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Important Stoichiometery & Mole concept Formulas for JEE and NEET

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STOICHIOMETRY
? Relative atomic mass (R.A.M) = 
atom carbon one of mass
12
1
element an of atom one of Mass
?
= Total Number of nucleons
? Y-map
Mole
× 22.4 lt
? 22.4 lt
Volume at STP
? N
A
× N
A
Number
× mol. wt.
× At. wt.
?
? At. wt.
 mol. wt.
Mass
Page 2


  
          
   
       
  
           
          
           
         
        
 
   
         
STOICHIOMETRY
? Relative atomic mass (R.A.M) = 
atom carbon one of mass
12
1
element an of atom one of Mass
?
= Total Number of nucleons
? Y-map
Mole
× 22.4 lt
? 22.4 lt
Volume at STP
? N
A
× N
A
Number
× mol. wt.
× At. wt.
?
? At. wt.
 mol. wt.
Mass
  
Density :
Specific gravity = 
C 4 at water of density
ce tan subs the of density
?
For gases :
Absolute density (mass/volume) = 
gas the of volume Molar
gas the of mass Molar
? ? =  
RT
PM
Vapour density      V.D.= 
2
H
gas
d
d
 = 
RT H
RT gas
2
PM
PM
 = 
2
H
gas
M
M
 = 
2
M
gas
M
gas
 = 2 V.D.
Mole-mole analysis :
Mass
??At. wt. / Mol. Wt.
Mole
Mole-mole 
relationship 
of equation
Mole
× 22.4 lt
Volume at STP
× mol. wt./At. wt.
Mass
Concentration terms :
Molarity (M) :
? Molarity (M) = 
inml
V ) solute of wt . Mol (
1000 w
?
?
Molality (m) :
Molality = 
1000
gram  in  solvent  of  mass
solute  of  moles  of  number
?
 = 1000 w
1
 / M
1
w
2
Mole fraction (x) :
? Mole fraction of solution (x
1
) = 
N n
n
?
? Mole fraction of solvent (x
2
) = 
N n
N
?
x
1
 + x
2
 = 1
Page 3


  
          
   
       
  
           
          
           
         
        
 
   
         
STOICHIOMETRY
? Relative atomic mass (R.A.M) = 
atom carbon one of mass
12
1
element an of atom one of Mass
?
= Total Number of nucleons
? Y-map
Mole
× 22.4 lt
? 22.4 lt
Volume at STP
? N
A
× N
A
Number
× mol. wt.
× At. wt.
?
? At. wt.
 mol. wt.
Mass
  
Density :
Specific gravity = 
C 4 at water of density
ce tan subs the of density
?
For gases :
Absolute density (mass/volume) = 
gas the of volume Molar
gas the of mass Molar
? ? =  
RT
PM
Vapour density      V.D.= 
2
H
gas
d
d
 = 
RT H
RT gas
2
PM
PM
 = 
2
H
gas
M
M
 = 
2
M
gas
M
gas
 = 2 V.D.
Mole-mole analysis :
Mass
??At. wt. / Mol. Wt.
Mole
Mole-mole 
relationship 
of equation
Mole
× 22.4 lt
Volume at STP
× mol. wt./At. wt.
Mass
Concentration terms :
Molarity (M) :
? Molarity (M) = 
inml
V ) solute of wt . Mol (
1000 w
?
?
Molality (m) :
Molality = 
1000
gram  in  solvent  of  mass
solute  of  moles  of  number
?
 = 1000 w
1
 / M
1
w
2
Mole fraction (x) :
? Mole fraction of solution (x
1
) = 
N n
n
?
? Mole fraction of solvent (x
2
) = 
N n
N
?
x
1
 + x
2
 = 1
  
% Calculation :
(i) % w/w = 
100
gm in  solution  of  mass
gm  in  solute  of  mass
?
(ii) % w/v = 
mass  of  solute  in  gm
100
Volume  of  solution  in  ml
?
(iii) % v/v = 
Volume  of  solute  in  ml
100
Volume  of  solution
?
Derive the following conversion :
1. Mole fraction of solute into molarity of solution M = 
2 2 1 1
2
x M M x
1000 x
?
? ?
2. Molarity into mole fraction x
2
 = 
2
1
MM 1000
1000 MM
? ? ?
?
3. Mole fraction into molality m = 
1 1
2
M x
1000 x ?
4. Molality into mole fraction x
2
 = 
1
1
mM 1000
mM
?
5. Molality into molarity M = 
2
mM 1000
1000 m
?
? ?
6. Molarity into Molality m = 
2
MM 1000
1000 M
? ?
?
M
1
 and M
2
 are molar masses of solvent and solute. ? is density of solution
(gm/mL)
M = Molarity (mole/lit.), m = Molality (mole/kg), x
1
 = Mole fraction of
solvent, x
2
 = Mole fraction of solute
Average/Mean atomic mass :
A
x
 = 
100
x a ..... x a x a
n n 2 2 1 1
? ? ?
Mean molar mass or molecular mass :
M
avg.
 = 
n 2 1
n n 2 2 1 1
n .... n n
M n ...... M n M n
? ?
? ?
    or M
avg.
 = 
?
?
?
?
?
?
n j
1 j
j
n j
1 j
j j
n
M n
Page 4


