CHE 101
BASIC ANALYTICAL
CHEMISTRY
Basic Analytical
Chemistry
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Basic Analytical Chemistry
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Introduction to Analytical Chemistry
Analytical Chemistry deals with;
identification and
determination of
components/chemical
species/analytes in a sample.
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Basic Analytical
Chemistry
Introduction…….continues
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• Analytical Chemistry intends to
gather and interpret chemical
information.
• The gathered information is of
value to society in a wide range of
contexts.
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Introduction…….continues
Chemical analysis can be divided into
two categories;
Qualitative analysis
 Quantitative analysis
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The role of Analytical Chemistry
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Chemistry
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Analytical Chemistry is applied in the following areas;
Ensure raw materials meet
the required specifications
Industry
Quality control – Ensure manufactured
products contains essential components within
predetermined range of composition.
Monitoring of pollutants from industrial
processes
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The role of Analytical Chemistry
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Chemistry
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Health
Diagnosis of illness
Monitoring the
patients’ conditions
Monitoring levels of chemical
species in foodstuffs,
pharmaceuticals and water
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The role of Analytical Chemistry
Basic Analytical
Chemistry
Agriculture
Determine the nature and
amount of fertilizers to be
used in a given soil and for a
particular crop.
Monitoring of pesticides and
herbicides levels in the
environment.
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Geology and mining
Determination of
composition of the
numerous rock and soil
samples collected in the
field.
Routine determination
of quantity of chemical
species found in soil
samples
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Analytical Methods
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Chemistry
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Analytical methods
The methods or techniques are based on;
• Suitable chemical reactions and measuring
either the amount of the reagent needed to
complete reaction or the amount of the product
obtained.
• Measurement of electrical quantities (I,R,V,Q)
• Measurement optical properties (e.g absorption
spectra)
• Combination of optical or electrical
measurements and quantitative chemical
reaction (potentiometric titration,
spectrophotometric determination)
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Basic Analytical Chemistry
Analytical Methods
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Classical methods of analysis include;
• Gravimetric analysis –in which mass is the final
measurement.
• Titrimetry analysis –involves measurement of
volume standard solutionneeded for complete
reaction. The common reactions are
Neutralization (acid-base) reactions
Complex-forming reactions
Precipitation reaction
Reduction oxidation reaction
• Volumetry- involves measuring the volume of gas
evolved or absorbed in a chemical reaction
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Analytical Methods
Instrumental methods of analysis include;
• Electrical methods –measurement of I,V, R, Q in
relation to the concentration of certain species in
solution. The techniques include;
Voltametry – measurement of current
Coulometry – measurement of current and time
Potentiometry - measurement of potential of an
electrode
Conductometry – measurement of electrical
conductivity of solution.
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Chemistry
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Analytical Methods
Instrumental methods of analysis include;
• Optical methods – depend on measurement of
amount of radiant energy either absorbed or
emission by a sample at a specific wavelength.
▫ Absorption methods are classified according to
the wavelength used. eg. Visible spectrophotometry
(colorimetry), Ultraviolet spectrophotometry,
Infrared spectrophotometry
▫ Emission methods involve the measurement the
intensity of the emitted energy. eg. Emission
spectroscopy, Flame photometry
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Analytical Methods
Basic Analytical Chemistry
Instrumental methods of analysis include;
• Chromatography – used in separation as well as in
qualitative and quantitative analysis of species in a
sample.
• X-ray methods,
• Radioactivity,
• Mass spectrometry,
• Kinetic methods,
• Thermal methods,
• Optical rotation, and
• Refractometry
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ERRORS IN CHEMICAL
ANALYSIS
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Measurement errors
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Chemistry
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• Error is any departure of a result from a true or
accepted value.
• Every experiment is accompanied by many
uncertainties which combine to produce a
scatter of results.
• Since the measurement uncertainties can not be
completely eliminated, a true value for any
quantity is always unknown.
• However it is possible to evaluate a probable
magnitude of error in a measurement
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Basic Analytical
Chemistry
Types of errors
• Systematic (determinate) errors
• Random (indeterminate) errors
• Gross errors
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Systematic errors
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Chemistry
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Systematic errors show consistent deviation of
analytical results from the true value, i.e are of the
same magnitude for replicate measurement. These are
classified into;
Instrumental
errors Caused by
faults in measuring
instruments and
instability in power
supply.
Method errors
Arise from non ideal
chemical or physical
conditions of reagents
and reactions on
which the analytical
method is based upon.
