Two oxides of a metal contains 50% and 40% of a metal respectively .The formula of the first oxide is MO. Then the formula of the second oxide is :-

Options

A. `MO_(2)`
B. `M_(2)O_(3)`
C. `M_(2)O`
D. `M_(2)O_(5)`

Answer

To solve the problem step by step, we will analyze the given information about the two oxides of a metal.

### Step 1: Determine the molecular weight of the metal from the first oxide (MO).
- The first oxide contains 50% metal and 50% oxygen.
- The formula of the first oxide is given as MO, where M is the metal and O is oxygen.
- The molar mass of oxygen (O) is 16 g/mol.

Using the percentage composition:
\[ \frac{x}{x + 16} \times 100 = 50 \]
where \(x\) is the molar mass of the metal.

### Step 2: Set up the equation and solve for \(x\).
\[ \frac{x}{x + 16} = \frac{50}{100} = \frac{1}{2} \]
Cross-multiplying gives:
\[ 2x = x + 16 \]
Subtracting \(x\) from both sides:
\[ x = 16 \text{ g/mol} \]
Thus, the molar mass of the metal (M) is 16 g/mol.

### Step 3: Analyze the second oxide.
- The second oxide contains 40% metal and 60% oxygen.
- Let's assume we have 100 g of the second oxide for easier calculation.
- Therefore, the mass of the metal is 40 g and the mass of oxygen is 60 g.

### Step 4: Calculate the moles of metal and oxygen in the second oxide.
- Moles of metal (M):
\[ \text{Moles of M} = \frac{40 \text{ g}}{16 \text{ g/mol}} = 2.5 \text{ moles} \]

- Moles of oxygen (O):
\[ \text{Moles of O} = \frac{60 \text{ g}}{16 \text{ g/mol}} = 3.75 \text{ moles} \]

### Step 5: Determine the mole ratio of metal to oxygen.
- The mole ratio of metal to oxygen is:
\[ \frac{2.5}{3.75} = \frac{2.5 \div 2.5}{3.75 \div 2.5} = \frac{1}{1.5} = \frac{2}{3} \]

### Step 6: Write the empirical formula based on the mole ratio.
- The ratio of metal (M) to oxygen (O) is 2:3.
- Therefore, the formula for the second oxide is \(M_2O_3\).

### Final Answer:
The formula of the second oxide is \(M_2O_3\).

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