An oxide of nitrogen contains `36.8%` by weight of nitrogen. The formula of the compound is

Options

A. `N_(2)O`
B. `N_(2)O_(3)`
C. `NO`
D. `NO_(2)`

Explanation

To determine the formula of the nitrogen oxide given that it contains 36.8% by weight of nitrogen, we can follow these steps:

### Step 1: Determine the percentage of oxygen
Since the total percentage must equal 100%, we can find the percentage of oxygen in the compound:
\[ \text{Percentage of Oxygen} = 100\% - \text{Percentage of Nitrogen} = 100\% - 36.8\% = 63.2\% \]

### Step 2: Convert percentages to grams
Assuming we have 100 grams of the compound, we can convert the percentages to grams:
- Mass of Nitrogen = 36.8 grams
- Mass of Oxygen = 63.2 grams

### Step 3: Calculate the number of moles of nitrogen and oxygen
Using the atomic masses (Nitrogen = 14 g/mol, Oxygen = 16 g/mol), we can calculate the number of moles:
\[ \text{Moles of Nitrogen} = \frac{36.8 \text{ g}}{14 \text{ g/mol}} = 2.62857 \text{ moles} \]
\[ \text{Moles of Oxygen} = \frac{63.2 \text{ g}}{16 \text{ g/mol}} = 3.95 \text{ moles} \]

### Step 4: Determine the simplest mole ratio
Next, we need to find the simplest ratio of moles of nitrogen to moles of oxygen:
- Divide both values by the smallest number of moles (2.62857):
\[ \text{Ratio of Nitrogen} = \frac{2.62857}{2.62857} = 1 \]
\[ \text{Ratio of Oxygen} = \frac{3.95}{2.62857} \approx 1.5 \]

### Step 5: Convert to whole numbers
Since we cannot have fractional atoms in a chemical formula, we multiply both ratios by 2 to eliminate the fraction:
\[ \text{Nitrogen} = 1 \times 2 = 2 \]
\[ \text{Oxygen} = 1.5 \times 2 = 3 \]

### Step 6: Write the empirical formula
Thus, the empirical formula of the compound is:
\[ \text{N}_2\text{O}_3 \]

### Final Answer
The formula of the nitrogen oxide is \( \text{N}_2\text{O}_3 \).

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