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TOPIC 5:
GRAVIMETRIC ANALYSIS
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Traditional - Quantitativeanalysis
Gravimetric analysis
A + B = C
An excess of B, at unknown [ ] , isadded to A so that A is completelytransformed in C.
C = Obtained by weighing solid product in apure form.
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Introduction Gravimetric analysis is the quantitative
determination of analyte through a processof precipitation/volatilization, isolation ofthe precipitate, and weighing the isolatedproduct.
Analyte + Reagent Precipitate
Precipitating pure
agent can be filtered,
well separated
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Divided into 2:
i. Precipitation gravimetry Analyte is separated from the solution of
the sample as a ppt & is converted to acompound of known composition that canbe weighed.
Eg: Cl- + AgNO3 AgCl + NO3-
ii. Volatilization gravimetry
Analyte is separated from otherconstituents of a gas then serves as ameasure of the analyte [ ].
Eg: NaHCO3 + H2SO4 CO2 + H2O +NaHSO
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CONTINUE
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CONTINUE
The ideal product should be;
very insoluble,
easily filtered, and
very pure and
posses a known and constant composition
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Steps of a Gravimetric Analysis which requirecondition optimization to eliminate systematicerror :
Preparation of the solution PrecipitationDigestion FiltrationWashingDrying or IgnitingWeighing Calculation
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Dissolution
Sample is dried, weighed, dissolved inacid, etc.
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Precipitation
Precipitation Techniques
Add precipitating reagent to sample solution.
Reacts with analyte to form insolublematerial.
Precipitate has known composition or can beconverted to known composition.
2 types of Precipitating agents:
Specific (react only with 1 chemical sp.) Selective (react with a limited number of sp.)
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Desirable properties of analyticalprecipitates:
Readily filtered and purified.
Low solubility, preventing losses duringfiltration and washing.
Stable final form (unreactive).
Known composition after drying orignition.
CONTINUE
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Ions in colloid Precipitate
Solution
(10-7cm) (10-7 - 10-4cm) (10-4cm)
During precipitation process, supersaturationoccurs, followed by nucleation and precipitation.
First supersaturation (soln. contains more of thedissolved salt than at equilibrium) occurs.
Nucleation particles come together to produce
microscopic nuclei. The the degree of supersaturation, the greater
the rate of nucleation.
Nucleation can be induced by scratching vessel
surface, dusts, etc
CONTINUE
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Nucleation
Formation of nucleus or primary particlesfrom ion, atoms or molecules, whichaggregates to form a stable second phase.
Growth of particles
Precipitation on top of formed nucleus.
The higher the degree of supersaturation,
the higher the rate of nucleation. Ifrate of nucleation < rate growth of
particles, big particles (crystals ) areformed (easy to filter, less impurities)
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Relative supersaturation = Q-S
S
Q : degree supersaturation ( [ ] Of mixedreagents before ppt)
S : solubility of ppts at equilibrium
If(Q-S)/S >, small crystals (surface area) If (Q-S)/S < , larger crystals ( surface area)
Therefore, we want to keep Q and S during pptn.
CONTINUE
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Favourable conditions for pptn
Keep Q by:Precipitate from dilute solution.
Add dilute ppt reagents slowly, withstirring.
Keep S
by:
Precipitate from hot solution.
Precipitate at as pH as possible.
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Digestion
Solution containing ppts are heated before filtering.
Heating the ppts within the mother liquor (or solution fromwhich it precipitated) for a certain period of time.
Done for crystalline ppts (e.g: BaSO4) diameter >10-4cm.
During digestion, small particles tend to dissolve andreprecipitate on larger ones.
Trapped impurities will be dissolved.
Objective: bigger and purer particles that are more
easily filtered from solution.
DT
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Filtering & Washing
Impurities on the surface can be removed bywashing the precipitate after filtering.
Water is not always a good choice. Washing with water will dilute the counter layer
and the primary layer charge causes theparticles to revert to the colloidal state(peptization).
So we wash with an electrolyte that can be
volatilized on heating (HNO3).
Test for completeness of washing ~ check thepresence of precipitation agent in filtrate.Impurities and excess ions are washed off the
ppt in this step.