Personal errors
Originating from the
individual analyst and
are not related to the
method or procedure;
result from carelessness,
personal limitations and
ignorance.
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Systematic errors
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Chemistry
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Systematic errors can either be constant or
proportional
Constant errors
▫ Magnitude is independent of the size of quantity
measured
▫ Become more serious as the size of the quantity
measured decreases.
▫ An effective way of minimizing constant error is to use
large sample sizes
Proportional errors
▫ Magnitude increases or decreases in proportion to the
size of the sample
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Correction of systematic errors
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Chemistry
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• Personalerrors
Instrumental errors
• Detected and corrected by
calibration
▫ Periodic calibration is important
because instruments
performance change over time
due to corrosion, misuse and
wear.
• Care and self discipline
• Careful select an analytical method
that you are competent in
• Get as much information or
knowledge as possible on the
method as well as reagents being
used.
Method errors
These can be detected by analyzing standard reference materialsmaterials containing exactly known concentration levels of one or more
analytes
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Random errors
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Chemistry
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• Arise from random fluctuations in measured
quantities, which always occur even under closely
controlled conditions.
• It is impossible to eliminate them entirely, but
they can be minimized by careful experimental
design and control.
• Experimental factors such as temperature,
pressure and humidity, and electrical properties
such as current, voltage and resistance are
susceptible to small continuous and random
variations.
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Gross errors
• Cause the result to be either too high or too low
i.e they lead to outlier.
• They are often caused by real blunders e.g
spilling of a solution required in a certain step or
accidentally skipping a step in a procedure.
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Reporting the experimental data
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Chemistry
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• A scientifically accepted manner of expressing a
quantity is;
Where q is the actual measurement and y is the
uncertainty or error associated with a particular
measurement.
• If y is not given, it is taken as ½ the unit of the last
significant figure. e.g if m = 1.2g, then m=1.2±0.05g
i.e the true value may lies between 1.15 to 1.25g.
• Similarly m = 1.00g, then m=1.00±0.005g and m
lies between 0.995 and 1.005g.
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Accuracy and Precision
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Accuracy
• Accuracy is the closeness of an
experimental measurement or result to the
true or accepted value.
• It is often more difficult to determine
because the true value is usually unknown.
• It is expressed in terms of either absolute
error or relative error.
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Absolute error
• Absolute error, E is the difference between
the measured quantity xi and the true value xt.
E= xi - xt
Example
If the true value of volume of a gas is 42.5cm3
and the measured values are 41.8cm3 and
43.7cm3, then their absolute errors will be -0.7
and +1.2 cm3 respectively.
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Relative error
Basic Analytical
Chemistry
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• Relative error is the absolute error divided by
the true value. It can be expressed in percent,
parts per thousand
or parts per million.
𝑥 −𝑥
𝐸𝑟 =
𝑖
𝑡
𝑥𝑡
× 100%
Thus, the relative errors for the gas volumes are;
41.8 − 42.5
𝐸𝑟 =
× 100% = −1.65%
42.5
43.7 − 42.5
𝐸𝑟 =
× 100% = 2.82%
42.5
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Precision
• Precision is the closeness or agreement
between replicated measurement or results
obtained under the same prescribed conditions.
• It is expressed by; standard deviation, variance
and coefficient of variance.
y
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Accuracy
• Measures the agreement
between the result and its
true value.
• Can never be exactly
determined since the true
value can not be known.
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Precision
• Describes the agreement
among several results which
have been measured by the
same procedure.
• Can be determined by
replicating a measurement.
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Measures of central tendency
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Chemistry
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• Mean, x is a quantity obtained by dividing the
sum of the measurements of data, xi by the
number of measurements, N.
1
𝑥=
𝑁
𝑁
𝑥𝑖
𝑖=1
• Median – middle value in ranked data.
• Mode – most frequent datum or value.
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Measures of spread/scatter
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a) Spread/range, w is the difference between the
largest value and the smallest value.
b) Deviation from the mean, d is the deviation of
the individual result xi from the mean.
𝑑 = 𝑥𝑖 − 𝑥
c) Variance, V is the sum of the deviations from
the mean, divided by N-1
𝑑2
𝑉=
=
𝑁−1
𝑥𝑖 − 𝑥
𝑁−1
2
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Measures of spread/scatter
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d) Standard deviation, s is the square root of
the variance.
