Gravimetric Analysis

Mrs. Jigna T. Patel M.Pharm., Ph.D. (Pursuing)

Assistant Professor

Department of Pharmaceutical Analysis and Pharmaceutical Chemistry

Saraswati Institute of Pharmaceutical Sciences

At. & Po.: Dhanap, Gandhinagar, Gujarat, India – 382355

Learning Outcomes
• Mention the main steps of gravimetric analysis.
• Define supersaturation.
• Identify types of impurities in precipitates.
• Define peptization.
• Define gravimetric factor.
• Apply gravimetric analysis to different samples.

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Gravimetric analysis
Gravimetric analysis is one of the most accurate and precise methods of macro-

quantitative analysis.

The analyte is selectively converted into an insoluble form (precipitate).
The separated precipitate is then dried or ignited, possibly to another form and is

accurately weighed.

From the weight of the precipitate and knowledge of its chemical composition, we
can calculate the weight of the analyte in the desired form.

Add a precipitating agent

Analyte in a soluble form Drying or Ignition Weighing
Precipitate Filtration

Properties of the Precipitate
The ideal product (precipitate) of a gravimetric analysis should be:

Sufficiently insoluble (the precipitate is of such low solubility that

losses from dissolution are negligible)

Easily filterable (crystals of large particle size so as not to pass through

the filtering system).

Very pure (less possibility that the precipitates carry some of the other

constituents of the solution with them).

Should possess a known composition (known chemical structure).

Steps of Gravimetric Analysis
1 • Precipitation

2 • Digestion

3 • Filtration

4 • Washing

5 • Drying or Ignition

6 • Weighing

7 • Calculations

1. Mechanism of Precipitation
When a solution of precipitating agent is added to a test solution to form a precipitate,

such as in the addition of AgNO3 to a chloride solution to precipitate AgCl.

The actual precipitation occurs in a series of steps:

1. Super Saturation 2. Nucleation 3. Particle Growth

Ionic product > Solubility product Nuclei join
A minimum number of

together to
Unstable state: solution contains a lot of particles come together to

produce microscopic form a crystal
dissolved ions more than it can

nuclei of the solid phase. of a certain
accommodate.

geometric
To become stable: Precipitation takes place. shape

1- Mechanism of Precipitation (cont.)

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1- Mechanism of Precipitation (cont.)

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1- Mechanism of Precipitation (cont.)
When a solution is super-

Important Notes A higher degree of supersaturation saturated, it is in an unstable
state and this favors rapid
nucleation to form a large

A greater rate of nucleation number of small particles.

A greater number of nuclei formed per unit time

Precipitate is in the form of a large number of small nuclei

Increase total surface area of
Precipitate is not of filterable

precipitate which increases the
size

possibility of entrapment of impurities

1- Mechanism of Precipitation (cont.)
Degree of supersaturation
Von Weimarn discovered that the particles size of precipitates is inversely
proportional to the relative supersaturation of the solution during the precipitation
process

Q is the concentration of the solute at any
Relative supersaturation = instant.

S is its equilibrium solubility.

HIGH RSS Many small crystals (Large Surface Area)

Low RSS Fewer large crystals (Small Surface Area)

Low RSS is favorable.
How to achieve it? During precipitation: Q S

1- Mechanism of Precipitation (cont.)

Favorable conditions for precipitaion

To decrease the value of Q
Precipitate from dilute solutions.

Add dilute precipitating agents slowly with constant stirring.

H+

H+ H+

To increase the value of S
Precipitate from hot solution.

Precipitate at as low pH as possible.

2. Digestion
 Digestion is keeping the precipitate formed in contact with the mother liquor for a specified amount

of time.
Mother liquor (the solution from which it was precipitated).

In case of Colloidal precipitates: In case of Crystalline precipitates:
Particle size (less than 100 m) Particle size (more than 100 m )

Digestion is performed by allowing the Digestion is performed by allowing the
precipitate to remain in contact with the precipitate to remain in contact with the
mother liquor at high temperature for a mother liquor for a long time.
couple of hours.

Why is it important?
1- The small particles tend to dissolve and re-precipitate on the surfaces of large crystal.

2- Individual particles tend to agglomerate together.

3- Imperfections of the crystals tend to disappear and adsorbed or trapped impurities

tend to escape into solution.

2. Digestion Cont.
For colloidal precipitate

• The surface of the precipitate tend to adsorb the ion of the
precipitated particle that is in excess in the solution, (Ag+ or
Cl- ).

• This primary adsorbed layer attracts oppositely charged ion
in a secondary or counter layer (they are less tightly held
than the primary adsorbed ion).

• The negatively charged ionic atmosphere of the particles
repel one another leading to the colloidal state.

How does digestion help solve this problem?
• The colloidal particles, therefore, must have enough kinetic energy to overcome electrostatic repulsion

before they can coalesce.
• Thus, digestion at high temperature increases the colloidal particles’ kinetic energies and can promotes

coalescence (coagulation).

Impurities encountered in Gravimetric Analysis
• 1. Occlusion
• This occurs when materials that are

not part of the crystal structure are
trapped within the crystal.

• For example, water or any counter
ion can be occluded in any
precipitate.

• This causes deformation in the
crystal.

• This type is hard to be removed,
digestion can decrease it to a certain
extent.

Impurities encountered in Gravimetric Analysis
• 2. Inclusion (isomorphous replacement)
• This occurs when a compound that is isomorphous to the

precipitate is entrapped within the crystal.
• Isomorphous means they have the same type of formula

and crystals in similar geometric form.
• This type of impurity doesn’t lead to deformation of the

crystals.
• Example, K + h as nearly the same size of NH + so it can

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replace it in Magnesium ammonium phosphate.

