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Module 3
Practical:
Observation and
Deduction
2002-7
NaCl
Find LEf
F Scullion. JustChemy.Com
1
2
Inorganic
Observation
and
Deduction
3
The observations and deductions come in pairs of
activities. One involves inorganic chemistry and
the other organic chemistry.
The inorganic exercise typically involves the
analysis of a so-called “Double Salt”. One of the
ions in these mixed salts is common to both salts
in most cases. Here are some examples: Sodium Sulphate and Sodium Chloride
+
Na
SO 2Cl4
Calcium Carbonate and Calcium Chloride
+
2
Ca
CO 2Cl3
4
Deductions.
Observations
A is Ammonium
Chloride NH4Cl
Identifying halide ions in solution
Na
The equations for the reactions are represented thus:
NaX(aq) + AgNO3(aq)
=
AgX(s) + NaNO3(aq)
10
Why
Can’t
You
hite
ream
ellow
NaCl(aq) + AgNO3(aq) = AgCl(s) + NaNO3(aq)
white ppt
NaBr(aq) + AgNO3(aq) = AgBr(s) + NaNO3(aq)
cream ppt
NaI(aq) + AgNO3(aq) = AgI(s) + NaNO3(aq)
yellow ppt
11
Both AgCl and
AgBr are light
sensitive.
They have
darkened
noticeably
after 5
minutes
AgBr
12
Solubility of Silver Halides in Ammonia solution.
Silver
halide
AgCl
Colour
White
Solubility in NH3(aq)
Dilute
√
Concentrated
√
AgBr
Cream
x
√
AgI
Yellow
x
x
13
Ammonium Compounds
warm
Ammonium + Alkali
Compound
 Salt + NH3 + H2O
warm
NH4Cl + NaOH  NaCl + NH3 + H2O
warm
(NH4)2SO4 + Ca(OH)2  CaSO4 + 2NH3 + 2H2O
14
AMMONIUM + ALKALI  SALT + AMMONIA + WATER
COMPD
NH4Cl
(ii)
+
Describe how you would carry out the test
state what you would observe.
NaOH 
NaCl + NH3
+
H2O
for hydrogen chloride gas and
White [1] fumes/smoke [1] glass rod [1]
dipped in conc [1] ammonia [1] (max [4])
16
Observations
Observations
19
It is Potassium Carbonate
20
Potassium flame test &
Emission spectrum
Lilac
21
Using HCl & Limewater to test for CO32-
Effervescence
Limewater
Sample of
dilute HCl
(aq)
Sign of
CO2 gas
22
Using Magnesium Nitrate to test for the presence of CO32-(aq)
CO32-(aq) + Mg2+(aq)  MgCO3(s)
23
Reactions of the Carbonate Ion.
Metal Carbonate + Acid  Salt + CO2 + H2O
Partial Ionic Equation
CO32- + H+  H2O + CO2
Balanced Symbol Equation
K2CO3 + 2HCl  2KCl + H2O + CO2
--------------------------------------------------------
Precipitation Reaction
AB(aq)
+
CD(aq)

AD(s)
+
CB(aq)
Partial Ionic Equation
CO32- + Mg2+  MgCO3
Balanced Symbol Equation
K2CO3 + MgCl2  MgCO3 + 2KCl
24
25
Observation
A is a “double salt”
A is a “double salt”
Colour is a feature of the compounds of TMs
Cobalt chloride above
Potassium
chromate
above
Above is from the Chemguide Website
Nickel chloride
opposite
28
Colour of Aqueous Ions
Copper(II)
ion, Cu2+(aq)
Chromium(III)
ion, Cr3+(aq)
Iron(II) ion,
Fe2+(aq)
Chromate ion,
CrO42-(aq)
Iron(III) ion,
Fe3+(aq)
Dichromate ion,
Cr2O72-(aq)
Cobalt(II) ion,
Co2+(aq)
Manganese(II)
ion, Mn2+(aq)
Nickel(II) ion,
Ni2+(aq)
Permanganate
ion, MnO4-(aq)
Colour of Aqueous Ions
Copper(II)
ion, Cu2+(aq)
blue
Chromium(III)
ion, Cr3+(aq)
deep green
Iron(II) ion,
Fe2+(aq)
green
Chromate ion,
CrO42-(aq)
yellow
Iron(III) ion,
Fe3+(aq)
Yellow
Dichromate ion,
Cr2O72-(aq)
orange
Cobalt(II) ion,
Co2+(aq)
pink
Manganese(II)
ion, Mn2+(aq)
very pale pink /
colourless
Nickel(II) ion,
Ni2+(aq)
deep
green
Permanganate
ion, MnO4-(aq)
deep purple
Ammonium Compounds
warm
Ammonium + Alkali
Compound
 Salt + NH3 + H2O
warm
NH4Cl + NaOH  NaCl + NH3 + H2O
warm
(NH4)2SO4 + Ca(OH)2  CaSO4 + 2NH3 + 2H2O
31
AMMONIUM + ALKALI  SALT + AMMONIA + WATER
COMPD
NH4Cl
(ii)
+
Describe how you would carry out the test
state what you would observe.
