Gravimetric analysis, P K MANI

Gravimetric Analysis and
Precipitation Equilibria
Analytical Chemistry,

ACSS-501

Class – I to VI

Dr. Pabitra Kumar Mani
Door meten tot weten
“Through measurement to knowledge.”
Heike Kamerlingh Onnes, NL
“gentleman of absolute zero”

A History of Gravimetric Analysis
Jons Jacob Berzelius (1779 - 1848),
considered the leading chemist of his time,
developed much of the apparatus and many
of the techniques of 19th century analytical
chemistry. Examples include the use of
ashless filter paper in gravimetry, the
use of hydrofluoric acid to decompose
silicates, and the use of the metric
system in weight determinations. He
performed thousands of analyses of pure
compounds to determine the atomic
weights of most of the elements known
then. Berzelius also developed our present
system of symbols for elements and
compounds.

A History of Gravimetric Analysis
Theodore W. Richards (1868 - 1928)
…and his graduate students at Harvard
developed or refined many of the techniques
of gravimetric analysis of silver and
chlorine. These techniques were used to
determine the atomic weights of 25 of the
elements by preparing pure samples of
the chlorides of the elements,
decomposing known weights of the
compounds, and determining the chloride
content by gravimetric methods. From this
work Richards became the first American to
receive the Nobel Prize in Chemistry in
1914.

Accuracy vs. Precision
• Accuracy: getting the correct answer
• Precision: being able to get the same answer
more than once
• We can have precision without accuracy and
accuracy without precision
• Both are important for valuable results

Chemistry 215 Copyright D
Sharma

4

Sharma

5
http://www.chem.ualberta.ca/courses/Chem211/Harrison/

Analytes: The constituents of interest in a sample.
Matrix: All other constituents in a sample except for the analytes.

How to Perform a Successful Gravimetric Analysis

• What steps are needed?
1.
2.
3.
4.
5.
6.
7.
8.
9.

Sampled dried, triplicate portions weighed
Preparation of the solution
Precipitation
Digestion
Filtration
Washing
Drying or igniting
Weighing
Calculation

Gravimetric Analysis
• Gravimetric Analysis – one of the most accurate and
precise methods of macro-quantitative analysis.
• Analyte selectively converted to an insoluble form.
• Measurement of mass of material
• Correlate with chemical composition
• Why?
• Simple
• Often required for high precision
Accuracy: A measure of the agreement between an experimental
result and its expected value.
Precision: An indication of the reproducibility of a measurement or
result.

Common Desiccants
Mechanism of Action
Hydration

ANHYDRONE® (Magnesium
Perchlorate anhydrous),
CaCl2, MgO, MgSO4, K2CO3,
KOH, Drierite, Na2SO4
(anhydrous), H2SO4, ZnCl2

Absorption and/ or
Adsorption

BaO, CaSO4, Molecular
Sieve, H3PO4, NaOH Pellets

Chemisorption

CaO, P2O5

• Silica gel goes from blue to • Anhydrous sodium sulfate
pink as it absorbs moisture
gets clumpy as it absorbs
Can be regenerated in oven
water
More Information about desiccants including common interferents
and regeneration temperature can be found at:

http://www.jtbaker.com/techlib/documents/3045.html

Principles of gravimetric Analysis
(i)Evolution or Volatalization method
(ii) Precipitation method

(iii) Miscellaneous method :
Magnetic separation,
Differential separation,
Electro gravimetry

Volatilization Gravimetry
(i)Direct Method
(ii)Indirect method

Direct Method
The desired constituent is directly weighed eg. estimation of water in
minerals
i) Talc H Mg (SiO ) Δ
water (absorbed in a suitable absorbant,
2

3

3 4

Anhydrous Mg (ClO4)2, P2O5
ii) Carbonates, bicarbonates

Δ or H+

CO2

ignited
iii) (Borates, Boric Acid ) + H2SO4 + CH3OH
(BOCH3)3
(weighed)
Methyl borate
Ca(BO3)2
Ca-Borate
(ester) absorbed in

Indirect Method
A) N in Nitrate, KNO3
2KNO3 + SiO2

heat

K2SiO3 + N2O5

(more volatile)

(residue in the crucible)
W1-W2 = wt of N2O5 ----- lost
heat
B) SiO2(Fe, Cu, Pb) + HF
Pt crucible

SiF4 + residue (impurities)

W1-W2 = wt of pure silica

For carbonate: carbon dioxide
Solid adsorbent: Ascarite II - NaOH on a nonfibrous silicate

Precipitation Technique:
To get a substance, sparingly soluble or very little soluble by
double decomposition reaction, in solution state
AB + CD
AD + BC
(soln) Reagent
Sparingly
soln
soluble

Due care should taken for selection of a suitable reagent for the
desired constituent to be analysed. The reagent must
(i)Produce a precipitate of low solubility product
(ii)Be selective or better specific for the desired constituent.
Other constituents should not produce a similar low solubility
product with the reagent

Desired
ion

Reagent

Precipitate
form

Weighable
form

Cl-

Ag+ as AgNO3 soln

AgCl

AgCl

Ag+

Cl- as KCL soln

AgCl

AgCl

SO4-2

Ba+2 as BaCl2 or BaNO3

BaSO4

BaSO4

Ba+2

SO4-2

BaSO4

BaSO4

Mg+2, Zn+2, Mn+2

HPO4-2 as (NH4)2HPO4
soln.

M(NH4)PO4.6H2O

M2P2O7
pyrophosphate

HPO4-2 , PO4-3

pH 5-6.5, MgCl2 soln
NH4Cl soln

Mg(NH4)PO4.6H2O

Mg2P2O7
pyrophosphate

K+

HClO4

KClO4

KClO4

Na+ [(C6H5)4 B]-

K [(C6H5)4 B]

K [(C6H5)4 B]

Na3 [ Co(NO2)6 ]

K2Na [ Co(NO2)6 ]

K2Na [ Co(NO2)6 ]

Desired
ion

Reagent

Precipitate
form

Ni+2

Dimethyl glyoxime (DMG)

Ni(DMG)2

Co+2

α-nitroso β-napthol

CoR3

N

O

Co3O4
N

OH

Al+3
pH 4.5-6.0
Mg+2
pH 7.0

8 hydroxy quinoline
at

Weighable
form

O
+

Co
O

AlR3

Desired
ion

Reagent

Be+2, Cu+2, Fe+3 N-nitroso phenyl hydroxyl
amine (Cupferron)

Precipitate
form

Weighable
form

BeR2
CuR2

Weighable
CuO

FeR3

N
N

Ca+2

O

Fe2O3
3+

Fe
O

CaC2O4.H2O

CaC2O4.H2O
Low heat

<400°C >600°C

C2O42-

CaO

Cu+2

CNS- or KCNS

CuCNS

Directly weighable

F-

PbCl2 satd. Soln.

