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Percent Composition Formula - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • 7 Questions around this concept.

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Percentage Composition And Equivalent Weight

Percentage Combination:

percentage combination of the compound is the relative mass of the each of the constituent element 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 which combines with or displaces 1.008 part by weight of H2 or 8 part by weight of O2, or 35.5 part by weight of Cl2.

  • 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 

Methods To Find Equivalent Weight: 

  • For Acids:
    E = (Molecular weight) / (Protocity or Basicity of Acid)
    For Example, for H3PO4, E = M/3
    For H2SO4  , E =M/2 
  • For Bases:
    E = (Molecular weight) / (Acidity or number of 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)
    For Example, for SO42-, E = M/2
    For PO43-, E = M/3 
  • For Compounds:
    E = (Molecular weight) / (Valency of cation or anion)
    For Example, for CaCO3, E = M/2
    For AlCl3, E =M/3 
  • For Redox Reactions:
    E = (Molecular weight) / (Change in oxidation number)
    For Example, for KMnO4
    (a) In acidic medium: E = M/5 

    \mathrm{\overset{+7}{KMnO_{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 
  • For Acidic Salt: 
    E = (Molecular weight) / (Number of replaceable H-atoms) 
    For example, for H3PO4 

\\ \mathrm{\\Ca(OH)_{2}\, +\, H_{3}PO_{4}\, \rightarrow CaHPO_{4}\, +\, 2H_{2}O}\\\\\mathrm{E\, =\, M/2}

  • Metal displacement method 
    E1 / E2 = W/ W2 
     

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Percentage Composition And Equivalent Weight

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Percentage Composition And Equivalent Weight

Chemistry Part I Textbook for Class XI

Page No. : 15

Line : 25

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