PRINCIPLE OF
GRAVIMETRIC
ANALYSIS
GROUP 1 :MIC 3A1
GRAVIMETRIC
ANALYSIS
Gravimetric
analysis is one of the
most accurate and precise method of
macroquantitative (large quantity)
analysis.
In this process the analyte is
selectively converted into insoluble
form
STEPS IN A GRAVIMETRIC
ANALYSIS
PREPARARION OF THE SOLUTION
PRECIPITATION
DIGESTION
FILTERATION
WASHING
DRYING AND IGNITING
CALCULATION
PREPARING OF THE SOLUTION
Factor that must considered during prepararion of
solution :
1.
2.
3.
4.
5.
Volume of the solution during precipitation
The concentration range of the test substance
The presence and concentration of other constituents
Temperature
pH
Some form of preliminary separation may be necessary
to eliminate interfering material.
 Adjust the solution condition to maintain low solubility
of the precipitate and to obtain it in a form suitable for
filtration.
 Proper adjustment of the solution conditions prior to
precipitation may also mask potential interferences.
 pH is important because it often influences both the
solubility of the analytical precipitate and the
possibility of interferences from other substances.

PRECIPITATION

Important conditions of substance that must be
taken into account before conducting
precipitation:
The precipitate should first be sufficiently insoluble
that the amount lost due to solubility will be
negligible.
 It should consist of consist of large crystals that can
be easily filtered.
 All precipitates tend to carry some of other
constituents of the solution with them.
 This contamination should be negligible. Keeping the
crystals large can minimize this contamination.

PRECIPITATION PROCESS:

After adding precipating agent such as AgNO3
,precipitation occur but in series(step by step)
1.
2.
SUPERSATURATION-The solution phase contain
more of the dissolved salt than occurs at
equilibrium.
NUCLEATION
A minimum number of partical must come
together to produce microscopic nuclei of the
solid phase.
 The higher the degree of supersaturation the
greater the rate of nucleation.

Von Weimarn ratio:
Relativesupersaturation 
Q -S
S
•Q - degree of supersaturation(the concentration of
the mixed reagent before precipitation occur)
•S - solubility of the precipitate at equilibrium
High relative supersaturation- many small crystal (high
surface area)
Low relative supersaturation- fewer, larger crystal (low
surface area)
STEPS TO MAINTAIN FAVOURABLE
CONDITIONS FOR PRECIPITATIONS
1.
2.
3.
4.
Precipitate from dilute solution. This keeps Q low.
Add dilute precipitating reagents slowly, with effective
stirring. This also keeps Q low. Stirring prevents local
excesses of the reagent.
Precipitate from hot solution. This increase S. the
solubility should not be too great or the precipitation will
not be quantitative (with less than 1 part per thousand
remaining). The bulk of the precipitation maybe
performed in the hot solution, and then the solution may
be cooled to make the precipitation quantitative.
Precipitate at as low a pH as is possible to maintain
quantitative precipitation. As we have seen, many
precipitates are more soluble in acid medium, and this
slows the rate of precipitation. They are more soluble
because the anion of the precipitate combines with
protons in the solution.
DIGESTION
To make precipitate become larger and more
pure crystals.
 Also known as Ostwald ripening (digestion).
 Ostwald ripening improves the purity and
crystallinity of the precipitate.
 In digestion, colloidal particles formed counter
layer(secondary layer) , primary absorptive layer
and colloidal(large number of small particle)
AgCl.
 There are 2 types of colloidal particles which is
hydrophilic and hydrophobic.

FILTRATION(IMPURITIES IN
PRECIPITATE)

Precipitates tend to carry down from the solution other
constituents, causing the precipitate to become
contaminated. This process of contamination is called
coprecipitation. Ways in which a foreign material may be
coprecipitate.
OCCLUSION AND INCLUSION.
1.
o
o
o
In the process of occlusion, material that is not part of
the crystal structure is trapped within a crystal.
For example, water may be trapped in pockets when
AgNO3 crystals are formed, and this can be driven off
by melting.
Inclusion occurs when ions, generally of similar size
and charge, are trapped within the crystal lattice
(isomorphous inclusion, as with K+ in NH4MgPO4
precipitation).
2. SURFACE ADSORPTION
The surface of the precipitate will have a primary adsorbed
layer of the lattice ions in excess.
 For example, after the barium sulfate is completely
precipitated, the lattice ion in excess will be barium, and
this will form the primary layer.
 Digestion reduces the surface area and the amount of
adsorption.
 Surface adsorption of impurities is the most common source
of error in gravimetry. It is reduced by proper precipitation
technique, digestion, and washing.

3. ISOMORPHOUS REPLACEMENT
Two compound are said to be isomorphous if they have
same type of formula and crystallize in similar geometric
forms.
 When the lattice dimensions are about the same, one ions
can replace another in the crystal, resulting in a mixed
crystal.
 For example, in the precipitation of Mg2+ as magnesium
ammonium phosphate, K+ has nearly the same ionic size as
NH4+ and can replace it to form magnesium potassium
phosphate.
 It is very serios and little can be done about it,so
precipitates in which it occurs are seldom used analytically.

4. POSTPRECIPITATION.
When the precipitate is allowed to stand in contact with the
mother liquor(the solution from it was precipitate), a
second substance will slowly form a precipitate with the
precipitating regent.
 For example, when calcium oxalate is precipitated in the
presence of magnesium ions, magnesium oxalate does not
immediately precipitate because it tends to form
supersaturated solutions.
 It will come down if the solution is allowed to stand too long
before being filtered.