NaOH 
NaCl + NH3
+
H2O
for hydrogen chloride gas and
White [1] fumes/smoke [1] glass rod [1]
dipped in conc [1] ammonia [1] (max [4])
NH3(aq)
A weak
alkali
JustChemy.Com
Dilute ammonia solution with UI: about pH 11
33
Lithium
Crimson Red
Calcium
Brick Red
Strontium
Red
34
Reactions of the Halide Ions with AgNO3(aq).
Silver + Sodium
Nitrate
Halide

Silver + Sodium
Halide
Nitrate
Partial Ionic Equation using X- for Halide Ions
Ag+ + X-  AgCl
Balanced Symbol Equation
AgNO3 + NaX
 AgX + NaNO3
AgNO3 + NaCl
 AgCl + NaNO3
AgNO3 + NaBr
 AgBr + NaNO3
AgNO3 + NaI
 AgI
+ NaNO3
35
36
Mixture of Ammonium Chloride and Li/Sr/Ca Chloride
37
38
39
Acidified Barium Chloride (or Nitrate) is used to test for SO
Ba2+(aq) + SO42-(aq) BaSO4(s)
2-
4
(aq)
40
Has Ammonium Sulphate and Sodium Sulphate
41
Ammonium Compounds
warm
Ammonium + Alkali
Compound
 Salt + NH3 + H2O
warm
NH4Cl + NaOH  NaCl + NH3 + H2O
warm
(NH4)2SO4 + Ca(OH)2  CaSO4 + 2NH3 + 2H2O
42
AMMONIUM + ALKALI  SALT + AMMONIA + WATER
COMPD
NH4Cl
(ii)
+
Describe how you would carry out the test
state what you would observe.
NaOH 
NaCl + NH3
+
H2O
for hydrogen chloride gas and
White [1] fumes/smoke [1] glass rod [1]
dipped in conc [1] ammonia [1] (max [4])
NH3(aq)
A weak
alkali
JustChemy.Com
Dilute ammonia solution with UI: about pH 11
44
Flame test for
Sodium
Orange-yellow
Na+
NH4+
?
45
Ba2+
Cl
Cl
Cu2+
SO42-
Barium nitrate
Copper(II) sulphate
BaCl2
CuSO4
Complete Formula Equation:
BaCl2(aq) + CuSO4(aq)  BaSO4(s) + CuCl2(aq)
Complete Ionic Equation:
Ba2+(aq) + 2 Cl-(aq) + Cu2+(aq) + SO42-(aq)  BaSO4(s) + Cu2+(aq) + 2 Cl-(aq)
Net Ionic Equation:
Ba2+ + 2 Cl- + Cu2+ + SO42-  BaSO4(s) + Cu2+ + 2 ClBa2+(aq) + SO42-(aq) BaSO4(s)
Na+
NH4+
2SO446
47
Observations
A is a mixture of 2 salts
48
Observations
A is a mixture of 2 salts
A is a White solid
49
50
51
52
Observations
X is a double salt
53
Observations
X is a double salt
X is a white solid
54
Has sodium chloride and ammonium chloride
55
56
61
A is a
double
salt
62
A is a
double
salt
A has
NaBr
and
Na2SO4
63
Oxidising ability of Concentrated Sulphuric Acid.
