An analysis of an oxide of nitrogen with a molecular weight of 92.02 amu has a percent composition of 69.57% oxygen and 30.43% nitrogen, what are the empirical and molecular formulas for this nitrogen oxide? Complete and balance the equation for its formation from the elements nitrogen and oxygen.
The Correct Answer and Explanation is:
Correct Answer:
- Empirical Formula: NO₂
- Molecular Formula: N₂O₄
- Balanced Formation Equation: N2(g)+2O2(g)→N2O4(g)N_2(g) + 2O_2(g) \rightarrow N_2O_4(g)
Explanation
To determine the empirical formula, we use the percent composition data provided:
- 30.43% nitrogen (N)
- 69.57% oxygen (O)
Assume we have 100 grams of the compound. This gives us:
- 30.43 g N → 30.43 g14.01 g/mol≈2.17 mol N\frac{30.43 \text{ g}}{14.01 \text{ g/mol}} ≈ 2.17 \text{ mol N}
- 69.57 g O → 69.57 g16.00 g/mol≈4.35 mol O\frac{69.57 \text{ g}}{16.00 \text{ g/mol}} ≈ 4.35 \text{ mol O}
Now, find the simplest whole-number ratio: 2.172.17:4.352.17=1:2\frac{2.17}{2.17} : \frac{4.35}{2.17} = 1 : 2
So, the empirical formula is NO₂.
Next, we calculate the molar mass of the empirical formula NO₂: 14.01+(2×16.00)=46.01 g/mol14.01 + (2 × 16.00) = 46.01 \text{ g/mol}
Now compare this with the given molecular weight of the compound: 92.02 amu 92.0246.01≈2\frac{92.02}{46.01} ≈ 2
So the molecular formula is 2 × (NO₂) = N₂O₄
To write the balanced formation reaction of N₂O₄ from its elements:
- Nitrogen gas: N₂(g)
- Oxygen gas: O₂(g)
In N₂O₄, there are 2 nitrogen atoms and 4 oxygen atoms per molecule. To form one molecule of N₂O₄: N2(g)+2O2(g)→N2O4(g)N_2(g) + 2O_2(g) \rightarrow N_2O_4(g)
This is already balanced with respect to both nitrogen and oxygen atoms.
Summary:
- Empirical Formula: NO₂
- Molecular Formula: N₂O₄
- Balanced Formation Equation: N2(g)+2O2(g)→N2O4(g)N_2(g) + 2O_2(g) \rightarrow N_2O_4(g)