𝑉=
𝑥𝑖 − 𝑥
𝑁−1
2
e) Relative standard deviation (RSD)
𝑠
𝑅𝑆𝐷 = (𝐼𝑡 𝑖𝑠 𝑜𝑓𝑡𝑒𝑛 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑎𝑠 %)
𝑥
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Significant figures in numerical
computation
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Chemistry
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• Significant figures in a number are all of
the certain digits plus the first uncertain digit.
• Care is to be taken to determine the appropriate
number of significant figures to be retained in a
result of computation of two or more numbers.
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Significant figures
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Chemistry
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• Addition and subtraction
▫ The result should contain the same number of
decimal places as the number with the smallest
number of decimal places.
Example:
3.4+0.020+7.31=10.730.
This is rounded to 10.7.
▫ When adding and subtracting numbers in scientific
notation, express the numbers to the same power of
10.
Example
7.234  10 6  1.215  10 4  7.234  10 6  0.01215  10 6  7.24615  10 6
This is rounded to 7.246  10 6
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Significant figures
Basic Analytical
Chemistry
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• Multiplication and division
It is suggested to round off the answer to the same
number of significant figures as the original
number with the smallest number of significant
figures.
• Logarithm and antilogarithm
The number of significant figures in the mantisa
is the same as the number of significant figures in
the original number.
Example;
log( 9.57  10 )  4.981
4
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Error propagation
• It is necessary to estimate the standard deviation
of a result that has been calculated from two or
more experimental data points, each of which
has a known sample standard deviation.
• Given; a±sa, b±sb, and c±sc ,the estimation of the
standard deviation of the value y is as shown in
the table in the next slide.
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Basic Analytical
Chemistry
Type of calculation
Example
Addition or subtraction
𝑦 =𝑎+𝑏+𝑐
Multiplication or division
Standard deviation of y
𝑎×𝑏
𝑦=
𝑐
Exponentiation
𝑦=𝑎
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𝑥
𝑠𝑎2 + 𝑠𝑏2 + 𝑠𝑐2
𝑠𝑦 =
𝑠𝑦
=
𝑦
𝑠𝑎
𝑎
2
𝑠𝑏
+
𝑏
𝑠𝑦
= 𝑥
𝑦
2
𝑠𝑐
+
𝑐
𝑠
𝑎
Logarithm
𝑦 = 𝑙𝑜𝑔10 𝑎
Antilogarithm
𝑦 = 𝑎𝑛𝑡𝑖 𝑙𝑜𝑔10 𝑎
𝑠𝑎
𝑠𝑦 = 0.434
𝑎
𝑠𝑦
= 2.303𝑠𝑎
𝑦
2
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Example
Consider the following
4.10( 0.02)  0.0050( 0.0001)
 0.010406(  ?)
1.97( 0.04)
compute the standard deviation of the result
where the numbers in parenthesis are absolute
standard deviation.
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• The standard deviation is calculated as follows;
2
2
2
 0.02 
 0.0001 
 0.04 
 






  0.0289
y
 4.10 
 0.0050 
 1.97 
sy
The calculation is completed by calculating the
absolute standard deviation
s y  y  (0.0289)  0.010406  (0.0289)  0.000301
We can write the answer with its uncertainty as
0.0104(±0.0003)
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Rounding results after error
propagation
• First,, round the error term (sy) to one
significant figure.
• Second, round the answer (y) to the same
number of decimal places as the error
term.
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Thank you for listening
CH 101;
BASIC ANALYTICAL
CHEMISTRY
EVALUATION OF ANALYTICAL
DATA
Introduction
• Experimental results seldom agree exactly
with those predicted from a theoretical
model.
• Scientists must judge whether a numerical
difference is a result of the random errors or
a result of systematic errors.
• The judgment is done by using the
significance tests such as Q-test, F-test and
t- test.
SIGNIFICANCE TESTS
• Significance testing involves a comparison
between a calculated experimental factor
and a tabulated factor determined by the
number of values in a set(s) of experimental
data or the degree of freedom and a selected
probability level.
OUTLIERS
• Outliers are measurements in a set of
replicate measurements or results that
appear to be considerably higher or lower
than the remainder and are suspected to be
outside the expected range.
Q-test
• It is devised to test suspected outliers in a
set of replicate results.
• It involves the calculation of a ratio, Qcalcu
(calculated value of Q), defined as the
absolute difference between the suspected
value and the value closest to it divided by
the spread of all values in the set.
