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

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.

image

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 = nimage 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

image= 3.33

image= 1

H

6.71%

1

image= 6.71

image= 2

O

53.28%

16

image= 3.33

image= 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

image

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 = image= 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 = image= 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 =image = 1

Simple ratio of H =image = 2

Simple ratio of O =image = 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

image

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.

image

Hence, the empirical formula becomes C3H4N.

Step 3:

Find molecular formula

For calculating the Molecular formula we need to

use the formula

image

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 = image = 0.050 and Y = image= 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

Related topics

Percentage composition

Atomic Number and Mass Number

mole

Atomic mass

Average atomic mass

Law of chemical reaction

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