An oxide of nitrogen contains 30.43% of nitrogen. The molecular weight of the compound is equal to 92 a.m.u. Calculate the molecular formula of the compound.

Explanation

To find the molecular formula of the nitrogen oxide compound, we will follow these steps:

### Step 1: Determine the percentage of oxygen in the compound.
Given that the compound contains 30.43% nitrogen, we can find the percentage of oxygen by subtracting the nitrogen percentage from 100%.

\[ \text{Percentage of Oxygen} = 100\% - 30.43\% = 69.57\% \]

### Step 2: Assume a sample mass.
For easier calculations, we assume a sample mass of 100 grams. This assumption allows us to directly convert percentages to grams.

\[ \text{Mass of Nitrogen} = 30.43 \text{ grams} \]
\[ \text{Mass of Oxygen} = 69.57 \text{ grams} \]

### Step 3: Calculate the number of moles of nitrogen and oxygen.
To find the number of moles, we divide the mass of each element by its atomic weight.

- Atomic weight of nitrogen (N) = 14 g/mol
- Atomic weight of oxygen (O) = 16 g/mol

Calculating moles of nitrogen:
\[ \text{Moles of Nitrogen} = \frac{30.43 \text{ g}}{14 \text{ g/mol}} \approx 2.17 \text{ moles} \]

Calculating moles of oxygen:
\[ \text{Moles of Oxygen} = \frac{69.57 \text{ g}}{16 \text{ g/mol}} \approx 4.35 \text{ moles} \]

### Step 4: Determine the simplest mole ratio.
Next, we will find the simplest ratio of moles by dividing each by the smallest number of moles calculated.

\[ \text{Ratio of Nitrogen} = \frac{2.17}{2.17} = 1 \]
\[ \text{Ratio of Oxygen} = \frac{4.35}{2.17} \approx 2 \]

Thus, the simplest mole ratio of nitrogen to oxygen is 1:2.

### Step 5: Write the empirical formula.
From the mole ratio, we can write the empirical formula of the compound:
\[ \text{Empirical Formula} = \text{NO}_2 \]

### Step 6: Calculate the empirical mass.
Now, we calculate the empirical mass of NO₂:
\[ \text{Empirical Mass} = (1 \times 14) + (2 \times 16) = 14 + 32 = 46 \text{ g/mol} \]

### Step 7: Determine the molecular formula.
Given that the molecular weight of the compound is 92 g/mol, we can find the ratio of the molecular mass to the empirical mass:
\[ N = \frac{\text{Molecular Mass}}{\text{Empirical Mass}} = \frac{92 \text{ g/mol}}{46 \text{ g/mol}} = 2 \]

### Step 8: Write the molecular formula.
To find the molecular formula, we multiply the subscripts in the empirical formula (NO₂) by N:
\[ \text{Molecular Formula} = N \times \text{Empirical Formula} = 2 \times \text{NO}_2 = \text{N}_2\text{O}_4 \]

Thus, the molecular formula of the compound is:
\[ \text{N}_2\text{O}_4 \]

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