Its reactions with halide ions in aqueous solution during the
formation of the Hydrogen Halides.
[CARE: CARRY OUT ALL OF THESE IN A FUME CUPBOARD WITH GREAT CARE]
1) NaCl(s)
+
H2SO4(l)

NaHSO4(aq) + HCl(g)
The concentrated sulphuric acid will not oxidise HCl
Dip a glass rod into concentrated
ammonia solution and then into the
test tube where you suspect the
presence of HCl. The HCl will form
a white smoke with NH3 if it is
present.
NH3 + HCl  NH4Cl
64
Oxidising ability of Concentrated Sulphuric Acid.
Its reactions with halide ions in aqueous solution during the
formation of the Hydrogen Halides.
[CARE: CARRY OUT ALL OF THESE IN A FUME CUPBOARD WITH GREAT CARE]
2a)
KBr(s)
+
H2SO4(l)

KHSO4(aq) +
HBr(g)
However, the concentrated sulphuric acid will
oxidise some of the HBr as follows:
-1
2b)
2HBr
An increase in O.N.
+
H2SO4
0

Br2
+
SO2 +
2H2O
Therefore, one will observe a Reddish Vapour due to some
bromine being present.
The SO2 is and acidic gas. It is also a reducing agent and
will, for example, decolourise purple potassium
65
permanganate solution
Oxidising ability of Concentrated Sulphuric Acid.
Its reactions with halide ions in aqueous solution
during the formation of the Hydrogen Halides.
[CARE: CARRY OUT ALL OF THESE IN A FUME CUPBOARD WITH GREAT CARE]
3a) KI(s)
+ H2SO4(l)
 KHSO4(aq)
+ HI(g)
The concentrated sulphuric acid will oxidise some of the HI as follows:
3b) H2SO4(l) + 2HI(g) + SO2 +
I2
+
2H20
3c)
H2SO4(l) +
6HI(g) + S
+
3I2 +
4 H20
3d)
H2SO4(l) +
8HI(g) + H2S +
4I2 +
4 H20
During the reaction one will observe: ·
·
·
·
·
Violet Iodine Vapour being evolved,
The violet vapour cooling and subliming to form dark solid iodine,
A smell of rotten eggs (H2S)
Some free yellow sulphur
Some HI(g) which could be identified in the way one shows the
66
presence of HCl. Use concentrated NH3 solution
SUMMARY
Oxidising ability of Concentrated Sulphuric Acid.
Its reactions with halide ions in aqueous solution during the formation of the
Hydrogen Halides.
[CARE: CARRY OUT ALL OF THESE IN A FUME CUPBOARD WITH GREAT CARE]
1) NaCl(s) +
H2SO4(l)
=
NaHSO4(aq) + HCl(g)
The concentrated sulphuric acid will not oxidise HCl
2a) KBr(s)
+
H2SO4(l)
=
KHSO4(aq) +
HBr(g)
The concentrated sulphuric acid will oxidise some of the HBr as follows:
2b) 2HBr +
H2SO4
+ Br2 +
SO2 +
2H2O
Therefore, one will observe a Reddish Vapour due to some bromine being present.
