Two oxides of a metal contain `27.6%` and `30.0%` of Oxygen, respecttively. If the formula of the first be `M_(3) O_(4)`. Find that of the second.

Options

A. XO
B. `XO_(2)`
C. `X_(2) O_(5)`
D. `X_(2) O_(3)`

Answer

To solve the problem, we need to determine the formula of the second oxide based on the given information about the first oxide and the percentages of oxygen in both oxides.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- First oxide: Contains 27.6% oxygen and has the formula \( M_3O_4 \).
- Second oxide: Contains 30.0% oxygen.

2. **Calculate the Percentage of Metal in Each Oxide**:
- For the first oxide:
\[ \text{Percentage of metal} = 100\% - 27.6\% = 72.4\% \]
- For the second oxide:
\[ \text{Percentage of metal} = 100\% - 30.0\% = 70.0\% \]

3. **Determine the Molar Mass of the First Oxide**:
- The formula of the first oxide is \( M_3O_4 \).
- Let the molar mass of the metal \( M \) be \( M \) g/mol.
- The molar mass of \( M_3O_4 \) can be calculated as:
\[ \text{Molar mass of } M_3O_4 = 3M + 4 \times 16 = 3M + 64 \]

4. **Calculate the Mass of Metal in the First Oxide**:
- The mass of metal in \( M_3O_4 \) is:
\[ \text{Mass of metal} = \frac{72.4}{100} \times (3M + 64) \]

5. **Calculate the Mass of Oxygen in the First Oxide**:
- The mass of oxygen in \( M_3O_4 \) is:
\[ \text{Mass of oxygen} = \frac{27.6}{100} \times (3M + 64) \]

6. **Set Up the Ratio for the Second Oxide**:
- For the second oxide, let its formula be \( M_xO_y \).
- The mass of metal in the second oxide is:
\[ \text{Mass of metal} = \frac{70.0}{100} \times (xM + y \times 16) \]

7. **Using the Relationship Between the Two Oxides**:
- From the first oxide, we know:
\[ \frac{3M}{4} \text{ (from } M_3O_4\text{)} \]
- For the second oxide, we need to find \( x \) and \( y \) such that:
\[ \frac{xM}{y} = \frac{3M}{4} \]
- Rearranging gives:
\[ \frac{x}{y} = \frac{3}{4} \]

8. **Calculate the Ratio of Metal to Oxygen in the Second Oxide**:
- Using the percentages, we can find the ratio of metal to oxygen:
\[ \text{For the first oxide: } \frac{72.4}{27.6} = \frac{3}{4} \]
- For the second oxide:
\[ \frac{70.0}{30.0} = \frac{7}{3} \]

9. **Finding the Empirical Formula**:
- The ratio of metal to oxygen in the second oxide can be simplified to find the empirical formula.
- If we scale the ratio \( 7:3 \) by multiplying by 2, we get \( 2M:3O \), which corresponds to \( M_2O_3 \).

10. **Final Answer**:
- Therefore, the formula of the second oxide is \( M_2O_3 \).