PbFCl

Directly weighable

Organic precipitating agents are chelating agents.
Organic precipitating agents are chelating agents.
They form insoluble metal chelates. .
They form insoluble metal chelates

©Gary Christian,
6th Ed. (Wiley)

Effect of particle size
Solubility of a precipitate increases with decreasing particle size. For
having diameter in the range less than 1-2 µ, the solubility

RT ln S r 2σ
=
M
S dr

Sr =solubility of particles having radius r
S= solubilty of macroparticles, solubility
M= molecular weight
σ = surface tension, d= density of the particle

For different substances, however, depending upon the nature of bond and
molecular structure, the ratio Sr/S becomes diffferent
For BaSO4

ppts, S0.02/S = 931 (at 25°C)
Ag2CrO4,
S0.02/S = 4 (at 25°C)
The particle size during pptn is influenced by the relative super
saturation of the medium. The initial velocity of formation of particles,
Velocity of formation,
V α
(Q-S)/S
where, Q=total concentration of the substance that is to precipitate
S= equilibrium solubility, (Q-S) denote the supersaturation at the moment pptn
commences.

This condition applies when the pptn. begins, by changing the value of relative
supersaturation, the particle size can be chnage. With very low value of it,
initially fewer no. of nuclei (particles) are formed, which later grow in size slowly
to produce larger particles of not only low solubility but also of perfect crystalline
set and easy filterability.
The rate of growth of initially precipitated particles is given by Noyes-Nernst

Equation

V= [ (D/l) x A] (QS)

D= diffusion coefficent
l= diffusion path, A= surface area

at a given temp. and stirring rate (D/l) x A= const, K

V= K(Q-S)
Particle size increases with decreasing super saturation. This is
due to two effects in combine . (i) The Aggregation velocity
(ii) The Orientation velocity
The final form of the precipitate particle depends upon the competition of these two
effects. On digestion and aging particularly at higher temp.(hot soln.) and dil.
soln (i.e. under condition of increasing solubility) , the crystals of precipitates
become stable and of perfect shape and size

Sometimes very finely dispersed particles of colloidal nature can be
readily coagulated to larger size by adding very small amount of an
organic substance behaving as a lyophillic colloid i.e., a dilute soln of
gelatin (1%) or gum arabic sol. The process is called sensitization.
Thus by adding a few ml of gelatin soln. quantitative pptn of colloidal
silica is obtained through digestion and evaporation of colloidal silicic
acid.

Important Factors for Gravimetric Analysis

• Nucleation

– Individual ions/atoms/molecules coalesce to form “nuclei”

• Particle Growth

– Condensation of ions/atoms/molecules with existing “nuclei”
forming larger particles which settle out

• Colloidal Suspension

– Colloidal particles remain suspended due to adsorbed ions
giving a net + or - charge

Coagulation, agglomeration
Suspended colloidal particles coalesce to form larger filterable

particles (inert electrolyte allows closer approach)

Peptization
Re-dissolution of coagulated colloids by washing and removing

inert electrolyte

Schematic model of the solid–solution interface at a particle of AgCl
in a solution containing excess AgNO3.

Mechanism of Precipitation
Digestion

• Heating the precipitate within the mother liquor
(or solution from which it precipitated) for a
certain period of time to encourage
densification of nuclei.
– During digestion, small particles dissolve
and larger ones grow (Ostwald ripening).
This process helps produce larger crystals
that are more easily filtered from solution
∆T

35

Conditions for Analytical Precipitation
• Precipitation from hot solution
– The molar solubility (S) of precipitates
increases with an increase in temperature
– An increase in S decreases the
supersaturation and increases the size of
the particle.
• Precipitation from dilute solution
– This keeps the molar concentration of the
mixed reagents low. Slow addition of
precipitating reagent and thorough stirring
keeps Q low. (Uniform stirring prevents high
local concentrations of the precipitating
agent.)
36

Conditions for Analytical Precipitation
• Precipitation at a pH near the acidic end of
the pH range in which the precipitate is
quantitative.
– Many precipitates are more soluble at the
lower (more acidic) pH values and so the rate
of precipitation is slower.
• Digestion of the precipitate.
– The digestion period can lead to
improvements in the organization of atoms
within the crystalline nuclei, such as expulsion
of foreign atoms (or other impurities).
37

During digestion at elevated temperature:
During digestion at elevated temperature:
Small particles tend to dissolve and reprecipitate on larger ones.
Small particles tend to dissolve and reprecipitate on larger ones.
Individual particles agglomerate.
Individual particles agglomerate.
Adsorbed impurities tend to go into solution.
Adsorbed impurities tend to go into solution.

©Gary Christian,
6th Ed. (Wiley)

Fig. 10.1. Ostwald ripening.

Cl- -adsorbs on the particles when in excess (primary layer).
Cl adsorbs on the particles when in excess (primary layer).
A counter layer of cations forms.
A counter layer of cations forms.
The neutral double layer causes the colloidal particles to
The neutral double layer causes the colloidal particles to
coagulate.
coagulate.
Washing with water will dilute the counter layer and the primary layer charge
Washing with water will dilute the counter layer and the primary layer charge
causes the particles to revert to the colloidal state (peptization).
causes the particles to revert to the colloidal state (peptization).
So we wash with an electrolyte
So we wash with an electrolyte
that can be volatilized on heating (HNO3).
that can be volatilized on heating (HNO ).
3

©Gary Christian,
6th Ed. (Wiley)

Representation of silver chloride colloidal particle and adsorptive layers when Cl is in excess.

This condition applies when the pptn. begins, by changing the value of relative
supersaturation, the particle size can be chnage. With very low value of it,
initially fewer no. of nuclei (particles) are formed, which later grow in size slowly
to produce larger particles of not only low solubility but also of perfect crystalline
set and easy filterability.
The rate of growth of initially precipitated particles is given by Noyes-Nernst

Equation

V= [ (D/l) x A] (Q-S)

D= diffusion coefficent
l= diffusion path, A= surface area

at a given temp. and stirring rate (D/l) x A= const, K

V= K(Q-S)
Particle size increases with decreasing super saturation. This is
due to two effects in combine . (i) The Aggregation velocity
(ii) The Orientation velocity
The final form of the precipitate particle depends upon the competition of these two
effects. On digestion and aging particularly at higher temp.(hot soln.) and dil.
soln (i.e. under condition of increasing solubility) , the crystals of precipitates
become stable and of perfect shape and size

Solubility product and Factors affecting solubility of a ppt.