• Digestion cannot handle this type and mixed crystals will
be formed.

Impurities encountered in Gravimetric Analysis
• 3. Surface adsorption
• Surface adsorption is very common especially in colloidal precipitates.
• Example, AgCl, BaSO 4, where each of them will have a primary adsorption

layer of the lattice ion present in excess followed by a secondary layer of
the counter ion of opposite charge.

• These adsorbed layers can often be removed by washing where they can
be replaced by ions that can be easily volatilized at the high temperature
of drying or ignition.

Adsorbed, occluded and included impurities are said to be coprecipitated.
That is, impurity is precipitated along with the desired product during its

formation.

Impurities encountered in Gravimetric Analysis
• 4. Post precipitation
• When the precipitate is allowed to stand in contact

with the mother liquor, a second substance will
slowly form a precipitate on the surface of the
original one.

• Examples, When calcium oxalate is precipitated in
the presence of magnesium ions, magnesium oxalate
may be if the solution is left without filtration for a
long time.

• Digestion will increase the extent of such type,
dissolution and reprecipitation will decrease the
extent of post precipitation.

3- Filtration and Washing of the Precipitate
Washing helps remove the co-precipitated impurities specially the occluded

and surface adsorbed.

The precipitate will also be wet with the mother liquor which is
also removed by washing.

Note that:
Colloidal precipitates can not be washed with pure water,

because peptization occurs. This is the reverse of coagulation.

3,4- Filtration and Washing of the Precipitate
Peptization
As we said before that colloidal precipitates have double layer of adsorbed primary and

counter ions.
The presence of another electrolyte during precipitation will cause the counter

ions to be forced into closer contact with primary layer, thus promoting
coagulation.

Washing with water will dilute and remove foreign ions and the counter ion
will occupy a larger volume, with more solvent molecules between it and the
primary layer.

This can be prevented by adding an electrolyte to the washing liquid. e.g. HNO 3
is used to wash AgCl.

This electrolyte should be volatile at the temperature of drying or ignition and must
not dissolve the precipitate.

5- Drying or Ignition

Ashless filter paper

Porcelain crucible Glass crucible

(Ignition up to 1000 C ) (Drying at 100-120 C )

5- Drying or Ignition
 After filtration, a gravimetric precipitate is heated until its mass becomes constant.
Drying at 110 to 120 °C for 1-2 hours is conducted If the collected precipitate is in a form

suitable for weighing (known, stable composition), it must be heated to remove water and to
remove adsorbed electrolyte from the wash liquid.

Ignition (strong heating) at much higher temperature is usually required if a precipitate must be
converted to a more suitable form for weighing.

In this case, the weighed form of the precipitate might be different from the precipitated
form.

Examples:

same
same
same

same

same

6- Weighing

7- Calculations
Gravimetric calculations relate moles of the product finally weighed to moles of

analyte.

Precipitating agent (B) Filtered
Washed
Dried

Analyte (A) 20 mls Weighed (P)

nA + B mP + S Convert moles into weights by
multiplying by the molecular

n Mwt m Mwt weight.

•Where W is t 2 1
2 he weight of W W

the analyte ion only
dissolved in 20 ml of
solution and W 1 is the
weight of precipitate (ppt). (nMwta nalyte/ mMwtp pt) is called the gravimetric factor.

Example 1

• Calculate the gravimetric factor for each of the following:
P Ag3 P O 4, K 2H PO 4 AgPO ,4 Bi S2 3 BaSO4
As2 O 3 Ag 3A sO ,4 K O2 KB(C H6 ) 5 4

• P/Ag 3P O 4 = atwt P/ Mwt Ag 3P O 4

• K 2H PO 4/ Ag 3P O =4 Mwt K H2 PO /4M wtAg P3O 4

• Bi 2S 3/ BaSO 4 = MWt Bi S2 /33 MWt BaSO 4

• As 2O 3/ Ag 3A sO 4= Mwt As O2 /32 MWt Ag A3s O 4

• K 2O /KB(C 6H 5) 4 = Mwt K O2 / 2MWt KB(C H6 ) 5 4

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Example 2
• Orthophosphate (PO 3

4
–

) is determined by weighing as ammonium
phosphomolybdate, (NH4 ) PO 4. 12MoO 3. Calculate the percent P in the sample and
the percent P2 O 5 if 1.1682g precipitate were obtained from a 0.2711 g sample.

• Remember: Wt of analyte = Wt of ppt x Gravimetric Factor

• Wt of P = 1.1682 X

% P = (0.0193/0.2711) x 100 = 7.11%

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Example 2 (cont.)

In general,

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Example 3
• An ore is analyzed for the manganese content by converting the manganese to

Mn3 O 4 and weighing it. If a 1.52 g sample yields Mn 3O 4 w eighing 0.126g, what
would be the percent Mn and Mn2 O3 in the sample?

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Example 3 (cont.)

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Example 4

The piperazine content of an impure commercial material can be
determined by precipitating and weighing the diacetate

In one experiment, 0.3126 g of the sample was dissolved in 25 mL of
acetone and 1 mL of acetic acid was added. After 5 min, the
precipitate was filtered, washed with acetone, dried at 110 C, and
found to weigh 0.7121 g. What is the percent of piperazine in the
commercial material?

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References
• D. A. Skoog, D.A. West, F.J. Holler, S.R. Crouch, Analytical

Chemistry, an introduction, 7th Edition, ISBN 0-03-020293-0 ,
(Chapter 8).

• Lecture 11 by Dr. Nesrine El Gohary, GUC, SS 2016

• Lecture 10 by Prof. Rasha Elnashar, GUC, SS 2015.

• Lecture 8 by Dr. Raafat Faraghly, GUC, SS 2007