Ch. 7.4 Determining Chemical Formulas

Ch. 7.4 Determining Chemical Formulas Objectives Define empirical formula, and explain how the term applies to ionic and molecular compounds. Determine an empirical formula from either a percentage or a mass composition. Explain the relationship between the empirical formula and the molecular formula of a given compound. Determine a molecular formula from an empirical formula.

Ch. 7.4 Determining Chemical Formulas An empirical formula consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole-number mole ratio of the different atoms in the compound. For an ionic compound, the formula unit is usually the compound’s empirical formula. For a molecular compound, however, the empirical formula does not necessarily indicate the actual numbers of atoms present in each molecule.

Everyone has 2 hands and 10 fingers: H2F10 Empirical Formula Example Everyone has 2 hands and 10 fingers: H2F10 Empirical formula: HF5​

More than one substance may reduce to the same empirical formula. Benzene Molecular Formula: C6H6 Proportion of C to H is 1:1 Empirical Formula: CH More than one substance may reduce to the same empirical formula. Benzene: C6H6 Acetylene: C2H2 

Ch. 7.4 Determining Chemical Formulas Calculation of Empirical Formulas To determine a compound’s empirical formula from its percentage composition, begin by converting percentage composition to a mass composition. Assume that you have a 100.0 g sample of the compound. Then calculate the amount of moles of each element in the sample. Simplest ratio of moles

Ch. 7.4 Determining Chemical Formulas Empirical Formula We have a sample that is 48.6% C, 8.16% H, and 43.2% O by weight. What is the empirical formula? Step 1: % Composition to Moles 48.6% C = 48.6g C; 48.6g / 12.01g/mol = 4.05 mol C 8.16% H = 8.16g H; 8.16g / 1.01g/mol = 8.08 mol H 43.2% O = 43.2g O; 43.2g / 16.00g/mol = 2.70 mol O

Ch. 7.4 Determining Chemical Formulas Empirical Formula Step 2: Mole Ratio Divide each element by the smallest number of moles C: 4.05 mol / 2.70 mol = 1.5 H: 8.08 mol / 2.70 mol = 2.99 RU = 3 O: 2.70 mol / 2.70 mol = 1

Ch. 7.4 Determining Chemical Formulas Empirical Formula Step 3: Atom Ratio Ratio of atoms MUST be a whole number! C = 1.5 x 2 = 3 H = 3 x 2 = 6 O = 1 x 2 = 2

Ch. 7.4 Determining Chemical Formulas Empirical Formula Step 4: Express as an Empirical Formula C3H6O2

Ch. 7.4 Determining Chemical Formulas Molecular Formula Given the composition percentages and the samples Molecular Mass. First: We use the percent composition to Empirical Formula Rules. Weight % → Mass → Mol → Ratio of Mol → Atoms = Empirical Formula

(EF)n = Molecular Formula Ch. 7.4 Determining Chemical Formulas Molecular Formula Second: Calculate the Empirical Mass from the Empirical Formula. Third: Divide the Molecular Mass (given) by Empirical Mass (MM/ EM) = n Last: Multiply Empirical Formula subscripts by n (EF)n = Molecular Formula

Ch. 7.4 Determining Chemical Formulas Molecular Formula A compound is 92.30% C and 7.80% H with a mass of 78.12g/mol. Find the empirical formula and the molecular formula of this compound.

Ch. 7.4 Determining Chemical Formulas Molecular Formula Mass to mol: C: 92.30g/ 12.01 = 7.69 mol H: 7.80g/ 1.01 = 7.72 mol Ratio of Mol: C: 7.69 / 7.69 = 1 H: 7.72/ 7.69 = 1 Atoms: C:H 1:1 Empirical Formula (EF): CH

Ch. 7.4 Determining Chemical Formulas Molecular Formula Empirical Mass (EM): (12.01) + (1.01)= 13.02 n = (MM/EM): 78.12 / 13.02 = 6 (EF)n: (CH) (6)= C6H6

Ch. 7.4 Determining Chemical Formulas Molecular Formula Sample Problem The empirical formula of a compound of phosphorus and oxygen was found to be P2O5. Experimentation shows that the molar mass of this compound is 283.89 g/mol. What is the compound’s molecular formula?