# Emperical Formula - Explanation and Worked Examples of Emperical Formula

### Empirical Formula:

Let's begin by considering two atoms, say Hydrogen (H) and oxygen (O). If these atoms are combined to form a compound, let's say (HO) for instance, we will not be able to tell the ideal formula of the compound if this was an experimental sample; as it also could be H2O, H2O2 or something else.

With regards to the uncertainty of the compound formed in the above instance, the empirical formula comes into play. It is the simplest formula of a compound because it is derived from experimental data or analysis.

It is the empirical formula that tells us the relative ratios the of different atoms in a compound.

The empirical formula of a compound may be defined as the formula that shows only the relative number of atoms of each element present in the compound.

Now let's consider water molecule (H2O) as a reference:

H2O = 2 atoms of hydrogen (H) and 1 atom of oxygen (O).

H2O = 1 mole of hydrogen molecule and 1 mole of oxygen.

Note: Always remember that when a number isn't written in front of a compound, then it means that the number "1" is actually in front of that compound. So, the H2O in our above example is actually 1H2O (1 mole of water).

The 1 (in front of the compound) is referred to as the Mole Ratio.

Therefore, if we know the amounts of mole of each element in a compound, then the emperical formula can be determined.

Note: Whenever we are working with the percentages of each element contained within a compound, we will need to convert the percentages into the mole ratio of the elements in order to get the empirical formula of the compound.

#### Worked Example of Emperical Formula:

A compound composed of iron (Fe) and oxygen (O) was analyzed and found to contain 69.94% iron and 30.06% oxygen. Find the empirical formula of the compound. ( Molar mass of Fe=55.85, O=16)

##### Solution

Step 1: Identify the given parameter from the question.

Fe = 69.94%,   O = 30.06%.

Empirical formula = Fe?O?

Step 2: Convert the percentages to gram. (just attribute grams to the %).

69.94% = 69.94g while 30.06% = 30.06g

Step 3: To get the mole ratio of each element, convert the gram to moles using the formula (mole = mass/molarmass). Please merorize this formula because we always work with moles in Emperical Formula.

Mole of Fe: 69.94/55.85 = 1.252mol

Mole of O: 30.06/16 = 1.879mol

Step 4: Divide both sides by the smallest mole ratio.

Iron has the smallest mole ratio in our case, therefore: 1.252/1.252 = 1,    1.879/1.252 = 1.5

We now have the formula = Fe1O1.5

Step 5: Multiply each of the moles by the smallest whole number that will convert each into a whole number. (In our case, the number "2" is the smallest whole number that will make "1.5" and "1" whole numbers when multiplied by it.

For iron (Fe), we will have 1 x 2 = 2

For oxygen (O), we will have 1.5 x 2 = 3

Step 6: Write the empirical formula.

The empirical formula= Fe2O3

Iron(III)tetraoxosulphate(VI) Alfred Ajibola is a Medical Biochemist, a passionate Academician with over 7 years of experience, a Versatile Writer, a Web Developer, a Cisco Certified Network Associate and a Cisco CyberOps Associate.

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