JEE Main Class 11 Syllabus 2025 PDF for Paper 1 and 2

Percent Composition Formula - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

18 Questions around this concept.

Solve by difficulty

A metal chloride contains 55.0% of chlorine by weight. 100 mL vapors of the metal chloride at STP weigh 0.57 g. The molecular formula of the metal chloride is: (Given the atomic mass of chlorine is 35.5u )

$\mathrm{MCl}$

$\mathrm{MCl}_3$

$\mathrm{MCl}_2$

$\mathrm{MCl}_4$

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Concepts Covered - 1

Percentage Composition And Equivalent Weight

Percentage Composition:

The percentage combination of the compound is the relative mass of each of the constituent elements in 100 parts of it.

$\mathrm{Mass\ \% \ of\ an\ element = \frac{\textrm{Mass of that element in one mole of the compound}}{\textrm{Molar mass of the compound}}\times100}$

Let us take an example of water (H₂O), it contains hydrogen and oxygen, the percentage composition of both these elements can be calculated as follow:

The molar mass of water = 18.02 g

$\mathrm{Mass\ \% \ of\ Hydrogen = \frac{2\times1.008}{18.02}\times100}=11.18 \%$

$\mathrm{Mass\ \% \ of\ Oxygen = \frac{16.00}{18.02}\times100}=88.79 \%$

One can check the purity of a given sample by analyzing percentage composition.

Equivalent Weight:

Equivalent weight is the weight of an element or a compound that combines with or displaces 1 gram of hydrogen or 8 grams of oxygen, or 35.5 part by weight of Chlorine.
Equivalent weight is a number and when it is denoted in grams, it is called gram equivalent.
It depends upon the nature of chemical reaction in which substance takes part

How To Find Equivalent Weight:

$\mathbf{Equivalent \:Weight = \frac {Molecular \:weight}{n- factor(x)}}$

n-Factor or Valence Factor:

It calculates the molar ratio of the species taking part in reactions that are, reactants. The reciprocal of the n-factor 's ratio of the reactants represents the molar ratio of the reactants. For example, If A (having n-factor = a) reacts with B (having n-factor = b) then its n-factor's ratio is a: b, so molar ratio of A to B is b: a.
It can be represented as follows:
$\\\mathrm{bA\,\: \: \: \: \: \: \: \: \:\: \: \: \: \: +\: \: \: \: \: \: \: \: \: \: \: \, aB\, \rightarrow \, Product}\\(\mathrm{n-factor=\, a)\:\: \: \: \: \: (n-factor=\, a)}$

Calculation of n-Factor
Before calculating the n-factor of any of the reactants in a given chemical reaction we must have a clear idea about the type of reaction. The reaction may be any of these types:

(i) Acid-base or neutralization reaction

(li) Redox reaction

Acid-Base or Neutralization Reactions:
As we know that according to the Arrhenius concept, "An acid provides H⁺ ion(s) while a base provides OH^- ion(s) in neutralization these H⁺ and OH^- ion/ions combines together".
The number of H⁺ ion(s) and OH^- ion(s) represent n-factor for acid and base respectively, that is, basicity and acidity respectively.
Example,
$\\ \mathrm{HCl\, \rightarrow \, H^{+}\, +\, Cl^{-} }\\ (\mathrm{n=1)\, that\, is,\, monobasic\, acid} \\\\ \mathrm{H_{2}SO_{4}\, \rightarrow \, 2H^{+}\, +\, SO_{4}^{2-}} \\ \mathrm{(n=2)\, that\, is,\, dibasic\, acid}$

Redox Reactions
These reactions involve oxidation and reduction simultaneously. Here the exchange of electrons occurs. To find the n-factor for Oxidizing or agent we must find out the change in the oxidation state of these species.

You will be learning the following in detail in the chapter of redox. For now, just look at the definition. Sufficient questions will be practiced later.

For Redox Reactions:
E = (Molecular weight) / (Change in oxidation number),

x= change in oxidation state
For Example, for KMnO₄
(a) In acidic medium: E = M/5

$\mathrm{\overset{+7}{2KMnO_{4}}\, +\, 3H_{2}SO_{4}\, \rightarrow K_{2}SO_{4} + \overset{+2}{2MnSO_{4}} + 3H_{2}O + 5 [O]}$
5 unit change in oxidation number.

(b) In basic medium: E = M/1

$\mathrm{\overset{+7}{2KMnO_{4}}\, +\, 2KOH\, \rightarrow \overset{+6}{2K_{2}MnO_{4}}\, +\, H_2O\, +\,[O]}$
one unit change in oxidation number

(c) In neutral medium: E = M/3

$\mathrm{\overset{+7}{2KMnO_{4}} + H_{2}O\, \rightarrow 2KOH + \overset{+4}{2MnO_{2}} + 3[O]}$
3 unit change in oxidation number

Formulae for calculation of Equivalent Weight:

For Acids:
E = (Molecular weight) / (Protocity or Basicity of Acid), x= number of furnishable protons
For Example, for H₃PO₄, E = M/3
For H₂SO₄ , E =M/2
For Bases:
E = (Molecular weight) / (Acidity or number of OH^- ions),x= number of furnishable OH^- ions
For Example, for Ca(OH)₂, E = M/2
For Al(OH)₃, E =M/3
For Ions:
E = (Molecular weight) / (Charge on ion), x= charge on ion
For Example, for SO₄^2-, E = M/2
For PO₄^3-, E = M/3
For Compounds:
E = (Molecular weight) / (total positive charge or negative charge present in compound),