𝑄𝑐𝑎𝑙𝑐𝑢
𝑠𝑢𝑠𝑝𝑒𝑐𝑡 𝑣𝑎𝑙𝑢𝑒 − 𝑛𝑒𝑎𝑟𝑒𝑠𝑡 𝑣𝑙𝑢𝑒
=
𝑟𝑎𝑛𝑔𝑒
Cont…
• The calculated value of Q (Qcalcu) for the
questionable datum or value is compared with a
critical value Qcritical from the table of critical
values. (see table 1).
• The result or datum is rejected if Qcalcu exceeds
Qcritical
Table 1:Critical values of Q at
95% confidence level
Sample size
Critical value
4
5
6
7
8
0.831
0.717
0.621
0.570
0.524
Example
Four replicate values were obtained for the
determination of pesticide in river water,
0.403, 0.410, 0.410, 0.380 µgdm-3.
▫ Inspect the obtained results;
▫ Is there a possible outlier? Identify it.
▫ Should it be rejected or retained?
to answer this you need to calculate the
value of Q
Cont…
0.380 µgdm-3 is a suspect value, the value of Q is
calculated as follows;
𝑄𝑐𝑎𝑙𝑐𝑢
0.380 − 0.403
=
= 0.767
0.410 − 0.380
Qcritical = 0.831 for four value at the 95%
probability level.
As Qcalcu is less than Qcritical, 0.380 µgdm-3
is not an outlier at 95% level and it should
be retained.
F-test
• This is used to compare the precision of;
▫ two sets of data which may originate from two
analysts in the same laboratory.
▫ two different methods of analysis for the same
analyte.
▫ results from two different laboratories.
• Statistical F is defined as the ratio of either the
population variance,
𝐹𝑐𝑎𝑙𝑐𝑢 =
𝜎12
𝜎22
• Or the sample variance of the two sets of data
𝐹𝑐𝑎𝑙𝑐𝑢 =
𝑠12
𝑠22
• The large variance is always taken as
numerator.
• When Fcalcu exceeds the critical value (see table 2),
there is a significant difference between the two
variances and hence between the precisions of the
two sets of data
Table 2:critical values of F at 95%
confidence level
Degrees of Degrees of freedom of the numerator
freedom of
the
denominator
5
6
7
5
7.146
5.285
4.484
6
6.853
4.995
4.197
7
6.681
4.823
4.026
Example
• A proposed new method for the determination of
sulphate in an industrial waste effluent ios compared
wih an existing method, giving the results in the table
below. Is there a significanmt difference between the
the precisions of the two methods?
Method
Mean
No. of
No. of
s (mgdm-3)
(gdm-3) replicate degrees of
freedom
Existing
72
8
7
3.38
New
72
8
7
1.06
Fcalcu 
2
sexisting
2
snew
(3.38) 2

 5.08
2
(1.50)
• The tabular value for F with 7 degrees of
freedom for both the numerator and the
denominator is Fcritical = 4.026 at 95%
probability level
• As Fcalcu is greater than Fcritical, the two
methods are giving significantly different
precisions
t-test
It
is used to compare;
• experimental means of two sets of data.
• the mean of one set of data with a known
reference value.
Comparison of two means
• When two experimental means are to be compared, the
value of t is obtained by the following equation, with
(M+N-2) degree of freedom
𝑡𝑐𝑎𝑙𝑐𝑢
𝑥−𝑦
=
𝑠𝑝𝑜𝑜𝑙𝑒𝑑
𝑀𝑁
𝑀+𝑁−2
1
2
• Where x is the mean of M determinations, y the mean of N
determinations and spooled the pooled standard deviation.
Spooled, pooled standard deviation
• The pooled standard deviation (spooled) for sets A and B is
given by the following equation;
𝑠𝑝𝑜𝑜𝑙𝑒𝑑 =
𝑁 − 1 𝑠𝐴2 + 𝑀 − 1 𝑠𝐵2
𝑀+𝑁−2
• Where sA and sB are the the standard deviations for sets A
and B respectively.
