Introduction: Analytical
Chemistry
NURSAKINAH MAT HAZIR
OUTLINE OF CHAPTER
 Types & Steps in analysis
 Review the terms: Moles, molarity and concentration.
 Review other forms of expressing concentration
i) ppm
ii) ppt
iii) ppb
iv) %(w/w)
v) %(w/v)
vi) %(v/v)
What is Analytical Chemistry?
“Analytical Chemistry is what analytical
chemists do.”
--- C. N. Reilley
 Chemical analysis is more than just detecting or
determining the general composition or a specific component
of a sample.
 It is the resolution or interpretation of a given problem.
What is Analytical Chemistry?
 Chemical analysis
determination of the chemical composition or
chemical make up as well as the quantity of each
composition presence in a sample
 E.g: The chemical analysis of a blood sample involves the
determination of its iron content, its alcohol content, or
perhaps its drug content as well as how much each of the
composition presence in the blood sample.
 E.g: The chemical analysis of a water sample involves the
determination of its mineral content, its pollutant content, and
its dissolved oxygen content and the percentage of each
constituents.
What is actually involve in
CHEMICAL ANALYSIS?
 Chemical analysis procedures involve two areas; the
GOAL of the analysis and the METHOD used in the
analysis.
 The GOAL of the analysis can be classified into
QUALITATIVE or QUANTITATIVE analysis:
 QUALITATIVE: Analysis conducted to identify what
are the constituents present in the sample
(identification of the sample component)
 QUANTITATIVE: Analysis to determine how much
of each constituent present in the sample.
 Qualitative Analysis
determination of chemical identity of the species
 Quantitative Analysis
determination of the amount of species or
analytes, in numerical terms. Hence, Math is
heavily involved.
In order to perform quantitative analysis,
typically one needs to complete qualitative
analysis.
One needs to know what it is and then select the
means to determine the amount
Qualitative analysis is what.
Quantitative analysis is how much.
©Gary Christian, Analytical Chemistry,
6th Ed. (Wiley)
WHAT HAPPEN TO OUR RIVER?
Why it is important?
 Analytical chemists must use a systematic approach
to obtain the required information from samples.
 This approach consists of a series of steps and
processes.
 Let we see the step in analytical analysis.
Step in chemical analysis….
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2005
Understanding and defining the problem
Familiarize with background of problem and
history of sample
Literature search
(selecting the analytical procedures)
Sampling
(obtain and store sample)
Preparation and pretreatment of sample
Analysis (perform measurement and compare
results with standards)
Apply required stastitical tecniques
Verify results
Reporting
1. Understanding and Defining Problem
 The goal of every chemical analysis is to obtain the required
information within a period of time acceptable to the customer.
 Many problem do not require complete identification and many
cases require only a general classification.
 It is important for the analyst to determine the information
required by the client.
 For example, in water analysis, only the total hardness is
required rather than the concentration of individual Ca2+ and
Mg2+ ion concentration is not necessary.
2. Sample History and Background
of the Problem
 Should know the origin of the sample.
 Sample history include information on how, where,
and when the sample were collected, transported and
stored.
 Analyst will use the sample history to maximize
advantage in solving the problem.
3. Literature Search
 Literature search is essential to find an appropriate
procedures.
 Patents or commercial literature usually help in
determination of the composition of industrial materials.
 Others:
- book
- review articles
- standard organization :
i) Association of official Analytical Chemists (AOAC)
ii) Food and Drug Administration (FDA)
- electronic media
OCT
2007
APR
2007
4. Sampling
 Definition: Process of selecting representative material





from the lot and storing the sample.
The suitable sampling method differ from one substance
to another depending on homogeneity.
Homogeneous: Substance that has the same
composition throughout the sample.
Heterogenous: Substance that has different composition
from one reagent to another reagent.
Proper procedure must be used to store both samples and
standards.
All sample must be properly labeled and recorded.
5. Sample Preparation and Pretreatment
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2007
 Step of sample preparation involved bulk materials:
1.
2.
3.
Bulk sample must be reduced in size to obtain a
laboratory sample of several grams.
From which a few grams to milligrams will be taken
to be analyzed (analysis sample).
The size reduction may require taking portions (eg,
two quarters) and mixing, in several steps, as well as
crushing and sieving to obtain a uniform powder
for analysis.
Why sample pretreatment is important?
Laboratory samples are often subjected to
physical or chemical pretreatment where it is
converted to a form that is suitable for the
measurement.
During pretreatment:
i) reduce and remove interferences
ii) adjust analyte concentrations to a range
suitable for measurement
iii) produce species from analyte that have
quantitatively measurable properties.
6. Analysis
 Analysis:
Incorporates the measurement of the
concentration of the analyte in replicates and comparing
with standards.
 Replicate measurement: the practice of taking multiple
readings.
 The replicate measurements are necessary to obtain the
measurement uncertainty.
 The uncertainty important as it indicates the reliability of
the measurements.
 Suitable standardization methods should be selected to
ensure the precision and accuracy of results.
i)
ii)
iii)
iv)
v)
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2006
vi)
CALIBRATION CURVE – A graph of detector
response as a function of analyte concentration.
EXTERNAL STANDARDIZATION
INTERNAL STANDARDIZATION
STANDARD ADDITION - Addition of a known
quantity of analyte to a real sample undergoing
analysis.
BLANKS – A blank contains the reagents and
solvents use in analysis, but no analyte.
