Law of conservation of mass :
“In all physical and chemical changes, the total mass of the reactants is equal to that of the products” or “Matter can neither be created nor destroyed”.
Law of constant composition or definite proportions :
“A chemical compound is always found to be made up of the same elements combined together in the same fixed ratio by weight”.
Law of multiple proportions :
“When two elements combine together to form two or more chemical compounds, then the weights of one of the elements which combine with a fixed weight of the other bear a simple ratio to one another”.
Law of Reciprocal Proportions :
The ratio of the weights of two elements A and B which combine with a fixed weight of the third element C is either the same or a simple multiple of the ratio of the weights of A and B which directly combine with each other.
Gay Lussac’s law of gaseous volumes :
“When gases react together, they always do so in volumes which bear a simple ratio to one another and to the volumes of the products, if gaseous, all measurements are made under the same conditions of temperature and pressure”.
Avogadro’s hypothesis :
“Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules”.
Atom :
It is the smallest particle of an element that takes part in a chemical reaction. It may or may not be capable of free existence.
Atomic mass unit (a.m.u) :
It is equal to \frac{1}{{12}}th of the mass of an atom of C-12. It is equal to 1.6606 × 10–24 g.
Atomic mass :
It is the average relative mass of its atoms as compared with an atom of carbon – 12 isotope taken as 12.
Calculation of average atomic mass :
If an element exists in two isotopes having atomic masses ‘a’ and ‘b’ in the ratio m: n, then average atomic mass = \frac{{{\text{m}}\,\, \times \,\,{\text{a}}\,\, + \,\,{\text{n}}\,\, \times \,\,{\text{b}}}}{{{\text{m}} + {\text{n}}}}.
Gram atomic mass (G.A.M.):
The atomic mass of an element expressed in grams is called gram atomic mass. This amount of the element is called one gram atom. This amount of the element can also be called one mole atom.
Method of determining atomic weight
(i) By application of the relation
At. wt. = Eq. wt. × Valency
Knowing the exact eq. wt. and approximate atomic weight, valency can be calculated (which is a whole number). Knowing exact eq. wt. and valency, exact atomic weight can be calculated.
(ii) Dulong and Petit’s method: According to Dulong and Petit’s law, for solid elements (except Be, B, C and Si)
At. wt. × Specific heat = 6.4 (approx.)
∴ Approx at. wt. = \frac{{6.4}}{{Sp.\,\,heat}}
= \frac{{{\text{Approx}}{\text{.}}\,\,{\text{At}}{\text{.}}\,\,{\text{wt}}{\text{.}}}}{{{\text{Eq}}{\text{.}}\,\,{\text{wt}}{\text{.}}}} = approx. Valency.
The approx. valency is converted to the nearest whole number which is called exact valence. Eq. wt. × exact valency = exact Atomic Weight.
(iii) Specific heat/molar heat capacity method: For gases, \gamma \, = \,\frac{{{C_p}}}{{{C_v}}} is 1.66 for monoatomic, 1.40 for diatomic and 1.30 for triatomic gases. Thus having known the atomicity of the gas from the value of g, Atomic wt. of the gaseous element = \frac{{Mol.\,\,wt.}}{{Atomicity}}
Molecular wt. = 2 × Vapour density.
(iv) Volatile chloride method: This method can be used for those elements whose chlorides are volatile so that their vapour densities can be determined. Then
Mol. wt. of the chloride = 2 × V.D.
If x is the valency of the element (M), then the formula of its chloride will be MClx. Hence
Mo. wt. of the Chloride MClx = At. wt. of M + x × 35.5
= Eq. wt. of M × Valency of M + x × 35.5
= E × x + x × 35.5
= x (E + 35.5)
∴ x (E + 35.5) = 2 × V.D. or x = \frac{{2\,\,\, \times \,\,V.D.}}{{E + 35.5}}
Knowing the eq. wt. E of the element, the valency x can be calculated. Then Atomic weight = eq. wt. × Valency.
(v) Isomorphism method: Compounds having similar molecular formulae and identical crystal structure are called isomorphs. The method is based upon the fact that elements in isomorphous compounds have same valencies e.g.,
(a) K2SO4, K2CrO4 and K2SeO4 are isomorphous. Hence valency of S, Cr and Se = 6.
(b) ZnSO4, 7H2O, FeSO4, 7H2O, MgSO4, 7H2O are isomorphous. Hence valency of Zn, Fe and Mg = 2.