Comparison of the mean with known
• When the set of data is compared to an accepted value, t is
value
computed from the following equation;
𝑡𝑐𝑎𝑙𝑐𝑢
𝑥−𝜇 1 2
=
𝑁
𝑠
Where x is the mean of the experimental set of data, µ is
the accepted mean, s is the experimental standard
deviation and N the number of results
• When tcalcu exceeds the critical value (see table 3)
for an appropriate number of degree of freedom
the difference between the means is said to be
significant
• When texp exceeds the critical value (see table 3)
for an appropriate number of degree of freedom
the difference between the means is said to be
significant
Table 3: Critical values of t at 95%
confidence level
Number of
degrees of
freedom
2
t values
5
2.57
10
2.23
18
2.10
4.03
•Example
A method of determination of mercury by
atomic absorption spectrometry gave values
of 400, 385 and 382 ppm for a standard
known to contain 400 ppm. Does the mean
value differ significantly from the true value
or is there any evidence of systematic
error(bias)?
Solution
• Mean =389 ppm
• s = 9.64ppm
• µ = 400 ppm
 x    12 (389  400) 12
tcalcu  
N


3


1
.
98

s
9.=1.98
64

• Ignore the negative sign, tcalcu
• For 2 degrees of freedom tcritical is 4.30 at 95%
probability level. As tcalcu is less than tcritical, the mean
is not significantly different from the true value.
Therefore there is no evidence of a systematic error
or bias
Some practical problems with relevant
statistical tests
Practical problems
Relevant
tests
One result in a replicate set differs
Examine for
widely from the rest. Is it a significant gross error.
result
Apply Q-test
Two operators analysing the same
Examine the
sample by the same method obtain data for
results with different spreads. Is
unreliable
there a significant difference in the
results. Apply
precisions between the results?
F-test
Practical problems
Relevant tests
A new method of analysis is Examine the data for
being tested by the analysis unreliable results.
of a standard sample with an Apply t-test
accurately known
composition. Is the
difference between the
experimental value and the
accepted value significant?
SAMPLING AND SAMPLE
PREPARATION
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INTRODUCTION
• Sampling is a process of obtaining a small
amount of a material that accurately
represent the bulk of the material being
analysed.
• Or sampling is the selection of some parts
of an aggregate or totality on the bases of
which judgment or inference about the
aggregate is made
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• Both sample collection and sample
preparation play an important role in
chemical analysis.
• If the samples collected are wrong, even the
best method of laboratory analysis will not
be able to correctly answer the questions
inspiring the analysis.
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SAMPLE COLLECTION
• Choosing appropriate samples can be difficult
especially where the material (population) is
heterogeneous.
• For example, consider a segregated or stratified
sample population. Should the whole sample be
taken or separate samples of each layer?
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Choosing appropriate samples can be difficult especially where the materia
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is heterogeneous. For example, consider a segregated or stratified sample p
Should the whole sample be taken or separate samples of each layer?
steps
involves the following
Sampling involves
• Sampling
the following
steps
Identify
population
Collect a gross
or bulk sample
Reduce the
gross sample
to a laboratory
sample
A gross sample is a miniature replica of the entire mass of the material to b
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• A gross sample is a miniature replica of the
entire mass of the material to be analysed. It
corresponds to the bulk material in chemical
composition and in particle size distribution if
composed of particles.
• A laboratory sample is a reduced gross
sample which is homogeneous.
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SAMPLE STORAGE
• Often samples are collected from places remote
from the analytical laboratory, therefore several
days or weeks elapse before they are received
and analysed.
• Moreover the workload of many laboratories is
such that the incoming samples are stored for a
period of time prior to analysis.
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• For this case the sample containers as well as
the storage conditions (e.g temperature,
humidity, and level of light) must be
controlled such that no significant changes
occur that could affect the validity of the
analytical data.
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• It should be ensured that the containers do not
allow contamination of samples. For instance
sodium potassium, boron, and silicates can be
leached from glass into sample solutions.
Therefore plastic containers should be used for
such samples.
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• Sample solution containing organic solvents
and other organic liquids should be stored in
glass containers because the base plastic may
be leached from the walls of plastic containers.
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SAMPLE PREPARATION
• Sample preparation refers to the ways in which a
sample is treated prior to its analysis.
Preparation is a very important step in most
analytical techniques, because the techniques
are often not responsive to the analyte in its insitu form, or the results are distorted by
interfering species.
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• Sample preparation may involve dissolution,
reaction with some chemical species, masking
filtering, dilution , sub-sampling or many
other techniques.
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General Chemistry:
An Integrated Approach
Hill, Petrucci, 4th Edition
Chapter 3
Stoichiometry:
Chemical Calculations
Mark P. Heitz
State University of New York at Brockport
© 2005, Prentice Hall, Inc.
79
Chapter 3: Stoichiometry
Molecular Mass
Molecular mass is the sum of the masses of the
atoms represented in a molecular formula.
