Transcript No Slide Title

Chapter 27
Gravimetric and
Combustion Analysis
Dr. S. M. Condren
Representative Gravimetric Analyses
Dr. S. M. Condren
Dr. S. M. Condren
Dr. S. M. Condren
Precipitation
AgNO3 + NaCl -----> AgCl + NaNO3
Dr. S. M. Condren
Precipitation
AgNO3 + NaCl -----> ?
AgNO3 + NaCl -----> AgCl + NaNO3
total ionic equation
Ag+ + NO3- + Na+ + Cl----->
AgCl + Na+ + NO3-
net ionic equation
Ag+ + Cl- -----> AgCl
Dr. S. M. Condren
Precipitation. WHY?
Solubility Rules
1. The common salts of the alkali metals
(Group IA) and the ammonium ion (NH4+)
are soluble.
2. Salts of nitrate (NO3-), chlorate (ClO3-),
perchlorate (ClO4-), and acetate (CH3COO-)
anions are soluble.
3. All chlorides, bromides, and iodides are
soluble, except those of Ag+, Pb+2, and
Hg2+2 (the form that Hg(I) exists in water).
Dr. S. M. Condren
Precipitation. WHY?
Solubility Rules
4. Sulfates (SO4-2) are soluble except those of
Ba+2, Pb+2, Hg+2, and Hg2+2.
Dr. S. M. Condren
Precipitation. WHY?
Solubility Rules
5. Most metal hydroxides are insoluble. The
exceptions are the hydroxides of the alkali
metals and the heavier alkaline earth metals
(Ca, Sr, Ba).
6. All carbonates (CO3-2) and phosphates
(PO4-3) are insoluble, except those of the
Group IA metals and NH4+ ions.
Dr. S. M. Condren
Crystal Growth
nucleation
particle growth
supersaturated
Dr. S. M. Condren
Crystal Growth
relative supersaturation = (Q - S)/S
where Q => actual concentration
S => equilibrium concentration
Dr. S. M. Condren
Crystal Growth
techniques to promote crystal growth
1.Raising the temperature to increase S and
thereby decrease relative supersaturation.
2.Adding precipitant slowly with vigorous
mixing, to avoid a local, highly supersaturated
condition where the stream of precipitant first
enters the analyte.
3.Keeping the volume of solution large so that
the concentration of analyte and precipitant are
low.
Dr. S. M. Condren
Dr. S. M. Condren
Precipitation in the Presence of
Electrolyte
coagulate => “to change or be changed from a
liquid into a thickened mass”
Dr. S. M. Condren
Precipitation in the Presence of Electrolyte
Dr. S. M. Condren
Precipitation in the Presence of
Electrolyte
adsorption => attached to surface
absorption => penetration beyond surface
Dr. S. M. Condren
Digestion
process of keeping mixture warm while the
size of the crystals increase
Dr. S. M. Condren
Purity
absorbed impurities
inclusions - impurity ions which randomly
occupy sites in the crystal lattice
occulusions - pockets of impurity trapped
inside growing crystal
coprecipitaion
Dr. S. M. Condren
Purity
gathering agent - precipitating agent used to
collect trace component (process gathering)
masking agent
post-precipitation
peptization
Dr. S. M. Condren
Product Composition
•
•
•
•
product must be of known composition
hygroscopic substance
ignition
thermogravimetric
Dr. S. M. Condren
Thermogravimetric curve for calcium salicylate
Dr. S. M. Condren
Example: What weight of Fe2O3
can be obtained from 1.63 g of
Fe3O4?
2 Fe3O4 + [O] -----> 3 Fe2O3
(1.63 g Fe2O3)
#g Fe2O3 = --------------------
Dr. S. M. Condren
Example: What weight of Fe2O3
can be obtained from 1.63 g of
Fe3O4?
2 Fe3O4 + [O] -----> 3 Fe2O3
(1.63 g Fe2O3)(1 mole Fe3O4)
#g Fe2O3 = -------------------------------------(231.54 g Fe3O4)
Dr. S. M. Condren
Example: What weight of Fe2O3
can be obtained from 1.63 g of
Fe3O4?
2 Fe3O4 + [O] -----> 3 Fe2O3
(1.63)(1 mole Fe3O4)(3 mole Fe2O3)
#g Fe2O3 = -----------------------------------------(231.54)(2 mole Fe3O4)
Dr. S. M. Condren
Example: What weight of Fe2O3
can be obtained from 1.63 g of
Fe3O4?
