An oxide of nitrogen contains 30.43% nitrogen. The molecular mass of the compound is 92 amu. Find the molecular formula of the given oxide.

Answer

To solve the problem of finding the molecular formula of the nitrogen oxide that contains 30.43% nitrogen and has a molecular mass of 92 amu, we can follow these steps:

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

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

### Step 2: Convert percentages to grams
Assuming we have 100 grams of the compound, we can directly convert the percentages to grams:
- Mass of Nitrogen = 30.43 g
- Mass of Oxygen = 69.57 g

### Step 3: Calculate the number of moles of each element
Next, we will calculate the number of moles of nitrogen and oxygen using their atomic masses:
- Atomic mass of Nitrogen (N) = 14 g/mol
- Atomic mass of Oxygen (O) = 16 g/mol

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

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

### Step 4: Determine the simplest mole ratio
To find the simplest ratio of moles of nitrogen to moles of oxygen, we divide both values by the smaller number of moles (2.17):

\[ \text{Ratio of Nitrogen} = \frac{2.17}{2.17} = 1 \]

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

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:
\[ \text{Empirical Formula} = NO_2 \]

### Step 6: Calculate the empirical formula mass
Now, we calculate the mass of the empirical formula (NO₂):
\[ \text{Mass of NO}_2 = 14 \text{ g/mol (N)} + 2 \times 16 \text{ g/mol (O)} = 14 + 32 = 46 \text{ g/mol} \]

### Step 7: Determine the value of N for the molecular formula
We know the molecular mass of the compound is 92 g/mol. We can find N using the formula:
\[ N = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}} = \frac{92 \text{ g/mol}}{46 \text{ g/mol}} = 2 \]

### Step 8: Write the molecular formula
Now, we can write the molecular formula by multiplying the empirical formula by N:
\[ \text{Molecular Formula} = N \times NO_2 = 2 \times NO_2 = N_2O_4 \]

### Final Answer
The molecular formula of the given oxide is:
\[ \text{Molecular Formula} = N_2O_4 \]

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