Purification Methods JEE Notes | EduRev

JEE : Purification Methods JEE Notes | EduRev

 Page 1


SIMPLE DISTILLATION TECHNIQUE
PURIFICATION METHODS
Typ e Conditions E xamp le s
(A) Simple distillation (i) When liquid sample has (i) Mixture of chloroform
non volatile impurities (BP = 334K) and Aniline
(BP = 457K)
(ii) When boiling point (ii) Mixture of Ether (BP = 308K) &
difference is 80° K or more. Toluene (BP = 384K)
(iii) Hexane
 
(342K) and Toulene(384K)
(B) Fractional distillation When BP difference is 10° (i) Crude oil in petroleum industry
(ii) Acetone (329) and  Methyl alcohol(338K)
(C) Distallation under When liquid boils at higher (i) Concentration of sugar juice
reduced pressure temperature and it may (ii) Recovery of glycerol from spent lye.
(Vacuum distillation) decompose before BP is attained. (iii) Glycerol
(D) Steam distillation When the substance is immiscible (i) Aniline is separated from water
 with water and steam volatile. (ii) Turpentine oil
(iii) Nitro Benzene
(iv) Bromo Benzene
(v) Naphthalene
(vi) o-Nitrophenol
? ? Distillation Techniques :
JEEMAIN.GURU
Page 2


SIMPLE DISTILLATION TECHNIQUE
PURIFICATION METHODS
Typ e Conditions E xamp le s
(A) Simple distillation (i) When liquid sample has (i) Mixture of chloroform
non volatile impurities (BP = 334K) and Aniline
(BP = 457K)
(ii) When boiling point (ii) Mixture of Ether (BP = 308K) &
difference is 80° K or more. Toluene (BP = 384K)
(iii) Hexane
 
(342K) and Toulene(384K)
(B) Fractional distillation When BP difference is 10° (i) Crude oil in petroleum industry
(ii) Acetone (329) and  Methyl alcohol(338K)
(C) Distallation under When liquid boils at higher (i) Concentration of sugar juice
reduced pressure temperature and it may (ii) Recovery of glycerol from spent lye.
(Vacuum distillation) decompose before BP is attained. (iii) Glycerol
(D) Steam distillation When the substance is immiscible (i) Aniline is separated from water
 with water and steam volatile. (ii) Turpentine oil
(iii) Nitro Benzene
(iv) Bromo Benzene
(v) Naphthalene
(vi) o-Nitrophenol
? ? Distillation Techniques :
JEEMAIN.GURU
FR ACTIONAL DISTILLATION
STEAM DISTILLATION
 DISTILLATION UNDER REDUCED PRESSURE
JEEMAIN.GURU
Page 3


SIMPLE DISTILLATION TECHNIQUE
PURIFICATION METHODS
Typ e Conditions E xamp le s
(A) Simple distillation (i) When liquid sample has (i) Mixture of chloroform
non volatile impurities (BP = 334K) and Aniline
(BP = 457K)
(ii) When boiling point (ii) Mixture of Ether (BP = 308K) &
difference is 80° K or more. Toluene (BP = 384K)
(iii) Hexane
 
