1. Ch. 7.4 Determining Chemical Formulas

2. 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.

3. 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.

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

5. 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 

6. 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 g sample of the compound.
Then calculate the amount of moles of each element in the sample.
Simplest ratio of moles

7. 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

8. 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 = RU = 3
O: 2.70 mol / 2.70 mol = 1

9. 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

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

11. 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

12. (EF)n = Molecular Formula
Ch 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

13. 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.

14. Ch. 7.4 Determining Chemical Formulas
Molecular Formula
Mass to mol:
C: 92.30g/ = 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

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

16. 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 g/mol. What is the compound’s molecular formula?