CONTINUE
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Type of filter will depend on the particle size ofppts.
filter paperGooch/glass sintered crucible
Colloidal ppts
E.g: AgCl.Mainly small particles. Cannot be filtered
using normal filtration as they are not pptfrom soln.
Bigger particles are not formed
Agglomeration of colloidal particles Is encouraged for formation of matter that
can be easily filtered, can be ppt fromsolutions.
Formation of larger particles
CONTINUE
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Drying/ Igniting & Weighing
Many ppt contain varying amounts of H2O.
Adsorbed from the air (i.e. hygroscopic).
Precipitates are dried for accurate, stable massmeasurements.
Heating removes the solvent & any volatile sp.carried down with ppt.
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Drying/ Igniting & Weighing
Drying T depends on ppt :
AgCl : > 110 oC Al2O3 : > 1000
oC
Ignition at higher temperature is required only if aprecipitate must be converted to a more suitable
form for weighing.
For example, Fe(HCO2)3.nH2O is ignited at 850oC
for 1 hour to give Fe2O3.
The compound finally weighed.
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Gravimetry Calculation
Generally express in units ofgram or %.
In gravimetry analysis, weight of the analyte isgiven by:
GF is gravimetry factor that is given by
a = mol of analyte
b = mol of ppt
xGF =
a
b
FW [analyte]
FW [precipitate]
weight of analyte (g) = weight of ppt (g) x GF
a & b must beequivalent
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CONTINUE
Examples of gravimetry factor (GF)
Analyte ppt GF
CaO CaCO3
Fe3O4 Fe2O3
Mg Mg2P2O7
1 Mg2P2O7
1 CaO
1 CaCO3
2 Fe3O43Fe2O3
2 Mg
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A general equation for calculating the % of
analyte (A) from gravimetric data;
Mass analyte= Mass ppt. x Gravimetric Factor (GF)
% Analyte = mass of ppt x GF x 100mass of sample
CONTINUE
% Analyte = mass of analyte (g) x 100 %
mass of sample (g)
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Example
An ore is analyzed for the manganesecontent by converting the manganese toMn3O4 & weighing it.
If a 1.52g sample yields Mn3O4 weighing0.126g, what would be the % Mn2O3 inthe sample? The % Mn?
(JMR: Mn= 54.94, O =16)
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Solution
%Mn2O3 =0.126g Mn3O4 x
3 Mn2O3
2 Mn3O4x 100%
1.52g sample
= 8.58 %
% Mn = 0.126g Mn3O4 x3 Mn
1 Mn3O4x 100%
1.52g sample
= 5.97 %
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A piece of impure marble (CaCO3) isanalyzed. It is found that 3.000 g ofthe marble react with Ba(NO3)2solution to produce 1.25 grams ofBaCO3. What % of the marble waspure CaCO3?
(Fw : CaCO3 = 100, BaCO3 = 197.3)
Ans: 21.1%
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Solution
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CaCO3 + Ba(NO3)2 BaCO3 + Ca(NO3)2
GF = 1 mole CaCO3 = 100 g CaCO31 mole BaCO3 197. 3 g BaCO3
= 0.507
% CaCO3 = 0.507 x 1.25 x 100% = 21.1%3.00 g
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Exercise
Treatment of a 0.4000 g sample ofimpure KCl with an excess of AgNO3
resulted in the formation of 0.7332 gof AgCl. Calculate the percentage ofKCl in the sample.
(JAR: Ag = 107.9, Cl= 35.5, K= 39.1, N= 14)
Ans: 95.35%
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A sample of ore weighing 1.2504 g containssulphur is treated with HNO3 and KClO3 toconvert sulphur to BaSO4. After treating with
HCL, chlorate and nitrate are removed andBaSO4 is precipitated, weighing 0.5473 g.calculate the percentage of sulphur in thesample .
(RAM S=32.064 & RMM BaSO4 = 233.40
Exercise
Ans: 6.010%
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Kuiz gantian akan diadakan semasa kelasganti khamis 24.5.2012, jam 8-10 mlm.
(kuiz pd 20.5.2012 - tidak diambil dikira)
Quiz