(c) Alums, M2SO4.M’2(SO4)3.24H2O in which M is monovalent and M’ is trivalent are isomorphous.
Knowing the valency, Atomic wt.= Eq. wt. × Valency
Molecules :
It is the smallest particle of an element or a compound that is capable of free existence.
Molecular mass :
Molecular mass of a substance is the average relative mass of its molecules as compared with an atom of C–12 isotope taken as 12.
Gram molecular mass or molar mass (G.M.M.):
The molecular mass of a substance expressed in gram is called gram molecular mass or molar mass. This amount of the substance is called one gram molecule. This amount of the substance is called one mole of the substance.
Methods of determining molecular weight
(i) Gram Molecular Volume (G.M.V.) method, 22.4 litres of every gas or vapour at STP weigh equal to molecular weight expressed in grams. This is the principle of Victor Meyer method used for volatile liquids.
(ii) Vapour density method
Molecular weight = 2 × Vapour Density
where Vapour Density = \frac{{Wt.\,\,of\,\,certain\,\,volume\,\,of\,\,the\,\,vapour}}{{Wt.\,\,of\,\,same\,\,volume\,\,of\,\,{H_2}\,\,under\,\,same\,\,conditions\,\,of\,\,temp.\,\,and\,\,pressure}}
(iii) Diffusion method: According to Graham’s law of diffusion, rates of diffusion of different gases under similar conditions of temperature and pressure are inversely proportional to the square root of their densities or molecular weights i.e.,
\frac{{{r_1}}}{{{r_2}}}\,\,\, = \,\,\sqrt {\frac{{{d_2}}}{{{d_1}}}} \,\,\, = \,\,\,\sqrt {\frac{{2\,\, \times \,\,{d_2}}}{{2\,\, \times \,\,{d_1}}}} \,\,\,\, = \,\,\,\sqrt {\frac{{{M_2}}}{{{M_1}}}}
Knowing the molecular weight of one gas, that of the other can be calculated.
(iv) Colligative property method: Discussed in the unit of ‘Solutions’.
Moles :
The collection of 6.023 × 1023 molecules of an element or ions or compounds constitute one mole of that element, ion or compound. This number, 6.023 × 1023 is known as Avogadro’s number. This is called mole-particle relationship.
Mole-Particle Relationship :
e.g. (i) one mole of Na contains 6.023 × 1023 atoms of Na.
(ii) one mole of oxygen i.e. one mole of O2 contains 6.023 × 1023 molecules of oxygen.
(iii) one mole of CCl4 carbontetrachloride contains 6.023 × 1023 molecule of CCl4 (carbon tetrachloride)
Mole-Weight Relationship :
(i) one mole of an element weighs equal to gram atomic weight of the element
e.g. one mole of Na weighs 23 gm (Gram atomic wt. of Na = 23)
(ii) one mole of Mg weighs 24 gms (GAM of Mg = 24)
(iii) one mole of substance (molecular state) weighs equal to the gram molecular weight of the substance.
e.g. one mole of O2 weighs 32 gms (GMM of O2 = 32)
one mole of SO2 weighs 64 gms.
Mole-Volume Relationship :
(i) one mole of every gas at STP occupies 22.4 litres of volume.
e.g. one mole of CO2 at STP occupies 22.4 litres of volume.
(ii) one mole of SO2 at STP occupies 22.4 litres of volume.
GRAM-ATOMS :
One gram-atom of an element means collection of 6.023 × 1023 atoms. This concept applies only to the elements which exists in polyatomic states (e.g. O, Cl, S, P etc.). It is meaningless for metals and compounds.
* The number of atoms in w gms of an element whose atomic mass is A is :
{\text{gm}} - {\text{atom}}\,\, = \,\,\frac{{\text{w}}}{{\text{A}}} = mole atom
* The number of atoms is given by:
No. of atoms = \frac{{\text{w}}}{{\text{A}}}\,\, \times \,\,{{\text{N}}_{\text{0}}}
Avogadro’s number :
It is the number of atoms present in one gram atom of an element or the number of molecules present in one gram molecule of the substance. In general, it is the number of particles present in one mol of the substance. Its value is 6.02 × 1023.