1 Oxygen atom
Example:
water - H2O
2(1.0079 u) + 15.9994 u
= 18.0152 u
2 Hydrogen atoms
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Molecular Mass
Chapter 3: Stoichiometry
Sulfur dioxide - SO2
= 32.066 u + 2(15.9994
= 64.065 u
Glucose - C6H12O6
= 6(12.0107 u) + 12(1.0079 u) + 6(15.9994 u)
= 180.1154 u
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Chapter 3: Stoichiometry
Formula Mass
Formula mass is the sum of the masses of the
atoms or ions present in a formula unit.
ClNa+
Na+
Cl-
Na+
Cl-
ClNa+
Crystal of
sodium chloride
One Na+ and one Cl– make a
formula unit for sodium
chloride
The mass of one formula unit
is:
= 22.9898 u + 35.4527 u
= 58.4425 u
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Chapter 3: Stoichiometry
Avogadro’s Number, NA
The number of elementary entities in one mole of
anything
NA = 6.02214199 x 1023 entities mol–1
-or- 1 mole  6.022 x 1023 entities
A mole (mol) is an amount of substance that
contains as many elementary entities as there are
atoms in exactly 12 g of the carbon-12 isotope.
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Chapter 3: Stoichiometry
Examples of NA
1 mole of stars in the universe
1 mole of pennies
1 mole of tennis balls
1 mole of glucose molecules
1 mole of helium atoms
1 mole of potassium ions (K+)
= 6.022 x 1023 stars
= 6.022 x 1023 pennies
(beats the lottery!)
= 6.022 x 1023 tennis balls
= 6.022 x 1023 molecules
= 6.022 x 1023 atoms
= 6.022 x 1023 ions
Clearly, NA is most practical for counting microscopic entities.
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Chapter 3: Stoichiometry
NA and Reactions
Example: consider the formation of carbon dioxide
At the molecular level ...
Problem: how does one mass out a single carbon
atom? Note that the mass in grams is ~2.00 x
–23 g!
10
Answer: one doesn’t!
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Chapter 3: Stoichiometry
The Solution ...
Use a measurable amount – molar quantities
For carbon, mass out:
2.0 × 10–23 g atom–1 × 6.0 × 1023 atoms mol–1
= 12 g C
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Making CO2
Chapter 3: Stoichiometry
same number
different entities
macroscopic
amounts
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Chapter 3: Stoichiometry
Molar Mass
The molar mass of a substance is the mass of
one mole of that substance.
numerically equal to the atomic mass,
molecular mass, or formula mass.
The units of molar mass are grams per mole (g
mol–1) Examples:
1 mol Na = 22.99 g mol–1
1 mol CO2 = 44.01 g mol–1
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Chapter 3: Stoichiometry
Calculations and Molar Mass
Molar mass is an equality that can be
expressed in fractional form
grams of substance
1mol
or
1mol
grams of substance
These are used as conversion factors
when problem solving …
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Chapter 3: Stoichiometry
Sample Calculation
Calculate the mass, in grams, of
0.250 moles of ammonia (NH3)
Molar mass = 14.00 g mol–1 + 3(1.008 g
mol–1)
= 17.024 g mol–1
17.024 g NH 3
0.250 mol NH 3 x
1 mol NH 3
= 4.256 g NH3
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Chapter 3: Stoichiometry
Percent Composition
A simple example:
calculate the percent
composition, by color, of the
collection of blocks below
Fraction of colors
4 orange blocks

6 blocks
 0.67 or 67%
2 green blocks

6 blocks
 0.33or 33%
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Chapter 3: Stoichiometry
Mass Percent Composition
The ratio of the total mass of a given element
in a compound to the mass of the compound
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Chapter 3: Stoichiometry
Mass Percent Composition
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Chapter 3: Stoichiometry
Molecular and Empirical Formulas
Empirical formula: the simplest whole
number ratio of elements in a compound
Example:
Molecular formula of glucose – C6H12O6
The elemental ratio C:H:O is 1:2:1, so the
empirical formula is CH2O.
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Chapter 3: Stoichiometry
Determining
Empirical
Formulas
A compound is comprised of 40.01% carbon, 6.72% hydrogen, and
53.27% oxygen. Calculate the empirical formula of the compound.
