Calculating Empirical, Molecular Formula and Percent Composition

There are two kinds of formulas: Empirical and Molecular

Empirical:
- Gives lowest-term ratio of atoms or moles in formula
- All Ionic compounds are empirical

Molecular:
- Gives all atoms that make up molecule
- Either Ionic or Covalent

Eg. C6H12O6 = Molecular Formula
CH2O = Empirical Formula

Example:
A gas has an Empirical Formula of CH2. What is the Molecular Formula if the mass of one mole is 42.0 g?

Have to find a whole number (N) = Molar Mass/Empiricial Mass = 42.0/14.0 = 3 (N)
(CH2) X 3 = C3H6

Determining Empirical Formula Given Mass


Example:

Determine the Empirical Formula of Fe and O given 10.87 g of Fe and 4.66 g of O.

First, grams must be converted to moles.
Fe - 10.87 g X 1 mole/55.8 g = 0.1948 moles
O - 4.66 g X 1 mole/16.0 g = 0.291 moles

Now, you must divide all of the derived numbers by the smallest amount of moles.
Fe - 0.1948/0.1948 = 1
O - 0.291/0.194 = 1.49, which can be rounded to 1.5

Lastly, you must multiply until you reach a whole number
Fe - 1 X 2 = 2
O - 1.5 X 2 = 3

Therefore, the Empirical Formula is Fe2O3.

Percent Composition:
- The % by mass of elements in a compound
- MUST ADD UP TO ABOUT 100% PERCENT (99.9 or 100.1)

Example:
What is the Percent Composition of CO2?

First, calculate the molar mass.
C - 12.0 g X 1 = 12.0 g
O - 16.0 g X 2 = 32.0 g
Therefore, the Molar Mass is 44.0 g.

Now, calculate each element's % of that total, rounding to one decimal place.
C - 12.0 g/44.0 g X 100 = 27.3 %
O - 32.0 g/44.0 g X 100 = 72.7 %

Percent Composition to Empirical Formula

Example:
A substance is 45.8% Sulphur and 54.2% Fluorine. What is the Empirical Formula?

First, assume 100.0 grams of material.
Then, convert to moles.
S - 45.8 g X 1 mole/32.1 g = 1.43 moles
F - 54.2 g X 1 mole/19.0 g = 2.85 moles

Then, divide by the smallest number.
S - 1.43 g/1.43 g = 1
F - 2.85 g/1.43 g = 2

Therefore, the Empirical Formula is SF2.