A Molecular Formula represents the actual number of atoms of each element present in one molecule of a compound. It tells us the exact composition of the molecule but does not give information about the arrangement of atoms. In simple terms, the molecular formula shows how many atoms of each element are present in a single molecule of a substance.
Examples:
Glucose: C6H12O6
This means one molecule of glucose contains 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.
Empirical Formula
- The Empirical Formula represents the simplest whole-number ratio of atoms of different elements in a compound.
- It does not show the actual number of atoms, but only the simplest ratio between them.
Example:
For glucose
Molecular formula: C6H12O6Divide each subscript by 6:
C6H12O6 →CH2O
So, the empirical formula of glucose is CH2O
This means the simplest ratio of C : H : O = 1 : 2 : 1.
There is a relationship between molecular formula and empirical formula, which is given by,
M = n × E
where,
M is the molecular formula
n is the ratio of molar mass and empirical formula mass
E is the empirical formula
Calculating Molecular Formula
To determine the Molecular Formula of a compound, we use its Empirical Formula and the molar mass of the compound. The molecular formula shows the actual number of atoms of each element present in a molecule, while the empirical formula shows the simplest ratio of those atoms.
Steps to Determine Molecular Formula:
Step 1: Find the Empirical Formula
First, determine the empirical formula of the compound. This gives the simplest whole-number ratio of the elements present in the compound.
Example:
Empirical formula = CH2 O
Step 2: Calculate the Empirical Formula Mass (EFM)
Find the mass of the empirical formula by adding the atomic masses of all atoms in it.
For CH2 O :
12 + ( 2 × 1 ) + 16 = 30
So, Empirical Formula Mass = 30
Step 3: Find the Value of n
Divide the molar mass of the compound by the empirical formula mass.
n = \frac{\text{Molar Mass of Compound}}{\text{Empirical Formula Mass}}
Step 4: Determine the Molecular Formula
Multiply the subscripts of the empirical formula by the value of n
Empirical formula: CH2O
(CH2O)6 = C6H12O6
So, the molecular formula is C6H12O6
Solved Examples
Example 1: A compound is composed of 82.78% nitrogen and 17.22% hydrogen. If its molecular weight is 17.031 g/mol, then find its molecular formula.
Solution:
The molecular weight of a compound = 17.031 g/mol
Nitrogen percentage in the given compound = 82.78%
Hydrogen percentage in the given compound = 17.22%
Step 1: Multiply the molecular weight with the given component percentage.
Nitrogen = 17.031 × (82.78/100) = 14.0982
Hydrogen = 17.031 × (17.22/100) = 2.9327
Step 2: Divide each value obtained by the atomic weight of that atom.
Nitrogen: 14.0982/14.0067 = 1.00654
Hydrogen: 2.9327/1.00794 = 2.90960
Step 3: Round off the obtained values to the closest whole number.
Nitrogen: 1
Hydrogen: 3
Thus, the molecular formula of the given compound is NH3 .
Example 2: The empirical formula of a compound of carbon, hydrogen, and oxygen is HO. If its molar mass is 60.052 g/mol , then determine the molecular formula of the compound.
Solution:
The empirical formula of a compound = HO
The molar mass of the compound = 60.052 g/mol
Step 1: First, let's calculate the empirical formula molar mass.
Empirical formula molar mass (EFM) = 1.00794 + 15.9994
= 17.007 g/mol
Step 2: Now, divide the molar mass of the given compound by the empirical formula mass.
n = Molar mass/EMF = 60.052/17.007 = 2
Step 3: Molecular Formula = n × (Empirical formula)
So, the molecular Formula of the given compound = 2 × (HO) = H2O2
Hence, the molecular formula of the compound is H2O2 .
Example 3: A compound is composed of 68.29% carbon, 12.02% hydrogen, and 21.69% oxygen. If its molecular weight is 86.136 g/mol, then find its molecular formula.
Solution:
The molecular weight of a compound = 86.136 g/mol
Carbon percentage in the given compound = 68.29%
Hydrogen percentage in the given compound = 12.02%
Oxygen percentage in the given compound = 21.69%
Step 1: Multiply the molecular weight with the given component percentage.
