Percent Composition Formula: Definition, Questions and Examples

Percentage Composition:

The percentage of any element or constituent in a compound is the number of parts by mass of that element or constituent present in 100 parts by mass of the compound.

Mass $\%$ of an element $=\frac{\text { Mass of that element in one mole of the compound }}{\text { Molar mass of the compound }} \times 100$

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

The molar mass of water = 18.02 g

Mass $\%$ of Hydrogen $=\frac{2 \times 1.008}{18.02} \times 100=11.18 \%$

Mass $\%$ of Oxygen $=\frac{16.00}{18.02} \times 100=88.79 \%$

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

Equivalent Weight:

• Equivalent weight of any substance is that weight which react or liberate 1g of hydrogen (11.2 litre of H₂ at STP) or 8g of oxygen or 35.5g of chlorine or 80g of bromine or 127g of iodine or 108g of silver or 1 mole of electron.

• Equivalent weight is a number and when it is denoted in grams, it is called gram equivalent.

• It depends upon the nature of the chemical reaction in which the substance takes part

How To Find Equivalent Weight:

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

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 the molar ratio of A to B is b: a.
It can be represented as follows: $\begin{array}{ll}\mathrm{bA} \quad+ & \mathrm{aB} \rightarrow \text { Product } \\ (\mathrm{n}-\text { factor }=\mathrm{a}) & (\mathrm{n}-\text { factor }=\mathrm{a})\end{array}$

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 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 the n-factor for acid and base respectively, that is, basicity and acidity respectively.
  Example,

  $\mathrm{HCl} \rightarrow \mathrm{H}^{+}+\mathrm{Cl}^{-}$ $(\mathrm{n}=1)$ that is, monobasic acid

  $$
  \mathrm{H}_2 \mathrm{SO}_4 \rightarrow 2 \mathrm{H}^{+}+\mathrm{SO}_4^{2-}
  $$

  $(\mathrm{n}=2)$ 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

$2 \mathrm{KMnO}_4^{+7}+3 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4^{+2}+3 \mathrm{H}_2 \mathrm{O}+5[\mathrm{O}]$
5 unit change in oxidation number.

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

$2 \mathrm{KMnO}_4^{+7}+2 \mathrm{KOH} \rightarrow 2 \mathrm{~K}_2 \stackrel{+6}{\mathrm{MnO}_4}+\mathrm{H}_2 \mathrm{O}+[\mathrm{O}]$
one unit change in oxidation number

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

$2 \mathrm{KMnO}_4^{+7}+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{KOH}+2 \stackrel{+4}{\mathrm{MnO}_2}+3[\mathrm{O}]$
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),

x= total positive charge or negative charge present in the compound
For Example, for CaCO₃, E = M/2
For AlCl₃, E =M/3

• For Acidic Salt:
  E = (Molecular weight) / (Number of replaceable H-atoms)
  For example, for H₃PO₄

$\begin{aligned} & 2 \mathrm{NaOH}+\mathrm{NaH}_2 \mathrm{PO}_4 \rightarrow \mathrm{Na}_3 \mathrm{PO}_4+2 \mathrm{H}_2 \mathrm{O} \\ & \mathrm{E}=\mathrm{M} / 2\end{aligned}$

• Metal displacement method

• E₁ / E₂ = W₁ / W₂

Also Read:

• Empirical And Molecular Formula
• Stoichiometric Calculations
• Gravimetric Analysis

Recommended topic video on (Percent Composition Formula)

Some Solved Examples

Que 1: Find the equivalent mass of H₃PO₄ in the reaction:

$\mathrm{Ca}(\mathrm{OH})_2+\mathrm{H}_3 \mathrm{PO}_4 \rightarrow \mathrm{CaHPO}_4+2 \mathrm{H}_2 \mathrm{O}$

1) 55

2) 43

3) 49

4) 37

Solution

In this given reaction only two hydrogen atoms are replaced so their equivalent mass will be given as follows:

Equivalent mass of H₃PO₄ = (Molecular Mass of H₃PO₄ ) / 2

=98 / 2

= 49

Hence, the answer is option (3).

Que 2: 74.5 g of metallic chloride contains 35.5 g of chlorine. The equivalent weight of the metal is

1) 19.5

2) 35.5

3) 39

4) 78

Solution

74.5 g of metallic chloride contains 35.5 g of chlorine.

Therefore, the weight of another element present in the chloride is

= Weight of Metal Chloride - Weight of Chloride

= 74.5 - 35.5 = 39 g

Thus, 39 g of the metal combines with 35.5 g of Chlorine

Mole of Chloride present = weight/molar mass = 35.5/35.5 = 1 mole

So, one mole of Cl is present and we know the charge on Cl is (-1).

So, MCl configuration will be there where M-metal.

MCl = M⁺+ Cl^-

So, one mole of Metal will be present and the charge would be (+1) on metal.

Molecular weight = mass of 1 mol metal = 39 g

Equivalent Weight = Molecular weight/ Valency.

Equivalent Weight = 39/1 =39

hence, the equivalent weight of the metal is 39.

Hence, the answer is the option (3).

Que 3: The oxide of a metal has 32% oxygen. Its equivalent weight would be

1) 17

2) 34

3) 32

4) 8

Solution

100g of the oxide contains 68g of metal and 32g of oxygen

$\therefore$ 8g of oxygen will be present with 17g of the metal.

$\therefore$ The equivalent weight of the metal is 17.

Hence, the answer is the option (1).

Que 4: What is the equivalent weight of Sulphuric Acid (H₂SO₄) when it behaves as a diprotic acid?

1) 98

2) 49

3) 24.5

4) 12.25

Solution

n-factor of H₂SO₄ = 2

$\therefore$ Equivalent weight $=\frac{98}{2}=49$

Hence, the answer is the option (2)

Que 5. A compound possesses 8% sulfur by mass. The least molecular mass is

1) 200

2) 400

3) 155

4) 355

Solution

Let, the molar mass of the compound be M.

The compound contains 8% by weight of Sulphur, mathematically it implies

$\frac{8}{100} \times M=32$

$\Rightarrow M=400 \mathrm{~g}$

Hence, the answer is the option (2).

Practice more Questions from the link given below:

Summary

Percentage composition is very important in chemistry because it helps in determining proportion of each element within a compound.This information is vital for understanding a compound's properties, identifying it, and calculating its empirical and molecular formulas.It’s also a analytical educational tool—serving as a bridge between chemical formulas and real-world applications. Its ability to authenticate purity and accuracy strengthens both lab precision and industrial credibility.