Quantitative estimation of `C,H` and extra elements (e.g., N.S.P., and halogens) is carried out by Liebig's combustion, Carius, Dumas, and Kjeldahl's method.
Ques. In the quantitative estimation of oxygen by using `I_2O_5`. The formula used is.............
Where W is the mass of `CO_2` and w is the mass of the compound.

Answer

To solve the question regarding the quantitative estimation of oxygen using iodine pentoxide (I2O5), we need to derive the formula that relates the mass of carbon dioxide (CO2) produced and the mass of the compound being analyzed.

### Step-by-Step Solution:

1. **Understanding the Components**:
- We know that when a compound containing oxygen is combusted, it produces carbon dioxide (CO2) and water (H2O).
- The question specifically asks for the estimation of oxygen using I2O5, which is a reagent that helps in the combustion process.

2. **Establishing the Relationship**:
- The mass of carbon dioxide produced (denoted as W) is directly related to the amount of oxygen in the original compound.
- In the combustion of carbon-containing compounds, each mole of CO2 produced corresponds to the presence of oxygen in the compound.

3. **Molar Mass Consideration**:
- The molar mass of CO2 is 44 g/mol (12 g/mol for carbon and 32 g/mol for two oxygen atoms).
- The mass of oxygen in one mole of CO2 is 32 g.

4. **Setting Up the Formula**:
- The percentage of oxygen in the compound can be calculated using the formula:
\[ \text{Percentage of Oxygen} = \left( \frac{\text{mass of oxygen}}{\text{mass of compound}} \right) \times 100 \]
- Since we have the mass of CO2 (W) and the mass of the compound (w), we can express the mass of oxygen in terms of W.

5. **Final Formula**:
- The formula for the percentage of oxygen using I2O5 is:
\[ \text{Percentage of Oxygen} = \left( \frac{32}{44} \times \frac{W}{w} \right) \times 100 \]
- Here, \( W \) is the mass of CO2 and \( w \) is the mass of the compound.

### Final Answer:
The formula used for the quantitative estimation of oxygen by using I2O5 is:
\[ \text{Percentage of Oxygen} = \frac{32}{44} \times \frac{W}{w} \times 100 \]