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

Options

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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the molar mass of metal X from the first oxide (XO₂).
We know that the first oxide, XO₂, contains 50% oxygen by mass.

Let the molar mass of metal X be \( M_X \).

The molar mass of oxygen (O) is approximately 16 g/mol. Since there are 2 oxygen atoms in XO₂, the total mass of oxygen in the oxide is:
\[ \text{Mass of O in XO₂} = 2 \times 16 = 32 \text{ g/mol} \]

According to the problem, the percentage of oxygen in XO₂ is given by:
\[ \frac{\text{Mass of O}}{\text{Mass of XO₂}} \times 100 = 50\% \]

This can be expressed as:
\[ \frac{32}{32 + M_X} \times 100 = 50 \]

### Step 2: Set up the equation and solve for \( M_X \).
From the equation above, we can simplify:
\[ \frac{32}{32 + M_X} = 0.5 \]

Cross-multiplying gives:
\[ 32 = 0.5(32 + M_X) \]

Expanding this:
\[ 32 = 16 + 0.5M_X \]

Subtracting 16 from both sides:
\[ 16 = 0.5M_X \]

Now, multiplying both sides by 2:
\[ M_X = 32 \text{ g/mol} \]

### Step 3: Analyze the second oxide.
Now, we have the molar mass of metal X, which is 32 g/mol.

The second oxide contains 60% oxygen. Let’s denote the formula of the second oxide as \( XO_n \), where \( n \) is the number of oxygen atoms.

The mass of oxygen in this oxide can be expressed as:
\[ \text{Mass of O in XO_n} = n \times 16 \text{ g/mol} \]

The percentage of oxygen in the second oxide can be expressed as:
\[ \frac{n \times 16}{n \times 16 + 32} = 0.6 \]

### Step 4: Set up the equation for the second oxide.
Cross-multiplying gives:
\[ n \times 16 = 0.6(n \times 16 + 32) \]

Expanding this:
\[ n \times 16 = 0.6n \times 16 + 19.2 \]

Rearranging gives:
\[ n \times 16 - 0.6n \times 16 = 19.2 \]

Factoring out \( n \):
\[ n \times (16 - 0.6 \times 16) = 19.2 \]

Calculating \( 16 - 0.6 \times 16 \):
\[ 16 - 9.6 = 6.4 \]

Thus, we have:
\[ n \times 6.4 = 19.2 \]

### Step 5: Solve for \( n \).
Dividing both sides by 6.4:
\[ n = \frac{19.2}{6.4} = 3 \]

### Conclusion: Determine the formula of the second oxide.
Since \( n = 3 \), the formula of the second oxide is:
\[ XO_3 \]

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

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