An element X forms two oxides. Formula of the first oxide is `XO_(2)`. The first contains 50% of oxygen. If the second oxide contains 60% oxygen, the formula of the second oxide is

Options

A. `XO_(3)`
B. `X_(2)O_(3)`
C. `X_(3)O_(2)`
D. `X_(2)O`

Answer

To solve the problem, we need to find the formula of the second oxide formed by element X, given that it forms two oxides: the first oxide is \( XO_2 \) and contains 50% oxygen, while the second oxide contains 60% oxygen.

### Step-by-Step Solution:

1. **Determine the Molar Mass of the First Oxide**:
The first oxide is \( XO_2 \). The molar mass of oxygen (O) is 16 g/mol. Therefore, the molar mass of \( XO_2 \) can be calculated as:
\[ \text{Molar mass of } XO_2 = m + 2 \times 16 = m + 32 \]
where \( m \) is the molar mass of element X.

2. **Calculate the Percentage of Oxygen in the First Oxide**:
The percentage of oxygen in \( XO_2 \) is given by:
\[ \text{Percentage of O} = \frac{\text{Mass of O}}{\text{Molar mass of } XO_2} \times 100 = \frac{32}{m + 32} \times 100 \]
According to the problem, this percentage is 50%. Therefore, we can set up the equation:
\[ \frac{32}{m + 32} \times 100 = 50 \]

3. **Solve for m**:
Rearranging the equation gives:
\[ 32 \times 100 = 50(m + 32) \]
\[ 3200 = 50m + 1600 \]
\[ 50m = 3200 - 1600 = 1600 \]
\[ m = \frac{1600}{50} = 32 \]
Thus, the atomic mass of element X is 32 g/mol.

4. **Determine the Formula of the Second Oxide**:
Let the formula of the second oxide be \( X_2O_n \). The molar mass of this oxide can be expressed as:
\[ \text{Molar mass of } X_2O_n = 2m + 16n = 2 \times 32 + 16n = 64 + 16n \]

5. **Calculate the Percentage of Oxygen in the Second Oxide**:
The percentage of oxygen in \( X_2O_n \) is given by:
\[ \text{Percentage of O} = \frac{16n}{64 + 16n} \times 100 \]
According to the problem, this percentage is 60%. Therefore, we can set up the equation:
\[ \frac{16n}{64 + 16n} \times 100 = 60 \]

6. **Solve for n**:
Rearranging the equation gives:
\[ 16n \times 100 = 60(64 + 16n) \]
\[ 1600n = 3840 + 960n \]
\[ 1600n - 960n = 3840 \]
\[ 640n = 3840 \]
\[ n = \frac{3840}{640} = 6 \]

7. **Write the Formula of the Second Oxide**:
The formula of the second oxide is \( X_2O_6 \). However, we need the empirical formula, which is obtained by simplifying the ratio:
\[ X_2O_6 \rightarrow XO_3 \]

### Final Answer:
The formula of the second oxide is \( XO_3 \).