(342K) and Toulene(384K)
(B) Fractional distillation When BP difference is 10° (i) Crude oil in petroleum industry
(ii) Acetone (329) and  Methyl alcohol(338K)
(C) Distallation under When liquid boils at higher (i) Concentration of sugar juice
reduced pressure temperature and it may (ii) Recovery of glycerol from spent lye.
(Vacuum distillation) decompose before BP is attained. (iii) Glycerol
(D) Steam distillation When the substance is immiscible (i) Aniline is separated from water
 with water and steam volatile. (ii) Turpentine oil
(iii) Nitro Benzene
(iv) Bromo Benzene
(v) Naphthalene
(vi) o-Nitrophenol
? ? Distillation Techniques :
JEEMAIN.GURU
FR ACTIONAL DISTILLATION
STEAM DISTILLATION
 DISTILLATION UNDER REDUCED PRESSURE
JEEMAIN.GURU
? LASSAIGNE METHOD :
1 . Nitrogen (N) : Na + C + N ? ? NaCN
(Sodium extract)
Test : NaCN + FeSO
4
 + NaOH
? ?boil & cool
FeCl
3
 + conc. HCl
?
Blue or green colour [Fe
4
[Fe(CN)
6
]
3
Prussian blue
2 . Sulphur (S) : 2Na + S 
?
? ? ?
 Na
2
S
(Sodium extract)
Test (1) Na
2
S
? Sodium nitro prusside
A deep voilet colour
Test (2) Na
2
S + CH
3
COOH + (CH
3
COO)
2
Pb
?
Black colour  (PbS)
3 . Halogen (X) : Na + Cl 
?
? ? ?
 NaCl
(Sodium extract)
Test : NaCl + HNO
3
 + AgNO
3
(a) White ppt soluble in aq. NH
3
 confirms Cl
(b) Yellow ppt partially soluble in aq. NH
3
 confirms Br
(c) yellow ppt insoluble in aq. NH
3
 confirms I
4 . Nitrogen and Sulphur together :
Na + C + N + S 
?
? ? ?
NaCNS
(Sodium thiocynate)
?
As in test of nitrogen, instead of green or blue colour,
blood red coloration confirms presence of N and S both.
5 . Phosphorous (P) :
P 
2 2
Na O
boils, ?
?? ? ? ?
 Na
3
PO
4
?
Solution in boiled with nitric acid and then treated
with ammonium molybdate (NH
4
)
2
 MoO
4
?
Formation of yellow ppt indicate presence of phosphate
hence phosphorous present in organic compounds.
? THE QUANTITATIVE ANALYSIS :
Quantitative analysis involves the estimation of percentage composition of various element by suitable method.
The molecular mass of the compound is also determined by suitable methods. The knowledge of molecular
mass and percentage composition help us to determine the molecular formula of the compound.
? ESTIMATION OF CARBON AND HYDROGEN OR LIEBIG COMBUSTION METHOD :
Organic compound  +  dry, black CuO
(w g) ? ?Heat strongly in combustion tube
mixture of oxide
? ? Pass through
(i) anhydrous CuSO
4
(ii) anhydrous CaCl
2
JEEMAIN.GURU
Page 4


SIMPLE DISTILLATION TECHNIQUE
PURIFICATION METHODS
Typ e Conditions E xamp le s
(A) Simple distillation (i) When liquid sample has (i) Mixture of chloroform
non volatile impurities (BP = 334K) and Aniline
(BP = 457K)
(ii) When boiling point (ii) Mixture of Ether (BP = 308K) &
difference is 80° K or more. Toluene (BP = 384K)
(iii) Hexane
 
