On analysis, a certain compound was found to contain iodine and oxygen in the ratio of 254:80. The formula of the compound is: (At. mass I=127,O=16)

Options

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

Answer

To find the formula of the compound containing iodine and oxygen in the ratio of 254:80, we can follow these steps:

### Step 1: Determine the mass ratio
The given mass ratio of iodine to oxygen is 254:80.

### Step 2: Convert the mass ratio to moles
To convert the mass ratio to moles, we will use the atomic masses of iodine and oxygen:
- Atomic mass of iodine (I) = 127 g/mol
- Atomic mass of oxygen (O) = 16 g/mol

Now, we can calculate the number of moles of each element:
- Moles of iodine = Mass of iodine / Atomic mass of iodine = 254 g / 127 g/mol
- Moles of oxygen = Mass of oxygen / Atomic mass of oxygen = 80 g / 16 g/mol

### Step 3: Calculate the moles
Calculating the moles:
- Moles of iodine = 254 / 127 = 2
- Moles of oxygen = 80 / 16 = 5

### Step 4: Determine the simplest mole ratio
Now we have the moles of each element:
- Moles of iodine = 2
- Moles of oxygen = 5

The simplest mole ratio of iodine to oxygen is 2:5.

### Step 5: Write the empirical formula
Based on the mole ratio, we can write the empirical formula of the compound as:
- I2O5

### Final Answer
The formula of the compound is I2O5.

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