PASSAGE-II : Empirical formula is the simplest formula of the compound which gives the atomic ratio of various elements present in one molecule of the compound. However, the molecular formula of the compound gives the number of atoms of various elements present in one molecule of the compound. Molacular formula = (Empirical formula) `xxn`
implies `n=("Molecular mass")/("Empirical formula mass")`
A compound may have same empirical and molecular formula. Both these formulae are calculated using percentage composition of constituent elements.
An oxide of iodine (Atomic mass of I = 127 u) contains 25.4 g of iodine and 8 g of oxygen. Its formula could be :

Options

A. `I_(2)O_(3)`
B. `I_(2)O`
C. `I_(2)O_(5)`
D. `I_(2)O_(7)`

Explanation

To determine the formula of the oxide of iodine based on the given masses of iodine and oxygen, we can follow these steps:

### Step 1: Calculate the moles of iodine and oxygen
To find the number of moles of each element, we use the formula:
\[ \text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \]

- For iodine (I):
- Mass of iodine = 25.4 g
- Molar mass of iodine = 127 g/mol
\[ \text{Moles of I} = \frac{25.4 \, \text{g}}{127 \, \text{g/mol}} \approx 0.2 \, \text{mol} \]

- For oxygen (O):
- Mass of oxygen = 8 g
- Molar mass of oxygen = 16 g/mol
\[ \text{Moles of O} = \frac{8 \, \text{g}}{16 \, \text{g/mol}} = 0.5 \, \text{mol} \]

### Step 2: Determine the mole ratio
Next, we need to find the simplest mole ratio between iodine and oxygen. We do this by dividing the number of moles of each element by the smallest number of moles calculated.

- The smallest number of moles is 0.2 (for iodine).
\[ \text{Ratio of I} = \frac{0.2}{0.2} = 1 \]
\[ \text{Ratio of O} = \frac{0.5}{0.2} = 2.5 \]

### Step 3: Convert to whole numbers
Since the ratio for oxygen is 2.5, we multiply both ratios by 2 to convert them into whole numbers:
\[ \text{I} : \text{O} = 1 \times 2 : 2.5 \times 2 = 2 : 5 \]

### Step 4: Write the empirical formula
From the mole ratio, we can write the empirical formula of the compound. The empirical formula is represented as:
\[ \text{I}_2\text{O}_5 \]

### Conclusion
The formula of the oxide of iodine based on the given masses is:
\[ \text{I}_2\text{O}_5 \]

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