S.

no Experiment Observation Inference

Preliminary tests
1 Action with sodium (i) Brisk effervescence (i) Presence of a carboxylic
bicarbonate: acid.
(ii) No brisk
Take 2 ml of saturated sodium effervescence (ii) Absence of a carboxylic
bi carbonate solution in a test acid.
tube. Add 2 or 3 drops (or a
pinch of solid) of an organic
compound to it.
2 Charring test: Charring takes place with Presence of carbohydrate
Take a small amount of an smell of burnt sugar
organic compound in a dry test
tube. Add 2 ml of conc H 2SO4
to it and heat the mixture.
3 Ignition test: (i) Burns with sooty (i) Presence of an aromatic
Take a small amount of the flame compound
organic compound in a nickel (ii) Presence of an aliphatic
spatula and burn it in bunsen (ii) Burns with non compound
flame. sooty flame
4 Test with KMnO4 solution: (i) Pink colour of (i) Substance is
Take small amount of the KmnO4 solution is unsaturated.
organic compound in a test decolourised
tube add 2 ml of distilled water (ii) No decolourisation (ii) Substance is saturated.
to dissolve it. To this solution takes place
add few drops of very dilute
alkaline KMnO4 solution and
shake it well.
SALT Esterification reaction: A pleasant fruity odour is Presence of carboxylic group.
2,3 Take 1 ml (or a pinch of solid) noted.
of an organic compound in
a clean test tube. Add 1 ml of
ethyl alcohol and 4 to 5 drops
of conc. sulphuric acid to it.
Heat the reaction mixture
strongly for about 5 minutes.
Then pour the mixture into a
beaker containing dil. Sodium
carbonate solution and note the
smell.

256

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SALT Tollen’s reagent test: Shining silver mirror is Presence of an aldehyde
1 Take 2 ml of Tollen’s reagent in formed.
a clean dry test tube. Add 3-4
drops of an organic compound
(or 0.2 g of solid) to it, and
warm the mixture on a water
bath for about 5 minutes.
SALT Legal’s test: Red colouration. Presence of a ketone.
3 A small amount of the
substance is taken in a test tube.
1 ml sodium nitro prusside
solution is added. Then sodium
hydroxide solution is added
dropwise.
SALT Biuret test: Violet colour is appeared. presence of a diamide
5 Take a small amount of an
organic compound in a test
tube. Heat strongly and then
allow to cool. Dissolve the
residue with 2 ml of water.
To this solution add 1 ml of
dilute copper sulphate solution
and few drops of 10% NaOH
solution drop by drop.
SALT Molisch’s test: Violet or purple ring is Presence of carbohydrate
6 Take a small amount of an formed at the junction of
organic compound in a test the two liquids.
tube. It is dissolved in 2 ml of
water. Add 3-4 drops of alpha
naphthol to it.Then add conc
H2SO4 through the sides of test
tube carefully.

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22-12-2022 19:48:33
 List of organic compounds for analysis:

1. Benzaldehyde 4. Benzoic acid 7. Glucose

2. Cinnamaldehyde 5. Cinnamic acid 8. Aniline
3. Acetophenone 6. Urea 9. Salicylic acid

REASONING

3. Action with sodium bicarbonate:

Carboxylic acids react with Sodium bicarbonate and liberate CO2. Evolution of
carbon dioxide gives brisk effervescence.
R-COOH+ NaHCO3 R-COONa+CO2   H2O

4. Action with Borsches reagent:

Borsches reagent is prepared by dissolving 2,4-dinitrophenylhydrazine in a solution
containing methanol and little of conc sulphuric acid.
Aldehydes and ketones react with borsches reagent to form yellow, orange or red
precipitate (dinitro phenylhydrazone)
Aliphatic carbonyl compounds give deep yellow precipitate.
Aromatic carbonyl compounds give red precipitate.
2,4-dinitrophenyl hydrazine can be used to qualitatively detect the carbonyl group
of an aldehyde or ketone. A positive result is indicated by the formation of an yellow or
orange-red precipitate of 2,4-dinitrophenyl hydrazone.