WASHING





Coprecipitated impurities, especially those on the surface,
can be removed by washing the precipitate after filtering.
The precipitate will be wet with the mother liquor, which is
also removed by washing.
Many precipitates cannot be washed with pure water,
because peptization(washing particles with water increases
the extend of solvent/water moleculer between the layers,
causing the secondary layer to be loosely bound,the
particles revert to colloidal state) occurs.
This is the reverse of coagulation
Coagulated particles have a neutral layer of adsorbed
primary and counterions.
presence of another electrolyte will cause the
counterions(an ion having a charge opposite to that of
the substance with which is associated).
 These foreign ions are carried along in the coagulation.
 Washing with water will dilute and remove foreign
ions, and the counterion will occupy a larger volume,
with more solvent molecules between it and the
primary layer.
 The repulsive forces between particles between
particles become strong again, and the particles
partially revert to the colloidal state and pass through
the filter.
 This can be prevented by adding an electrolyte to the
wash liquid. Eg: HNO3 or NH4NO3 for AgCl
precipitate.

DRYING OR IGNITING
This process is to removes the solvent and wash
electrolytes.
 If the collected precipitate is in a form suitable
for weighing, it must be heated to remove the
water and the absorbed electrolyte from the wash
liquid.
 The preferable temperature use for drying is by
drying it at 110°C to 120°C for 1 or 2 hours.
 Ignition at a higher temperature are required if a
precipitate must be converted to a more suitable
form for weighing.

CALCULATION

Gravimetric factor(GF)-the weight of analyte per
unit weight of precipitate
f wt analyte(g/mol)
a
GF 
 (molanalyte/mol precipitate)
f wt precipitate (g /mol) b
 g analyte/gprecipitate
So, if Cl2 in a sample is converted to chloride and
precipitated as AgCl, the weight of Cl2 that gives 1 g of
AgCl is
f wt Cl2 (g Cl2/molCl2)
1
g Cl2  g AgCl 
 (molCl2/mol AgCl)
f wt AgCl (g AgCl/molAgCl ) 2
 g AgCl  GF (g Cl2/g AgCl)
 g AgCl  0.24737 (g Cl2/g AgCl)
EXAMPLE 1
CALCULATE THE GRAMS OF ANALYTE PER GRAM OF
PRECIPITATE FOR THE FOLLOWING CONVERSION:
Analyte
Precipitate
P
Ag3PO4
K2HPO4
Ag3PO4
Bi2s3
3BaSO4
Solution
at wt P (g/mol)
1
 ( mol P /mol Ag3P O4)
g wt Ag3P O4 (g/mol) 1
30.97(g P /mol)
1


418.58( g Ag3P O4/mol) 1
 0.07399g P /g Ag3P O4  GF
g P /g Ag3P O4 
f wt K2HP O4 (g/mol) 1
 (mol K2HP O4/mol Ag3P O4)
f wt Ag3P O4 (g/mol) 1
174.18(g K2HP O4(g/mol ) 1


418.58( g Ag3P O4/mol) 1
 0.41612g K2HP O4/g Ag3P O4  GF
g K2HP O4/g Ag3P O4 
f wt Bi2S3 (g/mol) 1
g Bi2S3/g BaSO4 
 (mol Bi2S3/mol BaSO4)
f wt BaSO4 (g/mol) 3
514.15 (g Bi2S3/mol) 1


233.40(g BaSO4)
3
 0.73429g Bi2S3/g BaSO4  GF
•In order to calculate the percent composition by weight of analyte in
the sample, the weight of substance sought was divided with the
weight of sample and times with 100%.
•The weight of sample sought can be obtain from the weight of
precipitate and the corresponding weight/mole relationship, as in
equation below:
weightof substancesought(g)  weightof precipitate(g)
f wt substancesought(g/mol)

f wt precipitate (g/mol)
a
 ( mol substancesought/molprecipitate)
b
 weightof precipitate (g)  GF (g sought/gprecippitate)
•Thus, the general formula would be:
% sought
weightof precipitate (g)  GF (g sought/gprecipitate)
 100%
weightof sample (g)
EXAMPLE 2
An ore is analyzed for the manganese content by converting the
manganese
to Mn3O4 and weighing it. If a 1.52g sample yeilds Mn3O4 weighing
0.126g,what would be the percent Mn2O3 in the sample? The percent
Mn?
Solution
3Mn2O3
(g Mn2O3 / g Mn3O4)
2Mn3O4
% Mn2O3 
 100%
1.52g sample
0.126g  3(157.9)/2(228.8)

 100%  8.58%
1.52 g
0.126g Mn3O4 
3Mn
( g Mn/ g Mn3O4)
Mn3O4
% Mn 
 100%
1.52 g sample
0.126g  3(54.94)/228.8

 100%  5.97%
1.52 g
0.126g Mn3O4 
EXAMPLE 3:
What weight of pyrite ore (impure FeS2 ) must be taken for analysis
so that
the BaSO 4 precipitate weight obtained will be equal to one-half that
of the
percent S in the sample?
Solution:
gA
 100%
g sample
1
S
A(g BaSO4) 
(g S/g BaSO4)
BaSO4
A %S 2
 100%
g sample
1 32.064

1% S  2 233.40  100%
g sample
g sample  6.869g
%A
THE END….