Chemistry Notes Empirical & Molecular Formulas

Empirical Formula The empirical formula gives you the lowest, whole number ratio of elements in the compound. The empirical formula may or may not be the same as the molecular formula.

For example: For carbon dioxide, the molecular formula is CO 2, and the empirical formula is CO 2. One carbon and two oxygens are the lowest ratio of atoms. The molecular formula for dinitrogen tetrahydride is N 2 H 4, but the empirical formula is NH 2.

What is the empirical formula for: C 6 H 12 O 6

What is the empirical formula for: C 6 H 12 O 6 CH 2 O

What is the empirical formula for: C 6 H 12 O 6 CH 2 O C 6 H 12 O 2

What is the empirical formula for: C 6 H 12 O 6 CH 2 O C 6 H 12 O 2 C 3 H 6 O

What is the empirical formula for: C 6 H 12 O 6 CH 2 O C 6 H 12 O 2 C 3 H 6 O N 2 H 2

What is the empirical formula for: C 6 H 12 O 6 CH 2 O C 6 H 12 O 2 C 3 H 6 O N 2 H 2 NH

What is the empirical formula for: C 6 H 12 O 6 CH 2 O C 6 H 12 O 2 C 3 H 6 O N 2 H 2 NH CH 4

What is the empirical formula for: C 6 H 12 O 6 CH 2 O C 6 H 12 O 2 C 3 H 6 O N 2 H 2 NH CH 4 CH 4

To Find the Empirical Formula from % Composition: If given the percentages, assume there are grams of the compound. Convert the grams of each element to moles.

To Find the Empirical Formula from % Composition: Divide by the smaller amount of moles, then manipulate the ratio so that all numbers are whole.

Two Examples: What is the empirical formula of a compound that is 27.3% carbon and 72.7% oxygen?

Two Examples: What is the empirical formula of a compound that is 27.3% carbon and 72.7% oxygen? 27.3 g Carbon x 1 mole C = 2.28 mol C / 2.28 = 1 mole C 12 g C 12 g C 72.7 g Oxygen x 1 mole O = 4.54 mol O / 2.28 = 2 moles O 16 g O 16 g O The empirical formula would be: CO 2

Two Examples: What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen?

Two Examples: What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen? 25.9 g Nitrogen x 1 mole N = 1.85 mol N / 1.85 = 1 mole N x 2 = 2 moles N 14 g N 14 g N 74.1 g Oxygen x 1 mole O = 4.63 mol O / 1.85 = 2.5 moles O x 2 = 5 moles O 16 g O 16 g O The empirical formula would be: N 2 O 5

More to try… Calculate the empirical formula of a compound that is 94.1% oxygen, 5.9% hydrogen.

More to try… Calculate the empirical formula of a compound that is 94.1% oxygen, 5.9% hydrogen g Oxygen x 1 mole O = 5.88 mol O / 5.88 = 1 mole O 16 g O 16 g O 5.9 g Hydrogen x 1 mole H = 5.9 mol H / 5.88 = 1 mole H 1 g H 1 g H The empirical formula would be: OH

More to try… Calculate the empirical formula of a compound that is 79.8% carbon, 20.2% hydrogen.

More to try… Calculate the empirical formula of a compound that is 79.8% carbon, 20.2% hydrogen g Carbon x 1 mole C = 6.65 mol C / 6.65 = 1 mole C 12 g C 12 g C 20.0 g Hydrogen x 1 mole H = 20.2 mol H / 6.65 = 3 mole H 1 g H 1 g H The empirical formula would be: CH 3

From there, find it’s molecular formula: If given the molar mass (how many g/mol of the compound) then you can calculate the molecular formula from the empirical formula.

From there, find it’s molecular formula: Take the compound’s empirical formula mass and compare to the molecular mass. The molecular mass will be a multiple of the empirical formula’s mass.

Example: If the molecular mass of the first example (N 2 O 5 ) problem is 216 g/mol, then what is the molecular formula for the compound?

Example: 216 g /mol = g/mol The molecular formula is: N 4 O 10

Try these: What is the molecular formula of a compound whose molar mass is 60.0 g and whose empirical formula is CH 4 N?

Try these: What is the molecular formula of a compound whose molar mass is 60.0 g and whose empirical formula is CH 4 N? 60.0 g/mol = g/mol The molecular formula is: C 2 H 8 N 2

Try these: What is the molecular formula for a compound whose molar mass is 78 g and whose empirical formula is CH?

Try these: What is the molecular formula for a compound whose molar mass is 78 g and whose empirical formula is CH? 78 g/mol = 6 13 g/mol The molecular formula is: C 6 H 6