Empirical Formula

In chemistry, we represent every compound by some formula. The simplest formula for this representation is the empirical formula. It provides the lowest whole number ratio of the atoms existing in the compound. The relative number of atoms of every element in the compound is available by this formula. This article will explain the concept of empirical formula with examples. Let us learn it!

Empirical Formula

Concept of Empirical Formula

The empirical formula of any compound is the simplest whole-number ratio of each individual type of atom in that compound. It may be the same as the compound’s molecular formula sometimes. But it is not possible always. We may calculate the empirical formula from information about the mass of each element in that compound or also from the percentage composition.

Steps for Determining an Empirical Formula:

To calculate the empirical formula, we have to first determine the relative masses of the various elements present. We may either use mass data in grams or percent composition. Also, for the percentage composition, we may assume the total percent of a compound like 100% and the percentage composition in grams.

The steps to determine the empirical formula are as follows:

Step 1: Find out the mass of each element present in grams

m = Element percentage = mass in gram

Step 2: Obtain the number of moles of each type of atom present

M = $\frac {m} {atomic mass}$ = Molar amount

Step 3: Now, divide the number of moles of each element by the smallest number of moles

R = $\frac {M}{ least M value}$ = Atomic Ratio

Step 4: Finally, convert numbers to the whole numbers. This set of whole numbers will be the subscripts in the empirical formula.

i.e. R  $\times whole number$ = Empirical Formula

Solved Examples on Empirical Formula

Q.1: A compound contains 88.79% oxygen element and 11.19% hydrogen element. Compute the empirical formula of this compound.

Solution:

Assume 100.0g of the compound. Then, we see that the percentage of each element matches the grams of each element, as

11.19 gram of H and 88.79 gram of O

Convert the grams of each element to moles as:

H: (11.19g) divided by 1 mole H i.e. 1.008 g H = 11.10 mol H atoms

O: (88.79g) divided by I mole O i.e. 16.00 g O = 5.549 mol O atoms

The formula will be determined as:

$H_{11.10} O_{ 5.549}$

However, it’s usual to use the smallest whole number ratio of atoms.

Thus by dividing the lowest number alter the numbers to whole numbers, we get

H = 2.000

O = 1.000

So, the simplest ratio of H to O is 2:1

Therefore the Empirical formula of the compound is $H_2 O$ .

Q.2: An oxide of Aluminum is formed by the reaction of 4.151 grams of Aluminum with 3.692 grams of Oxygen. Find out the empirical formula for this compound.

Solution:

Step 1: Determine the masses

4.151 gram of Al and 3.692 gram of O

Step 2: Determine the number of moles

4.151 g Al $\times  \frac {1 mol \;Al} { 26.98 g \; Al}$ = 0.1539 mol Al atoms

3.692 g O $\times \frac {1 mol \;O} {16.00 g\; O}$   = 0.2398 mol O atoms

Step 3: Divide the number of moles

$\frac {0.1539 mol Al} { 0.1539 }$ = 1.000 mol Al atoms

$\frac{ 0.2398 mol O } { 0.1539 }$ = 1.500 mol O atoms

Step 4: Convert numbers to whole numbers

The compound contains 2 Al and 3 O atoms

So, the Empirical Formula = $Al_2O_3$ .