JEE Exam  >  JEE Notes  >  Important Formulas for JEE Mains & Advanced  >  Important Formulas: Stoichiometry & Mole concept

Important Stoichiometery & Mole concept Formulas for JEE and NEET

Download, print and study this document offline
Please wait while the PDF view is loading
 Page 1


  
          
   
       
  
           
          
           
         
        
 
   
         
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.
 
  
  
      
      
   
  
  
    
  
            
   
  
           
Read More
79 docs

Top Courses for JEE

79 docs
Download as PDF
Explore Courses for JEE exam

Top Courses for JEE

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

video lectures

,

Previous Year Questions with Solutions

,

shortcuts and tricks

,

practice quizzes

,

Free

,

mock tests for examination

,

MCQs

,

Important Stoichiometery & Mole concept Formulas for JEE and NEET

,

Summary

,

Extra Questions

,

Viva Questions

,

study material

,

ppt

,

Sample Paper

,

Important questions

,

Important Stoichiometery & Mole concept Formulas for JEE and NEET

,

pdf

,

Objective type Questions

,

past year papers

,

Semester Notes

,

Exam

,

Important Stoichiometery & Mole concept Formulas for JEE and NEET

;