Given:
Molecular weight of compound=136 g.
Percentage composition is C=70.54%, H=5.87% and O=23.52%
Step-1: Calculate the empirical formula
- Empirical formula: Empirical formula is the simplest formula of the compound which gives the simplest whole number ratio of the atoms of the various elements present in one molecule of the compound.
| Element | % Of Element | Atomic mass (g) | Atomic ratio | Simplest ratio |
| Carbon (C) | 70.54 | 12 | 70.5412=5.87 | 5.871.47=3.99≈4 |
| Hydrogen (H) | 5.87 | 1 | 5.871=5.87 | 5.871.47=3.99≈4 |
| Oxygen (O) | 23.52 | 16 | 23.5216=1.47 | 1.471.47=1 |
As the simplest ratio of Carbon is 4, Hydrogen is 4 and Oxygen is 1, therefore the empirical formula comes out to be C4H4O.
Step-2: Calculate the empirical and molecular mass
Empirical formula weight: Empirical formula weight is obtained by the addition of the atomic weight of the various atoms present in the empirical formula.
Atomic weight of Carbon= 12 g
Atomic weight of Hydrogen= 1 g
Atomic weight of Oxygen= 16 g
Empirical formula weight= 4(Atomic weight of Carbon) + 4(Atomic weight of Hydrogen) + Atomic weight of Oxygen
Empirical formula weight= 4(12) + 4(1) + 16= 68 g
Molecular formula weight: Molecular formula weight is obtained by the addition of the atomic weight of the various atoms present in the molecular formula.
Molecular weight =136 g (given)
Step-3: Calculate the molecular formula
- Molecular formula: Molecular formula is the actual formula of the compound which gives the actual number of the atoms present in the molecule of the compound.
- Relationship between empirical formula and molecular formula is given as follows: Molecular formula = n × Empirical formula, where n is an integer such as 1,2,3, etc.
- n=MolecularweightofthecompoundEmpiricalformulamass
- n=13668=2
As stated above, Molecular formula = n× Empirical formula, and empirical formula is C4H4O.
Therefore, the molecular formula is (C4H4O)2 or C8H8O2.
