Empirical formula from mass data: A compound contains nitrogen and oxygen in masses of 28 g and 80 g, respectively. What is its formula?

Introduction / Context:
Deriving a chemical formula from elemental mass data is a basic but essential skill for process chemists and engineers. The method uses molar masses to convert mass to moles, then simplifies to the smallest whole-number ratio.

Given Data / Assumptions:

• Mass of nitrogen = 28 g; atomic mass N ≈ 14 g/mol.
• Mass of oxygen = 80 g; atomic mass O ≈ 16 g/mol.
• Assume the given masses correspond to one formula unit ratio (empirical composition).

Concept / Approach:
Compute moles of each element, divide by the smallest mole count to obtain a whole-number ratio, and map to the simplest empirical formula. If the empirical formula matches a known molecular formula under typical valences, it is likely the molecular formula as well (nitrogen pentoxide is a well-known compound).

Step-by-Step Solution:
n_N = 28 g / 14 g·mol^-1 = 2.00 mol.n_O = 80 g / 16 g·mol^-1 = 5.00 mol.Mole ratio N:O = 2:5 → empirical formula N2O5.

Verification / Alternative check:
Check valence consistency: nitrogen can reach +5 in N2O5; oxygen is −2, satisfying charge neutrality with known oxide N2O5.

Why Other Options Are Wrong:
N2O4 implies a 1:2 N:O ratio; N2O3 implies 2:3; NO2 is 1:2; none match 2:5.

Common Pitfalls:
Forgetting to reduce to the simplest whole-number ratio; rounding errors when mole ratios are near halves or thirds.

Final Answer:
N2O5