At first note 2.37 g + 3.04 g = 5.41 g, which implies that the example was a compound of just N, H and O. Change over the majority of N2 and H2O into number of moles by utilizing the molar masses of each. 1) N moles = mass of N2/molar mass of N2 = 2.37 g/( 14.0 g/mol) = 0.169 mol of N 2) H2O moles = mass of H2)/molar mass of H2O = 3.04 g/(18.0 g/mol) = 0.169 mol 0.169 moles of H2O => 2 * 0.169 moles of H and 0.169 moles of O 3) Emipirical equation N: 0.169/0.169 = 1 H: 2* 0.169/0.169 = 2 O: 0.169/0.169 = 1 => NH2O 4) Molar mass of the observational equation: 14.0 g/mol + 18 g/mol = 32 g/mol 5) Number of times that the observational equation is contained in the sub-atomic recipe: 64 g/mol/32 g/mol = 2 6) Molecular equation = 2 * Empirical formula = N2H4O2 Finally The chemical formula of the compound is N2H4O2