Find empirical formula of the compound if M=68%
(atomic mass=34) and remaining 32% oxygen.

Options

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

Explanation

To find the empirical formula of the compound given the percentages of the elements, we can follow these steps:

### Step 1: Identify the elements and their percentages
We have:
- Metal (M): 68%
- Oxygen (O): 32%

### Step 2: Convert percentages to grams
Assuming we have 100 grams of the compound:
- Mass of M = 68 grams
- Mass of O = 32 grams

### Step 3: Calculate the number of moles of each element
To find the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

For Metal (M):
- Molar mass of M = 34 g/mol
\[ \text{Moles of M} = \frac{68 \text{ g}}{34 \text{ g/mol}} = 2 \text{ moles} \]

For Oxygen (O):
- Molar mass of O = 16 g/mol
\[ \text{Moles of O} = \frac{32 \text{ g}}{16 \text{ g/mol}} = 2 \text{ moles} \]

### Step 4: Determine the simplest mole ratio
Now, we have:
- Moles of M = 2
- Moles of O = 2

To find the simplest ratio, we divide both by the smallest number of moles:
\[ \text{Ratio of M to O} = \frac{2}{2} : \frac{2}{2} = 1 : 1 \]

### Step 5: Write the empirical formula
From the mole ratio, we can write the empirical formula:
\[ \text{Empirical formula} = \text{MO} \]

### Conclusion
The empirical formula of the compound is **MO**.

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