BmAn (solid) ⇋ BmAn (soln.) ⇋ m Bx+ + n AySparingly soluble

almost complete dissociation since the resulting electrolyte

substance

[ B ] = ms moles/L [ A ] = ns moles/L

soln. is very dilute

Solubility = s moles/L
(having ionic lattice)

[B ] [ A ]
Ke =
x+ m

y− n

[ Bm An ]

[ ] [A ]

S=B

x+ m

y− n

x+

y−

But, [BmAn] is constant because the soln is
in contact with large excess of solid BmAn
Ke X [BmAn] = S (Solubility product )

= (ms)m (ns)n

Precipitation Equilibria:The Solubility Product
• Solubility of Slightly Soluble Salts:
• AgCl(s)(AgCl)(aq) Ag+ + Cl• Solubility Product KSP = ion product
• KSP = [Ag+][Cl-]
• Ag2CrO4(s)  2 Ag+ + CrO42• KSP = [Ag+]2[CrO42-]

The molar solubility depends on the stoichiometry of the salt.
The molar solubility depends on the stoichiometry of the salt.
A 1:1 salt is less soluble than a nonsymmetric salt with the same K sp..
A 1:1 salt is less soluble than a nonsymmetric salt with the same K sp

©Gary Christian, Analytical Chemistry,

6th Ed. (Wiley)

Let’s consider silver chloride
AgCl (s)  Ag+ + Cls

s

s

• What is the equilibrium constant?
= s2
Ksp = (Ag+) (Cl-)
• Now what about the value of the Ksp?
Ksp = 1.6 x 10-10
• Can we calculate 1.3 xsolubility, s?
s = (Ksp)1/2 = the 10-5 M
PGCC CHM 102 Sinex

What is the pH of a bottle of Milk of
Magnesia, a common antacid?
Milk of Magnesia is Mg(OH)2 with Ksp = 8.9 x 10-12
Mg(OH)2 (s)  Mg++ + 2OHs
s
2s
Ksp = (Mg++)(OH-)2 = s(2s)2 = 4s3
s = (Ksp/4)1/3 = 1.3 x 10-4 M
(OH-) = 2s = 2.6 10-4 M
PGCC CHM 102 Sinex

pOH = 3.58 pH = 10.42

Problem: A saturated solution of silver chromate was to found
contain 0.022 g/L of Ag2CrO4. Find Ksp
Eq. Expression:

Ag2CrO4 (s) ⇔ 2Ag+ + CrO42–

Ksp = [Ag+]2[CrO42–]

So we must find the
concentrations of each
ion and then solve
for Ksp.

52

Problem: A saturated solution of silver chromate was to contain
0.022 g/L of Ag2CrO4. Find Ksp
Eq. Expression:

Ag2CrO4 (s) ⇔ 2Ag+ + CrO42–

Ksp = [Ag+]2[CrO42–]
Ag+: 0.022 g Ag2CrO4
L

2 Ag +
mol
332g 1 Ag 2CrO 4

2
1 CrO 40.022g Ag2CrO4 mol
CrO4 :
L
332 g 1 AgCrO 4

–2

[

K sp = 1.33 x 10

mol Ag +
= 1.33 x 10–4
L

= 6.63 x 10–5

2
mol CrO 4L

] [6.63 x 10 ] = 1.16 x 10

-4 2

-5

-12

53

Problems working from Ksp values.
Given: Ksp for MgF2 is 6.4 x 10–9 @ 25 oC
Find: solubility in mol/L and in g/L
Ksp = [Mg2+][F–]2
MgF2(s) ⇔ Mg2+ + 2F–
I.
C.
E.

N/A
N/A
N/A

0
+x
+x

0
+2x
+2x

Ksp= [x][2x]2 = 4x3
6.4 x 10–9 = 4x3

[

]

6.4 x 10 -9
x=3
= 1.2 x 10-3 = Mg 2+ = [ MgF2 ]
4

now for g/L:

1.2 x 10-3 mol MgF2
L

62.3 g
mol

= 7.3

x 10–2

g MgF2
L
54

The common ion effect:
A salt is generally less soluble in a solution containing an ion which is the same as one
of the constituent ions of that salt. This is known as the COMMON ION EFFECT.

We can see that this must necessarily occur if we apply Le Chatelier's
Principle to an equilibrium such as
Adding an excess Cl- (or Na+) to a saturated solution of NaCl imposes a stress on the
equilibrium, which will adjust in order to oppose the stress. A shift to the left will use
up Na+ or Cl- to form solid NaCl.
Le Chatelier's Principle
"If

a system in equilibrium is subjected to a stress the
equilibrium will shift in the direction which tends to relieve
that stress."

The degree of ionization of an electrolyte is suppressed by the addition of a strong
electrolyte containing common ion. This effect is known common ion effect.
The phenomenon of lowering the degree of ionization of a weak electrolyte by
adding a solution of a strong electrolyte having a common ion is called common
ion effect.
APPLICATION OF COMMON ION EFFECT
Knowledge of common ion effect is very useful in analytical chemistry. It is frequently
applied in qualitative analysis.
An electrolyte is precipitated only when the concentration of its ions
exceeds the solubility product (KSP)..
PRECIPITATION OF THE CATIONS OF GROUP II
Sulphides of basic radicals of groups II are precipitated by passing H2S gas through
the acidified solution by HCl.
Ionization of H2S:
H2S ⇋ 2H+ + S-2
Here HCl provides common ion H+ which shifts the above equilibrium to the left as
given by Le-Chatelier's principle.
HCl ⇋ H+ + ClAddition of HCl suppresses the ionization of H2S and lowers the concentration of S-2
ions, just enough to exceeds the KSP of II group sulphides. In this way only cations
of gr. II are precipitated as CuS, PbS, CdS etc. but precipitation of the sulphides of gr.
IV is prevented because they have high KSP values as compared to the sulphides of
gr. II.

PRECIPITATION OF THE CATIONS OF GROUP IV (Co,Ni, Zn)
Cations of groups IV are precipitated as sulphides by passing H2S gas
through the solution in the presence of NH4OH.
Ionization of NH4OH:

NH4OH ⇋

NH4+ + OH-

In this analysis NH4OH provides OH- ions which combines with H+ ions of H2S
to form H2O.

