- Therefore, mass of oxygen = Total mass of oxide - Mass of element M
\[ \text{Mass of oxygen} = 4 \, \text{g} - 0.8 \, \text{g} = 3.2 \, \text{g} \]

### Step 3: Write the ratios of the masses of oxygen
Now we have:
- In the first oxide: Mass of M = 0.5 g, Mass of O = 0.5 g
- In the second oxide: Mass of M = 0.8 g, Mass of O = 3.2 g

To find the ratio of the masses of oxygen that combine with a fixed mass of element M, we will take the mass of M from both oxides as a reference.

### Step 4: Normalize the mass of element M
To compare the two oxides, we need to express the masses of oxygen in terms of the same mass of element M.

For the first oxide, we can consider the mass of M as 0.5 g:
- Mass of O in the first oxide = 0.5 g

For the second oxide, we need to express the mass of oxygen in terms of 0.5 g of M:
- Mass of M in the second oxide = 0.8 g
- To find the mass of oxygen that would combine with 0.5 g of M, we can set up a proportion:
\[ \frac{0.8 \, \text{g of M}}{3.2 \, \text{g of O}} = \frac{0.5 \, \text{g of M}}{x \, \text{g of O}} \]
Cross-multiplying gives:
\[ 0.8x = 3.2 \times 0.5 \]
\[ 0.8x = 1.6 \]
\[ x = \frac{1.6}{0.8} = 2 \, \text{g of O} \]

### Step 5: Establish the ratio of the masses of oxygen
Now we have:
- From the first oxide: 0.5 g of O
- From the second oxide: 2 g of O

Now we can write the ratio of the masses of oxygen:
\[ \text{Ratio of masses of O} = \frac{0.5 \, \text{g}}{2 \, \text{g}} = \frac{1}{4} \]

### Conclusion
The ratio of the masses of oxygen that combine with a fixed mass of element M (0.5 g) in the two oxides is 1:4. This confirms Dalton's law of multiple proportions, as the masses of oxygen are in the ratio of small whole numbers.

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