To find formula of compound composed of A & B which is given by `A_(X)B_(Y)` ,it is strongly heated in oxygen as per reaction
`A_(X)B_(Y) + O_(2) to AO` + oxide of B
If 2.5 gm of `A_(X)B_(Y)` on oxidation gives 3 gm oxide of A, Find empirical formula of `A_(X)B_(Y)`,
[Atomic mass of A = 24 & B = 14 ]

Answer

To find the empirical formula of the compound \( A_XB_Y \) based on the information provided, we will follow these steps:

### Step 1: Determine the mass of A in the oxide
We know that 2.5 g of \( A_XB_Y \) produces 3 g of the oxide of A, which is \( AO \). The molar mass of \( AO \) can be calculated as follows:
- Atomic mass of A = 24 g/mol
- Atomic mass of O = 16 g/mol
- Molar mass of \( AO = 24 + 16 = 40 \) g/mol

### Step 2: Calculate the moles of \( AO \) produced
Using the mass of \( AO \):
\[ \text{Moles of } AO = \frac{\text{mass}}{\text{molar mass}} = \frac{3 \text{ g}}{40 \text{ g/mol}} = 0.075 \text{ mol} \]

### Step 3: Relate moles of A in \( A_XB_Y \) to moles of \( AO \)
From the reaction, we know that the moles of A in \( A_XB_Y \) will equal the moles of A in \( AO \):
\[ \text{Moles of A in } A_XB_Y = \text{Moles of } AO = 0.075 \text{ mol} \]
Since \( A_XB_Y \) contains \( x \) moles of A, we can express this as:
\[ \frac{2.5 \text{ g}}{(24x + 14y)} \times x = 0.075 \]

### Step 4: Set up the equation
From the previous step, we can rearrange to find:
\[ 2.5x = 0.075(24x + 14y) \]
Expanding this gives:
\[ 2.5x = 1.8x + 1.05y \]
Rearranging leads to:
\[ 2.5x - 1.8x = 1.05y \implies 0.7x = 1.05y \]

### Step 5: Simplify the equation
Dividing through by 0.7 gives:
\[ x = \frac{1.05}{0.7}y = 1.5y \]

### Step 6: Relate x and y
To express \( x \) and \( y \) in whole numbers, we can multiply both sides by 2:
\[ 2x = 3y \]
This implies:
\[ \frac{x}{y} = \frac{3}{2} \]

### Step 7: Assign values to x and y
From \( x = 3 \) and \( y = 2 \), we find:
\[ A_XB_Y = A_3B_2 \]

### Conclusion
The empirical formula of the compound \( A_XB_Y \) is \( A_3B_2 \).

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