Carbon = 86.136 × (68.29/100) = 58.8223
Hydrogen =86.136 × (12.02/100) = 10.3535
Oxygen = 86.136 × (21.69/100) = 18.6828
Step 2: Divide each value obtained by the atomic weight of that atom.
Carbon: 58.8223 /12.0107 = 4.8973
Hydrogen: 10.3535/1.00794 = 10.2719
Oxygen: 18.6828/15.9994 = 1.1677
Step 3: Round off the obtained values to the closest whole number.
Carbon: 5
Hydrogen: 10
Oxygen: 1
Thus, the molecular formula of the given compound is C5H10O .
Example 4: Oxalic acid is composed of 27.42% carbon, 2.33% hydrogen, and 70.25% oxygen. If its molecular weight is 90.035 g/mol, then find its molecular formula.
Solution:
The molecular weight of oxalic acid = 90.035 g/mol
Carbon percentage in oxalic acid = 27.42%
Hydrogen percentage in oxalic acid = 2.33%
Oxygen percentage in oxalic acid = 70.25%
Step 1: Multiply the molecular weight with the given component percentage.
Carbon = 90.035 × (27.42/100) = 24.6875
Hydrogen = 90.035 × (2.33/100) = 2.0978
Oxygen = 90.035 × (70.25/100) = 63.2496
Step 2: Divide each value obtained by the atomic weight of that atom.
Carbon = 24.6875/12.0107 = 2.0554
Hydrogen: 2.0978/1.00794 = 2.08127
Oxygen: 63.2496/15.9994 = 3.9532
Step 3: Round off the obtained values to the closest whole number.
Carbon: 2
Hydrogen: 2
Oxygen: 4
Thus, the molecular formula of the given compound is C2H2O4.
Example 5: The empirical formula of a compound of carbon, hydrogen, and oxygen is CH2O. If its molar mass is 60.052 g/mol, then determine the molecular formula of the compound.
Solution:
The empirical formula of a compound = CH2O
The molar mass of the compound = 60.052 g/mol
First, let's calculate the empirical formula molar mass.
Empirical formula molar mass (EFM) = 12.0107 + 2 × 1.00794 + 15.9994
= 30.026 g/mol
Now, divide the molar mass of the given compound by the empirical formula mass.
n = Molar mass/EMF = 60.052/30.026 = 2
Molecular Formula = n × (Empirical formula)
So, the molecular Formula of the given compound = 2 × (CH2O) = C2H4O2
Hence, the molecular formula of the compound is C2H4O2.
Example 6: Boric acid is composed of 21.14% boron, 4.65% hydrogen, and 74.21% oxygen. If its molecular weight is 61.83 g/mol, then find its molecular formula.
Solution:
The molecular weight of boric acid = 61.83 g/mol
Boron percentage in boric acid = 21.14%
Hydrogen percentage in boric acid = 4.65%
Oxygen percentage in boric acid = 74.21%
Step 1: Multiply the molecular weight with the given component percentage.
Boron = 61.83 × (21.14/100) = 13.0709
Hydrogen = 61.83 × (4.65/100) = 2.8750
Oxygen = 61.83 × (74.21/100) = 45.8840
Step 2: Divide each value obtained by the atomic weight of that atom.
Boron: 13.0709/10.811 = 1.209
Hydrogen: 2.2568/1.00794 = 2.8524
Oxygen: 45.8840/15.9994 = 2.8679
Step 3: Round off the obtained values to the closest whole number.
Boron: 1
Hydrogen: 3
Oxygen: 3
Thus, the molecular formula of the given compound is H3BO3.
Example 7: The empirical formula of a compound that is composed of hydrogen, chlorine, and oxygen is HClO. If its molar mass is 52.46 g/mol, then determine the molecular formula of the compound.
Solution:
The empirical formula of a compound = HClO
The molar mass of the compound = 52.46 g/mol
First, let's calculate the empirical formula molar mass.
Empirical formula molar mass (EFM) = 1.00794 + 35.453 + 15.9994
= 52.460 g/mol
Now, divide the molar mass of the given compound by the empirical formula mass.
n = Molar mass/EMF = 52.460/52.46 = 1
Molecular Formula = n × (Empirical formula)
So, the molecular Formula of the given compound = 1 × (HClO) = HClO
Here, the empirical formula and the molecular formula of the given compound are the same.
Hence, the molecular formula of the compound is HClO.