Initially, chemical formulas were obtained by determination of masses of all the elements that are combined to form a molecule and subsequently we come up with two important types of formulas in chemistry : molecular formula and empirical formula.
The empirical formula of a compound gives the simplest ratio of the number of different atoms present, whereas the molecular formula gives the actual number of each different atom present in a molecule. If the formula is simplified then it is an empirical formula. The molecular formula is commonly used and is a multiple of the empirical formula.
The general statement relating molecular formula and the empirical formula is
What is Molecular formula?
• The molecular formula is the formula derived from molecules and is representative of the total number of individual atoms present in a molecule of a compound.
• A molecular formula uses a subscript that reports the actual number of each type of atom in a molecule of the compound.
• Molecular formulas are associated with gram molecular masses that are simple whole number multiples of the corresponding empirical formula mass.
What is Empirical formula?
• The empirical formula is the simplest formula for a compound which is defined as the ratio of subscripts of the smallest possible whole number of the elements present in the formula. It is also known as the simplest formula.
• An empirical formula for a compound is the formula of a substance written with the smallest integer subscript.
• The empirical formula gives information about the ratio of numbers of atoms in the compound. The percent composition of a compound directly leads to its empirical formula. Empirical Formula Vs Molecular Formula.
Empirical Formula |
Molecular Formula |
An empirical formula represents the simplest whole-number ratio of various atoms present in a compound |
The molecular formula shows the exact number of different types of atoms present in a molecule of a compound. |
Example : The empirical formula for Acetylene is CH . |
Example : The molecular formula for Acetylene is C2H2 . |
Let’s take the example of glucose. The molecular formula of glucose is C6H12O6 and the empirical formula of glucose is CH2O. We can derive a relation between the Molecular formula and the empirical formula of glucose.
C6H12O6 = 6 × CH2O
We can derive a general expression as,
Molecular formula = n x empirical formula where n is a whole number
Sometimes, the empirical formula and molecular formula both can be the same.
Question-1 : The empirical formula of Boron Hydride is BH3. Calculate the molecular formula when the measured mass of the compound is 27.66.
Solution :
The atomic mass is given by = B + 3(H) = 10.81 + 3(1) = 13.81 u
But, the measured molecular mass for Boron atom is given as 27.66 u. By using the expression,
Molecular formula = n x empirical formula n
= molecular formula/empirical formula
= 27.66 / 13.81
= 2
Putting the value of n = 2 in the empirical formula we get molecular formula as Molecular formula = 2(BH3) = B2H6.
Question-2 : The empirical formula of a compound is COCI2 and its molecular mass is 90.00u. Find out the molecular formula of that compound.
Solution :
COCI2 = C + O + 2(CI) = 12 + 16 + 2(35.5) = 99u
Empirical formula is the same as molecular mass as n=1, this means molecular formula is COCI2.
Question-3 : What is the molecular formula of a compound which has an empirical formula of CH2 and a relative molecular mass of 70?
Solution :
Relative molecular mass = 70
Empirical formula mass = 12 + 2 = 14
The relative molecular mass is 5 x the relative empirical formula mass
The molecular formula is 5 x the empirical formula The molecular formula is C5H10 .
Question-4 : A compound of iron and oxygen is analysed and found to contain 69.94% iron and 30.06% oxygen. Find the empirical formula of the compound.
Solution :
Steps for Problem Solving |
Find the empirical formula of a compound of 69.94% iron and 30.06% oxygen. |
Identify the “Given information and what the problem is asking you to find.” |
Given : % of Fe = 69.94% % of O = 30.06% Find : Empirical Formula = Fe?O? |
Calculate |
|
a. Assume a 100g sample, convert the same % value to grams. |
69.94g Fe 30.06g O |
b. Covert to moles. |
69.94g Fe x 1 mol Fe = 1.252 mol Fe 55.85g Fe
30.06g O x 1 mol O = 1.879 mol O 16.00g O |
c. Divide both moles by the smallest of the results. |
Fe x 1 .252 mol 1.252
O x 1.879 mol 1.252 The “non-whole” empirical formula of the compound is Fe1O1.5
|
d. Multiply each of the moles by the smallest whole number that will convert each into a whole number. |
Since the moles of O is still not a whole number, both moles can be multiplied by 2, while rounding to a whole number. |
e. Write the empirical formula |
The empirical of the compound is Fe2O3 |