Equivalent weight :
The equivalent weight of a substance is the number of parts by weight of the substance that combine with or displace directly or indirectly 1.008 parts by weight of hydrogen or 8 parts by weight of oxygen or 35.5 parts by weight of chlorine.
Eq. wt. of an acid = \frac{{Mol.\,\,wt.\,\,of\,\,the\,\,\,acid}}{{Basicity}}
Eq. wt. of a base = \frac{{Mol.\,\,wt.\,\,of\,\,the\,\,\,base}}{{acidity}}
Gram equivalent weight :
The equivalent weight of a substance expressed in grams is called gram eq. wt. or one gram equivalent.
Now we can define gram-equivalent (gm eq) in g gms of a substance whose equivalent weight is E as follows:
No. of e q = \frac{g}{E}
* The number of milliequivalents (meq) in g gms is given by:
m e q = \frac{g}{E} × 1000
Methods of determining equivalent weight
(i) Hydrogen displacement method: This method is used for metals which react with an acid to evolve hydrogen gas.
Eq. wt. of the metal is the weight of the metal which displace 1.008 g of H2 or 11200 c.c. of H2 at STP.
Eq. wt. of metal = \frac{{Wt.\,\,of\,\,metal}}{{Wt.\,\,of\,\,{H_2}\,\,displaced}}\,\, \times \,\,1.008
or = \frac{{Wt.\,\,of\,\,metal}}{{Vol.\,\,of\,\,{H_2}\,\,in\,\,ml\,\,displaced\,\,at\,\,STP}}\,\,\, \times \,\,\,11200
(ii) Oxide formation or reduction of the oxide method: In this method a known weight of the metal is converted into its oxide directly or indirectly. Knowing the weight of the metal oxide formed, the weight of oxygen combined can be calculated. Alternatively, a known weight of the metal oxide may be reduced to metal whose weight is determined.
Eq. wt. of metal = \frac{{Wt.\,\,of\,\,metal}}{{Wt.\,\,of\,\,oxygen\,\,combined}}\,\, \times \,\,8
or = \frac{{Wt.\,\,of\,\,metal}}{{Vol.\,\,of\,\,{O_2}\,\,displaced\,/\,combined\,\,in\,\,ml\,\,at\,\,STP}}\,\, \times \,\,5600
Eq. wt. of metal =\frac{{Wq.\,\,of\,\,metal}}{{Wt.\,\,of\,\,chlorine\,\,combined}}\,\,\, \times \,\,35.5
or \frac{{Wt.\,\,of\,\,metal}}{{Vol.\,\,of\,\,C{l_2}\,\,combined\,\,in\,\,ml\,\,at\,\,STP}}\,\,\, \times \,\,11200
(iii) Double decomposition method: For a reaction of the type
AB + CD → AD ↓ + BC
(e.q. AgNO3 + NaCl → AgCl ↓ + NaNO3)
\frac{{Weight\,\,of\,\,AB\,\,taken}}{{Weight\,\,of\,\,AD\,\,formed}} = \frac{{Eq.\,\,wt.\,\,of\,\,AB}}{{Eq.\,\,wt.\,\,of\,\,AD}} = \frac{{Eq.\,\,wt.\,\,of\,\,A\,\, + \,Eq.\,\,wt.\,\,of\,\,B}}{{Eq.\,\,wt.\,\,of\,\,A\,\, + \,\,Eq.\,\,wt.\,\,of\,\,D}}
Knowing the equivalent weights of any two radicals out of A, B and D, that of the third can be calculated.
(iv) Electrolytic method (based on Faraday’s laws of electrolysis)
1 Faraday (96500 coulombs) deposit one gram equivalent of the substance.
Electrochemical equivalent of a substance is the weight of the substance deposited by one coulomb.
When the same quantity of electricity flows through solutions of different electrolytes,
\frac{{Wt.\,\,of\,\,X\,\,deposited}}{{Wt.\,\,of\,\,Y\,\,deposited}} = \frac{{Eq.\,\,wt.\,\,of\,\,X}}{{Eq.\,\,wt.\,\,of\,Y}}
(v) Neutralization method: This is based on the fact that acids and bases react in equivalent amounts. Hence equivalent weight of an acid is the weight of the acid which is neutralized by 1000 cc of 1N base solution (which contains 1g eq. of the base) Likewise equivalent weight of a base can be determined.