(342K) and Toulene(384K)
(B) Fractional distillation When BP difference is 10° (i) Crude oil in petroleum industry
(ii) Acetone (329) and  Methyl alcohol(338K)
(C) Distallation under When liquid boils at higher (i) Concentration of sugar juice
reduced pressure temperature and it may (ii) Recovery of glycerol from spent lye.
(Vacuum distillation) decompose before BP is attained. (iii) Glycerol
(D) Steam distillation When the substance is immiscible (i) Aniline is separated from water
 with water and steam volatile. (ii) Turpentine oil
(iii) Nitro Benzene
(iv) Bromo Benzene
(v) Naphthalene
(vi) o-Nitrophenol
? ? Distillation Techniques :
JEEMAIN.GURU
FR ACTIONAL DISTILLATION
STEAM DISTILLATION
 DISTILLATION UNDER REDUCED PRESSURE
JEEMAIN.GURU
? LASSAIGNE METHOD :
1 . Nitrogen (N) : Na + C + N ? ? NaCN
(Sodium extract)
Test : NaCN + FeSO
4
 + NaOH
? ?boil & cool
FeCl
3
 + conc. HCl
?
Blue or green colour [Fe
4
[Fe(CN)
6
]
3
Prussian blue
2 . Sulphur (S) : 2Na + S 
?
? ? ?
 Na
2
S
(Sodium extract)
Test (1) Na
2
S
? Sodium nitro prusside
A deep voilet colour
Test (2) Na
2
S + CH
3
COOH + (CH
3
COO)
2
Pb
?
Black colour  (PbS)
3 . Halogen (X) : Na + Cl 
?
? ? ?
 NaCl
(Sodium extract)
Test : NaCl + HNO
3
 + AgNO
3
(a) White ppt soluble in aq. NH
3
 confirms Cl
(b) Yellow ppt partially soluble in aq. NH
3
 confirms Br
(c) yellow ppt insoluble in aq. NH
3
 confirms I
4 . Nitrogen and Sulphur together :
Na + C + N + S 
?
? ? ?
NaCNS
(Sodium thiocynate)
?
As in test of nitrogen, instead of green or blue colour,
blood red coloration confirms presence of N and S both.
5 . Phosphorous (P) :
P 
2 2
Na O
boils, ?
?? ? ? ?
 Na
3
PO
4
?
Solution in boiled with nitric acid and then treated
with ammonium molybdate (NH
4
)
2
 MoO
4
?
Formation of yellow ppt indicate presence of phosphate
hence phosphorous present in organic compounds.
? THE QUANTITATIVE ANALYSIS :
Quantitative analysis involves the estimation of percentage composition of various element by suitable method.
The molecular mass of the compound is also determined by suitable methods. The knowledge of molecular
mass and percentage composition help us to determine the molecular formula of the compound.
? ESTIMATION OF CARBON AND HYDROGEN OR LIEBIG COMBUSTION METHOD :
Organic compound  +  dry, black CuO
(w g) ? ?Heat strongly in combustion tube
mixture of oxide
? ? Pass through
(i) anhydrous CuSO
4
(ii) anhydrous CaCl
2
JEEMAIN.GURU
say increase in weight = a g
? ? Pass through
(i) KOH (aq)
(ii) Saphnolite resin
say increase in weight = b g
For % of H - The increase in weight is 'a' g due to the formation of H
2
O
? ? 18 g H
2
O has H = 2 g
?? a g H
2
O has  
2a
H g
18
?
Since w g organic compound has =  
2a
18
 g hydrogen.
? ? 100 g organic compound has= 
2a 100
18 w
?
?
 g hydrogen
?? 
2
 of H O formed 2 weight
% H 100
18 weight of subs tan ce
? ? ?
For % of C -
The increase in wt is b g due to formation of CO
2
? ? 44 g CO
2
 has C = 12 g
?? b g CO
2
 has C = 
12
b
44
? g
Since w g organic compound has 
12
b
44
? ?
?
? ?
? ?
 g carbon
?? 100 g organic comound has 
12 100
b
44 w
? ?
 g carbon
?? 
2
weight of CO formed 12
% C 100
44 weight of subs tan ce
? ? ?
Note : This method is suitable for estimation if organic compound contains C and H only. In case if other
elements e.g., N,S, halogens are also present the organic compound will also give their oxides which on
being absorbed in KOH will increase the percentage of carbon and therefore following modification
should be made.
(a) If N is also present : The gaseous oxide mixture is first passed over heated Cu gauze which reduces all
oxides of N to N
2
 which are neither absorbed by H
2
O nor by KOH
2NO  +  2Cu  ??? N
2
  + 2CuO
(b) If halogens are present : A roll of silver gauze is placed at the exit end of combustion tube which retains
all the halogens on itself.
2Ag  + X
2
 ??? 2AgX
2Ag  +  CuX
2
 ??? 2AgX  +  Cu
(c) If S is present : A layer of fused lead chromate is placed near the exit end, all the oxides of sulphur are
oxidised to PbSO
4
 & left in tube it self.
Cr
+6
  + 3e
–
  ?? Cr
+3
S
+4
 ?? S
+6
  + 2e
–
(in SO
2
) (in SO
4
–2
)
JEEMAIN.GURU
Page 5


SIMPLE DISTILLATION TECHNIQUE
PURIFICATION METHODS
Typ e Conditions E xamp le s
(A) Simple distillation (i) When liquid sample has (i) Mixture of chloroform
non volatile impurities (BP = 334K) and Aniline
(BP = 457K)
(ii) When boiling point (ii) Mixture of Ether (BP = 308K) &
difference is 80° K or more. Toluene (BP = 384K)
(iii) Hexane
 
