1. Empirical and Molecular Formulas

2. We can determine its percent composition
If we are given an unknown substance how could we determine its chemical formula?
We can determine its percent composition
Then you can determine its chemical formula
There are two types of chemical formula:
Empirical Formula
Molecular Formula

3. Empirical Formula
Empirical formula: the simplest whole-number ratio of atoms in a compound
This is the true formula for ionic compounds, but not for necessarily for molecular

4. 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, CH3, CO2

5. How to determine the Empirical Formula
You are given the % composition of a compound
Ex: % Carbon % Hydrogen
Assume that you have a g sample of the compound:
Ex: g Carbon 7.7 g Hydrogen

6. 3. Use the atomic mass of each element to determine the number of moles of each element
Covert mass to moles
nC = g
12.0 g/mol
nC = 7.69 mol C
Covert mass to moles
nH = 7.7 g
1.01 g/mol
nH = 7.70 mol H
4. To obtain the simplest molar ratio (empirical formula), divide both by the smaller number of moles present
C:H mol C : 7.70 mol H
7.69
1 mol C : 1 mol H, therefore C1H1 or CH is the empirical formula

7. 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
Covert mass to moles
nC = 40.0 g
12.0 g/mol
nC = 3.33 mol
nO = 53.3 g
16.0 g/mol
nO = 3.33 mol
 The empirical formula of the substance is CH2O
nH = 6.70 g
1.01 g/mol
nH = 6.63 mol

8. Calculating Molecular Formula
Molecular formula: shows the actual number of atoms of each element in a molecule
A true chemical formula for molecular compounds
There are two different ways to find a molecular formula, depending on what type of information you are given…

9. Type 1: Given empirical formula and molar mass
Given: Empirical formula: CH2O
molar mass: 180 g/mol
Calculate the empirical formula molar mass
M of CH2O = g/mol + 2(1.01 g/mol) g/mol
M of CH2O = g/mol
2. Calculate the number of formula units
# of formula units = molar mass (given)
empirical molar mass (step 1)
# of formula units = g/mol = 6 formula units g/mol

10. Molecular formula = # of formula units x empirical formula
6 x (CH2O) = C6H12O6
Therefore the molecular formula is C6H12O6 (glucose)

11.  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

12.  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

13. Type 2: Given the % Composition and the Molar Mass
Step 1: Using the % composition, find the empirical formula
21.9 g Na g C 1.9 g H g O
22.99g/mol g/mol g/mol g/mol
= 0.95 mol Na = 3.81 mol C =1.88 mol H =1.91 mol O
0.95 mol Na mol C mol H mol O
= 1 mol Na = 4 mol C = 2 mol H = 2 mol O
Given:
% Na = 21.9 % % C = 45.7% % H = 1.9%
% O = 30.5 % MM = g/mol

14. Empirical formula: NaC4H2O2
Step 2: Continue to solve for the molecular formula as for Type 1
M of NaC4H2O2 = g/mol + 4(12.01g/mol) + 2(1.01 g/mol) + 2(16.00g/mol)
M of NaC4H2O2 = g/mol
# of formula units = g/mol
g/mol
# of formula units = 2 formula units
Molecular Formula = 2 x (NaC4H2O2)
Molecular Formula = Na2C8H4O4

15.  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 = assume 100 g
6.87 mol C 17.3 mol H
6.87 mol mol
= 1 mol C = 2.5
*** multiply both by 2 to get full numbers
Empirical formula = C2H5
M of C2H5 = 2(12.01g/mol) + 5(1.01)
M of C2H5 =29.07g/mol
mc = 82.5 g
nc = g
12.01 g
= 6.87 mol
mH = 17.5 g
nH = 17.5 g
1.01 g
= 17.3 mol
# of formula units = g/mol
g/mol
# of formula units = 2 formula units
Molecular formula = 2 x (C2H5)
 The molecular formula is C4H10

16. Homework
P. 211 #
P. 214 # 1-3
P. 218 # 17-20