The chemical formula of a compound gives the exact number of moles of each element which have combined to form the compound. This gives the identity of a compound and also, its nature. Some elements, say carbon, can combine with hydrogen in many ratios. So, another type of formula, used to predict a chemical compound is the empirical formula. It is an important entity in organic chemistry.
Empirical Formula Definition
Empirical formula is the simplest ratio of all the elements present in a compound. Empirical formula only gives the ratio in which the elements are present in a compound.
The empirical formula of a compound is the “Simplest formula which expresses its percentage composition”.
The empirical formula of a compound is the “Simplest formula which expresses its percentage composition”.
So, any number of compounds can have the same empirical formula, since many compounds can be made with the same ratio of elements in them.
Determining empirical formula is explained in detail below :
1. Some chemists carry out chemical reactions called synthesis to produce new compounds for many different uses.
2. Chemists must be able to determine the identity of the substance produced by these synthesis reactions.
3. One procedure that the chemists use to identify an unknown substance is the elemental analysis.
4. This technique tells the chemists about the elements present in the compound that is synthesized and the percent by mass of these elements.
5. The information gained from such an analysis is the “percent composition” of the substance synthesized.
6. For example, a compound that is 88.7% Oxygen and 11.3% Hydrogen has a percentage composition that matches that of water, H2O. A compound that is 60.7% chlorine and 39.3% sodium has a percentage composition that matches Sodium chloride, NaCl.
7. These data along with the further tests are used to identify an unknown compound
Note: For a compound, the empirical formula is the same as the molecular formula. Such is the case of Water, H2O and Nitric acid, HNO3. But, for some compounds, molecular formula is the whole number multiple of empirical formula.
For example,
For example,
Compound
|
Compound Empirical formula
|
Molecular formula
|
Acetic acid
|
CH2O
|
(CH2O) x 2 = C2H4O2
|
Glucose
|
CH2O
|
(CH2O) X 6 =C6H12O6
|
So, molecular formulas are calculated using empirical formula and molar mass data. The empirical formula is a multiple of the molecular formula, as shown in the relationship:
n(empirical formula) = Molecular formula
where, ‘n’ will always be a whole number and it is occasionally a very large number. With acetic acid, it is 2, but with glucose, it is 6. Likewise, true molar mass should be a multiple of the empirical formula mass.
Example: 1
A hydrocarbon was found to contain the following percentage composition.
C = 92.3%, H = 7.69%. Calculate its empirical formula (C=12, H=1)
Solution:
The given problem is solved in the following table
Element
|
Symbol
|
Relative atomic mass
|
% Composition
|
Relative number of atoms
|
Simple ratio
|
Simple whole number ratio
|
Carbon
|
C
|
12
|
93.3
|
92.312 = 7.69
|
7.697.69
|
1
|
Hydrogen
|
H
|
1
|
7.69
|
7.691 = 7.69
|
7.697.69
|
1
|
Empirical formula = C1H1 = CH
Example: 2
Example: 2
A substance was found to have the following percentage composition;
Ca = 40% C =12%, O = 48%. Calculate its empirical formula. (Relative atomic mass of Ca=40, C=12, O=16)
Ca = 40% C =12%, O = 48%. Calculate its empirical formula. (Relative atomic mass of Ca=40, C=12, O=16)
Solution:
The given problem is solved in the following table.
The given problem is solved in the following table.
Element
|
Symbol
|
Relative atomic mass
|
% composition
|
Relative number of atoms
|
Simple ratio
|
Simple whole number ratio
|
Calcium
|
Ca
|
40
|
40
|
4040 = 1
|
1
|
1
|
Carbon
|
C
|
12
|
12
|
1212 = 1
|
1
|
1
|
Oxygen
|
O
|
16
|
48
|
4816 = 3
|
3
|
3
|
Empirical formula = Ca1C1O3 = CaCO3
Example: 3
An organic compound was found to contain the following percentage composition:
Example: 3
An organic compound was found to contain the following percentage composition:
H = 2.22%, C = 26.67%, O = 71.11%. Find its empirical formula. (Relative atomic mass of H=1, C=12, O=16)
solution:
The given problem is solved in the following table.
Element
|
Symbol
|
Relative atomic mass
|
% composition
|
Relative number of atoms
|
Simple ratio
|
Simple whole number ratio
|
Hydrogen
|
H
|
1
|
2.22
|
2.221 = 2.22
|
2.222.22 = 1
|
1
|
Carbon
|
C
|
12
|
26.67
|
26.6712 = 2.22
|
2.222.22 = 1
|
1
|
Oxygen
|
O
|
16
|
71.11
|
71.1116 = 4.44
|
4.442.22 = 2
|
2
|
Empirical formula = H1C1O2 = HCO2.
Example: 4
A compound was found to have the following percentage composition: Nitrogen = 21.21%, Hydrogen= 6.06%, Sulfur = 24.24% and Oxygen = 48.48%
A compound was found to have the following percentage composition: Nitrogen = 21.21%, Hydrogen= 6.06%, Sulfur = 24.24% and Oxygen = 48.48%
(N=14, H=1, S=32, O=16). Find its empirical formula.
Solution:
The given problem is solved in the following table.
Solution:
The given problem is solved in the following table.
Element
|
Symbol
|
Relative atomic mass
|
% composition
|
Relative number of atoms
|
Simple ratio
|
Simple whole number ratio
|
Nitrogen
|
N
|
14
|
21.21
|
21.2114 = 1.15
|
1.50.75 = 2
|
2
|
Hydrogen
|
H
|
1
|
6.06
|
6.061 = 6.06
|
6.060.75 = 8.08
|
8
|
Sulfur
|
S
|
32
|
24.24
|
24.2432 = 0.75
|
0.750.75 = 1
|
1
|
Oxygen
|
O
|
16
|
48.48
|
48.4816 = 3.03
|
3.030.75 = 4.04
|
4
|