JEE Main & Advanced Chemistry Atomic, Molecular and Equivalent masses

Atomic, Molecular and Equivalent masses

         (1) Atomic mass : It is the average relative mass of atom of element  as compared with an atom of carbon –12 isotope taken as 12.

           \[\text{Atomic mass }=\frac{\text{Average mass of an atom}}{\text{1/12}\times \text{Mass of an atom of }{{C}^{12}}\,}\]

         Average atomic mass : If an elements exists in two isotopes having atomic masses ‘a’ and ‘b’ in the ratio   m : n, then average atomic mass =\[\frac{(m\times a)+(n\times b)}{m+n}.\] Since the atomic mass is a ratio, it has no units and is expressed in amu, 1 amu = \[1.66\times {{10}^{-24}}g\]. One atomic mass unit (amu) is equal to \[\frac{1}{12}th\] of the mass of an atom of carbon-12 isotope.

         Gram atomic mass (GAM) : Atomic mass of an element expressed in grams is called Gram atomic mass or gram atom or mole atom.

           (i) Number of gram atoms =\[\frac{\text{Mass of an element}}{\text{GAM}}\]

           (ii) Mass of an element in gm. = No. of gm. atom \[\times \] GAM

           (iii) Number of atoms in 1 GAM = 6.02 \[\times \]\[{{10}^{23}}\]

           \[\therefore \,\,\,\,\]Number of atoms in a given substance

           = No. of  GAM \[\times \]6.02 \[\times \]\[{{10}^{23}}\] = \[\frac{\text{Mass }}{\text{GAM}}\,\]\[\times \]6.02 \[\times \]\[{{10}^{23}}\]

         (iv) Number of atoms in 1gm of element  =\[\frac{6.02\,\times \,{{10}^{23}}}{\text{Atomic mass}}\]

           (v) Mass of one atom of the element (in gm.) \[=\frac{\text{GAM}}{\text{6}\text{.02}\,\,\times \,\text{1}{{\text{0}}^{\text{23}}}}\]

           Methods of determination of atomic mass

         (i) Dulong and Pettit's method : According to Dulong and Pettit's law

           Atomic mass \[\times \]Specific heat = 6.4 (approx.)

           Atomic mass (approx.)  = \[\frac{6.4}{\text{Specific heat (in cals}\text{.)}}\]

         This law is applicable to solid elements only except Be, B, C and Si because their specific heat is variable with temperature.

           Atomic mass = Equivalent mass \[\times \]Valency

                  \[\text{Valency }=\frac{\text{Approximate atomic mass}}{\text{Equivalent mass}}\]

           (ii) Vapour density method : It is suitable for elements whose chlorides are volatile.

           Valency of the element =\[\frac{\text{Molecular mass of chloride}}{\text{Equivalent mass of chloride}}\]

                                                     =\[\frac{2\,\times \,\text{Vapour density of chloride}}{\text{Equivalent mass of metal }+\text{ 35}\text{.5}}\]

           Atomic mass = Equivalent mass of metal \[\times \]Valency

           (iii) Specific heat method : It is suitable only for gases. The two types of specific heats of gases are C_P (at constant pressure) and \[{{C}_{v}}\] (at constant volume). Their ratio is known as g whose value is constant (1.66 for monoatomic, 1.40 for diatomic and 1.33 for triatomic gases).

           \[\text{Atomic mass of a gaseous element }=\frac{\text{Molecular mass}}{\text{Atomicity}}\]

         (iv) Isomorphism method : It is based on law of isomorphism which states that compounds having identical crystal structure have similar constitution and chemical formulae.

           Example : \[{{K}_{2}}S{{O}_{4}},\,\,{{K}_{2}}CrO{}_{4}\] and \[{{K}_{2}}Se{{O}_{4}}\]

(valency of S, Cr, Se = 6),

               \[ZnS{{O}_{4}}.\,7\,{{H}_{2}}O,\,\,MgS{{O}_{4}}.7\,{{H}_{2}}O,\,\,FeS{{O}_{4}}.\,7\,{{H}_{2}}O\]

(valency of Zn, Mg, Fe = 2).

           (2) Molecular mass : Molecular mass of a molecule, of an element or a compound may be defined as a number which indicates how many times heavier is a molecule of that element or compound as compared with \[\frac{1}{12}\]of the mass of an atom of carbon–12. Molecular mass is a ratio and hence has no units. It is expressed in a.m.u.

           \[\text{Molecular mass }=\text{ }\frac{\text{Mass of one molecule of the substance}}{1/12\text{ }\times \text{ Mass of one atom of C-12}}\]

         Actual mass of one molecule = Mol. mass \[\times \,1.66\,\,\times \,{{10}^{-24}}\,gm.\]

           Molecular mass of a substances is the additive property and can be calculated by adding the atomic masses present in one molecule.

