Empirical and Molecular

Formulas

Empirical Formula

• What are we talking about???

• Empirical Formula represents the

smallest ratio of atoms in a

formula.

• In other words it represents the

simplest chemical formula for a

particular compound.

Steps to Solve!

• You need to follow these basic steps to

determine your answer.

• 1) Make sure you are starting in grams

• 2) Convert all grams to moles (Molar Mass)

• 3) Divide each by the smallest number of

moles.

• 4) Create whole number ratio of subscripts

• 5) Multiply if necessary

Example 1

• In an unknown molecule, there are

4.15 grams of Carbon and 1.38

grams of Hydrogen. Determine the

empirical formula for the substance!

• Step 1) Make sure you are starting in

grams

• Step 2) Convert each mass in grams

to moles.

Now What?• Step 3) Divide each by the smallest

number of moles. Set up the Ratio of

elements

• The 1 and the 4 represent the whole

number ratio of the elements in the

chemical formula.

• The 1 represents that there is only one

Carbon atom

• The 4 represents that there are four

Hydrogen atoms

• The Solution: The empirical formula is

CH4

Solution!

• Phenol, a general disinfectant, is 76.57

% Carbon, 6.43 % Hydrogen, and 17.00

% Oxygen. Determine it’s empirical

formula.

• Step 1) Make sure you are starting in

grams…….

• How can we go from percent to grams?

• Easy! Assume we have a 100 gram

sample so there would be how much of

each element?

• Step 2) Convert each mass in grams to moles.

Try this!

• A compound has 78.1% B and 21.9% H. Determine the empirical formula.

• 78.1g B X 1mol/10.81g = 7.22 mol B

• 21.9g H X 1mol/1.01g = 21.7 mol H

• 7.22mol ÷ 7.22mol =1

• 21.7mol ÷ 7.22mol = 3.01

• Ratio of 1:3

• BH3

Molecular Formula

• The molecular formula is related, but

different to the empirical formula.

• Remember that Empirical Formula

represents the lowest ratio of the

atoms in a compound.

• The Molecular Formula is the actual,

or true ratio of the elements in the

compound.

Needs and Steps!

• 1. In order to solve you need the molar mass of the molecular formula for a compound in g/mol.

• 2. Determine the empirical formula.

• 3. Calculate the molar mass of the empirical formula

• 4. Divide the molar mass of the molecular formula by the molar mass of the empirical formula. Apply ratio to all subscripts.

Example

• The empirical formula for hydroquinone, a chemical used in photography, is C3H3O. The molecular weight of the compound Is 110 g/mol. Determine the molecular formula.

• Step 1) Determine the molecular weight of the compound: Given at 110 g/mol.

• Step 2) Determine the empirical

formula: Given C3H3O.

• Step 3) Determine the molar mass of

the empirical formula.

Think…I Know it’s hard…

• Step 4) Think about the molar

mass of the molecular formula

compared to the molar mass of

the empirical formula.

• Mol. Formula Emp. Formula

• 110 g/mol 55.06 g/mol

• Divide and get the ratio

• 110 g/mol / 55.06 g/mol = 2

So….

• That 2 represents what you will

multiply all the subscripts by to

determine the correct molecular

formula.

• C3H3O x 2 = C6H6O2

Molecular Formula

Try this

• The empirical formula for a compound

is C3H7. If the molecular weight is 86

g/mol, then what is the molecular

formula?

• 1. The molecular weight is given:

– 86 g/mol

• 2. The empirical formula is given:

- C3H7

• 3. Calculate the molar mass for

the empirical formula.

Compare the two….

• The Molecular weight was 86

g/mol

• The Empirical weight was 43.1

g/mol

• Divide Molecular by Empirical to

determine the ratio!

• 86 g/mol / 43.1 g/mol = 2

• So 2 is the ratio, multiply all

the subscripts by 2

• C3H7 x 2 = C6H14 is the

Mol. Formula

Practice!

A) Empirical Formula is S

Molecular weight is 256 grams per mol

B) Empirical Formula is NO2

Molecular Weight is 46 g/mol