H2S ⇋

2H+ + S-2

:

H+ + OH-

⇋ H2 O

Removal of H+ ions from product side shifts the equilibrium to right and the
concentration of S-2 increases which is enough to exceed the KSP of the
sulphides of group IV. In this way CoS, NiS or ZnS can easily be precipitated.
PRECIPITATION OF THE CATIONS OF GROUP III

(Al,Fe,Cr)

Cations of groups III are precipitated as hydroxides by passing NH4OH in the
presence of NH4Cl. Here, NH4Cl provides common ion NH4+ which suppresses the
ionization of NH4OH.

NH4OH ⇋

NH4+ + OH-

:

NH4Cl ⇋

NH4+ + Cl-

Common ion NH4+ shifts the equilibrium to left side and the concentration of OH- ions
decreases. Under these circumstances, the KIP of the hydroxides of Al, Fe and Cr
is only exceeded and they are precipitated as Al (OH)3, Fe (OH)3 and Cr (OH)3 but the

COMMON ION EFFECT
Consider a simple sparingly soluble precipitate like (BA) e.g. AgCl
(1:1 electrolyte)
B+ or A- is a common ion with ppt BA.
In a soln. Containing ion as B+ as its soluble salt in C moles/L
BA ⇋
B+ + As = solubility
S = [B+] [A- ] = s x s = s2
In terms of common ion B+ , the solubilty of BA changes to and
becomes s′
∴ [B+] = s′ + C
[A- ] = s′
∴ S = (s′ + C )(s′) = s′ 2 + s′C
∴ s′ < s
Hence, solubilty of a ppt or a sparingly soluble electrolyte decreases due to
common ion effect. This is why during pptn from a soln the reagent is added in a
little excess of the theoretical value

The common ion effect “Le Chatelier”
Why is AgCl less soluble in sea water than in fresh water?
AgCl(s) ⇔ Ag+ + Cl–

Seawater contains
NaCl

60

Problem: The solubility of AgCl in pure water is 1.3 x 10–5 M.
What is its solubility in seawater where the [Cl–] = 0.55 M?
(Ksp of AgCl = 1.8 x 10–10)
I.
C.
E.

AgCl(s) ⇔ Ag+ + Cl–
N/A
0
0.55
N/A
+x
+x
N/A
+x
0.55 + x

Ksp= [x][0.55 + x]

Ksp= [Ag+][Cl–]

try dropping this x

Ksp = 0.55x
1.8 x 10–10 = 0.55x
x = 3.3 x 10–10 = [Ag+]=[AgCl]
“AgCl is much less soluble in seawater”
61

more Common ion effect:
a. What is the solubility of CaF2 in 0.010 M Ca(NO3)2?
Ksp(CaF2) = 3.9 x 10–11
CaF2(s) ⇔ Ca2+ + 2F–
[Ca2+] [F–]
I. 0.010
0
C.
+x
+2x
2x
E. 0.010 + x
Ksp= [0.010 + x][2x]2

Ksp=[Ca2+][F-]2

≅ [0.010][2x]2 = 0.010(4x2)

3.9 x 10–11 = 0.010(4x2)
x = 3.1 x 10–5 M Ca2+ from CaF2 so = M of CaF2
Now YOU determine the solubility of CaF2 in 0.010 M NaF.
62

Answer:

3.9 x 10–7 M Ca2+

CaF2(s) ⇔ Ca2+ + 2F–
0
0.010
+x
2x
x
0.010 + 2x
Ksp = [x][0.010 + 2x]2 ≅ x(0.010)2
3.9 x 10-11 =x(0.010)2
x = 3.9 x 10-7
What does x tell us
63

Effect of Temperature on solubility
Raising the temp of the soln. usually increases the solubility of a
ppt. or of the sparingly soluble substance. Thus a ppt is usually
more soluble in hot soln than in the cold. But difft. ppts show
different trend in respect of increase in solubility with temp.
ppt

10ºC

50ºC

100ºC

AgCl

0.89

5.23

21.1

BaSO4

2.2

3.36

3.9

Presence of a little excess of the precipitant (peptising reagent),
however, diminishes the solubility of the ppt to such an extent
that the temp. effect on solubility often becomes negligible

Factors affecting solubility
product

ACSS-501

Class –IV

Heike Kamerlingh Onnes , NL

Effect of solvent:
As a rule solubility of inorganic precipitates in water is decreased by
addition of organic solvents like methanol, ethanol, acetones etc.
CaSO4 ---- a so much insoluble in 50% ethanol that quantitative
separation of Ca or ppt is possible in their way.
comps like LiCl----- is soluble in alcohols whereas NaCl or KCl is
insoluble in alcohol. LiCl---- Li low ionic size and covalent character
Thus, by using Amyl alcohol(C5-alcohol), Li can be separated from
Na+ or K+ in the form of their chlorides. In the organic solvents LiCl is
completely soluble.
Solubility (at 25ºC , g/100g)
Water
KClO4
NaClO4

Methanol

Ethanol

N-Butanol

Ethyl acetate

2.02

0.105

0.012

0.0045

0.0015

Soluble in all the solvents

Separation of KClO4 from NaClO4 is possible by using ethyl
acetate or n-butanol

Effect of Electrolytes:
Presence of other ions added in the form of electrolytes or already
present in the soln. as foreign ions increase the ionic strength of the
medium.
Ionic strength, μ = ½ ∑ Ci Zi2

Ci = concn of ion, Zi= charge or valence of ion

As ionic strength increases activity coeffcicent of an ion including that of the
ion derived from ppt itself decreases

μ = -A logfi

A = constant , fi = activity coefficient

ai = ci.fi

a= activity of any ion, ci = concn of ion,

From thermodynamics it is found that the solubility pdt as defined in terms of
molar concn is not a trur constant. Instead the activity product is true constant

S = aB+ . aA
a

-

= (cB+ x cA-) (fB+ x fA-)

= Sc.(f±)2

Thus, in-order to keep the activity pdt const at a
particular temp. as the activity coeffcient value
decreases, the molar concn must increase i.e., the
substance(ppt) becomes more soluble. Hence,
solubility of a precipitae increases by adding
large amount of electrolytes in the solution.

