# Calculating Compound Empirical & Molecular Formula

## Steps of Determining Empirical and Molecular Formulas

The empirical formula of a chemical compound is a representation of the simplest whole number ratio between the elements comprising the compound. The molecular formula is the representation of the actual whole number ratio between the elements of the compound. This step by step tutorial shows how to calculate the empirical and molecular formulas for a compound.

### Empirical and Molecular Problem

A molecule with molecular weight of 180.18 g/mol is analyzed and found to contain 40.00% carbon, 6.72% hydrogen and 53.28% oxygen.

What are the empirical and molecular formulas of the molecule?

### How To Find the Solution

Finding the empirical and molecular formula is basically the reverse process used to calculate mass percent.

Step 1: Find the number of moles of each element in a sample of the molecule.

Our molecule contains 40.00% carbon, 6.72% hydrogen and 53.28% oxygen. This means a 100 gram sample contains:

40.00 grams of carbon (40.00% of 100 grams)
6.72 grams of hydrogen (6.72% of 100 grams)
53.28 grams of oxygen (53.28% of 100 grams)

Note: 100 grams is used for a sample size just to make the math easier. Any sample size could be used, the ratios between the elements will remain the same.

Using these numbers we can find the number of moles of each element in the 100 gram sample. Divide the number of grams of each element in the sample by the atomic weight of the element (from the periodic table) to find the number of moles.

moles C = 40.00 g x 1 mol C/12.01 g/mol C = 3.33 moles C

moles H = 6.72 g x 1 mol H/1.01 g/mol H = 6.65 moles H

moles O = 53.28 g x 1 mol O/16.00 g/mol O = 3.33 moles O

Step 2: Find the ratios between the number of moles of each element.

Select the element with the largest number of moles in the sample.

In this case, the 6.65 moles of hydrogen is the largest. Divide the number of moles of each element by the largest number.

Simplest mole ratio between C and H: 3.33 mol C/6.65 mol H = 1 mol C/2 mol H
The ratio is 1 mole C for every 2 moles H

Simplest ratio between O and H: 3.33 moles O/6.65 moles H = 1 mol O/2 mol H
The ratio between O and H is 1 mole O for every 2 moles of H

Step 3: Find the empirical formula.

We have all the information we need to write the empirical formula. For every 2 moles of hydrogen, there is one mole of carbon and one mole of oxygen.

The empirical formula is CH2O.

Step 4: Find the molecular weight of the empirical formula.

We can use the empirical formula to find the molecular formula using the molecular weight of the compound and the molecular weight of the empirical formula.

The empirical formula is CH2O. The molecular weight is

molecular weight of CH2O = (1 x 12.01 g/mol) + (2 x 1.01 g/mol) + (1 x 16.00 g/mol)
molecular weight of CH2O = (12.01 + 2.02 + 16.00) g/mol
molecular weight of CH2O = 30.03 g/mol

Step 5: Find the number of empirical formula units in the molecular formula.

The molecular formula is a multiple of the empirical formula. We were given the molecular weight of the molecule, 180.18 g/mol.

Divide this number by the molecular weight of the empirical formula to find the number of empirical formula units that make up the compound.

Number of empirical formula units in compound = 180.18 g/mol/30.03 g/mol
Number of empirical formula units in compound = 6

Step 6: Find the molecular formula.

It takes six empirical formula units to make the compound, so multiply each number in the empirical formula by 6.

molecular formula = 6 x CH2O
molecular formula = C(1 x 6)H(2 x 6)O(1 x 6)
molecular formula = C6H12O6

Solution:

The empirical formula of the molecule is CH2O.
The molecular formula of the compound is C6H12O6.

### Limitations of the Molecular and Empirical Formulas

Both types of chemical formulas yield useful information. The empirical formula tells us the ratio between atoms of the elements, which can indicate the type of molecule (a carbohydrate, in the example).

The molecular formula lists the numbers of each type of element and can be used in writing and balancing chemical equations. However, neither formula indicates the arrangement of atoms in a molecule. For example, the molecule in this example, C6H12O6, could be glucose, fructose, galactose, or another simple sugar. More information than the formulas is needed to identify the name and structure of the molecule.