Transcript Chapter8

Chapter12
Gravimetric Methods of Analysis
Gravimetric methods of analysis
12A Precipitation gravimetry
12A-1 Properties of precipitates and precipitating reagents
12A-2 Particle Size and Filterability of Precipitates
12A-3 Colloidal Precipitates
12A-4 Crystalline Precipitates
12A-5 Coprecipitateion
12A-6 Precipitation from Homogeneous Solution
12A-7 Drying and Ignition of Precipitates
12B Calculation of results from Gravimetric data
Example12-1,12-2,12-3
12C Applications of Gravimetric methods
12c-1 inorganic Precipitating Agents
12c-2 Reducing Agents
12c-3 organic precipitating Agents
12c-4 Organic functional group analysis
12c-5 Volatilization gravimetry
There are Several analytical method
on mass measurements
1.Precipitation Gravimetry methods
The analyte is separated from a solution of the sample as a
precipitate and is converted to a compound of known
composition that can be weight.
2.Volatilization Gravimetry methods
The analyte is separated from other constituents of a sample
by conversion or a gas of known chemical composition.
3.Electrogravimetry
4.Gravimetry titrimetry
5.Atomic mass spectrometry
12A PERCIPITAION GRAVIMETRY
•
•
•
•
2HN3+H2C2O4→2NH4++C2O42Ca2+(aq)+C2O42-(aq)→CaC2O4(s)
CaC2O4(s)→CaO(s)+CO (g)+CO2(g)
Example12-1
Ex12-1：The calcium in a 200.0ML sample of a natural water was determined
by precipitating the cation as CaC2O4.The precipitate was filtered, washed,
and ignited in a crucible with an empty mass of 26.6002g.the mass of the
crucible plus Cao（56.077g/mol）was 26.7134 g. Calcualte the
concentration of Ca（40.078g/mol）in water in units of grams per 100mL of
the water.
Ans：
The mass of CaO is 26.7134-26.6002＝0.1132g
The number of moles Ca in the sample is equal to the number of moles CaO
amount of Ca
1 molCaO
1 molCa
＝ 0.1132gCaO × 56 .077 gCaO
× molCaO
＝2.0186×10-3
Conc.Ca＝
2 . 0186  10
3
molCa  40 . 07 gCa / molCa
200 MLSample
＝0.04045g/100ML
 100 ml
12A-1 Properties of precipitates
and precipitating reagents
• Specific reagents, which are rare, react
only with a single chemical species.
• Selective reagents, which are more
common, react with only a limited number
of species. P. 315 note
The ideal precipitating reagent
1. Readily filtered and washed free of
contaminants
2. Of sufficiently low solubility so that no
significant loss of the solid occurs during
filtration and washing
3. Unreactive with constituents of the
atmosphere
4. Of known composition after it is dried or,
if necessary, ignited
12A-2 Particle size and Filterability of
Precipitates
• Precipitates consisting of large particles because these particles are
easy to filter and wash free of impurities
1. Factors that determine the particle size of precipitates
(1) Colloidal suspensions (10-7 to 10-4 cm in diameter)
show no tendency to settle from solution and are not easily filtered
(2) Crystalline suspension
Tend to settle spontaneously and are easily filtered.
• Particle size of a precipitate is influenced by Precipitate
solubility, temp., reactants conc., the rate at which the
reactants are mixed
relateve
sup ersaturati on 
Q S
(12  1)
S
Q is the concentration of the solute at any instant
and S is its equilibrium solubility
p. 316 note: supersaturated
(Q-S)/S large: colloid;
(Q-S)/S small: crystalline
2.Mechanism of Precipitates formation
(1) Nucleation
is a process in which a minimum number of atoms, ions, or
molecules join together to produce a stable solid.
(2) Particle growth
•
If nucleation predominates, a precipitate containing a
large number of small particles results；if growth
predominates, a smaller number of large particles is
produced.
3.Controlling particle size
(1)Elevated temperatures to increase the solubility of the
precipitate(2) dilute solution (3) slow addition of the
precipitating agent with good stirring.
12A-3 Colloidal Precipitates
1.Coagulation of colloids:
by heating, stirring, adding electrolyte
(1) Why colloidal suspensions are stable and do not
coagulate spontaneously.
(a) Charge (cations or anions that are bound to the
surface of the particles .Check)
(b) Adsorption is a process in which a substance (gas,
liquid, or solid) is held on the surface of a solid..
(c) Absorption involves retention of a substance within
the pores of a solid.