Sa = activity pdt., Sc = solubilty pdt in
terms of molar concn , f± = mean activity coeffcient

Influence of complex ion formation
Many ppts show a tendency to form a complex ion with an excess of their
own ion resulting in the increase in solublty of the precipitate

Ag+ KCN AgCN

KCN

[Ag(CN)2]Complex ion formation

AgCl HCl

H[AgCl2]

S AgCN = 4 x 10-12 , when CN- is added in equivalent amount to Ag+ in soln
[CN- ] = √4 x 10-12 = 2 x10-6 g-ion/ L
Complex formation rreaction.
AgCN + CN- ⇋ [Ag(CN)2]-

K⇋

[AgCN] + [CN-]
[Ag(CN) ]2
long as. AgCN

As
∴ [CN-]

[Ag(CN) ]2

=

K= dissociation const /instability constant

remains in solid phase, [AgCN]= const

K′ ...........10-10 or, [CN-] = 10-10 [Ag(CN)2]-

eq.....1
Now, suppose

excess of [CN-] is added corresponding to 0.01 molar soln (10-2)
[CN-] + [Ag(CN)2]- = 10-2
eq........2
free
complexed ion

[Ag(CN)2]- {10-10 +1 } = 10-2
or,
[Ag(CN)2]- ≈ 10-2
Hence, practically all the [CN-] added is transformed into the complex ion and it
can be said that KCN dissolves an equiv amount of AgCN .
Complex formation as above is quite common during pptn analysis
particularly when the resulting complex ion has a very high stability.

Hg+2 + I- → HgI2
HgI2 + 2I- → [HgI4]-2

( insoluble)
(highly stable)

As a rule, solubility for many precipitates at first decreases due to common ion
effect and a min. solubilty may be found at a certain excess of the common
ion and the solubility then increases with further addition of the reagent
providing the common ion but forming the complex ion at same time
The complex ions may also be formed with constituents other than the common –
ion present in the soln. Solubilty of AgOH or Cu(OH) 2 in ammonia solution.

AgOH + 2NH3 ⇋ [Ag(NH3)2]+ + OHCu(OH)2 + 4NH3 ⇋ [Cu(NH3)4]+2 + 2OH-

Influence of [H+] or pH
If the ppt. is salt of a weak acid, strong acids will have a solvent effect on it
CaF2 + 2HCl → Ca+2 + 2Cl- + 2HF
Salt

strong acid

weak acid

Thus all carbonates dissolved in mineral acid.
Phosphates, Aresnates, Sulphides and Cyanide except AgCN which is in effect the
salt of relatively strong acid

H[Ag(CN)2]

Ag+

Ag[Ag(CN)2] ≡ 2AgCN HNO3 Insoluble

Argento cyanic acid (strong acid)

The solvent effect of acid is of importance even incase of slightly soluble sulfate.
Among the hydroxides and hydrous oxides precipitates pH ha s agreat role on their
formation. Fractional precipitation of hydrous oxide can be carried out by
controlling the pH of the medium.
pH

Metal ions
Pptd as
Hydrous
oxides

3

4

5

6

7

8

9

11

Fe+3

Th+4

Al+3

Be+2

Fe+2

Co+2

Ag+

Mg+2

U+4

Cu+2

Ni+2

Mn+2

Zn+2
Cr+3

U+6
Cd+2

Hg+2

Zr+4
Sn+2

H+R- + M+n → MRn + H+ pH has effect in all cases of pptn
pptant

ppt

where reagent is an weak acid

Co-precipitation and Post precipitation
Co-precipitation means incorporation of foreign ions into a ppts having a
common ion or an ion of identical size or charge with any ion of the ppt.
Co-pptn may take place due to mainly 3 reasons
(i)Due to adsorption at the surface of the particles exposed to the soln. having
ionic lattice. Colloidal particles like Silver halide of hydrated oxides are
highly contaminated in this way.
(ii)Due to adsorption at the interior or occlusion of foreign ions during growth
of primary particles. In this case ions may or may not fit into the lattice and even
solvent molecules are often held in the ppt in this way.
(iii)Due to formation of chemical comp. with the unsatisfied valency of any
ions of the lattice. e.g.

Ba Cl- When SO4-2 is pptd with BaCl2 soln

Ba

+

SO4

SO4
Ba+ Cl-

Ba

+

Ba+ K+ When Ba+2 is pptd by K2SO4 soln
SO4
Ba+ K+
lattice

The degree of contamination of a ppt. by Co-pptn depends
on specific adsorption, specific surface, decreasing
solubility of the contaminant, increasing concn of the
contaminant, higher valence, and smaller size of the
foreign ion, lower temp. and greater no. of particles/
unit mass of the ppts

Co-pptn can be minimised by the following techniques
i)

By pptn from hot soln so that initially a few particles are formed and relative
supersaturation is increased
ii) For ppts having appreciable solubility, however the soln must be cooled and
the relative super saturation is increased by the control of pH or any other
device
iii) By pptn from dilute soln to decrease the concn. of foreign ions
iv) By adding reagent as a solution , slowly with stirring in order to reduce the
local concn. (Q)
v) By prior removal of the foreign ion as ppt by rn with a difft. Reagent or
by conversion of the foreign ion to a difft. valency state or sometimes to a
complex spp.
vi) E.g.
Fe+2 during BaSO4 pptn
N

N

[Fe(O-pn)3]
Fe(OH)3. xH2O
Phenanthroline
i) By allowing the ppt. to satnd (digestion and aging) for perfect growth of
crystals
ii) By thorough and proper washing of the ppt.
iii) By double pptn.
+2

Ist ppt.

Fe+3, Ca+2

Fe(OH)3.xH2O (Ca+2)
HCl

+ Ca+2 in soln

Included as co-ppt.

Fe+3, Ca+2 (less)
OHFe(OH)3.xH2O + Soln (Ca+2)
2nd ppt.

Three examples of impurities that may form during precipitation. The
cubic frame represents the precipitate and the blue marks are
impurities: (a) inclusions, (b) occlusions, and (c) surface adsorbates.
Inclusions are randomly distributed throughout the precipitate.
Occlusions are localized within the interior of the precipitate and
surface adsorbates are localized on the precipitate’s exterior. For ease
of viewing, in (c) adsorption is shown on only one surface.

Post precipitation
Post –pptn involves incorporation of impurities when the supernatant
liquid is super saturated w.r.t some phase which crystalline slowly.
Eg. thus in case of metal sulfide pptn from 0.1N acid medium
(HCl), ZnS is not normally precipitated
A soln saturated with H2S, however, remains supersaturated with
ZnS, which slowly separates as a ppt which is accelerted in
presence of the fine particles of HgS or CuS.
This is known as post precipitation of ZnS
 Similar situation arises when Mg remains in a solution containing
large excess of oxalate after Ca-Oxalate pptn. Mg2C2O4
precipitate slowly from the supernatant liquid by post-pptn

Post precipitation differs from Co-pptn in the following respects

Contamination increases with aging or digestion (in case
of post pptn), but the reverse is true in case of Co-pptn
Contamination increases with faster agitation or
stirring of the supernatant liquid in case of post pptn.,
but the reverse is true for co-pptn.
Degree of contamination or its magnitude is usually
much greater due to post pptn in comparison to Co-pptn.
To avoid post –pptn, a shorter period of digestion
particularly under warm condition is to be carried out and
the ppt. should be filtered as quickly as possible so that
the ppt does not remain in contact with a mother–liquor
for a long time

Washing of the precipitate
The precipitate after filtration must be washed thoroughly
with a proper liquid.