1) Given the percent composition, assume a mass of sample
100.00 g for convenience
6.72%
H
40.01%
C
- use
6.72 g
H
53.27%
O
40.01 g
C
53.27 g
O
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Chapter 3: Stoichiometry
Determining Empirical Formulas
2) Convert masses to mole
mol C  40.01 g C x
1 mol C
12.01 g C
 3.331 mol C
mol H  6.72 g H x
1 mol H
1.008 g H
 6.667 mol H
1 mol O
mol O  53.27 g O x
15.99 g O
 3.331 mol O
3) Divide by the least number of
moles
3.331
C
1
3.331
6.667
H
2
3.331
3.331
O
1
3.331
= CH2O
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Chapter 3: Stoichiometry
Stoichiometry of Chemical Reactions
For any chemical conversion of reactants,
Reactants

Products
stoichiometry expresses the mathematical connections between all
reactants and products involved in the reaction
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Chapter 3: Stoichiometry
Writing Chemical Equations
•
•
•
•
•
A chemical equation is a shorthand description of a chemical
reaction, using symbols and formulas to represent the elements and
compounds involved.
Starting substances are called reactants.
The substances formed from the reaction are called products.
A plus sign (+) is used between each reactant and between each
product.
An arrow () points from the reactants to the products.
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Chapter 3: Stoichiometry
Equation Notation
• Extra notation added to each substance:
–
–
–
–
(g) = gas
(l) = liquid
(s) = solid
(aq) = aqueous (water) solution
• Extra notation added above the arrow:
–  = heat
– sometimes the actual temperature is used instead
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Chapter 3: Stoichiometry
Chemical Equations
Example: consider the formation of water
H2(g) + O2(g)  H2O(g)
Law of Conservation of Mass must be obeyed …
therefore, equations must be balanced.
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Chapter 3: Stoichiometry
Balancing Equations
Chemical “bookkeeping” of atoms involved in the reaction
H2(g) + O2(g) 
H–2 O–2
Reactants
H2O(g)
H–2 O–1
Products
Note the imbalance in oxygen atoms
Whole units of reactants or products must be added to equate
numbers of reactant atoms with numbers of product atoms
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Chapter 3: Stoichiometry
Balancing Equations
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Chapter 3: Stoichiometry
Stoichiometric Equivalents
Coefficients from a balanced chemical equation
show molar equivalents of reactants and
products
In the formation
of water:
==> form
conversion
2 mol H2 = 1 mol O2
2 mol H2 = 2 mol H2O
1 mol O2 = 1 mol H2O
factors
2 mol H 2
1 mol O2
or
1 mol O2
2 mol H 2
2 mol H 2
2 mol H 2O
or
2 mol H 2O
2 mol H 2
1 mol O2
2 mol H 2O
or
2 mol H 2O
1 mol O2
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Chapter 3: Stoichiometry
Limiting Reagents
Chemical reactant that is completely consumed in a reaction and
therefore limits the quantity of product formed.
**Depends on stoichiometry of reaction
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Chapter 3: Stoichiometry
Limiting Reagents (cont.)
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Chapter 3: Stoichiometry
Limiting Reagents (cont.)
How many meals can be made from 105 sandwiches, 202 cookies, and
107 oranges?
105 sandwiches x
1 meal
 105 meals
1 sandwich
202 cookies x
1 meal
 101 meals
2 cookies
107 oranges x
1 meal
 107 meals
1orange
Cookies limit the total
number of whole meals
with excess
sandwiches and
oranges
VideoClip
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Reaction Yields
Chapter 3: Stoichiometry
Theoretical yield – predicted amount of
product formed from the limiting reagent,
based only on the stoichiometry of the
reaction
2H2(g) + O2(g)  2 H2O(g)
If all worked
perfectly ...
Example: 1 mol H2 will
produce 1 mol of water
Actual yield – amount of product produced
In practice, actual < theoretical: errors, poor technique, etc. ...
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Chapter 3: Stoichiometry
Reaction Yields (cont.)
Reaction yields are expressed as a ratio in the form of a percentage
actual
percent yield 
x 100
theoretical
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Chapter 3: Stoichiometry
Solution
Components
Solute – the substance that is dissolved;
usually present in lesser quantities than the solvent
Solvent - the substance that does the dissolving
(examples: water, hexane, methanol)
Making Solutions
NaCl
NaCl
NaCl
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Chapter 3: Stoichiometry
Concentrations
• Substances enter into chemical reactions
according to certain molar ratios.
• Volumes of solutions are more convenient to
measure than their masses.
