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Important Stoichiometery & Mole concept Formulas for JEE and NEET
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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.
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