The washed liquid should dissolve out the foreign
substances only and it should have no peptising or disperse
action on the ppt.

It should not form any volatile or insoluble pdt. with
the ppt and should be itself easily removed during drying.

It should contain no substance which interferes for
the determination of any constituents in the filtrate.

The washed liquid should not interact chemically
with the ppt during drying or ignition.

Nature of the wash liquid
To prevent peptisation the wash liquid must contain an electrolyte.
Usually dilute ammonia, dil. Acids and some salts particularly
ammonium salts are used as electrolytes. The wash liquid often
contains a common ion with the ppt to decrease the solubility during
washing. A few drops of reagent solution is usually added to
washed liquid for this purpose
During AgCl pptn wahing with dil HNO3
For Hydrous oxide pptn…. dil NH3 containing small amount of
NH4NO3 is used
For washing of PbSO4 ppt…… a saturated soln of PbSO4 is used
as first wash liquid, the ppt was then washed with 40% alcohol
The washing must be carried out by using a very small amount of
wash liquid at a time and the number of washing must be as large
as possible.

The common ion effect is used to decrease the solubility.
The common ion effect is used to decrease the solubility.
Sulfate concentration is the amount in equilibrium and is equal to the
Sulfate concentration is the amount in equilibrium and is equal to the
BaSO4 solubility.
BaSO solubility.
4

In absence of excess barium ion, solubility is 10-5 M.
In absence of excess barium ion, solubility is 10-5 M.

©Gary Christian,
6th Ed. (Wiley)

Fig. 10.3. Predicted effect of excess barium ion on solubility of BaSO 4.

Ksp = Ksp0/fAg+fSO42Ksp = Ksp0/fAg+fSO42Solubility increases with increasing ionic strength as activity coefficients decrease.
Solubility increases with increasing ionic strength as activity coefficients decrease.

©Gary Christian,
6th Ed. (Wiley)

Predicted effect of increased ionic strength on solubility of
BaSO4. Solubility at zero ionic strength is 1.0 x 10-5 M.

Gravimetric Calculation

ACSS-501

Class –V

Heike Kamerlingh Onnes

Find moles of precipitate.

Find moles of sought substance(Analyte)

Find weight of sought substance (Analyte).

mol ratio=mol analyte/ mol ppt

Calculate the percentage of the sample that is the sought substance

Gravimetric Calculations

The point here is to find the weight of analyte from the weight of
precipitate. We can use the concepts discussed previously in stoichiometric
calculations but let us learn something else. Assume Cl 2 is to be precipitated as
AgCl, then we can write a stoichiometric factor reading as follows: one mole of Cl 2
gives 2 moles of AgCl. This is in fact what is called the gravimetric factor (GF)
where we can substitute for the number of moles by grams to get:

GF for Cl2 = 1 mol Cl2/2 mol AgCl = FW Cl2/2 FW AgCl
= x g analyte/y g precipitate
Weight of substance sought = weight of precipitate x GF
One can also consider the problem by looking at the number of mmoles of analyte in
terms of the mmoles of the precipitate where for the precipitation of Cl 2 as AgCl, we
can write
(analyte) Cl2 = 2 AgCl (ppt)
mmol Cl2 = 1/2 mmol AgCl
(mg Cl2/FW Cl2) = 1/2 (mg AgCl/FW AgCl)

Q1. Calculate the grams analyte to mg precipitate for the following:
P (at wt =30.97) in Ag3PO4 (FW = 711.22), Bi2S3 (FW 514.15) in BaSO4
(FW = 233.40)

Solution: 1

P = Ag3PO4
mmol p = mmol Ag3PO4
Mg P/30.97 = mg Ag3PO4/711.22
Mg P/mg Ag3PO4 = 30.97/711.22 = 0.04354 (GF)

Bi2S3 = 3 BaSO4
mmol Bi2S3 = 1/3 mmol BaSO4
mg Bi2S3/FW Bi2S3 = 1/3 mg BaSO4/FW BaSO4
mg Bi2S3/514.15 = 1/3 mg BaSO4/233.40
mg Bi2S3/ BaSO4 = 1/3 (514.15/ 233.40) = 0.73429 (GF)

(NH4)2PO4.12 MoO3 (FW = 1876.5). Find percentage P (at wt = 30.97) and
percentage P2O5 (FW = 141.95) in the sample.

Solution: 2.
First we set the mol relationship between analyte and precipitate
P = (NH4)2PO4.12 MoO3
mmol P = mmol (NH4)2PO4.12 MoO3

mg P/at wt P = mg (NH4)2PO4.12 MoO3/ FW (NH4)2PO4.12 MoO3
mg P = at wt P x (mg (NH4)2PO4.12 MoO3/ FW (NH4)2PO4.12 MoO3)

mg P = 30.97 (1.1682x103 / 1876.5) = 19.280 mg
% P = (19.280/271.1) x 100 = 7.111%
The same procedure is applied for finding the percentage of P 2O5
P2O5 = 2 (NH4)2PO4.12 MoO3
mmol P2O5 = 1/2 (NH4)2PO4.12 MoO3
mg P2O5/FW P2O5 = 1/2 (mg (NH4)2PO4.12 MoO3/ FW (NH4)2PO4.12 MoO3)
mg P2O5 = 1/2 x FW P2O5 x (mg (NH4)2PO4.12 MoO3/ FW (NH4)2PO4.12 MoO3)
mg P2O5 = 1/2 x 141.95 (1.1682x103/1876.5) = 44.185 mg

Q.3. Manganese in a 1.52 g sample was precipitated as Mn3O4 (FW = 228.8)
weighing 0.126 g. Find percentage Mn2O3 (FW = 157.9) and Mn (at wt = 54.94) in
the sample.