           Gram molecular mass (GMM) and Gram molar volume : Molecular mass of an element or compound when expressed in gm. is called its gram molecular mass, gram molecule or mole molecule.

           Number of gm molecules =\[\frac{\text{Mass of substances}}{\text{GMM}}\]

           Mass of substances in gm = No. of gm. molecules \[\times \]GMM

         Volume occupied by one mole of any gas at STP is called Gram molar volume. The value of gram molar volume is 22.4 litres. Volume of 1 mole of any gas at STP = 22.4 litres    

         Expression for mass and density

           Mass of \[11.2L\] of any gas at STP = V.D. of that gas in gm.

           Density of a gas at NTP = \[\frac{\text{Mol}\text{. mass in }gm.}{22400\,\,ml}\]

           Important generalisations

         Number of atoms in a substance

               = Number of GMM \[\times \]6.02 \[\times \]10²³ \[\times \] Atomicity 

           Number of electrons in given substance

= Number of GMM \[\times \]6.02 \[\times \]10²³ \[\times \]Number of electrons

         Methods of determination of molecular mass

           Following methods are used to determine molecular mass,

           (i) Diffusion method (For gases) : The ratio of rates of diffusion of two gases is inversely proportional to the square root of their molecular masses.

                               \[\frac{{{r}_{1}}}{{{r}_{2}}}\,\,=\,\,\sqrt{\frac{{{M}_{2}}}{{{M}_{1}}}}\]

           (ii) Vapour density method (For gases only) : Mass of a fixed volume of the vapour is compared with the mass of the same volume of hydrogen under same conditions. The ratio of these masses is called Vapour density or Relative density.

                \[\text{Molecular mass }=\text{ }2\,\,\times \,\text{Vapour desity}\]

         (iii) Victor Meyer method (For volatile liquids or solids)

It is based on Dalton's law of partial pressure and Avogadro's hypothesis (gram molar volume).

           22400 ml of vapours of a substance = Molecular mass of that substance

           (iv) Colligative property method (For non-volatile solids)

Discussed in colligative properties of solutions.

         Average atomic mass and molecular mass

           \[\overline{A}\,\](Average atomic mass) =\[\frac{\sum {{A}_{i}}{{X}_{i}}}{\,\sum {{X}_{\text{total}}}}\]

           \[\overline{M}\] (Average molecular mass) = \[\frac{\sum {{M}_{i}}{{X}_{i}}}{\,\sum {{X}_{\text{total}}}}\]

           Where  \[{{A}_{1}},\,{{A}_{2}},\,{{A}_{3}}.....\] are atomic mass of species 1, 2, 3,…. etc. with % ratio as \[{{X}_{1}},\,{{X}_{2}},\,{{X}_{3}}.....\] etc. Similar terms are for molecular masses.

         (3) Equivalent mass : The number of parts by mass of a substance that combines with or displaces 1.008 parts by mass of hydrogen or 8.0 parts of oxygen or 35.5 parts of chlorine is called its equivalent mass (EM). On the other hand quantity of a substance in grams numerically equal to its equivalent mass is called its gram equivalent mass (GEM) or gram equivalent.

           \[\text{Number of GEM }=\text{ }\frac{\text{Mass of the substance in grams}}{\text{GEM of the substance}}\]

         Expressions for equivalent mass (EM)

           (i) EM of an element \[=\,\frac{\text{Atomic mass}}{\text{Valency}}\]

• (ii) EM of an acid \[=\,\frac{\text{Molecular mass}}{\text{Basicity}}\]
• (Basicity of acid is the number of replaceable hydrogen atoms in one molecule of the acid).

           (iii) EM of a base \[=\,\frac{\text{Molecular mass}}{\text{Acidity}}\] 

           (Acidity of a base is the number of replaceable–\[OH\]groups in one molecule of the base).

           (iv) EM of a salt  = \[\frac{\text{Formula mass}}{\text{Total positive or negative charge}}\]

           (v) EM of an oxidising agent

\[=\,\frac{\text{Formula mass }}{\text{Number of electrons gained per molecule or Total change in O}\text{.N}\text{.}}\]

         Equivalent mass of common oxidising agent changes with the medium of the reaction.

           Methods of determination of equivalent mass

           (i) Hydrogen displacement method : The mass of metal which displaces 11200 ml of hydrogen at NTP from an acid, alkali or alcohol is the equivalent mass of the metal.

           (a) Equivalent mass of metal

\[=\,\frac{\text{Mass of metal}}{\text{Mass of }{{\text{H}}_{\text{2}}}\,\text{displaced}}\,\times \,1.008\]\[=\frac{W}{M}\times 1.008g\]

           (b) Equivalent mass of metal

           \[=\,\frac{\text{Mass of metal}}{\text{Vol}\text{. (ml) of }{{\text{H}}_{\text{2}}}\,\text{displaced at STP}}\text{  }\times \,\text{11200}\]= \[\frac{W}{V}\,\times \,11200\,\]

           This method is useful for metals which can displace hydrogen from acids or can combine with hydrogen \[(Mg,\,Zn,\,Na,\,Ca\,\text{etc}\text{.)}\]

           (ii) Oxide formation method : The mass of the element which combines with 8 grams of oxygen is the equivalent mass of the element.