O2N O2N

R R
H H
C O + H N NO C N NO
2 N 2 N 2

H H

Aldehyde 2,4 dinitrophenylhydrazine Aldehyde 2,4 dinitrophenylhydrazone

(Yellow or orange)
O2N O2N

R R
H H
C O + H N NO C N NO
2 N 2 N 2

R R

Ketone 2,4 dinitrophenylhydrazine Ketone 2,4 dinitrophenylhydrazone

(Yellow or orange or red)

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22-12-2022 19:48:33
 1. Estimation of Ferrous Sulphate (Fe2+)
Aim :
To estimate the amount of ferrous sulphate dissolved in 750 ml of the given unknown
solution volumetrically. For this you are given with a standard solution of ferrous
ammonium sulphate (FAS) of normality 0.1102 N and potassium permanganate solution
as link solution. 4

Short procedure:

S.no Content Titration-I Titration-II

1 Burette solution KMnO4 KMnO4
2 Pipette solution 20 ml of standard FAS 20 ml of unknown FeSO4
3 Acid to be added 20ml of 2N H2SO4 (approx) 20ml of 2N H2SO4 (approx)
4 Temperature Lab temperature Lab temperature
5 Indicator Self-indicator (KMnO4) Self-indicator (KMnO4)
Appearance of permanent Appearance of permanent pale
6 End point
pale pink colour pink colour
7 Equivalent weight of FeSO4 = 278

Titration –I
(Link KMnO4 )Vs (Standard FAS)

Burette readings Concordant value

Volume of
S.no standard FAS Initial Final (Volume of KMnO4)
(ml)
(ml) (ml) (ml)
1 20
2 20
3 20
Calculation :
Volume of KMnO4 (link) solution (V1) =-------------ml
Normality KMnO4 (link) solution (N1) =------------ N
Volume of standard FAS solution (V2) = 20 ml
Normality of standard FAS solution (N2) = 0.1102

According to normality equation: V1× N1 = V2 X N2

N1 = (V2 X N2 )\ V1
Normality of KMnO4 (link) solution (N1) = X N

Calculation :
 Volume of KMnO4 (link) solution V1 = ml
Normality KMnO4 (link) solution N1 = ?N
Volume of standard FeSO4 solution V2 = 20 ml

Normality of standard = N
FeSO4 solution N2

According to normality equation:

According to normality equation: V1× N1 = V2 × N2
N1 = (V2 x N2)\V1

Normality of KMnO4 (link) solution (N1) =

Titration –II
(Link KMnO 4)Vs (Unknown FeSO 4 solution)

Volume of Burette readings Concordant value

s.no Unknown FeSO4 Initial Final (Volume of KMnO4 )
(ml) (ml) (ml) (ml)

1 20
2 20
3 20

Calculation :
Volume of Unknown FeSO4 solution V1 = 20 ml
Normality of Unknown FeSO4 solution N1 = ?N
Volume of KMnO4 (link) solution V2 = ml RESULT:
Normality KMnO4 (link) solution N2 = X N The amount of FeSO4
According to normality equation: V1× N1 = V2 × N2 dissolved in 750 ml of the
solution = g

N1=(V2 X N2)/V1
N1 = ------------ N
The normality of unknown FeSO 4 solution = N
Weight calculation:

The amount of FeSO4 dissolved in 1 lit of the

solution = (Normality) x (equivalent weight)
The amount of FeSO4 dissolved in 750 ml of the Normality x equivalentweight x 750
solution =
1000
Y  278  3
N1 
4
 2. Estimation of Ferrous Ammonium Sulphate (FAS)
Aim :
To estimate the amount of ferrous ammonium sulphate (FAS) dissolved in 1500 ml of
the given unknown solution volumetrically. For this you are given with a standard solution of
ferrous sulphate ( FeSO4 )of normality 0.1024 N and potassium permanganate solution as link
solution.
SHORT PROCEDURE :

s.no Content Titration-I Titration-II

4

1 Burette solution KMnO4 KMnO4

2 Pipette solution 20 ml of standard FeSO4 20 ml of unknown FAS

3 Acid to be added 20ml of 2N H2SO4(approx)

20ml of 2N H2SO4 (approx)
4 Temperature Lab temperature Lab temperature
5 Indicator Self-indicator ( KMnO4 ) Self-indicator ( KMnO4 )

6 End point Appearance of permanent Appearance of permanent

pale pink colour pale pink colour
7 Equivalent weight of FAS = 392

Titration –I
(Link KMnO4)Vs (Standard FeSO4 )