.
Fig 12-1 Colloidal silver chloride particle in a solution that
contains an excess of silver nitrate.
The primary adsorption layer
attach directly to the solid surface
Ex：consists mainly of adsorbed silver ions.
The counter–ion layer
surrounding the charged particle is a layer of
solution.
Ex：which contains sufficient excess of negative ion
（principally nitrate）to balance the charge on the
surface of the particle.
Electrical double layer
The primarily adsorbed and the counter-ion layer
constitute .This double later exerts an electrostatic
repulsive force that prevents particles from colliding
and adhering.
Figure 12-2 Effect of AgNO3 and electrolyte concentration on the
thickness of the double layer surrounding a colloidal AgCl particle in a
solution containing excess AgNO3
Upper portion
The effective charge on the
particles prevents them from
approaching one another more
closely than about 2d1
Lower part
In the dilute silver nitrate
solution .the two particles can
within 2d2of one another the
distance between particles
becomes small enough for the
forces of agglomeration to take
effect and a coagulated
precipitate to appear.
short period of heating
heating decreases the number
of adsorbed ions and the
thickness. The particles may
gain enough kinetic energy at
the higher temper.
Coagulation of a colloidal suspension can often
be brought
1.by short period of heating
heating decreases the number of adsorbed ions and the
thickness. The particles may gain enough kinetic energy at
the higher temper.
2.Increase the electrolyte concentration of the solution.
If we add ionic compound to a colloidal suspension. The
concentration of counter – ions increases in the vicinity of each to
balance the charge of the primary adsorption layer decreases.
The net effect of adding an electrolyte is thus a shrinkage of
the counter-ions layer
2.Peptization of colloid
Peptization:A coagulated colloid returns to its dispersed state
(1) When a coagulated colloid is washed some of the electrolyte
responsible for its coagulation is leached from the internal liquid
in contact with the solid particles.
(2)Washing is needed to minimize contamination on the other,
there is a risk of losses resulting from peptization if pure water
is used.
(3)Solved by washing the precipitate with a solution containing an
electrolyte that volatilizes when the precipitate is dried or ignited.
commonly
3.Practical treatment of Colloidal precipitates
Digestion：It is a process in which a precipitate is
heated for an hour or more in the solution from which
it was formed（the mother liquor）
12A-4 Crystalline Precipitates
1、Easily filtered and purified than are coagulated colloids.
2、Method of improving particle size and Filterability
QS
relateve
sup ersaturati on 
S
(1) minimization of Q
(using dilute solution and adding the precipitating reagent slowly and with
good mixing.)
(2) maximizing S
(precipitating from hot solution or adjusting the pH of the precipitation
medium.)
12A-5 Coprecipitation
Coprecipitation
It is a process in which normally soluble compounds
are carried out of solution by precipitate.
There are four types of coprecipitation:
1.surface adsorption,
2.mixed-crystal formation,
3.occlusion,
4. mechanical entrapment.
1.surface adsorption,
1.Adsorption is often the major source of
contamination in coagulated colloids but is of no
significance in crystalline precipitates.
The net effect of surface adsorption is therefore the carrying
down of an otherwise soluble compound as a surface
contaminant.
（1）Minimizing Adsorbed Impurities on Colloids
(A) coagulated colloids is improved by digestion
(B) Washing a coagulated colloid containing a volatile
electrollyte
（2）Reprecipitation：A drastic but effective way to
minimize the effects of adsorption is Preprecipitation.
2.Mixed-crystal formation
(1)Mixed-crystal formation ,one of the in the crystal
lattice of a solid is replaced by an ion of another
element. It is necessary that the two ions have
the same charge and that their sizes differ by no
more than about 5%. The same crtstal class.
Ex (Pb ion replace some of the barium ion)
(2)The interfering ion may have to be separated
before the final precipitation step.
Precipitation reagent.
3.Occlusion is a type of coprecipitation in
which a compound is trapped within a
pocket formed during rapid crystal growth.
4.Mechanical entrapment occurs when
crystals lie close together during growth.
Here, several crystals grow together and
in so doing trap a portion of the solution in
a tiny pocket.
Coprecipitation Errors
It impurities may cause either negative or
positive errors in an analysis .
12A-6 Precipitation from Homogeneous Solution
• Homogeneous precipitation is a process in
which a precipitate is formed by slow
generation of a precipitating reagent
homogeneously throughout a solution.
