Some Basic Concepts of Chemistry

Mole and Equivalent Weight

If we ask you to count the total number of stars in the sky, can you do that? No. Similarly, scientists can not count the exact number of atoms and molecules as they are very tiny and present in a large number. Therefore, we have the concepts of mole and equivalent weight to make this calculation easy. In this chapter, we will study these two concepts and look at a few examples for the same.

Suggested Videos

previous arrow
next arrow
previous arrownext arrow


What Do You Mean By A Mole?

As we already mentioned, atoms and molecules are very tiny. That is why they are very difficult to be counted in exact figures. Well, this problem is solved by using the Avogadro’s number. This number is expressed as NA = 6.023 x 1023. We define a mole as the number equal to Avogadro’s number. This is just like you say dozen is equivalent to 12, a score is 20 and a century is 100.

In other words, a mole is a unit that we use to represent 6.023 x 1023 particles of the same matter. Describing further, a mole is the total amount of substance that contains as many atoms, molecules, ions, electrons or any other elementary entities as there are carbon atoms in exactly 12 gm of it. There are various kinds of moles of a substance that we can calculate. They are as follows:

  • The number of moles of molecules
  • Number of moles of atoms
  • The number of moles of gases (Standard molar volume at STP = 22.4 lit)
  • Number of moles of particles e.g. atoms, molecules ions etc

We calculate the mole fraction of a substance as the fraction of the substance in the mixture expressed in terms of mol. It is represented as X.

Browse more Topics under Some Basic Concepts Of Chemistry

Equivalent Weight

We define the equivalent weight of a substance (element or compound) as:

“The number of parts by weight of it, that will combine with or displace directly or indirectly 1.008 parts by weight of hydrogen, 8 parts by weight of oxygen, 35.5 parts by weight chlorine or the equivalent parts by weight of another element”.

It is important to know that the equivalent weight of any substance is dependent on the reaction in which it takes part. Equivalent weight is a relative quantity so it does not have any unit. When we express the equivalent weight of a substance in grams, we call it Gram Equivalent Weight (GWE).

How to Calculate Equivalent Weight?

Equivalent weight  = Molar mass / Valence factor

(The Valence factor for a base = acidity of the base, the Valence factor for an acid = basicity of the acid and the Valence factor for an element = valency)

Equivalent Weight


Titration is a procedure in which we can determine the concentration of an unknown solution with the help of solution, the concentration of which we know. In this procedure, we determine the concentration of solution A by adding carefully measured volumes of a solution of known concentration B. We continue adding these solutions until the reaction of A with B is just complete.

Law of Equivalence

Titration is based on the Law of Equivalence. This law states that,

“At the endpoint of a titration, volumes of the two titrants reacted have the same number of equivalents or milliequivalents”

Acid-Base Titration

One gram equivalent of acid is neutralized by one gm equivalent of base. It means,

One equivalent of Acid = One Equivalent of Base

Acid [N1V1] = Base [N2V2]

[Gram equivalent = Normality x Volume]

Let us understand this with the help of a simple example. Let us calculate the number of milliequivalents of H2SO4 present in 10 mL of N/2 H2SO4 solution. Milliequivalents = Normality x Volume (mL) = ½ × 10 = 5 milliequivalent of H2SO4

Limiting Reactant

The reactant that is completely consumed in the course of the reaction is the limiting reactant. When it is fully consumed, the reaction itself stops. However, what we must note is that the concept of limiting reactant is applicable to reactions other than monomolecular reactions. These are the reactions where more than one type of reactant is involved.

To determine the limiting reagent, first, we must be aware of the amount of all reactants and mole ratio of reactants. If the ratio of moles of reactant A with respect to reactant B is greater than the ratio of the moles of A to moles of B for a balanced chemical equation then B is the limiting reactant.

We can use the concept of stoichiometry to determine the other terms like left (unused) mass of other reactants, amount of formed product once we know the limiting reactant.

Solved Example for You

Q: Calculate the number of moles in each of the following:

  1. 392 grams of sulphuric acid
  2. 44.8 litres of carbon dioxide at STP
  3. 6.022 x 1023 molecules of oxygen


1) 1 mole of H2SO4= 98 g
Thus, 98 g of H2SO4= 1 mole of H2SO4
392 g of H2SO4 = 4 moles of H2SO4

2) 1 mole of CO2 = 22.4 litres at STP
i.e. 22.4 litres of CO2 at STP = 1 mole
44.8 litres of CO2 at STP = 2 moles CO2

3) 1 mole of O2 molecules = 6.022 x 1023 molecules.
6.022 x 1023 molecules = 1 mole of oxygen molecules.

Share with friends

Customize your course in 30 seconds

Which class are you in?
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.
Ashhar Firdausi
IIT Roorkee
Dr. Nazma Shaik
Gaurav Tiwari
Get Started

One response to “Mole and Equivalent Weight”

  1. says:

    Where are Equivalent weight video in chapter

Leave a Reply

Your email address will not be published. Required fields are marked *

Download the App

Watch lectures, practise questions and take tests on the go.

Customize your course in 30 seconds

No thanks.