Percent Composition Formula: Definition, Questions and Examples

To determine the molecular formula of a compound, we begin by measuring the mass or mass percentage of each element present in the compound. The composition is generally expressed as the mass percentage composition. The mass percentage gives the mass of each element expressed as the percentage of the total mass. In other words, it gives the number of grams of the element present in 100 g of the compound.

This Story also Contains

1. Percentage Composition:
2. n-Factor or Valence Factor:
3. Calculation of n-Factor
4. Some Solved Examples
5. Summary

Percentage Composition:

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

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 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 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

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 option is the 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).

Summary

Percentage composition is very important and it is in terms of chemistry it refers to as the percentage by mass of every atom present in the compound. There are various steps to find it first determine the molecular formula by calculating the molar mass of the compound, find the mass of each atom in the compound, and after doing all that find the percentage composition. percentage composition has various significances in the field of the mole concept.