Manganese forms non-stoichiometric oxides having the gereral formula formula `MnO_x`. The value of `x` for the compound that analyzed `64%` by mass mn :

Options

A. `1.16`
B. `1.83`
C. `2`
D. `1.93`

Answer

To find the value of \( x \) in the non-stoichiometric oxide \( \text{MnO}_x \) that is 64% manganese by mass, we can follow these steps:

### Step 1: Understand the Composition
We know that the compound has the general formula \( \text{MnO}_x \). The percentage by mass of manganese (Mn) in this compound is given as 64%.

### Step 2: Set Up the Equation
The percentage by mass of manganese can be expressed as:
\[ \text{Percentage of Mn} = \left( \frac{\text{mass of Mn}}{\text{mass of MnO}_x} \right) \times 100 \]
Given that the mass of Mn is 55 g/mol and the mass of oxygen (O) is 16 g/mol, the molar mass of the compound \( \text{MnO}_x \) can be calculated as:
\[ \text{Molar mass of } \text{MnO}_x = 55 + 16x \]

### Step 3: Substitute Values into the Equation
Substituting the known values into the percentage equation:
\[ 64 = \left( \frac{55}{55 + 16x} \right) \times 100 \]

### Step 4: Rearrange the Equation
To eliminate the fraction, we can rearrange the equation:
\[ 64(55 + 16x) = 5500 \]
Expanding this gives:
\[ 3520 + 1024x = 5500 \]

### Step 5: Solve for \( x \)
Now, isolate \( x \):
\[ 1024x = 5500 - 3520 \]
\[ 1024x = 1980 \]
\[ x = \frac{1980}{1024} \approx 1.9375 \]

### Step 6: Round to Appropriate Value
Since \( x \) represents the number of oxygen atoms in the compound, we can round \( x \) to two decimal places:
\[ x \approx 1.94 \]

### Final Answer
Thus, the value of \( x \) for the compound \( \text{MnO}_x \) is approximately \( 1.94 \).

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