Solution. 3
(analyte) 3 Mn2O3 = 2 Mn3O4 (ppt)
mmol Mn2O3 = 3/2 mmol Mn3O4
mg Mn2O3 / FW Mn2O3 = 3/2 (mg Mn3O4/FW Mn3O4)
mg Mn2O3 = 3/2 FW Mn2O3 (mg Mn3O4/FW Mn3O4)
mg Mn2O3 = 3/2 x 157.9 ( 126/228.8) = 130 mg
% Mn2O3 = (130/1520) x 100 = 8.58%

The same idea is applied for the determination of Mn in the sample

3 Mn = Mn3O4
mmol Mn = 3 mmol Mn3O4
mg Mn / at wt Mn = 3 (mg Mn3O4/FW Mn3O4)
mg Mn = 3 at wt Mn (mg Mn3O4/FW Mn3O4)
mg Mn2O3 = 3 x 54.94 ( 126/228.8) = 90.8 mg

% Mn = (90.8/1520) x 100 = 5.97%

Q.4. A mixture containing only FeCl3 (FW = 162.2) and AlCl3 (FW = 133.34)
weighs 5.95 g. The chlorides are converted to hydroxides and ignited to Fe2O3 (FW
= 159.7) and Al2O3 (FW = 101.96). The oxide mixture weighs 2.26 g. Calculate the
percentage Fe (at wt = 55.85) and Al (at wt = 26.98) in the sample.

Solution 4.
Fe = FeCl3
1 mol Fe = 1 mol FeCl3
g Fe/at wt Fe = g FeCl3/ FW FeCl3
Rearrangement gives
g FeCl3 = g Fe (FW FeCl3/at wt Fe)
In the same manner
g AlCl3 = g Al ( FW AlCl3/at wt Al)
g FeCl3 + g AlCl3 = 5.95
g Fe (FW FeCl3/at wt Fe) + g Al ( FW AlCl3/at wt Al) = 5.95
assume g Fe = x,
g Al = y
then:
x (FW FeCl3/at wt Fe) + y ( FW AlCl3/at wt Al) = 5.95
x (162.2/55.85) + y (133.34/26.98) = 5.95

2.90 x + 4.94 y = 5.95

................ (1)

Soln 4:
The same treatment with the oxides gives
2 Fe =Fe2O3
mol Fe = 2 mol Fe2O3
g Fe/at wt Fe = 2 (g Fe2O3/FW Fe2O3)
g Fe2O3 = 1/2 g Fe (FW Fe2O3/at wt Fe)
In the same manner
g Al2O3 = 1/2 g Al (FW Al2O3/at wt Al)
g Fe2O3 + g Al2O3 = 2.26
1/2g Fe(FW Fe2O3/at wt Fe) + 1/2 g Al (FW Al2O3/at wt Al) = 2.26
1/2 x (159.7/55.85) + 1/2 y (101.96/26.98) = 2.26
1.43 x + 1.89 y = 2.26
...............2)
from (1) and (2) we get
x = 1.07
y = 0.58
% Fe = (1.07/5.95) x 100 = 18.0%
% Al = (0.58/5.95) x 100 = 9.8%

Q : A 0.4960 g sample of a CaCO3 is dissolved in an acidic solution.
The calcium is precipitated as CaC2O4.H2O and the dry precipitate is
found to weigh 0.6186 g. What is the percentage of CaO in the sample?
Unknown: % CaO in the 0.4960 g sample of CaCO3
Knowns: wt of CaCO3 sample = 0.4960 g, wt of CaC2O4.H2O = 0.6186 g
Concepts: gravimetric factor, moles, mole ratio, FW

wt of CaO = 0.2374 g

x= 1.07g, y=0.58 g
%Fe = (1.07/5.95) x100

=18.0%
%Al = (0.58/5.95) x100
=9.8.0%

Solubility in Presence of Diverse Ions
Find the solubility of AgCl (ksp = 1.0 x 10-10) in 0.1 M NaNO3. The activity
coefficients for silver and chloride are 0.75 and 0.76, respectively.

Solution
AgCl(s) = Ag+ + ClWe can no longer use the thermodynamic equilibrium constant (i.e in absence of diverse
ions ) and we have to consider the concentration equilibrium constant or use activities instead
of concentration :

Ksp = aAg+ aClKsp = [Ag+] fAg+ [Cl-] fCl1.0x10-10 = s x 0.75 x s x 0.76
s = 1.3x10-5 M
We have calculated the solubility of AgCl in pure water to be 1.0x10-5 M, if we
compare the this value to that obtained in presence of diverse ions we see
% increase in solubility = {(1.3x10-5 – 1.0x10-5)/1.0x10-5}x 100 = 30%
Therefore, once again we have an evidence for an increase in dissociation or a shift
of equilibrium to right in presence of diverse ions .

Q. A certain barium halide exists as the hydrated salt BaX2.2H2O,
where X is the halogen. The barium content of the salt can be
determined by gravimetric methods. A sample of the halide
(0.2650 g) was dissolved in water (200 cm3) and excess sulfuric acid
added. The mixture was then heated and held at boiling for 45
minutes. The precipitate (barium sulfate) was filtered off, washed
and dried. Mass of precipitate obtained = 0.2533 g. Determine the
identity of X.

Answer:
The precipitate is barium sulfate. The first stage is to determine the number of moles of
barium sulfate produced, this will, in turn give us the number of moles of barium in the
original sample.
Relative Molecular Mass(RMM) of barium sulfate
= 137.34 (Ba) + 32.06 (S) + (4 x 16.00) (4 x O)
= 233.40
Number of moles
= mass / RMM
= 0.2533 / 233.40
= 1.09 x 10 -3
This is the number of moles of barium present in the precipitate and, therefore, the number
of moles of barium in the original sample. Given the formula of the halide, (i.e. it contains
one barium per formula unit), this must also be the number of moles of the halide. From this
information we can deduce the relative molecular mass of the original halide salt:
RMM = mass / number of moles
= 0.2650 / 1.09 x 10-3
= 244.18
The relative atomic mass (RAM) of 2 X will be given by the RMM of the whole salt - that of the
remaining components; So RAM of 2 X = 244.18 - 173.37 = 70.81
2 X = 70.81, so X = 35.41.
The RAM of chlorine is 35.45 which is in good agreement with the result obtained and hence
the halide salt is hydrated barium chloride and X = Chlorine

Other Gravimetric Techniques
Gravimetric Combustion Analysis
Determines the carbon and hydrogen content of organic
compounds burned in excess O2.