(342K) and Toulene(384K)
(B) Fractional distillation When BP difference is 10° (i) Crude oil in petroleum industry
(ii) Acetone (329) and  Methyl alcohol(338K)
(C) Distallation under When liquid boils at higher (i) Concentration of sugar juice
reduced pressure temperature and it may (ii) Recovery of glycerol from spent lye.
(Vacuum distillation) decompose before BP is attained. (iii) Glycerol
(D) Steam distillation When the substance is immiscible (i) Aniline is separated from water
 with water and steam volatile. (ii) Turpentine oil
(iii) Nitro Benzene
(iv) Bromo Benzene
(v) Naphthalene
(vi) o-Nitrophenol
? ? Distillation Techniques :
JEEMAIN.GURU
FR ACTIONAL DISTILLATION
STEAM DISTILLATION
 DISTILLATION UNDER REDUCED PRESSURE
JEEMAIN.GURU
? LASSAIGNE METHOD :
1 . Nitrogen (N) : Na + C + N ? ? NaCN
(Sodium extract)
Test : NaCN + FeSO
4
 + NaOH
? ?boil & cool
FeCl
3
 + conc. HCl
?
Blue or green colour [Fe
4
[Fe(CN)
6
]
3
Prussian blue
2 . Sulphur (S) : 2Na + S 
?
? ? ?
 Na
2
S
(Sodium extract)
Test (1) Na
2
S
? Sodium nitro prusside
A deep voilet colour
Test (2) Na
2
S + CH
3
COOH + (CH
3
COO)
2
Pb
?
Black colour  (PbS)
3 . Halogen (X) : Na + Cl 
?
? ? ?
 NaCl
(Sodium extract)
Test : NaCl + HNO
3
 + AgNO
3
(a) White ppt soluble in aq. NH
3
 confirms Cl
(b) Yellow ppt partially soluble in aq. NH
3
 confirms Br
(c) yellow ppt insoluble in aq. NH
3
 confirms I
4 . Nitrogen and Sulphur together :
Na + C + N + S 
?
? ? ?
NaCNS
(Sodium thiocynate)
?
As in test of nitrogen, instead of green or blue colour,
blood red coloration confirms presence of N and S both.
5 . Phosphorous (P) :
P 
2 2
Na O
boils, ?
?? ? ? ?
 Na
3
PO
4
?
Solution in boiled with nitric acid and then treated
with ammonium molybdate (NH
4
)
2
 MoO
4
?
Formation of yellow ppt indicate presence of phosphate
hence phosphorous present in organic compounds.
? THE QUANTITATIVE ANALYSIS :
Quantitative analysis involves the estimation of percentage composition of various element by suitable method.
The molecular mass of the compound is also determined by suitable methods. The knowledge of molecular
mass and percentage composition help us to determine the molecular formula of the compound.
? ESTIMATION OF CARBON AND HYDROGEN OR LIEBIG COMBUSTION METHOD :
Organic compound  +  dry, black CuO
(w g) ? ?Heat strongly in combustion tube
mixture of oxide
? ? Pass through
(i) anhydrous CuSO
4
(ii) anhydrous CaCl
2
JEEMAIN.GURU
say increase in weight = a g
? ? Pass through
(i) KOH (aq)
(ii) Saphnolite resin
say increase in weight = b g
For % of H - The increase in weight is 'a' g due to the formation of H
2
O
? ? 18 g H
2
O has H = 2 g
?? a g H
2
O has  
2a
H g
18
?
Since w g organic compound has =  
2a
18
 g hydrogen.
? ? 100 g organic compound has= 
2a 100
18 w
?
?
 g hydrogen
?? 
2
 of H O formed 2 weight
% H 100
18 weight of subs tan ce
? ? ?
For % of C -
The increase in wt is b g due to formation of CO
2
? ? 44 g CO
2
 has C = 12 g
?? b g CO
2
 has C = 
12
b
44
? g
Since w g organic compound has 
12
b
44
? ?
?
? ?
? ?
 g carbon
?? 100 g organic comound has 
12 100
b
44 w
? ?
 g carbon
?? 
2
weight of CO formed 12
% C 100
44 weight of subs tan ce
? ? ?
Note : This method is suitable for estimation if organic compound contains C and H only. In case if other
elements e.g., N,S, halogens are also present the organic compound will also give their oxides which on
being absorbed in KOH will increase the percentage of carbon and therefore following modification
should be made.
(a) If N is also present : The gaseous oxide mixture is first passed over heated Cu gauze which reduces all
oxides of N to N
2
 which are neither absorbed by H
2
O nor by KOH
2NO  +  2Cu  ??? N
2
  + 2CuO
(b) If halogens are present : A roll of silver gauze is placed at the exit end of combustion tube which retains
all the halogens on itself.
2Ag  + X
2
 ??? 2AgX
2Ag  +  CuX
2
 ??? 2AgX  +  Cu
(c) If S is present : A layer of fused lead chromate is placed near the exit end, all the oxides of sulphur are
oxidised to PbSO
4
 & left in tube it self.
Cr
+6
  + 3e
–
  ?? Cr
+3
S
+4
 ?? S
+6
  + 2e
–
(in SO
2
) (in SO
4
–2
)
JEEMAIN.GURU
? GENER AL CONCEPT FOR DRYING AND DEHYDR ATING AGENTS
(1) Acidic oxides ( non metals oxides such as CO
2
, NO
2
, SO
2
 P
2
O
5
 etc.) are absorbed by alkali (KOH,
NaOH, Ca(OH)
2
 etc.)
(ii) Basic oxides (Na
2
O, K
2
O etc.) are dissolved in acids (H
2
SO
4
, HCl etc.)
(iii) Acidic oxides can not be dried by basic dehydrating agent (e.g CaO) but they are dried by acid dehydrat-
ing agent (e.g H
2
SO
4
, P
2
O
5
 etc). However H
2
S can be dried by H
2
SO
4
 because it reacts with H
2
SO
4
.
H
2
S  +  H
2
SO
4
 ?? 2H
2
O  + S  + SO
2
(iv) Basic substances can not be dried by acidic dehydrating agent but are dried by basic dehydrating agent
(e.g CaO). It is because NH
3
 is dried by CaO but not by H
2
SO
4
.
? ESTIMATION OF NITROGEN
(a) Duma's method :
Organic compound 
CuO
heated strongly
?? ? ? ? ? ? Mixture of oxides 
Pass over Cu gauze
? ? ? ? ? ? ? ? ?
? N
2
+ Other oxides
   (w g) ?
Dry N
2
  
