2. Qualitative Analysis of Organic
• The analysis and identification of unknown
organic compounds constitutes a very important
aspect of experimental organic chemistry.
• In order to deduce the identity of your two
unknowns, combine one qualitative test.
3. General Scheme of Analysis
• A. Preliminary Test
physical characteristics: solid, liquid, color, and odor.
• Amines often have a fish-like odor, while esters usually have a
pleasant fruity or floral smell.
• B. Physical Constants
Determine the boiling point or melting point. Distillation is
used for liquids. It serves the dual purpose of determining the
boiling point as well as purifying the liquid for subsequent
• C. Solubility Tests
The solubility of the unknown in the following reagents
provides very useful information. In general, about 1 mL of
the solvent is used with approximately 0.1 g or 0.2 mL (2-3
drops) of the unknown compound.
4. Solubility table
REAGENT AND TEST CLASS
Soluble in cold or hot
Neutral, acidic or basic.
(Test with litmus or
Neutral, e.g. alcohols;
acids,phenols; Basic, e.g.
Soluble in dil. HCl Basic Most amines
Soluble in dil. NaOH Acidic
Most acids, most
Soluble in NaHCO3 Strongly acidic Most carboxylic acids.
Insoluble in water, acid
Hydrocarbons, alkyl or
aryl halides, esters and
ethers. Higher molecular
aldehydes and ketones
5. Group test
• Test for –OH Group
• Soluble in NaOh
• FeCl3 Test
• Alc. Sol. Of sample + 2-3 drops of neu.FeCl3
• violet p-nitrophenol
• pink a-napthol
• Reimer-tiemann test
• Comp.+2mlCHCl3+3ml NaOh Heat
CHCl3 layer turns blue to green
• Soluble in NaOH
• Comp.+Br2 water white ppts of tribromo
• comp.+ NaNO2+ conc.H2SO4
Dark blue color appears
• Comp.+ neutral FeCl3 Violet colour
• Based on the principle of Separation of
Thin layer chromatography
10. UV and IR
• Basic techniques
• Based on the principle of absorption of light
• UV=Ultra violet Spectroscopy
• IR =Infrared spectroscopy
• 10 cm-1 -14000 cm-1
• AAS= Atomic Absorption Spectroscopy
• Based on the principle of absorption of light
• Mostly elemental analyzed
• AES=Atomic Emission Spectroscopy
• Based on the principle of emission of light
• Minute elements analyzed
12. Mass spectrometry
• The highest tool to identify and quantify a
• Measures Molecular weight of a molecule
13. What is an empirical formula?
• A chemical formula in which the ratio of the
elements are in the lowest terms is called an
• The empirical formula for a glucose molecule
(C6H12O6) is CH2O. All the subscripts are
divisible by six.
C6 H12 O6
6 6 6
C H2 O
16. Finding empirical formulas using molecular
• Find mass (or %) of each element.
• Find moles of each element.
• Divide moles by the smallest # to find
• When necessary, multiply subscripts by 2, 3, or
4 to get whole #’s.
100g of CH4:
74.9 g of C & 24.9g of H.
- Dividing each quantity by the proper atomic weight gives
the no. of moles of each element.
C = 74.9/12.01
= 6.24 moles
H = 24.9/1.008
= 24.7 moles
Since a mole of one element contains the same no. of
atoms as a mole of any other element.
C = 6.24/6.24
H = 24.7/6.24
= 3.96 =4
18. EXAMPLE: Find the empirical formula for a
sample of 25.9% N and 74.1% O.
• 25.9g = 1.85 mol N 1.85mol N = 1 N x 2 = N2
14g/mol 1.85 mol
• 74.1 g = 4.63 mol O 4.63 mol O = 2.5 O x 2 = O5
16 g/mol 1.85 mol
19. PRACTICE PROBLEMS:
1. A substance is 36.1% by weight calcium
and 63.9% chlorine. What is the empirical
formula of this compound?
Atomic wt of calcium = 40.07
Atomic wt of chlorine = 35.45
20. 2. A compound is 43.4%Na, 11.3%C, and
45.3% O. What is the empirical formula for
Na = 22.98
C = 12.01,
O = 15.99
21. Molecular formula
• A molecular formula is the “true formula” of a
• The chemical formula for a molecular
compound shows the actual number of atoms
present in a molecule.
22. To find the molecular formula
from the empirical formula:
• Find the empirical formula.
• Determine the empirical formula mass.
• Divide the molecular mass by the empirical
formula mass to determine the multiple.
• Multiply the empirical formula by the multiple
to find the molecular formula.
MF mass = n
(EF)n = molecular formula
The empirical formula for ethylene is CH2.