Volume of Burette readings Concordant value

s.no standard FeSO4 Initial Final (Volume of KMnO4 )
(ml) (ml) (ml) (ml)
1 20
2 20
3 20

CALCULATION :

Volume of KMnO4 (link) solution V1 = ml

Normality of standard FeSO4 =0.1024N
Volume of KMnO4 (link) solution N2 = N

Volume of standard FeSO4 solution V2 = 20ml

 Titration –II
(Link KMnO4)Vs (Unknown FAS)
Burette readings Concordant value
Volume of
s.no Unknown FAS Initial Final (Volume of KMnO4 )
(ml) (ml) (ml) (ml)
1 20
2 20
3 20

Calculation :
Volume of Unknown FAS solution V1 = 20ml
Normality of Unknown FAS solution N1 = ?N
Volume of KMnO4 (link) solution V2 = ml
Normality KMnO4 (link) solution N2 = N
According to normality equation: V1× N1 = V2 × N2
N1 = (V2 X N1)\V1
N1 = -----------------N

The normality of unknown FAS solution = Y N

Weight calculation:

The amount of FAS dissolved in 1 lit of the = (Normality) x (equivalent weight)

solution
The amount of FAS dissolved in 1500 ml of the Normality  equivalentweight  1500
=
solution 1000

Y  392  1500
=
1000
= g

Report :
The amount of FAS dissolved in 1500 ml of the solution = g
 3.Estimation of oxalic acid
Aim :
To estimate the amount of oxalic acid dissolved in 500 ml of the given solution
volumetrically. For this you are given with a standard solution of ferrous ammonium sulphate
(FAS) of normality 0.1 N and potassium
4 4
permanganate solution as link solution.

4
Short procedure:
s.no Content Titration-I Titration-II
1 Burette solution KMnO4 KMnO4

2 Pipette solution 20 ml of standard FAS 20 ml of unknown oxalic acid

3 Acid to be added 20ml of 2N H2SO4(approx) 20ml of 2N H2SO4 (approx)
4 Temperature Lab temperature 60 – 70 0C
5 Indicator Self-indicator ( KMnO4 ) Self-indicator (KMnO4)
Appearance of permanent Appearance of permanent
6 End point
pale pink colour pale pink colour
7 Equivalent weight of oxalic acid = 63

Titration –I
(Link KMnO4 )Vs (Standard FAS solution)
Volume of standard FAS solution
1
V1
Volume of Burette readings Concordant value
s.no standard FAS (Volume of KMnO4 )
Normality KMnO4 (link) solution N1 = Initial Final
N
(ml) (ml)
(ml) (ml)

Titration–II
1 20
(Unknown2 oxalic acid
20 ) Vs (Link KMnO 4 )
Burette is washed with water, rinsed with KMnO4 solution and filled with same KMnO4
3 20
solution up to the zero mark. Exactly 20 ml of unknown oxalic acid solution is pipetted out
into the clean, washed conical flask. To this oxalic acid solution approximately 20ml of 2N
Calculation
sulphuric acid is added. :This mixture is heated to 60 – 700C using Bunsen burner and that
hot solutionVolume
is titrated
of against Link solution from
(link) 4solution
KMnO4 KMnO V1 the
= burette.
ml KMnO 4 is added drop
Normality KMnO4 (link) solution N1 = ?N
Volume of standard FeSO4 V2 = 20ml
Normality of standard FeSO4 N2 = 0.1024 N
Titration –II
(Link KMnO4 )Vs (Unknown oxalic acid)

Volume of Burette readings Concordant value

s.no Unknown oxalic Initial Final (Volume of KMnO4 )
acid (ml) (ml) (ml) (ml)
1 20
2 20
3 20
Calculation :
Volume of Unknown oxalic acid solution V1 = 20 ml
Normality of Unknown oxalic acid solution N1 = ?N
Volume of KMnO4 (link) solution V2 = ml
Normality KMnO4 (link) solution N2 = N
According to normality equation:

V1 x N1 = V2 x N2

V2 x N2
N1 =
V1

Normality of Unknown oxalic acid solution N1 = Y N

Weight calculation:
The amount of oxalic acid dissolved in 1 lit
=(Normality) x (equivalent weight)
of the solution
The amount of oxalic acid dissolved in 500 Y  63  500
=
ml of the solution 1000
x 63 x 500
=
1000
= g
Report :
The amount of oxalic acid dissolved in 500 ml of given the solution = g