3a) KI(s)
+ H2SO4(l) à KHSO4
+
HI(g)
The concentrated sulphuric acid will oxidise some of the HI as follows:
3b) H2SO4(l) + 2HI(g) + SO2 +
I2
+
2H20
3c)
H2SO4(l) +
6HI(g) + S
+
3I2 +
4 H20
3d)
H2SO4(l) +
8HI(g) + H2S +
4I2 +
4 H20
During the reaction one will observe: ·
·
·
·
·
Violet Iodine Vapour being evolved,
The violet vapour cooling and subliming to form dark solid iodine,
A smell of rotten eggs (H2S)
Some free yellow sulphur
Some HI(g) which could be identified in the way one shows the
presence of HCl. Use concentrated NH3 solution
67
68
Organic
Observation
and
Deduction
69
We are informed that B is a
mixture of hydrocarbons
71
72
B has a C=C functional group
73
Cyclohexene
Only a very weak permanent dipole.
Essentially non-polar
Molecules held together by only van der Waals forces
Memo: “Like dissolves Like.”
74
Alcohols, carboxylic acids, aldehydes and
ketones are miscible with water. In other
words, they are soluble in water.
As the organic molecule increases in length,
masking of the functional group by the
hydrocarbon chain occurs. This reduces
solubility.
H-bonding between alcohol and water
molecules is shown to the left.
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
Ethanal
H-bonded
to water
75
The hydrogen/carbon ratio has an influence on how
cleanly a fuel burns. In general, the higher this
ratio, the cleaner the flame.
Ethanol
C2H5OH (Ratio = 6/2 = 3.00)
Propane
C3H8 (Ratio = 8/3 = 2.67)
Large alkane
C30H62 (Ratio = 62/30 = 2.07)
Cyclohexane
C6H12 (Ration =
Ethene
C2H4 (Ration = 4/2 = 2.00)
Cyclohexene
C6H10 (Ration =
12/
10/
6
6
= 2.00)
= 1.67)
Methylbenzene C6H5CH3 (Ration = 8/6 = 1.34)
Benzene
C6H6 (Ration = 6/6 = 1.00)
The black smoke and soot is caused by unburnt carbon
76
ratio and more smoke and soot
C2H6 (Ratio = 6/2 = 3.00)
carbon
Ethane
hydrogen/
CH4 (Ratio = 4/1 = 4.00)
Decreasing
Methane
Increasing carbon/hydrogen ratio = Increasing smoke and soot
Ethanol
Hexane
Cyclohexane
Cyclohexene
77
Ethene + Bromine  1,2-dibromoethane
Br
+ Br2

Br
Cyclohexene + Bromine  1,2-dibromocyclohexane
Clear
Clear
Clear
colourless
brown
colourless
Liquid
liquid
liquid
78
79
George Bush
gets the smell of
this chemical
Spirit burner
82
83
A primary or secondary alcohol
Alcohols, carboxylic acids, aldehydes and
ketones are miscible with water. In other
words, they are soluble in water.
As the organic molecule increases in length,
masking of the functional group by the
hydrocarbon chain occurs. This reduces
solubility.
H-bonding between alcohol and water
molecules is shown to the left.
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
Ethanal
H-bonded
to water
84
Alcohols undergo combustion in air, burning with
a clean blue flame.
C2H6 (Ratio = 6/2 = 3.00)
Ethanol
C2H5OH (Ratio = 6/2 = 3.00)
The alcohol has its on
“inbuilt oxygen” that helps
it to burn.
Propane
C3H8 (Ratio = 8/3 = 2.67)
Large alkane
C30H62 (Ratio = 62/30 = 2.07)
ROH + O2 = CO2 + H2O
Cyclohexane
C6H12 (Ration = 12/6 = 2.00)
C2H5OH + 3O2 
Ethene
C2H4 (Ration = 4/2 = 2.00)
Benzene
C6H6 (Ration = 6/6 = 1.00)
2CO2 + 3H2O
Soot is unburnt carbon
ratio and more smoke and soot
Ethane
carbon
CH4 (Ratio = 4/1 = 4.00)
hydrogen/
Methane
Decreasing
The hydrogen/carbon ratio has an influence on how
cleanly a fuel burns. In general, the higher this
ratio, the cleaner the flame.
85
Potassium dichromate (acidified) is an oxidising agent.
Its formula is K2Cr2O7.