(vi) Silver salt method: (for organic acids only) The organic acid is converted into its silver salt (RCOOAg), a known weight of which is ignited to give residue of Ag which is weighed. Then
\frac{{Eq.\,\,wt.\,\,of\,RCOOAg}}{{Eq.\,\,wt.\,\,of\,\,Ag\,\,(108)}} = \frac{{Wt.\,\,of\,\,silver\,\,salt}}{{Wt.\,\,of\,\,silver}}
Eq. wt. of acid (RCOOH) = Eq. wt. of RCOOAg–107.
(vii) Conversion method: When one compound of a metal is converted into another compound of the same metal (e.g. metal carbonate → metal oxide), then
\frac{{Weight\,\,of\,\,compound\,\,I}}{{Weight\,\,of\,\,compound\,\,II}} = \frac{{Eq.\,\,wt.\,\,of\,\,metal\,\, + \,\,Eq.\,\,wt.\,\,of\,\,anion\,\,of\,\,compound\,\,I}}{{Eq.\,\,wt.\,\,of\,\,metal\,\, + \,\,Eq.\,\,wt.\,\,of\,\,anion\,\,of\,\,compound\,\,II}}
Relationship between eq. wt. atomic wt. and valency of an element: Eq. wt. = \frac{{Atomic\,\,\,weight}}{{Valency}}
Equivalent weight of oxidizing/reducing agent = \frac{{Molecular\,\,weight\,\,of\,\,the\,\,substance}}{{No.\,\,of\,\,electrons\,gained/lost\,\,by\,\,one\,\,molecule}}
Expressing concentration of solutions
Solution is a homogenous mixture of two or more components in which intermingling particles are of atomic or molecular dimensions. A solution consists of a dissolved substance known as solute and the substance in which the solute is dissolved is known as solvent. The concentration of a solution means the quantity of solute dissolved per unit volume of solution, or per unit quantity of solvent.
concentration of solution = \frac{{quantity\,\,of\,\,solute}}{{quantity\,\,of\,\,solvent\,\,or\,\,solution}}
Note: While discussing various methods for expressing concentration, we have taken solute as B dissolved in solvent A and gA as grams of solute and eB as grams of solvent.
Mass fraction is the fractional part of a component that is contributed by it to the total mass of solution.
mass fraction of B = \frac{{{{\text{w}}_{\text{B}}}}}{{{{\text{w}}_{\text{A}}} + {{\text{w}}_{\text{B}}}}}
mass fraction of A = \frac{{{{\text{w}}_B}}}{{{{\text{w}}_{\text{A}}} + {{\text{w}}_{\text{B}}}}}
Mole fraction is the fractional part of the moles that is contributed by each component to the total number of moles that comprises the solution. In containing nA moles of solvent and nB moles of solute:
mole fraction of B = xB = \frac{{{n_B}}}{{{n_A} + {n_B}}}
mole fraction of A = xB = \frac{{{n_A}}}{{{n_A} + {n_B}}}
Molality (m) is expressed as number of moles of solute dissolved in 1000 gms (1 Kg) of solvent. It is denoted by m.
{\text{m}}\,\, = \,\,\frac{{{{\text{n}}_{\text{B}}}}}{{{{\text{w}}_{\text{A}}}}}\,\,\, \times \,\,\,{\text{1000}}Molarity (m) is expressed as moles of solute contained in one litre of solution or it is also taken as millimoles of solute in 1000 cc (ml) of solution. It is denoted by M.
molarity (M) = \frac{{moles\,\,of\,\,solute}}{{litres\,\,of\,\,solution}} = \frac{{{\text{millimoles}}\,\,{\text{of}}\,\,{\text{solute}}}}{{{\text{millilitres}}\,\,{\text{of}}\,\,{\text{solution}}}}
{\text{M}}\,\,\, = \,\,\,\frac{{{{\text{n}}_{\text{B}}}}}{{{{\text{V}}_{{\text{lt}}}}}} = \frac{{{g_B}/{m_B}}}{{{V_{lt}}}}
Normality (N) is expressed as the number of gram equivalents (gm eq) of solute contained in one litre of solution or it can also be taken as number of milliequivalents (meq) in 1000 cc (ml) of solution. It is denoted by N.
