Determination of Empirical Formulae - Determination of Molecular Formulae - Definitions, Examples, practice problems & FAQs

During research you have discovered a chemical species which has some very useful properties now how can you write its exact chemical formula (molecular formula).

First, We test this chemical in the lab and find its constituents and percentage composition of constituents. Then with suitable techniques (like osmotic pressure), find its molecular mass with minimum error. Then we can easily write its exact molecular formula.

Table of contents

Empirical formula
Rules to finding empirical formula of any compound if percentage composition is known
Practice problems
Frequently asked questions-FAQs

Empirical formula

It is very easy to calculate the percentage of elements present in a molecule if the molecular formula is known but if the percentage of elements are known then to find the molecular formula is not a simple task.

First, we have to calculate the relative number of atoms of each element in the particular compound. Then, we can find the empirical formula of the compound. From some other technique, we have to find the molecular mass of the compound then we are able to calculate the exact molecular mass of the compound.

The empirical formula is a chemical formula presenting the simplest ratio of elements in a compound (nor exact number).

The molecular formula gives the exact number of different atoms present in a compound.

E.g- if empirical formula id CH then the molecular formula can be C₂H₂, C₆H₆

if empirical formula id CH₂ then the molecular formula can be C₂H₄, C₃H₆ etc.

E.g- Molecular formula of Glucose is C₆H₁₂O₆ but empirical formula is CH₂

Molecular formula of benzene is C₆H₆ but empirical formula is CH

With the help of empirical formula, the exact molecular formula can be calculated only if the molecular mass of the mentioned compound is known.

So, Molecular formula = n empirical formula

Rules to finding empirical formula of any compound if percentage composition is known

Step 1: Relative mole of atoms

If a compound on analysis gave the following results: C = 39.99%, H = 6.71% and O = 53.28%.

Element	Percentage	Molar weight (g)	No. of moles
C	39.99%	12	3.33
H	6.71%	1	6.71
O	53.28%	16	3.33

Step 2: Find the simplest ratio of moles of atoms

Element	Percentage	Molar weight (g)	No. of moles	Simplest whole no. ratio
C	39.99%	12	= 3.33	= 1
H	6.71%	1	= 6.71	= 2
O	53.28%	16	= 3.33	= 1

Thus, CH₂O this is the empirical formula of the above compound.

Step 3: Find empirical formula then multiply the empirical formula by n to find the molecular formula.

Empirical formula mass for CH₂O = 12 + 2 + 16 = 30

Molecular formula = n × Empirical formula

= 6 × (CH₂O)

= C₆H₁₂O₆

Practice problems

Q1. Empirical formula if acetylene and benzene are

A. CH
B. CH₂
C. CH & CH₂
D. C₂H₂ & C₆H₆

Answer: (A)

Solution: molecular formula of acetylene and benzene are C₂H₂ & C₆H₆ respectively.

Empirical formula is a chemical formula presenting the simplest ratio of elements in a compound (nor exact number).

So, the empirical formula of both compounds is CH.

Q2. A compound on analysis gave the following results: C = 39.99 %, H = 6.71 %, and O = 53.28 %. Determine the molecular formula of the compound if the molar mass is 180 g mol^-1.

A. C₆H₁₂O₆
B. CH₂O
C. C₆H₁₀O₆
D. C₅H₁₂O₆

Solution: (A)

Step 1: Find the moles of each element

Given: Mass % of C = 39.99 %, H = 6.71 % and O = 53.28 %.

Assuming mass of compound to be 100 g

Molar mass of C = 12 g mol^-1, H = 1 g mol^-1l and O = 16 g mol^-1.

Given Mass of C = 39.99 g, moles of C = = 3.33 mol

Given Mass of H = 6.71 g, moles of H = 6.71 mol

Given Mass of O = 53.28 g, moles of O = = 3.33 mol

Step 2: Find the empirical formula

For making the elements in a simple whole number ratio, we divide by the element having the least moles (by carbon or oxygen)

Simple ratio of C = = 1

Simple ratio of H = = 2

Simple ratio of O = = 1

Carbon : Hydrogen : Oxygen is 1 : 2 : 1

Empirical formula becomes = CH₂O

Step 3: Find the molecular formula

Empirical formula mass = 12 + (1 x 2) + 16 = 30 g

Given: Molar mass = 180 g mol^-1

So, Molecular formula = n empirical formula

Molecular formula = n × Empirical formula= 6 × (CH₂O) = C₆H₁₂O₆

Q3. In an organic compound of molar mass 108 g mol-1, C, H, and N atoms are present in 9: 1: 3.5 ratio by weight. The molecular formula can be

A. C₆H₈N₂
B. C₇H₁₀N
C. C₅H₁₆N₃
D. C₄H₁₈N₃

Answer: (A)

Solution: Step 1: Converting the given ratio into simple whole numbers ratio

Given: The mass ratio of C: H: N = 9: 1: 3.5 = 36: 4: 14

Step 2:

Finding the mole ratio and empirical formula

Dividing individual masses by respective molar mass.

Hence, the empirical formula becomes C₃H₄N.

Step 3:

Find molecular formula

For calculating the Molecular formula we need to

use the formula

Given: Molecular mass = 108 g mol^-1

Hence, the molecular Formula is C₆H₈N₂

Q4. The simplest formula of a compound containing 50 % by mass of element X (atomic weight 10 amu) and 50 % by mass of element Y (atomic weight 20 amu) is

A. XY₂
B. X₂Y
C. X₂Y₂
D. XY₃

Answer: Step 1: Find the number of moles in 1 g of the sample

Given: 50% by mass of both X and Y

Molar mass of X = 10 g mol^-1

Molar mass of Y = 20 g mol^-1

Assuming that we have 1 g of the compound.

So, we have 0.5 g of X and 0.5 g of Y.

Now, find the number of moles,

Moles of X = = 0.050 and Y = = 0.025.

Step 2:

Find the simple whole-number ratio

So, X: Y = 0.050: 0.025

X = 2 and Y = 1

Hence, the formula is X₂Y

Frequently asked questions-FAQs

Question 1. Which methods are generally used for calculation of the molecular mass of unknown compounds?
Answer: Generally, the osmotic pressure technique is used to calculate the molecular mass of unknown compounds, elevation in boiling points, and depression in freezing points also can be used.

Question 2. Can any compound have the same empirical and molecular formula?
Answer: Yes, compounds have the same molecular formula and empirical formula

E.g- formaldehyde

Molecular formula = HCHO, Empirical formula = HCHO

Question 3. Can molecular formula apply to ionic compounds?
Answer: No, the Molecular formula is not defined for the ionic compounds. To find the formula of an ionic compound, first, identify the cation and write down its symbol and charge. Then, identify the anion and write down its symbol and charge. Finally, combine the two ions to form an electrically neutral compound.

Question 4. Can two or more two compounds have the same empirical formula?
Answer: yes, e.g- The empirical formula is acetylene and benzene is CH.

Their molecular formula is C₂H₂& C₆H₆

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