The following equation shows it accepting electrons: Cr(VI)
Cr2O72- + 6 e- = 2 Cr3+
Cr (III)
The role of the acid in “mopping up the
oxygens” is seen in this next equation:
Cr2O72- + 6 e- + 14H+ = 2 Cr3+ + 7H2O
At this stage (AS) one only need to learn the following: H
I
H–C–
I
H
H
I
Warm
C – O-H + [O]

I
H ethanol
H
I
H – C – C = O + H2 O
I I
H H ethanal
86
87
Magnesium reacting with 2M Ethanoic Acid
2CH3COOH + Mg  Mg(CH3COO)2 + H2
90
Deduction
B is a carboxylic acid
RCOOH
Vinegar is a dilute
aqueous solution of
ethanoic acid.
It is approximately
5% CH3COOH
2CH3COOH + Na2CO3 = 2CH3COO-Na+ + CO2 + H2O
2CH3COOH + Mg = Mg(CH3COO)2 + H2
93
94
96
97
98
99
THE TRIIODOMETHANE
(IODOFORM) REACTION
Gives a positive result for two groupings: -
Iodoform
molecule HCI3
H
Triiodomethane
I
Yellow crystals with
WITH
CH3 C - R
ALCOHOL
an antiseptic smell.
I
OH where R = H, CH3, C2H5, etc
-------------------------------------O
You will see that the alcohol above is oxidised
II
to this carbonyl structure in step 1 of 3 steps!
WITH CH3 C - R
Ethanal and Methyl Ketones
therefore also give +ve tests
where R = H, CH3, C2H5, etc
100
Which of these alcohols would give a +ve iodoform test?
H
I
H – C – OH
I
H
H H
I
I
H – C – C - OH
I
I
H H
H H H
I I
I
H – C – C – C – OH
I
I I
H H H
Ethanol is the only primary
alcohol to give the iodoform
If "R"is a hydrocarbon group,
then you have a secondary
alcohol. Lots of secondary
alcohols give this reaction,
but those that do all have a
methyl group attached to the
carbon with the -OH group.
H H H
I
I I
H–C–C–C-H
I
I I
H HO H
No tertiary alcohols can contain this group because no tertiary alcohols
can have a hydrogen atom attached to the carbon with the -OH group.
No tertiary alcohols give the triiodomethane (iodoform) reaction.
101
Which of these carbonyls would give a +ve iodoform test?
H–C=O
I
H
H H
I I
H–C–C–C=O
I
I I
H H H
H
I
H–C–C=O
I
I
H H
Ethanal is the
only aldehyde
that gives a
+ve iodoform
test
H
H
I
I
H–C–C–C-H
I II I
H O H
All the 2-ones of the ketones will give a +ve iodoform test
102
Version 1 of the Iodoform Test
Sample
Iodine in
KI(aq)
Pure iodoform
103
Version 1 of the Iodoform Test I2/NaOH
To about 5 cm3 of a saturated solution of iodine in
potassium iodide in a test-tube add 5 drops of
alcohol.
Add alcohol
Iodine in KI(aq)
104
Version 1 of the Iodoform Test I2/NaOH
To about 5 cm3 of a saturated solution of iodine in
potassium iodide in a test-tube add 5 drops of
ethanol.
Add sodium hydroxide solution carefully until the
colour has almost gone.
The cloudiness is a sign of
precipitation. Iodoform is a
pale yellow solid with an
antiseptic smell
105
Version 1 of the Iodoform Test I2/NaOH
To about 5 cm3 of a saturated solution of iodine in
potassium iodide in a test-tube add 5 drops of
ethanol.
Add sodium hydroxide solution carefully until the
colour has almost gone.
3 REACTIONS or STEPS occur:
1. OXIDATION: Alcohol + [O]  Carbonyl
106
Version 1 of the Iodoform Test I2/NaOH
To about 5 cm3 of a saturated solution of iodine in
potassium iodide in a test-tube add 5 drops of
ethanol.