CONTROL CHARTS
7. Reporting and Interpretation
 Deliver a clearly written, complete report and their
limitations.
HOMEWORK
Turn to page 17 (Analytical Chemistry, 5th Ed. Gary
D. Christian, Wiley, New York, 1994.) and please answer
all questions.
This is your tutorial question !!! It not required you to
Submit this homework to me…. But if I ask you to explain
Later on.. I do hope you can… thank you.
PREFIX NOTATION
 Main measures such as gram and meter are given prefix
number for the power of ten.
Exponential
Prefix name
Symbol
1012
tera
T
109
giga
G
106
mega
M
103
kilo
k
10
-
-
10-1
deci
d
10-2
centi
c
10-3
milli
m
10-6
micro
µ
10-9
nano
n
10-12
pico
p
CONCENTRATION UNITS
 Concentration : the quantity of solute in a specified volume
or mass of solution or solvent.
 The concentration of solutes can be expressed in various
ways.
Name
Component Total sample
measured by measured by
Abbreviation
Units
used
Molarity
Number
Volume
M
Mol L-1
Molality
Number
Weight
m
Mol kg-1
Percent composition
Weight-toweight
Weight
Weight
w/w
%(w/w)
Weight-tovolume
Weight
Volume
w/v
%(w/v)
Volume-tovolume
Volume
Volume
v/v
%(v/v)
DEFINITION
 Molarity (M) : No of moles of the solute in 1L of
solution.
M = no.mol solute
no.L solution
 Molality (m) : No of moles of the substance per
kilograms of the solvent.
m = no.mol solute
kg solution
Continued….
Analytical molarity
 There are 2 versions of molarity
Species molarity
 Analytical Molarity: The total number of moles of the solute
in 1L of solution.
 For example, a HCl solution that has an analytical
concentration of 1.0 M can be prepared by dissolving 1.0 mol
or 36.5 g of HCl in water and diluting to exactly 1.0 L.
 Species Molarity (Equilibrium molarity) : The number of
moles of a particular species in 1L of a solution at equilibrium.
 The concentration unit is usually written in bracket,
[H+],means the concentration of H+.
Continued…. Species Molarity
 For example, the species molarity of HCl in a solution
with an analytical molarity of 1.0M is 0.0 M because
HCl is entirely ionized into H+ and Cl- ions and
essentially no HCl molecules as such present in the
solution.
CHCl =
[HCl] =
[H+ ] =
[Cl-] =
1.0 M
0.0 M
1.0 M
1.0 M
You should calculate this…….
 What is the molarity of a solution of 0.60 g NaCl
in 100 mL of solution?
Given, MW NaCl = 58.5g/mol.
Example: Saline water concentration
Typical seawaters contain sodium chloride, NaCl as
much as 2.7 g per 100 mL.
(a) What is the molarity of NaCl in the saline water ?
(b) The MgCl2 content of the saline water is 0.054 M.
Determine the weight (grams) of MgCl2 in 50 mL of the
saline water ?
Solution:
(a) What is the molarity of NaCl in the saline water ?
 MW of NaCl = 22.99 + 35.54 = 58.44 g/mol
 No.of mol of NaCl in 100 mL saline water = 2.7 g
58.44 g/mol
= 0.046 mol
 Molarity of saline water = no.mol = 0.046 mol = 0.46 M
L
(100/1000) L
Continued……….
(b) The MgCl2 content of the saline water is 0.054
M. Determine the weight (grams) of MgCl2 in 50
mL of the saline water ?


MW of MgCl2 = 24.30 + 2(35.45) = 95.20 g/mol
Weight of in 50 mL of saline water
= (M x V) x MW
= 0.054 M x (50/1000) x 95.20 g/mol
= 0.26 g
DEFINITION
 Weight percent, %w/w =
mass of solute x
weight of solution
100
 Volume percent, %v/v =
volume of solute x
volume of solution
100
 Volume percent, %w/v =
mass of solute
x
volume of solution
100
DEFINITION
 Parts per million (ppm) is grams of solute per million
grams of total solution.
ppm = mass of solute x
mass of solution
106
 Parts per billion (ppb) is grams of solute per billion
grams of total solution.
 Parts per trillion (ppt) is grams of solute per trillion
grams of total solution.
Common unit to express content
Expression
Abbreviation
Units
w/v
Parts per million
ppm
w/w
µg/g
mg/kg
µg/mL
mg/L
v/v
nL/mL
µL/L
Parts per billion
ppb
ng/g
ng/mL
nL/L
µg/kg
µg/L
Example: Percent Composition
 A solution contains 118.5 KI per liter solution.
Calculate the concentration in
(a) % (w/v)
(b) % (w/w)
Given the density of the solution at 25oC is 1.078
g/mL.
Solution……
 (a) Volume percent, % (w/v) = mass of solute x 100
volume of solution
= 118.5 g
x
1000 mL
= 11.85% (w/v)
100
(b) weight percent, % (w/w) = mass of solute x 100
weight of solution
= 118.5 g x 1mL x 100
1000 mL
1.078 g
= 10.99% (w/w)
Example: Conversion of ppb to molarity
 An aqueous solution contains 56 ppm SO2. Calculate
the molarity of the solution.
 Solution:
MW of SO2 = 32 + 2(16) = 64 g/mol
56 ppm = 56 mg/L = (56 x 10-3 g / 64 gmol-1)/L
solution
= 8.75 x 10-4 M