Empirical and Molecular Formulas

• 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

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

Molecular Formula Empirical Formula

C2H2

CHC4H4

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

How to determine the Empirical Formula

1. You are given the % composition of a compound

Ex: 92.3 % Carbon 7.7% Hydrogen

2. Assume that you have a 100.0 g sample of the compound:

Ex: 92.3 g Carbon 7.7 g Hydrogen

3. Use the atomic mass of each element to determine the number of moles of each elementCovert mass to moles

nC = 92.3 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 7.69 mol C : 7.70 mol H 7.69

1 mol C : 1 mol H, therefore C1H1 or CH is the empirical formula

Calculating Empirical FormulaE.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

The empirical formula of the substance is CH2O

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 molnH = 6.70 g

1.01 g/mol

nH = 6.63 mol

Divide each element by the lowest quantity

3.33 mol C = 1 6.63 mol H = 1.99 3.33 mol O = 1

3.33 3.33 3.33

Mole ratio of C:H:O = 1:2:1

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…

Type 1: Given empirical formula and molar mass

Given: Empirical formula: CH2O

molar mass: 180 g/mol

1. Calculate the empirical formula molar mass

M of CH2O = 12.01 g/mol + 2(1.01 g/mol) + 16.00 g/mol

M of CH2O = 30.03 g/mol

2. Calculate the number of formula units

# of formula units = molar mass (given)

empirical molar mass (step 1)

# of formula units = 180 g/mol = 6 formula units 30.03 g/mol

3. Molecular formula = # of formula units x empirical formula

6 x (CH2O) = C6H12O6

Therefore the molecular formula is C6H12O6 (glucose)

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

The molecular formula of the compound is H2O2

Molecular Formula = MMcmpd

MMHO

= 34 g/mol

17.01 g/mol

= 2

MMHO = 1.01 + 16.00

= 17.01 g/mol

E.g. What is the molecular formula of a compound that has a molar mass of 45.00 g/mol and the empirical formula CH3?

Given:

Empirical Formula: CH3

MMcmpd = 45.00 g/mol

The molecular formula of the compound is C3H9

Molecular Formula = MMcmpd

MMCH3

= 45.00 g/mol

15.04 g/mol

= 3

MMCH3 = 12.01 + 3(1.01)

= 15.04 g/mol

Type 2: Given the % Composition and the Molar Mass

Step 1: Using the % composition, find the empirical formula21.9 g Na 45.7 g C 1.9 g H 30.5 g O

22.99g/mol 12.01 g/mol 1.01 g/mol 16.00 g/mol

= 0.95 mol Na = 3.81 mol C =1.88 mol H =1.91 mol O

0.95 mol Na 3.81 mol C 1.88 mol H 1.91 mol O

0.95 0.95 0.95 0.95

= 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 = 210.00 g/mol

Empirical formula: NaC4H2O2

Step 2: Continue to solve for the molecular formula as for Type 1

M of NaC4H2O2 = 22.99 g/mol + 4(12.01g/mol) + 2(1.01 g/mol) + 2(16.00g/mol)M of NaC4H2O2 = 105.05 g/mol

# of formula units = 210 g/mol 105. 05 g/mol

# of formula units = 2 formula units

Molecular Formula = 2 x (NaC4H2O2)

Molecular Formula = Na2C8H4O4

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

The molecular formula is C4H10

mc = 82.5 g

nc = 82.5 g

12.01 g

= 6.87 mol

mH = 17.5 g

nH = 17.5 g

1.01 g

= 17.3 mol

6.87 mol C 17.3 mol H 6.87 mol 6.87 mol= 1 mol C = 2.5 *** multiply both by 2 to get full numbersEmpirical formula = C2H5

M of C2H5 = 2(12.01g/mol) + 5(1.01)M of C2H5 =29.07g/mol

# of formula units = 58.0 g/mol 29. 07 g/mol

# of formula units = 2 formula units

Molecular formula = 2 x (C2H5)

Homework

• P. 211 # 13- 16

• P. 214 # 1-3

• P. 218 # 17-20