           (a) Equivalent mass of metal =\[\frac{\text{Mass of metal}}{\text{Mass of oxygen}}\,\times \,8\]  

           (b) Equivalent mass of metal

= \[\frac{\text{Mass of metal}}{\text{Vol}\text{. of }{{O}_{\text{2}}}\,\text{at S}\text{.T}\text{.P}\text{. in ml}}\,\times 5600\]

           (iii) Chloride formation method : The mass of an element which reacts with 35.5 gm. of chlorine is the equivalent mass of that element.

           (a) Equivalent mass of metal \[=\,\frac{\text{Mass of metal}}{\text{Mass of chlorine}}\,\times \,35.5\]         

           (b) Equivalent mass of metal 

\[=\,\frac{\text{Mass of metal}}{\text{Vol}\text{. of }C{{l}_{2}}\,\text{(in ml}\text{.) at STP}}\,\times \,11200\]

           (iv) Neutralisation method : (For acids and bases).

         \[\text{Equivalent mass of acid (or base) }=\,\,\frac{W}{V\times \ N}\]

           Where ,

           W = Mass of acid or base in gm.,

           V = Vol. of base or acid in litre required for neutralisation

            N is Normality of base or acid

           (v) Metal displacement method : It is based on the fact that one gm. equivalent of a more electropositive metal displaces one gm equivalent of a less electropositive metal from its salt solution.

\[\frac{\text{Mass of metal added}}{\text{Mass of metal displaced}}\,=\,\frac{\text{Eq}\text{. mass of metal added}}{\text{Eq}\text{. mass of metal displaced}}\]

                                       \[\frac{{{W}_{1}}}{{{W}_{2}}}\,\,\,=\,\,\frac{{{E}_{1}}}{{{E}_{2}}}\]

           (vi) Electrolytic method : The quantity of substance that reacts at electrode when 1 faraday of electricity is passed is equal to its gram equivalent mass.

           Gram equivalent mass = Electrochemical equivalent \[\times \]96500

           The ratio of masses of two metals deposited by the same quantity of electricity will be in the ratio of their equivalent masses.                

                                        \[\frac{{{W}_{1}}}{{{W}_{2}}}\,\,\,=\,\,\frac{{{E}_{1}}}{{{E}_{2}}}\]

           (vii) Double decomposition method 

         \[AB\,\,+\,\,CD\,\xrightarrow{{}}\,AD\,\,\downarrow \,\,+\,\,CB\]

           \[\frac{\text{Mass of compound }AB}{\text{Mass of compound }AD}\,=\,\frac{\text{Eq}\text{. mass of }A\text{ }+\text{ Eq}\text{. mass of }B}{\text{Eq}\text{. mass of }A\text{ }+\text{ Eq}\text{. mass of }D}\]

           or \[\frac{\text{Mass of salt taken (}{{W}_{\text{1}}})}{\text{Mass of ppt}\text{. obtained (}{{\text{W}}_{\text{2}}})}\,=\,\frac{\text{Eq}\text{. mass of salt (}{{E}_{\text{1}}})}{\text{Eq}\text{. mass of salt in ppt}\text{. (}{{E}_{\text{2}}})}\]

           (viii) Conversion method : When one compound of a metal is converted to another compound of the same metal, then

          \[\frac{\text{Mass of compound I (}{{W}_{1}})}{\text{Mass of compound II (}{{W}_{2}})}\,=\,\frac{E\,+\text{ Eq}\text{. mass of radical I}}{E\,+\,\text{Eq}\text{. mass of radical II}}\]   

(E = Eq. mass of the metal)

             (ix) Volatile chloride method

           Valency of metal \[=\,\frac{\text{2}\,\times \,\text{V}\text{.D}\text{. of Chloride}}{\text{Eq}\text{. mass of metal chloride}}\]\[=\,\,\frac{\text{2}\,\times \,\text{V}\text{.D}\text{.}}{\text{E}+\text{35}\text{.5}}\]

          \[\therefore \] E \[=\,\frac{\text{2}\,\times \,\text{V}\text{.D}\text{. of Chloride}}{\text{Valency}}\,-\,35.5\]

           (x) Silver salt method (For organic acids)

           Equivalent Mass of acid =\[\frac{108\,\,\times \,\text{Mass of silver salt }}{\text{Mass of }Ag\,\text{metal}}-107\]

                Molecular mass of acid = Equivalent mass of acid \[\times \]Basicity