Answer :
To determine the empirical formula of the nitrogen oxide compound given that it contains 30.4% nitrogen and the rest is oxygen, let's follow the steps for empirical formula calculation:
1. Determine the percentage of each element:
- Percentage of nitrogen (N): [tex]\( 30.4\% \)[/tex]
- Since the rest of the compound is oxygen (O), the percentage of oxygen is [tex]\( 100\% - 30.4\% = 69.6\% \)[/tex].
2. Convert these percentages to masses:
- For ease of calculation, assume we have a 100 g sample of the compound.
- Mass of nitrogen in the sample: 30.4 g.
- Mass of oxygen in the sample: 69.6 g.
3. Convert the masses to moles:
- Molar mass of nitrogen (N): 14 g/mol.
- Moles of nitrogen: [tex]\( \frac{30.4 \text{ g}}{14 \text{ g/mol}} = 2.1714285714285713 \text{ mol} \)[/tex].
- Molar mass of oxygen (O): 16 g/mol.
- Moles of oxygen: [tex]\( \frac{69.6 \text{ g}}{16 \text{ g/mol}} = 4.35 \text{ mol} \)[/tex].
4. Determine the simplest whole number ratio of moles:
- Find the smallest number of moles between nitrogen and oxygen to determine the ratio:
- Smallest mole quantity: [tex]\( 2.1714285714285713 \)[/tex] (moles of nitrogen).
- Ratio of nitrogen:
[tex]\[ \frac{2.1714285714285713 \text{ mol}}{2.1714285714285713 \text{ mol}} = 1.0 \][/tex]
- Ratio of oxygen:
[tex]\[ \frac{4.35 \text{ mol}}{2.1714285714285713 \text{ mol}} = 2.0032894736842106 \approx 2 \][/tex]
This simplifies to a ratio of 1 nitrogen atom to 2 oxygen atoms.
5. Write the empirical formula:
- Using the mole ratio, the empirical formula is [tex]\( NO_2 \)[/tex].
Therefore, the empirical formula of the nitrogen oxide compound is [tex]\( NO_2 \)[/tex].
The correct answer from the given choices is [tex]\( NO_2 \)[/tex].
1. Determine the percentage of each element:
- Percentage of nitrogen (N): [tex]\( 30.4\% \)[/tex]
- Since the rest of the compound is oxygen (O), the percentage of oxygen is [tex]\( 100\% - 30.4\% = 69.6\% \)[/tex].
2. Convert these percentages to masses:
- For ease of calculation, assume we have a 100 g sample of the compound.
- Mass of nitrogen in the sample: 30.4 g.
- Mass of oxygen in the sample: 69.6 g.
3. Convert the masses to moles:
- Molar mass of nitrogen (N): 14 g/mol.
- Moles of nitrogen: [tex]\( \frac{30.4 \text{ g}}{14 \text{ g/mol}} = 2.1714285714285713 \text{ mol} \)[/tex].
- Molar mass of oxygen (O): 16 g/mol.
- Moles of oxygen: [tex]\( \frac{69.6 \text{ g}}{16 \text{ g/mol}} = 4.35 \text{ mol} \)[/tex].
4. Determine the simplest whole number ratio of moles:
- Find the smallest number of moles between nitrogen and oxygen to determine the ratio:
- Smallest mole quantity: [tex]\( 2.1714285714285713 \)[/tex] (moles of nitrogen).
- Ratio of nitrogen:
[tex]\[ \frac{2.1714285714285713 \text{ mol}}{2.1714285714285713 \text{ mol}} = 1.0 \][/tex]
- Ratio of oxygen:
[tex]\[ \frac{4.35 \text{ mol}}{2.1714285714285713 \text{ mol}} = 2.0032894736842106 \approx 2 \][/tex]
This simplifies to a ratio of 1 nitrogen atom to 2 oxygen atoms.
5. Write the empirical formula:
- Using the mole ratio, the empirical formula is [tex]\( NO_2 \)[/tex].
Therefore, the empirical formula of the nitrogen oxide compound is [tex]\( NO_2 \)[/tex].
The correct answer from the given choices is [tex]\( NO_2 \)[/tex].
