1. Empirical and Molecular Formulas

2. Empirical Formula
If we are given an unknown substance how could we determine its chemical formula?
We can determine its percent composition
Then we can determine its empirical formula: the simplest whole-number ratio of atoms in a compound

3. However, the empirical formula does not necessarily give information about the number of atoms in a molecule (molecular formula).
Molecular Formula
Empirical Formula
C2H2 CH
C4H4
C6H6
The empirical formula gives the combining ratio in its simplest form
The molecular formula gives the same ratio but with the actual number of atoms
Note: The empirical formula and the molecular formula can be the same in some cases. E.g. H2O, CH4, CO2

4. Calculating Empirical Formula
E.g. What is the empirical formula of a substance that is 40.0% carbon, 6.70% hydrogen, and 53.3% oxygen by mass?
Given:
Assume 100 g
mC = 40.0 g
mH = 6.70 g
mO = 53.3 g
Divide each element by the lowest quantity
3.33 mol C = mol H = mol O = 1
Mole ratio of C:H:O = 1:2:1
Convert mass to moles
nC = 40.0 g X 1 mol
12.01 g
nC = 3.33 mol
nO = 53.3 g X 1 mol
16.00 g
nO = 3.33 mol
 The empirical formula of the substance is CH2O
nH = 6.70 g X 1 mol
1.01 g
nH = 6.63 mol

5. Calculating Molecular Formula
The empirical formula tells us the simplest ratio of atoms in a molecule, but it does not tell us the actual number of atoms in a molecule
However, if we know molar mass and the empirical formula of a compound we can determine its molecular formula
Divide the molar mass of the compound by the molar mass obtained from the empirical formula
Multiply the subscripts by this number

6.  The molecular formula of the compound is H2O2
E.g. What is the molecular formula of a compound that has a molar mass of 34 g/mol and the empirical formula HO?
Given:
Empirical Formula: HO
MMcmpd = 34 g/mol
MMHO =
= g/mol
Molecular Formula = MMcmpd
MMHO
= 34 g/mol
17.01 g/mol
= 2
 The molecular formula of the compound is H2O2

7.  The molecular formula of the compound is C3H9
E.g. What is the molecular formula of a compound that has a molar mass of g/mol and the empirical formula CH3?
Given:
Empirical Formula: CH3
MMcmpd = g/mol
MMCH3 = (1.01)
= g/mol
Molecular Formula = MMcmpd
MMCH3
= g/mol
15.04 g/mol
= 3
 The molecular formula of the compound is C3H9

8. Calculating Molecular Formula
We can calculate molecular formula if empirical formula and molar mass is given
We can also calculate molecular formula if percent composition and molar mass is given.

9.  The molecular formula is C4H10
E.g. What is the molecular formula of a compound with a molar mass of 58.0 g/mol and a percent composition of 82.5% C and 17.5% H?
Given:
% C = 82.5%
% H = 17.5%
MM = 58.0 g/mol
m = 58.0 g (Assume 1 mole)
mH = X 58.0 g
= 10.2 g
nH = 10.2 g X 1 mol
1.01 g
= 10.1 mol
mc = X 58.0 g
= 47.8 g
nc = 47.8 g X 1 mol
12.01 g
= 3.98 mol
Mole ratio C:H = 4:10
 The molecular formula is C4H10

10. Homework
P. 186 #2-6
P. 188 #1-6
P. 193 # 8,9

11. Hydrates
Solid ionic compounds are formed from dehydrating ionic solutions
Some of the liquid water molecules may remain between molecules of the solids, creating a hydrate
When a hydrate is heated and water is evaporated it is referred to as anhydrous

12. Hydrate Example
A 50.0 g sample of a hydrate of barium hydroxide was heated. The anhydrous solid weighed 27.2 g.
Determine the percent composition of water in the hydrate.
Calculate the molecular formula of the hydrate. Name the hydrate.
Hint: Start with a BCE
Ba(OH)2xH2O Solve for x