  
          
   
       
  
           
          
           
         
        
 
   
         
STOICHIOMETRY
? Relative atomic mass (R.A.M) = 
atom carbon one of mass
12
1
element an of atom one of Mass
?
= Total Number of nucleons
? Y-map
Mole
× 22.4 lt
? 22.4 lt
Volume at STP
? N
A
× N
A
Number
× mol. wt.
× At. wt.
?
? At. wt.
 mol. wt.
Mass
  
Density :
Specific gravity = 
C 4 at water of density
ce tan subs the of density
?
For gases :
Absolute density (mass/volume) = 
gas the of volume Molar
gas the of mass Molar
? ? =  
RT
PM
Vapour density      V.D.= 
2
H
gas
d
d
 = 
RT H
RT gas
2
PM
PM
 = 
2
H
gas
M
M
 = 
2
M
gas
M
gas
 = 2 V.D.
Mole-mole analysis :
Mass
??At. wt. / Mol. Wt.
Mole
Mole-mole 
relationship 
of equation
Mole
× 22.4 lt
Volume at STP
× mol. wt./At. wt.
Mass
Concentration terms :
Molarity (M) :
? Molarity (M) = 
inml
V ) solute of wt . Mol (
1000 w
?
?
Molality (m) :
Molality = 
1000
gram  in  solvent  of  mass
solute  of  moles  of  number
?
 = 1000 w
1
 / M
1
w
2
Mole fraction (x) :
? Mole fraction of solution (x
1
) = 
N n
n
?
? Mole fraction of solvent (x
2
) = 
N n
N
?
x
1
 + x
2
 = 1
  
% Calculation :
(i) % w/w = 
100
gm in  solution  of  mass
gm  in  solute  of  mass
?
(ii) % w/v = 
mass  of  solute  in  gm
100
Volume  of  solution  in  ml
?
(iii) % v/v = 
Volume  of  solute  in  ml
100
Volume  of  solution
?
Derive the following conversion :
1. Mole fraction of solute into molarity of solution M = 
2 2 1 1
2
x M M x
1000 x
?
? ?
2. Molarity into mole fraction x
2
 = 
2
1
MM 1000
1000 MM
? ? ?
?
3. Mole fraction into molality m = 
1 1
2
M x
1000 x ?
4. Molality into mole fraction x
2
 = 
1
1
mM 1000
mM
?
5. Molality into molarity M = 
2
mM 1000
1000 m
?
? ?
6. Molarity into Molality m = 
2
MM 1000
1000 M
? ?
?
M
1
 and M
2
 are molar masses of solvent and solute. ? is density of solution
(gm/mL)
M = Molarity (mole/lit.), m = Molality (mole/kg), x
1
 = Mole fraction of
solvent, x
2
 = Mole fraction of solute
Average/Mean atomic mass :
A
x
 = 
100
x a ..... x a x a
n n 2 2 1 1
? ? ?
Mean molar mass or molecular mass :
M
avg.
 = 
n 2 1
n n 2 2 1 1
n .... n n
M n ...... M n M n
? ?
? ?
    or M
avg.
 = 
?
?
?
?
?
?
n j
1 j
j
n j
1 j
j j
n
M n
  
Calculation of individual oxidation number :
Formula : Oxidation Number  =  number of electrons in the valence shell
? number of electrons left after bonding
Concept of Equivalent weight/Mass :
For elements, equivalent weight (E) =  
factor - Valency
weight Atomic
For acid/base,        
Acidity / Basicity
M
E?
Where M = Molar mass
For O.A/R.A,        
lost / gained e of moles of . no
M
E
?
?
Equivalent weight (E) = 
v.f.
weight moleculear or Atomic
(v.f. = valency factor)
Concept of number of equivalents :
No. of equivalents of solute = 
M/n
W
E
W
wt. Eq.
Wt
? ?
No. of equivalents of solute = No. of moles of solute × v.f.
Normality (N) :
Normality (N) = 
litres) (in solution of Volume
solute of s equivalent of Number
Normality  = Molarity × v.f.
Calculation of valency  Factor :
n-factor of acid =  basicity = no. of H
+
 ion(s) furnished per molecule of the
acid.
n-factor of base = acidity = no. of OH
?
 ion(s) furnised by the base per
molecule.
At equivalence point :
N
1
V
1
 = N
2
V
2
n
1
M
1
V
1
 = n
2
M
2
V
2
Page 5


  
          
   
       
  
           
          
           
         
        
 
   