drying agent
? ? ? ? ? ? ?
?N
2
  + Moisture  
Pass through
KOH (aq.)
? ? ? ? ? ? ?
(Let Vml at N.T.P.)
?? 22400 ml N
2
 weigh at N.T.P. = 28 g
?? V ml N
2
 weigh at NTP = 
28
224 0 0
? V
 g
since w g organic compound gives 
28
22 40 0
10 0 ? ?
V
w
 g
??  
2
Volume of N at NTP 28
% N 100
22400 weight of the compound
? ? ?
Note :   This method can be used to estimate nitrogen in all types of organic compounds.
? KJELDAHL'S METHOD :
This method is based on the principle that a nitrogenous organic compound on heating with conc. H
2
SO
4
converts all its nitrogen quantitatively to ammonium sulphate ;
organic compound  +  conc. H
2
SO
4
+ K
2
SO
4
+ CuSO
4
(w g)      (raises the B.P. of H
2
SO
4
) (catalyst)
? ?
?
?
?
Heating in Kjeldahl flask
4 2 4 3
add KOH
drop by drop
(NH ) SO NH is formed ? ? ? ? ? ? ?
Pass this gas through V
1
 ml of N
1
 H
2
SO
4
Now titrate the H
2
SO
4
 by another alkali say of N
1
 H
2
SO
4
 (NH
3
 is absorbed here)
V
2
 ml of N
2
 ,KOH  are used
Meq of H
2
SO
4
 taken = N
1
V
1
Meq. of KOH  used for H
2
SO
4
 = N
2
V
2
?? Meq. of H
2
SO
4
 for NH
3
  = N
1
V
1
 – N
2
V
2
?? Meq. of NH
3
 formed = (N
1
V
1
 – N
2
V
2
) or say NV)
(the difference obtained or V ml of N H
2
SO
4
 are used to absorb NH
3
)
JEEMAIN.GURU
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