Chapter12

Gravimetric Methods of Analysis

Gravimetric methods of analysis12A 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 p

recipitate 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++C2O4

2-

• Ca2+(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

＝ 0.1132gCaO × × ＝ 2.0186×10-3

Conc.Ca ＝ ＝ 0.04045g/100ML

gCaOmolCaO077.561

molCaOmolCa1

mlMLSample

molCagCamolCa100

200

/07.40100186.2 3

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 reagent1. Readily filtered and washed free of conta

minants

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

Q is the concentration of the solute at any instant and S is its equilibrium solubility

)112(sup

S

SQonersaturatirelateve

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 preci

pitate(2) dilute solution (3) slow addition of the precipitating agent with good stirring.

12A-3 Colloidal Precipitates1.Coagulation of colloids: by heating, stirring, adding electrolyte(1) Why colloidal suspensions are stable and do not c

oagulate spontaneously. (a) Charge (cations or anions that are bound to the su

rface 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..

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 solutio

n. 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 con

stitute .This double later exerts an electrostatic repulsive force that prevents particles from colliding and adhering.

Fig 12-1 Colloidal silver chloride particle in a solution that contains an excess of silver nitrate.

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

(1) minimization of Q (using dilute solution and adding the precipitating reagent slowly and with go

od mixing.)

(2) maximizing S (precipitating from hot solution or adjusting the pH of the precipitation mediu

m.)

S

SQonersaturatirelateve

sup

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.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 electr

ollyte

（ 2 ） Reprecipitation ： A drastic but effective way to minimize the effects of adsorption is Preprecipitation.

1.surface adsorption,

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 p

ositive errors in an analysis .

• 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-

12A-6 Precipitation from Homogeneous Solution

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 Precipitates12A-7 Drying and lgnition of PrecipitatesWeighing formWeighing formSome precipitates are also ignited to decompose the solid aSome precipitates are also ignited to decompose the solid and form a compound of Known compositionnd 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 amount of Ca

＝ 0.1132gCaO × × ＝ 2.0186×10-3

Conc.Ca ＝ ＝ 0.04045g/100ML

gCaOmolCaO077.561

molCaOmolCa1

mlMLSample

molCagCamolCa100

200

/07.40100186.2 3

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 (55.847

g/mol ) and (b) The ％ Fe3O4(231.54g/mol) in the sample.Ans ：amount Fe2O3 ＝ 0.5394g Fe2O3×

（ a ） The number of moles of Fe is twice the number of moles of Fe2O3,and Mass Fe

＝ 3.3778×10-3× ＝ 0.37728Fe

％ Fe ＝ ％＝ 33.32 ％(b) massFe3O4 ＝ 3.3778×10-3

％ Fe3O4 ＝ ％＝ 46.04 ％

323

69.1591 103778.3

32

32 OmolFeOgFeOmolFe

molFegFe

OmolFemolFe 847.55

32

2 1001324.1

37728.0 gsamplegFe

OgFeOmolFeOgFe

OmolFeOmolFe

354.231

32 52140.0

3

3

32

43

1001324.15140.0 43 gsample

OgFe

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＝ xgNaCl× ＝ 0.017111xmol AgclMass AgCl from NaCl ＝ 0.017111xmol AgCl×143.32 g AgCl/molAgCl

＝ 2.4524gProceeding in the same way, wi can write that the number of moles

of AgCl from the BaCl2 is given by

Amount AgCl from BaCl2

＝ ygBaCl2×Amount AgCl from BaCl2 ＝ 9.605×10-3×143.32 ＝ 1.3766yg AgCl2.4524x ＋ 1.3766y ＝ 0.4637Y ＝ 0.2356-x ； 2.4524x ＋ 1.3766(0.2356-x) ＝ 0.46371.0758x ＝ 0.13942 ； x ＝ mass NaCl ＝ 0.12960gNaCl

％ NaCl ＝％ BaCl2 ＝ 100.00%-55.01% ＝ 44.99%

molNaClmolAgCl

gNaClmolNaCl 144.58

1

molAgClmolBaClmolAgCl

gBaClmolBaCl 3223.208

1 10605.922

2

%01.55%1002356.012956.0 gsample

NaCl

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