Add sodium hydroxide solution carefully until the
colour has almost gone.
3 REACTIONS or STEPS occur:
1. OXIDATION: Alcohol + [O]  Carbonyl
2. All 3 of the H atoms of the methyl group are substituted by I atoms
107
Version 1 of the Iodoform Test I2/NaOH
To about 5 cm3 of a saturated solution of iodine in
potassium iodide in a test-tube add 5 drops of
ethanol.
Add sodium hydroxide solution carefully until the
colour has almost gone.
3 REACTIONS or STEPS occur:
1. OXIDATION: Alcohol + [O]  Carbonyl
2. All 3 of the H atoms of the methyl group are substituted by I atoms
3. The CI3COR formed then goes on to form HCI3 and RCOO-Na+
Stand the test-tube in water at about 70 oC for two or three
minutes, then remove and allow to cool.
108
Version 1 of the Iodoform Test I2/NaOH
To about 5 cm3 of a saturated solution of iodine in
potassium iodide in a test-tube add 5 drops of
ethanol.
Add sodium hydroxide solution carefully until the
colour has almost gone.
3 REACTIONS or STEPS occur:
1. OXIDATION: Alcohol + [O]  Carbonyl
2. All 3 of the H atoms of the methyl group are substituted by I atoms
3. The CI3COR formed then goes on to form HCI3 and RCOO-Na+
Stand the test-tube in water at about 70 oC for two or three
minutes, then remove and allow to cool.
Yellow crystals of iodoform separate out on standing
and the smell is like that of antiseptic
109
Reactions taking place.
H
I
CH3 C - OH + I2 + 2OH-  CH3 C = O + 2I- + 2H2O
I
I
R
R
CH3 C = O + 3I2 + 3OHI
R
CI3 C = O
I
R
+ NaOH
 CI3 C = O
I
R
+ 3 I- + 3H2O
 R – C = O + HCI3
I
O-Na+
110
111
Iodoform.
Triiodomethane
CHI3
A fine yellow
precipitate
Antiseptic smell
112
Version 2. KI/NaClO
Using potassium iodide and sodium chlorate(I) solutions
Sodium chlorate(I) is also known as sodium hypochlorite.
Sodium chlorate is the active ingredient in most household
bleaches. It is an oxidising agent and will oxidise Iodide ions
to Iodine (I2).
NaClO
Potassium Iodide
Iodine liberated
113
Version 2. KI/NaClO
Using potassium iodide and sodium chlorate(I) solutions
Sodium chlorate(I) is also known as sodium hypochlorite.
Potassium iodide solution is added
to a small amount of organic sample,
Sample
+ KI
114
Version 2. KI/NaClO
Using potassium iodide and sodium chlorate(I) solutions
Sodium chlorate(I) is also known as sodium hypochlorite.
Potassium iodide solution is added
to a small amount of organic sample,
This is followed by sodium chlorate(I) soln.
Sample
+ KI
+ NaClO
Iodoform
NaClO is alkaline (source of OH-) and oxidises I- to I2
As a result KI/NaClO is equivalent to using I2/NaOH
115
Summary of Version 2. KI/NaClO
Using potassium iodide and sodium chlorate(I) solutions
Sodium chlorate(I) is also known as sodium hypochlorite.
Potassium iodide solution is added
to a small amount of organic sample,
Sample
+ KI
+ NaClO
This is followed by sodium chlorate(I) soln.
NOTE THAT THE NaClO OXIDISES IODIDE (I-) TO IODINE (I2)
So as well as any possible yellow precipitate, you will also see the typical
reddish-brown colour of iodine solution being formed during the reaction.
Note also, that sodium chlorate(I) solution is alkaline and contains a sufficietly
high [OH-] to carry out the second half of the reaction.
In effect you are making I2 “in situ” so the tests are essentially the same.
If no precipitate is formed in the cold, it may be
necessary to warm the mixture very gently.
Look for the formation of a pale yellow precipitate
with antiseptic smell
116