A compound contains 0.130 g of nitrogen and 0.370 g of oxygen. What is its empirical formula, using N = 14.007 and O = 16.00 g/mol?

1. Convert nitrogen to moles: 0.130 g / 14.007 g/mol = 0.009281 mol N. 2. Convert oxygen to moles: 0.370 g / 16.00 g/mol = 0.023125 mol O. 3. Divide both by the smaller amount, 0.009281 mol. 4. Nitrogen gives 1.000 and oxygen gives 0.023125 / 0.009281 = 2.492, which rounds to 2.5. 5. The tentative formula is NO2.5, so a decimal subscript still remains. 6. Multiplying every subscript by 2 gives N2O5. 7. Truncating 2.5 to 2 gives NO2 and rounding to 3 gives NO3, both of which change the atom ratio. _Source: OpenStax Chemistry (CC BY 4.0), Ch 3 "Composition of Substances and Solutions", section 3.2 Determining Empirical and Molecular Formulas_