A compound is 49.47% C, 5.201% H, 28.84% N and 16.48% O by mass, with a molecular mass of 194.2 amu. What is its molecular formula? Use C = 12.01, H = 1.008, N = 14.007, O = 16.00.

1. Take a 100 g sample so each percentage becomes a mass in grams. 2. Carbon gives 49.47 g / 12.01 g/mol = 4.119 mol, and hydrogen gives 5.201 g / 1.008 g/mol = 5.160 mol. 3. Nitrogen gives 28.84 g / 14.007 g/mol = 2.059 mol, and oxygen gives 16.48 g / 16.00 g/mol = 1.030 mol. 4. Divide all four by the smallest, 1.030 mol: that yields 4.00 C, 5.01 H, 2.00 N and 1.00 O. 5. The empirical formula is therefore C4H5N2O, with an empirical formula mass of 48.04 + 5.04 + 28.01 + 16.00 = 97.09 amu. 6. Divide the molecular mass by it: 194.2 amu / 97.09 amu = 2 formula units per molecule. 7. Multiplying every subscript by 2 gives the molecular formula C8H10N4O2. 8. C4H5N2O is the trap of stopping at the empirical formula and ignoring the 194.2 amu molecular mass. 9. C8H10N2O4 doubles only some subscripts, breaking 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_