2 Fe3O4 + [O] -----> 3 Fe2O3
(1.63)(1)(3moleFe2O3)(159.69gFe2O3)
#g Fe2O3 = -----------------------------------------------(231.54)(2)
(1 mol Fe2O3)
Dr. S. M. Condren
Example: What weight of Fe2O3
can be obtained from 1.63 g of
Fe3O4?
2 Fe3O4 + [O] -----> 3 Fe2O3
(1.63)(1)(3)(159.69gFe2O3)
#g Fe2O3 = ----------------------------------- = 1.69 g
(231.54)(2)(1)
Dr. S. M. Condren
EXAMPLE: What mass of PbI2 will
precipitate if 2.85 g Pb(NO3)2 is
added to 225 mL of 0.0550 M KI(aq)?
Pb(NO3)2 + 2 KI ------> PbI2 + 2 KNO3
Dr. S. M. Condren
EXAMPLE: What mass of PbI2 will
precipitate if 2.85 g Pb(NO3)2 is
added to 225 mL of 0.0550 M KI(aq)?
Pb(NO3)2 + 2 KI ------> PbI2 + 2 KNO3
if use all of the 2.85 g Pb(NO3)2
Dr. S. M. Condren
EXAMPLE: What mass of PbI2 will
precipitate if 2.85 g Pb(NO3)2 is
added to 225 mL of 0.0550 M KI(aq)?
Pb(NO3)2 + 2 KI ------> PbI2 + 2 KNO3
if use all of the 2.85 g Pb(NO3)2
(2.85 g Pb(NO3)2
#g PbI2 = -----------------------Dr. S. M. Condren
EXAMPLE: What mass of PbI2 will
precipitate if 2.85 g Pb(NO3)2 is
added to 225 mL of 0.0550 M KI(aq)?
Pb(NO3)2 + 2 KI ------> PbI2 + 2 KNO3
if use all of the 2.85 g Pb(NO3)2
(2.85 g Pb(NO3)2)(1 mol Pb(NO3)2
#g PbI2 = -------------------------------------------(331 g Pb(NO3)2)
Dr. S. M. Condren
EXAMPLE: What mass of PbI2 will
precipitate if 2.85 g Pb(NO3)2 is
added to 225 mL of 0.0550 M KI(aq)?
Pb(NO3)2 + 2 KI ------> PbI2 + 2 KNO3
if use all of the 2.85 g Pb(NO3)2
(2.85 g Pb(NO3)2)(1 mol Pb(NO3)2
#g PbI2 = -------------------------------------------(331 g Pb(NO3)2)
Dr. S. M. Condren
EXAMPLE: What mass of PbI2 will
precipitate if 2.85 g Pb(NO3)2 is
added to 225 mL of 0.0550 M KI(aq)?
Pb(NO3)2 + 2 KI ------> PbI2 + 2 KNO3
if use all of the 2.85 g Pb(NO3)2
(2.85)(1 mol Pb(NO3)2)(1mol PbI2)
#g PbI2 = -------------------------------------------(331) (1 mol Pb(NO3)2)
Dr. S. M. Condren
EXAMPLE: What mass of PbI2 will
precipitate if 2.85 g Pb(NO3)2 is
added to 225 mL of 0.0550 M KI(aq)?
Pb(NO3)2 + 2 KI ------> PbI2 + 2 KNO3
if use all of the 2.85 g Pb(NO3)2
(2.85)(1)(1 mol PbI2)(461 g PbI2)
#g PbI2 = -------------------------------------------(331) (1)
(1 mol PbI2)
Dr. S. M. Condren
EXAMPLE: What mass of PbI2 will
precipitate if 2.85 g Pb(NO3)2 is
added to 225 mL of 0.0550 M KI(aq)?
Pb(NO3)2 + 2 KI ------> PbI2 + 2 KNO3
if use all of the 2.85 g Pb(NO3)2
(2.85)(1)(1 mol PbI2)(461 g PbI2)
#g PbI2 = -------------------------------------------(331) (1)
(1 mol PbI2)
Dr. S. M. Condren
EXAMPLE: What mass of PbI2 will
precipitate if 2.85 g Pb(NO3)2 is
added to 225 mL of 0.0550 M KI(aq)?
Pb(NO3)2 + 2 KI ------> PbI2 + 2 KNO3
if use all of the 2.85 g Pb(NO3)2
(2.85)(1)(1)(461 g PbI2)
#g PbI2 = ------------------------------- = 3.97 g
(331)(1)(1)
Dr. S. M. Condren
Dr. S. M. Condren
Dr. S. M. Condren
Dr. S. M. Condren
Dr. S. M. Condren