normality of solution (N) = \frac{{gm - eq\,\,of\,\,solute}}{{litres\,\,of\,\,solution}} = \frac{{meq\,\,of\,\,solute}}{{millilitres\,\,of\,\,solution}}
N = \frac{{{\text{gm}} - {\text{eq}}}}{{{{\text{V}}_{{\text{lt}}}}}}
Note : The following results should be remembered and used directly
(1) moles = M Vlt = \frac{{{{\text{w}}_{\text{B}}}}}{{{{\text{M}}_{\text{B}}}}}
milimoles = M Vcc = \frac{{{{\text{w}}_{\text{B}}}}}{{{{\text{M}}_{\text{B}}}}}\,\, \times \,\,{\text{1000}}
(2) gm-eq = N Vlt = \frac{{{{\text{w}}_{\text{B}}}}}{{{{\text{E}}_{\text{B}}}}}
meq = N Vcc = \frac{{{{\text{w}}_{\text{B}}}}}{{{{\text{E}}_{\text{B}}}}}\,\, \times \,\,{\text{1000}}
(3) Molarity (M) = \frac{{10\,\,x\,d}}{{{M_0}}} Normality (N) = \frac{{10\,\,x\,d}}{E}
d : density of solution in g/cc x :% age strength of solution (by weight)
(4) N = z × M
(5) m = \frac{{1000\,\, \times \,\,{x_B}}}{{{M_A}\, \times \,{x_A}}}
x : acidity for base or basicity for acid or electron transfer / mole for O.A. and R.A.
Strength of a solution is generally expressed as grams of solute in one litre of solution.
strength = \frac{{{{\text{w}}_{\text{B}}}}}{{{\text{litres}}\,\,{\text{of}}\,\,{\text{solution}}}}
strength = Normality × Eq. wt. = Molarity × Mole wt.
Diluting a solution:
Whenever a given solution of known concentration i.e. normality & molarity (known as standard solution) is diluted by adding more of solvent, the number of millimoles (or millequivalents) of solutes remain unchanged. The concentration of solution however changes.
In such cases M1 V1 = M2 V2 (M1 & V1a re molarity and volume of original solution)
(M2 & V2 are molarity and volume of diluted solution)
Stoichiometric Calculations:
(Quantitative analysis of a balance chemical equation)
Consider a balanced chemical reaction of equation:
mA + nB → pC + qD
A and B are reactants; C and D are products; m, n, p, q are the stiochiometric coefficients.
The above balanced reaction is analysed as:
m moles of A react with n moles of B to produce p moles of C plus q moles of D.
This can be represented as:
m moles of a ≡ n moles of B ≡ p moles of C ≡ q moles of d
It is also observed that when a substance takes part in a chemical reaction then the amount of the substance taking part is proportional to its equivalent weight
i.e. 1 equivalent of every substance combines (displaces, double displaces, exidises, reduces, neutralizes) are equivalent of another substance. Hence in a chemical reaction of the type
aA + bB → mC + nD
1 eq. of A combines with 1 eq. of B to form 1 eq of C and 1 eq of D or
neq of A ≡ neq of B ≡ neq of C ≡ neq of D.
In general, there are two types of stiochiometric reactions (which are most common):
Neutralisation Reactions and Redox Reactions
The analysis of two types of reactions is generally carried out in the form of mass of reactants (or products) taking part in a given reaction (gravimetric analysis) or in terms of concentrations of reactants (or products) taking part in a given reaction (volumetric analysis).
Neutralisation:
A reaction in which an acid (or a base) completely reacts with a base (or an acid) to form salt and water. If HA be the acid, BOH be the base and BA be the salt, then neutralisation reaction can be represented as follows:
HA + BOH → BA + H2O
As we known that an acid may be monobasic (HCl, HNO3, etc.), dibasic (H2SO4, H2C2O4 etc.) or tribasic (H3PO4 etc.) and similarly a base may be monoacidic (NaOH, NaHCO3 etc.), diacidic (Ca(OH)2, Na2CO3 etc.) or triacidic (Al(OH)3 etc.), so it is better to define the neutralisation reaction in the following manner.
A reaction in which 1 gram equivalent (or 1 meq of an acid (or a base) completely reacts with 1 gram equivalent (or 1 meq) of a base (or an acid) to form 1 gram equivalent (or 1 meq) of corresponding salt.
A stage at which the process of neutralisation is complete is known as end point or equivalence point. At equivalence point:
gm eq. (or meq) of acid = gm eq. (or meq) of base.
NaVa = Nb Vb (N : normality, V: volume in ml of lt, at acid, b:)