Empirical and Molecular Formulas Chemical Compounds Unit Notes #5

Writing a Formula What if we wanted to be able to give a name/formula to an unknown compound, based on data? Ex. A compound weighs 298.12 g. It is 72.2% magnesium and 27.8% nitrogen by mass. How could we determine its formula?

Empirical Formula A compound’s empirical formula is the smallest whole number ratio of atoms in the compound. The ratio is shown in the subscripts. Ex) NO2 vs. N2O2 It does not necessarily tell us how many atoms are in the compound.

Some Empirical Formulas Actual (Molecular) Formula HO HO, H2O2, H3O3… CH2O CH2O, C2H4O2, C6H12O6 … CH3O CH3O, C2H6O2 …

Calculating the Empirical Formula Ex. A compound weighs 298.12 g. It is 72.2% magnesium and 27.8% nitrogen by mass. Change the percents into masses. Mg: 72.2%  72.2 g N: 27.8%  27.8 g

Calculating the Empirical Formula Find the moles of each element. Mg: 72.2 g x 1 mole = 2.97 moles 24.3 g N: 27.8 g x 1 mole = 1.97 moles 14.01 g

Calculating the Empirical Formula Divide both numbers of moles by the smaller number. We have 2.97 moles Mg and 1.97 moles N Mg : 2.97 moles / 1.97 moles = 1.50 N: 1.97 moles / 1.97 moles = 1.00

Calculating the Empirical Formula Multiply both quotients by the same number until they are both whole numbers. Mg : 1.50 x 2 = 3.00 N: 1.00 x 2 = 2.00 The final answers become the subscripts in the empirical formula. Mg3N2

Try it! A sample is 25.9% nitrogen and 74.1% oxygen. In grams: In moles: Divided: Multiplied: Final: 25.9 g N; 74.1 g O 1.85 moles N; 4.63 moles O N: 1.00; O: 2.50 N: 1.00 x 2 = 2.00; O: 2.50 x 2 = 5.00 N2O5

Actual (Molecular) Formula The molecular formula is the actual ratio of atoms in the compound. Many molecular formulas can have the same empirical formula. Empirical Formula Actual (Molecular) Formula HO HO, H2O2, H3O3… CH2O CH2O, C2H4O2, C6H12O6 … CH3O CH3O, C2H6O2 …

Calculating the Molecular Formula A compound has the empirical formula CH2 and a molecular mass of 28.1 g/mol.

Calculating the Molecular Formula Find the empirical formula CH2 Find the molar mass of the empirical formula 1(12.01) + 2(1.01) = 14.03 g/mol

Calculating the Molecular Formula Divide the empirical mass by the molecular mass 28.1 g/mol = 2.00 14.03 g/mol Multiply each subscript by that quotient. CH2 x 2.00 = C2H4

Try it! A sample is 25.9% nitrogen and 74.1% oxygen, and has a molecular mass of 216 g/mol. Empirical formula: Empirical mass: Molecular mass divided by empirical mass: Subscripts multiplied: N2O5 108 g/mol 216 / 108 = 2 (N2O5) * 2 = N2O5