• (H2N)2CO+3H2O→CO2+2NH4++2OH-
Fig 12-5 shows hydrous oxide precipitates of aluminum formed by
direct addition of base and by homogeneous precipitates with urea.
12A-7 Drying and lgnition of Precipitates
Weighing form
Some precipitates are also ignited to decompose the solid
and form a compound of Known composition
Fig 12-6 Effect of temperature of precipitate mass.
12BCalculation of results from
gravimetric data
Ex12-1：The calcium in a 200.0ML sample of a natural water was determined by precipitating
the cation as CaC2O4.The precipitate was filtered, washed, and ignited in a crucible with an
empty mass of 26.6002g.the mass of the crucible plus Cao（56.077g/mol）was 26.7134 g.
Calcualte the concentration of Ca（40.078g/mol）in water in units of grams per 100mL of
the water.
Ans：
The mass of CaO is 26.7134-26.6002＝0.1132g
The number of moles Ca in the sample is equal to the number of moles CaO
＝ 0.1132gCaO × 1 molCaO
× 1 molCa ＝2.0186×10-3
56 . 077 gCaO
Conc.Ca＝
2 . 0186  10
3
molCaO
molCa  40 . 07 gCa / molCa
200 MLSample
＝0.04045g/100ML
 100 ml
amount of Ca
Example 12-2：Anirom ore was analyzed by dissolving a 1.1324g sample in
concentrated HCL .The resulting solution was diluted with water, and the iron
（Ⅲ）was precipitated as the hydrous oxide Fe2O3 .xH2O by the addition of
NH3. After filtration and washing, the residue was ignited at a high temperature
to give 0.5394 g of pure Fe2O3 (159.69g/mol.). Calculate (a) a the ％Fe
(55.847g/mol ) and (b) The ％ Fe3O4(231.54g/mol) in the sample.
Ans：
amount Fe2O3＝0.5394g Fe2O3×
1 molFe 2 O 3
159 . 69 gFe 2 O 3
 3 . 3778  10
3
molFe 2 O 3
（a）The number of moles of Fe is twice the number of moles of
Fe2O3,and Mass Fe
2 molFe
-3
＝3.3778×10 ×
％Fe ＝
molFe 2 O 3
0 . 37728 gFe
1 . 1324 gsample
 55 . 847
 100
(b) massFe3O4＝3.3778×10-3
％Fe3O4＝
0 . 5140 gFe 3 O 4
1 . 1324 gsample
gFe
＝0.37728Fe
％＝33.32％

molFe
2 molFe 3 O 4
3 molFe 2 O 3

231 . 54 gFe 3 O
molFe 3 O
 100 ％＝46.04％
 0 . 52140 gFe 3 O
Ex12-3 A 0.2356g sample containing only Nacl and BaCl2
Yielded 0.4637 g of dried Ag Cl. Calculate the percent of each
halogen compound in the sample.
Sol：X＋y ＝0.2356g sample ；X= NaCl y= BaCl2
Amount AgCl from NaCl
1 molAgCl
1 molNaCl

＝xgNaCl×
＝0.017111xmol Agcl
58 . 44 gNaCl
molNaCl
Mass AgCl from NaCl＝0.017111xmol AgCl×143.32 g AgCl/molAgCl
＝2.4524g
Proceeding in the same way, wi can write that the number of
moles of AgCl from the BaCl2 is given by
1 molBaCl
2 molAgCl
3
Amount AgCl from BaCl22 
 9 . 605  10 molAgCl
＝ygBaCl2×
208 . 23 gBaCl
2
molBaCl
2
Amount AgCl from BaCl2 ＝9.605×10-3×143.32＝1.3766yg AgCl
2.4524x＋1.3766y＝0.4637
Y＝0.2356-x；2.4524x＋1.3766(0.2356-x) ＝0.4637
1.0758x＝0.13942；x＝mass NaCl＝0.12960gNaCl
％NaCl＝
0 . 12956 NaCl
0 . 2356 gsample
 100 %  55 . 01 %
％BaCl2＝100.00%-55.01%＝44.99%
12C Applications of gravimetric
methods
12C-1 Inorganic Precipitating Agents
12C-2 Reducing Agents
12C-3 Organic precipitating
Agents
12C-5 Voltilization Gravimetry
The two most common gravimetric methods
based on volatilization are those for determining
water and carbon dioxide
（1）water
A. Direct determination B. Indirect method
（2）CO2 （It is the determination of the sodium
hydrogen carbonate content of antacid tablets）
2NaOH＋CO2 → Na2CO3＋H2O