Find the molecular formula if the molecular mass
C = 1 x 12 = 12
H = 2 x 1 = +2
14g/mol = empirical formula mass
28.1 g/mol = 2
24. Practice Problems:
1. Find the molecular formula for a compound
with a mass of 78 amu and the empirical formula
2. Find the molecular formula for a compound
with a mass of 82 amu and the empirical
3. Find the molecular formula for a compound
with a mass of 90 amu and the empirical
25. 4. Find the molecular formula for a compound
with a mass of 112 amu and the empirical
5. Find the molecular formula for a compound
with a mass of 40 amu and the empirical
26. More Practice Problems:
Writing Empirical Formulas:
1.Determine the empirical formula of a compound
containing 2.644 g of gold and 0.476 of chlorine.
2. Determine the empirical formula of a
compound containing 0.928 g of gallium and
0.412 of phosphorus.
27. Qualitative elemental analysis:
• The detection of various elements
present in an organic compound is called
• Carbon and hydrogen are present in almost all
the organic compounds.
• Other commonly present elements in organic
compounds are oxygen, nitrogen, halogens,
sulphur and sometimes phosphorus.
28. Ignition test
• This simple test will indicate the presence of
oxygen atoms, and also the presence of aromatic
rings or metals.
• A small amount of the unknown is burned in the
blue part of a Bunsen Burner flame.
• The combustion flame and residue are
examined. A blue flame indicates the presence of
an oxygen atom in the unknown.
• A yellow, sooty flame indicates the presence of
29. Beilstein Test
• The presence of bromine, chlorine or iodine in
organic compounds can be detected by the
• This test depends on the production of a volatile
copper halide (CuX2) produced when an organic
halide is strongly heated with a copper oxide.
• The copper halide will give off blue-green light
when excited in the high temperature of a burner
30. Sodium Fusion Test
• This is the most useful test for the presence of N,
S, Cl, Br and I. An organic unknown is degraded
under high temperature in the presence of
• Depending on the elements present
the following sodium salts will be formed:
Nitrogen: NaCN, Sulfur: Na2S and halogens:
NaX (X = Cl,Br, I).
31. Quantitative elemental analysis: C,H,X
• In quantitative combustion, a weighed sample of org.
compound is passed through a combustion train: a tube packed
with copper oxide heated to 600-800°C, a tube containing a
drying agent ( dehydrite, magnesium percolate) & strong base.
• Water : absorbed by drying agent,
• CO2 : absorbed by base
• For ex: Methane – 9.67 mg produced 26.53 mg of CO2 &
21.56mg of H2O.
• C/CO2 = 12.01/44.01 of the carbon dioxide is C.
• 2H/H2O = 2.016/18.02 of the water is H.
32. • Therefore,
• wt. of C = 26.53 x 12.01/44.01
• = 7.24 mg
• wt. of H = 21.56 x 2.016/18.02
• = 2.41 mg
• % composition:
• %C = (7.24/9.65) x 100 = 74.9
• % H = (2.41/9.65) x 100 = 24.9
• Total of H & C = 100%, Oxygen must be absent.
33. Estimation of halogens (Carius
• • In this method a known weight of organic compound is heated
with fuming nitric acid in the presence of AgNO3 in a hard glass
tube called Carius tube.
• • Carbon and hydrogen of the compound are oxidized to CO2 and
H2O. Halogen forms Silver halide (AgX).
• It is filtered, washed dried & weighed.
• % of Halogen(X) = At.wt of X × Wt.of AgX ×100
• Mol.wt of AgX x Wt of Organic compound
• % of Cl = 35.5 × Wt. of AgCl formed ×100
• 143.5 Wt of Organic compound
34. • % of Br= 80 × Wt.of AgBr formed ×100
188 Wt of Oraganic compound
• % of I = 127 Wt.of Ag I formed 100
235 Wt of Oraganic compounds
35. Estimation of nitrogen
• Two method :
1.Duma’s method, 2.Kjeldahl’s method
• In this method a known weight of organic compound is heated strongly
with dry cupric oxide.
• Carbon and hydrogen get oxidized to carbon dioxide and water vapour.
• Nitrogen, if present is converted to N2 is collected over KOH solution
and its volume is determined at S.T.P.
CxHyNz + (2x + y/2)CuO xCO2 +y/2 . H2O +z/2.N2 + (2x+y/2)Cu
• % of N = 28 × Volume of Nitrogen at S.T.P ×100
• 224 × Weight of organic compound
36. 2. Kjeldahl’s method:
• organic compound is treated with Conc.H2SO4 in the presence
of small amounts of CuSO4 to convert nitrogen into
• Ammonium sulphate is treated with excess of NaOH to
liberate NH3 gas.
• The ammonia evolved is neutralized with excess of Conc.
H2SO4, which is relatively more in amount than that is
required to neutralize NH3 gas.
• Now, the excess of acid is titrated with standard alkali
• From this the amount of H2SO4 used to neutralise NH3
formed is calculated and from that percentage of nitrogen is
37. Organic compound + H2SO4 (NH4)2SO4
(NH4)2SO4 + 2NaOH Na2SO4 + 2H2O + 2NH3
2NH3 + H2SO4 (NH4)2SO4
• % of N = 14 x N x V
Wt. of organic compound
Where N = Normality of acid,
V = volume of acid neutralized by ammonia