         
STOICHIOMETRY
? Relative atomic mass (R.A.M) = 
atom carbon one of mass
12
1
element an of atom one of Mass
?
= Total Number of nucleons
? Y-map
Mole
× 22.4 lt
? 22.4 lt
Volume at STP
? N
A
× N
A
Number
× mol. wt.
× At. wt.
?
? At. wt.
 mol. wt.
Mass
  
Density :
Specific gravity = 
C 4 at water of density
ce tan subs the of density
?
For gases :
Absolute density (mass/volume) = 
gas the of volume Molar
gas the of mass Molar
? ? =  
RT
PM
Vapour density      V.D.= 
2
H
gas
d
d
 = 
RT H
RT gas
2
PM
PM
 = 
2
H
gas
M
M
 = 
2
M
gas
M
gas
 = 2 V.D.
Mole-mole analysis :
Mass
??At. wt. / Mol. Wt.
Mole
Mole-mole 
relationship 
of equation
Mole
× 22.4 lt
Volume at STP
× mol. wt./At. wt.
Mass
Concentration terms :
Molarity (M) :
? Molarity (M) = 
inml
V ) solute of wt . Mol (
1000 w
?
?
Molality (m) :
Molality = 
1000
gram  in  solvent  of  mass
solute  of  moles  of  number
?
 = 1000 w
1
 / M
1
w
2
Mole fraction (x) :
? Mole fraction of solution (x
1
) = 
N n
n
?
? Mole fraction of solvent (x
2
) = 
N n
N
?
x
1
 + x
2
 = 1
  
% Calculation :
(i) % w/w = 
100
gm in  solution  of  mass
gm  in  solute  of  mass
?
(ii) % w/v = 
mass  of  solute  in  gm
100
Volume  of  solution  in  ml
?
(iii) % v/v = 
Volume  of  solute  in  ml
100
Volume  of  solution
?
Derive the following conversion :
1. Mole fraction of solute into molarity of solution M = 
2 2 1 1
2
x M M x
1000 x
?
? ?
2. Molarity into mole fraction x
2
 = 
2
1
MM 1000
1000 MM
? ? ?
?
3. Mole fraction into molality m = 
1 1
2
M x
1000 x ?
4. Molality into mole fraction x
2
 = 
1
1
mM 1000
mM
?
5. Molality into molarity M = 
2
mM 1000
1000 m
?
? ?
6. Molarity into Molality m = 
2
MM 1000
1000 M
? ?
?
M
1
 and M
2
 are molar masses of solvent and solute. ? is density of solution
(gm/mL)
M = Molarity (mole/lit.), m = Molality (mole/kg), x
1
 = Mole fraction of
solvent, x
2
 = Mole fraction of solute
Average/Mean atomic mass :
A
x
 = 
100
x a ..... x a x a
n n 2 2 1 1
? ? ?
Mean molar mass or molecular mass :
M
avg.
 = 
n 2 1
n n 2 2 1 1
n .... n n
M n ...... M n M n
? ?
? ?
    or M
avg.
 = 
?
?
?
?
?
?
n j
1 j
j
n j
1 j
j j
n
M n
  
Calculation of individual oxidation number :
Formula : Oxidation Number  =  number of electrons in the valence shell
? number of electrons left after bonding
Concept of Equivalent weight/Mass :
For elements, equivalent weight (E) =  
factor - Valency
weight Atomic
For acid/base,        
Acidity / Basicity
M
E?
Where M = Molar mass
For O.A/R.A,        
lost / gained e of moles of . no
M
E
?
?
Equivalent weight (E) = 
v.f.
weight moleculear or Atomic
(v.f. = valency factor)
Concept of number of equivalents :
No. of equivalents of solute = 
M/n
W
E
W
wt. Eq.
Wt
? ?
No. of equivalents of solute = No. of moles of solute × v.f.
Normality (N) :
Normality (N) = 
litres) (in solution of Volume
solute of s equivalent of Number
Normality  = Molarity × v.f.
Calculation of valency  Factor :
n-factor of acid =  basicity = no. of H
+
 ion(s) furnished per molecule of the
acid.
n-factor of base = acidity = no. of OH
?
 ion(s) furnised by the base per
molecule.
At equivalence point :
N
1
V
1
 = N
2
V
2
n
1
M
1
V
1
 = n
2
M
2
V
2
  
Volume strength of H
2
O
2 
:
20V H
2
O
2 
 means one litre of this sample of H
2
O
2 
on decomposition
gives 20 lt. of O
2
 gas at S.T.P.
Normality of H
2
O
2
 (N) = 
2 2
Volume,strengthof H O
5.6
Molarity of H
2
O
2 
(M) = 
2 . 11
O H of strength Volume
2 2
Measurement of Hardness :
Hardness in ppm  =  
6
3
10
water of mass Total
CaCO of mass
?
Calculation of available chlorine from a sample of bleaching powder :
% of Cl
2
 = 
) g ( W
) mL ( V x 55 . 3 ? ?
 where x = molarity of hypo solution
and v = mL. of hypo solution used in titration.
 
  
  
      
      
   
  
  
    
  
            
   
  
           
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