In chemistry, a chemical formula is a group of symbols of the same or different elements which represents one molecule of the substance. It basically represents chemical composition of the substance.
There are three types of formulae in the case of chemical compounds. They are as:
- Empirical formula
- Molecular formula
- Structural formula
In this tutorial, we will understand the basics of empirical formula, how to find or calculate it with the help of some important examples.
Table of Contents
What is Empirical Formula in Chemistry?
An empirical formula is a simplest formula of a substance that represents the relative whole number ratio of atoms of each element present in the molecule of that substance. In contrast, a molecular formula shows the actual number of atoms of each element present in a molecule or compound. Let’s understand it with the help of simple examples.
The molecular formula of benzene is C6H6. In benzene, there are six carbon atoms and six hydrogen atoms in its molecule. The simplest whole number ratio between the number of carbon and hydrogen atoms is 1 : 1, as 6 : 6 can be simplified to 1 : 1. Thus, the empirical formula of benzene having the molecular formula of C6H6 is CH.
Let us take another example to make it more clearly. The molecular formula of glucose is C6H12O6. In glucose, there are six carbon atoms, twelve hydrogen atoms, and six oxygen atoms in its molecule. The simplest whole number ratio between the number of carbon, hydrogen and oxygen atoms is 1 : 2 : 1, as 6 : 12 : 6 can be simplified to 1 : 2 : 1. Thus, the empirical formula of glucose having the molecular formula of C6H12O6 is CH2O.
Diamond has a three-dimensional giant structure in which each carbon atom is bound with four different carbon atoms. We can represent this giant structure by the formula Cn, where n may contain any value from a few hundreds to many thousands depending on the size of the sample. Hence, the empirical formula of diamond is C.
In this way, an empirical formula of a compound provides the lowest whole number ratio between the number of atoms of all elements present in the molecule of that compound.
How to find Empirical Formula of a Compound?
There are the following steps or rules by which we can find the empirical formula of any compound. They are as:
(1) Determine the percentage composition or mass percent of each element present in the compound.
(2) Divide the percentage of each element by its atomic mass of the respective element that gives an atomic ratio of the elements present in the compound.
(3) Divide the atomic ratio of each element by the minimum value of atomic ratio to get the simplest ratio of elements present in the compound.
(4) Convert these simplest ratios into the nearest whole numbers. These numbers give the simplest whole number ratio of the number of atoms of the various elements present in the compound.
(5) To get the empirical formula of the compound, write down symbols of all elements present in the compound side by side with their respective whole number ratio as a subscript to the lower right-hand comer of the symbol.
Using the above rules, you can easily calculate the empirical formula of any compound in the chemistry.
Important Examples of Empirical Formula
Let us look at some examples of empirical formula calculations:
Example 1:
Calculate the empirical formula for a compound that contains 9.76% magnesium, 13.01% sulphur, 26.01% oxygen, 51.22% H2O. The atomic masses for elements Mg, S, O, H are 24, 32, 16, and 1, respectively.
Solution:
Step 1: First, convert the mass per cent into grams
As we have mass percent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g of compound, 9.76g magnesium, 13.01g sulphur, 26.01g oxygen, and 51.22g H2O are present.
Step 2: Convert into the relative number moles of each element:
To find out the number of moles of each element in the compound, just divide the masses obtained above by respective atomic masses of various elements. This will give the atomic ratio of each element present in the compound.
No. of moles of magnesium = 9.76 / 24 = 0.406
No. of moles of sulphur = 13.01 / 32 = 0.406
No. of moles of oxygen = 26.01 / 16 = 1.625
No. of moles of water = 51.22 / 18 = 2.846
Step 3: Determine the simplest ratio:
To get the simplest ratio, divide each of the mole values obtained above by the smallest number amongst them. In other words, divide the atomic ratio of each element by the minimum value of atomic ratio to get the simplest ratio.
Simplest ratio of Mg = 0.406 / 0.406 = 1
Simplest ratio of S = 0.406 / 0.406 = 1
Simplest ratio of O = 1.625 / 0.406 = 4
Simplest ratio of H2O = 2.846 / 0.406 = 7
Step 4: Determine the simplest whole number ratio.
Simplest whole number ratio of Mg = 1
Simplest whole number ratio of S = 1
Simplest whole number ratio of O = 4
Simplest whole number ratio of H2O = 7
Hence, the empirical formula of above compound is MgSO4.7H2O.
Example 2:
A compound has the following composition: K = 26.6%, Cr = 35.4%, and O = 38.1%. What is its empirical formula of this compound? [Given K 39.1; Cr= 52; 0 = 16]
Solution:
| Element | Mass Per cent | Atomic mass | Relative number of moles | Simplest ratio | Simplest whole number ratio |
|---|---|---|---|---|---|
| Potassium (K) | 26.6 | 39.1 | 26.6 / 39.1 = 0.68 | 0.68 / 0.68 = 1 | 1 * 2 = 2 |
| Chromium (Cr) | 35.4 | 52.0 | 35.4 / 52.0 = 0.68 | 0.68 / 0.68 = 1 | 1 * 2 = 2 |
| Oxygen (O) | 38.1 | 16.0 | 38.1 / 16 = 2.38 | 2.38 / 0.68 = 3.5 | 3.5 * 2 = 7 |
Thus, the empirical formula of the above compound is K2Cr2O7.
Example 3:
Calculate the empirical formula of an organic compound that contains 57.82% carbon, 3.60% hydrogen, and the rest oxygen.
Solution:
| Element | Percentage | Atomic mass | Relative number of moles | Simplest ratio | Simplest whole number ratio |
|---|---|---|---|---|---|
| Carbon | 57.82 | 12 | 57.82 / 12 = 4.80 | 4.80 / 2.40 = 2 | 2 * 2 = 4 |
| Hydrogen | 3.60 | 1 | 3.60 / 1 = 3.60 | 3.60 / 2.40 = 1.5 | 2 * 1.5 = 3 |
| Oxygen | 100 – (57.82 + 3.60) = 38.58 | 16 | 38.58 / 16 = 2.40 | 2.40 / 2.40 = 1 | 1 * 2 = 2 |
Thus, the empirical formula of the above compound is C4H3O2.
How to Calculate Empirical Formula Mass?
Empirical formula mass or simply formula mass of a substance (i.e. molecule or compound) is equal to the sum of the atomic masses of atoms present in the empirical formula. For example, the empirical formula of benzene (C6H6) is CH. So, formula mass of CH = 12 + 1 = 13u (or 13 g/mol).
Example 4:
Calculate the empirical formula mass of a compound that contains 34.8% oxygen, 52.2% carbon and 13.0% hydrogen.
Solution:
| Element | Percentage | Atomic mass | Atomic ratio | Simplest ratio | Simplest whole number ratio |
|---|---|---|---|---|---|
| Oxygen | 34.8 | 16.0 | 34.8 / 16 = 2.175 | 2.175 / 2.175 = 1 | 1 |
| Carbon | 52.2 | 12 | 52.2 / 12 = 4.35 | 4.35 / 2.175 = 2 | 2 |
| Hydrogen | 13.0 | 1 | 13.0 / 1 = 13.0 | 13.0 / 2.175 = 6 | 6 |
Thus, the empirical formula of the above compound is C2H6O.
Empirical formula mass of C2H6O = (2 x 12) + (6 x 1) + 16 = 46
In this tutorial, we have discussed what is an empirical formula in chemistry and how to find an empirical formula of a compound. Hope that you will have understood the basic concepts of finding the empirical formula and practiced all important example numerical.
Thanks for reading!!!