CO2 trap

Guard tube prevents
H2O and CO2 entering
from the air

H2O trap
103

Combustion Analysis, or Elemental Analysis
C, H, N, S  → CO (g) + H O(g) + N (g) + SO (g) + SO (g)

1050  C

2

2

2

2

3

Determines the C, H, N, and S content in a single operation by using GC
and thermal conductivity measurements

104

Measuring Principle
The sample under test is weighed in using a tin capsule. The required amount is 2 to
3 mg of organic material and can hardly exceed 10 mg, if inorganic matter with little
carbon content is investigated. After folding the capsule (looking rather like wrapped
tin foil) the sample is placed in the autosampler.
The tin capsule enclosing the sample falls into the reactor chamber where
excess oxygen is introduced before. At about 990 °C the material is "mineralized".
Formation of CO is probable at this temperature even under these conditions of excess
O2 .
The complete oxidation is reached at a tungsten trioxide (WO3) catalyst
which is passed by the gaseous reaction products. The resulting mixture should thus
consist of CO2, H2O und NOx. But also some excess O2 passes the catalyst.
The product gas mixture flows through a silica tube packed with copper
granules. In this zone held at about 500 °C remaining oxygen is bound and
nitric/nitrous oxides are reduced. The leaving gas stream includes the analytically
important species CO2, H2O und N2. Eventually included SO2 or hydrohalogenides
are absorbed at appropiate traps.
High purity helium (Quality 5.0) is used as carrier gas. Finally the gas
mixture is brought to a defined pressure/volume state and is passed to a gas
chromatographic system. Separation of the species is done by so called zone
chromatography. In this technique a staircase type signal is registrated. Step height

Working Range

Mineralization and detection covers every species of the analyte elements
in the sample. Beside purely organic samples various metal organic
compunds and even inorganic samples as carbides and nitrides have
successfully been characterized.
The detection limit for carbon and nitrogen at sample amounts of 2 to 3 mg
was found to be at about 0,05 w-% (500 ppm) in what case the uncertainty
stays at about 0,02 w-%.
Problems and Interferences
The weighing of oily or fluid substances is impossible using the thin walled
tin capsules. For this purpose alumina pans with a lid are available. These
pans are tightly closed by cold welding to prevent loss of sample by
spillage and evaporation. As the blank value for nitrogen is dramatically
increased be the enclosed volume of air the determination limit for N in
liquid samples is increased to 0,1 to 0,2 w-%.
With highly viscous or even glassy materials elemental analysis is even
impossible with the above method.

It has long been known that phosphorus can interfere in the mineralization
of organic material. In literature the formation of glassy P2O5 xH2O yC has
been described. This effect can be controlled by the addition of vanadium
pentoxide (V2O5).
Fluorine is mineralized to form HF which reacts at the wall of the silica tubes
which form the main part of the reaction zone. The gaseous products, such as SiF4
and relatives, can cause systematic errors which rarely become significant with
respect to the 0,3 w-% tolerance.
The mineralization of metal containing samples can also be affected by
interferences. By modification of the method most of these can be compensated
for.
Percentage Composition
CxHyOz (9.83 mg) + excess O2 x CO2 (23.26 mg) + y/2 H2O (9.52 mg)
millimoles of CO2 = 23.26 mg/ 44.01mg/mmol = 0.5285 mmoles of CO2
mmoles of CO2 = mmoles of C in original sample
(0.5285 mmoles of C)(12.01mg/mmol C) = 6.35 mg of C in original sample
mmoles of H2O = 9.52 mg/ 18.02 mg/mmol = 0.528 mmoles of H2O
mmoles of H2O = 1/2 mmoles of H in original sample
(0.528 mmoles of H)(2)(1.008mg/mmol H) = 1.06 mg of H in original sample

Weight Percentage Composition
%C = 6.35 mg/9.83 mg x 100 = 64.6%
%H = 1.06 mg/9.83 mg x 100 = 10.8%
%O = 100-(64.6 + 10.8) = 24.6%
Calculation of Empirical Formula
assume for example a 100g sample
64.6% of C: 64.6 g/12.01 g/mol = 5.38 moles of C
10.8% of H: 10.8 g/1.008 g/mol = 10.7 moles of H
24.6% of O: 24.6 g/16.0 g/mol = 1.54 moles of O
Thus: C5.38 H10.7 O1.54
converting to simplest ratio:
C5.38/1.54; H10.7/1.54; O1.54/1.54
C3.50 H7.00 O1.00 = C7H14O2
What are possible structures?

Thermal Gravimetric Analysis (TGA)
• Precisely monitoring weight loss of a
sample in a given atmosphere as a
function of temperature and/or time
• Atmospheres: N2, O2, air, or He

• Temperature: ambient to 1000 °C
• Records the first derivative of the mass loss
• Evaluate the thermal decomposition and
stability of materials
– Polymers, resins, rubbers, explosives
• Information on bulk composition of
compounds
– Thermal oxidation, heat resistance
– Residual water or solvents
– Compositional analysis
– Ash content in a sample
– Quantity of inorganic filler in a polymer

109

TGA cont.
• The percent weight loss of a test sample is recorded while the
sample is being heated at a uniform rate in an appropriate
environment.
• The loss in weight over specific temperature ranges provides
an indication of the composition of the sample, including
volatiles and inert filler, as well as indications of thermal
stability.
• The gas environment is pre-selected for either a thermal
decomposition (inert – He or N2 gas), an oxidative
decomposition (air or O2), or a combination therein.

Sharma

110

Thermogravimetry
One method for determining the products of a thermal decomposition is to
monitor the sample’s mass as a function of temperature, a process called
thermogravimetry.

From 100–250 oC the sample loses 17.61 mg – 15.44 mg, or 2.17 mg,
which is
of the sample’s original mass. In terms of CaC2O4•H2O, this corresponds
to a loss of
The product’s molar mass and the temperature range for the
decomposition, suggest that this is a loss of H2O(g), leaving a residue
of CaC2O4. The loss of 3.38 mg from 350–550 oC is a 19.2% decrease
in the sample’s original mass, or a loss of
which is consistent with the loss of CO(g) and a residue of CaCO3.
Finally, the loss of 5.30 mg from 600-800 oC is a 30.1% decrease in
the sample’s original mass, or a loss of
This loss in molar mass is consistent with the release of CO2(g),
leaving a final residue of CaO.

TGA cont.
TGA tests may be run in a
heating mode at some
controlled heating rate, or
isothermally. Typical weight
loss profiles are analyzed
for the amount or percent
of weight loss at any given
temperature, the amount or
percent of noncombusted
residue at some final
temperature, and the
temperatures of various
sample degradation
processes.

TGA cont.

H2O, C7H3O6, C6H4, CO2
are consecutively lost.

Sharma

114

Gravimetric analysis, P K MANI

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