Molar concentration expresses the
amount of solute in one liter of
molarity ( M ) 
solution
mol solute
L of solution
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Chapter 3: Stoichiometry
Examples of Concentration
Dilute solution
NaCl
Concentrated solution
NaCl
NaCl
NaCl
NaCl
NaCl
e.g., 0.1 M NaCl solution
e.g., a 5 M NaCl solution
(0.05 mol NaCl dissolved in 500 mL of
water)
(5 mol NaCl dissolved in 1.0 L
of water)
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Chapter 3: Stoichiometry
Dilutions
• Dilution is the process of preparing a more
dilute solution by adding solvent to a more
concentrated one.
• Addition of solvent does not change the
amount of solute in a solution but does
change the solution concentration.
Dilution Illustration
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Chapter 3: Stoichiometry
Same number of
solute molecules
after dilution
Added solvent
volume, therefore
less concentrated
Concentrated
solution
Dilute
solution
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Chapter 3: Stoichiometry
Summary of Concepts
•
•
•
•
•
Molecular and formula masses relate to the masses of molecules and
formula units.
A mole is an amount of a substance containing Avogadro’s number of
elementary entities.
Avogadro’s number (NA) = 6.022 × 1023.
The molar mass of a substance is the mass in grams of one mole of that
substance.
Formulas and molar masses can be used to calculate mass percent
composition.
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Chapter 3: Stoichiometry
Summary (cont.)
•
•
•
•
A chemical equation uses symbols and formulas for the elements
and/or compounds involved in a reaction.
Calculations involving reactions use stoichiometric factors based on
stoichiometric coefficients in the balanced equation.
The limiting reactant determines the amounts of products in a reaction.
The molarity of a solution is the number of moles of a solute per liter of
solution.
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CONCENTRATIONS OF SOLUTIONS
CONCENTRATIONS OF SOLUTIONS
An intimate homogeneous mixture of two or more substances
Components of Solutions
• Solvent- a component which its physical state is retained (or substance
present in a greater amount)
• Solute – Substances being dissolved (with lesser amount).
Solution expression.
The concentration of the Solution is given by a measure of the amount of
solute in specific quantity of solvent/solution.
• Mass percent, Volume percent and mass/volume percent
– Percent by mass (%w/w) is widely used in industrial chemical
applications.
CONCENTRATIONS OF SOLUTIONS
• Calculate the mass of nickel(II) sulfate, NiSO4 contained in
200. g of a 6.00% solution of NiSO4 .
Solution
• The percentage information tells us that the solution
contains 6.00 grams of NiSO4 per 100. grams of solution.
The desired information is the mass of NiSO4 in 200.
grams of solution.
• A unit factor is constructed by placing 6.00 grams NiSO4
over 100. grams of solution.
• Multiplication of the mass of the solution, 200. grams, by
this unit factor gives the mass of NiSO4 in the solution.
CONCENTRATIONS OF SOLUTIONS
Also the percentage concentration can be expressed as
• percent by volume (%v/v)
• Mass/Volume percent (%m/v). This is commonly
used in medicines and Pharmacy
CONCENTRATIONS OF SOLUTIONS; MOLARITY
• Molarity: the ratio between the moles of dissolved
substance (solute) and the volume of the solution (in
liters or cubic decimeters)
Ammount of a Solute in moles 
Molarity 
Volume of Solution in litres 
Example: If 0.455mol. of Urea (CO(NH2)2 is dissolved
in enough water to make exactly 1 L of solution, what is
the molarity of that solution?
No. of Moles of Urea
0.455 mol.
Molarity 

 0.455 M .Urea.
Volume of Solutionof urea 1.00 L of a So ln .
The SI unit for Molarity, M is mol. Per Litre.
CONCENTRATIONS OF SOLUTIONS; MOLARITY Contd.
Sample Problem:
1 Calculate the molarity of the following
• 145 g (NH4)2C4H4O6 in 500 mL of solution
• 13.2 g MnSeO4 in 500 mL of solution
• 45.1 g cobalt (II) sulfate in 250 mL of solution
• 41.3 g iron (II) nitrate in 100 mL solution
2 Consider the following reaction
CaCO3(s) + 2HCl(aq) → Ca2+(aq) + 2Cl1-(aq) + H2O(l) + CO2(g).
▫ Calculate the molarity of C a2+ and Cl1- if 10 dm3 of CO2 was produced at STP,.
▫ What is the mass of CaCO3 required to produce 10 Litres of CO2 gas at STP in
excess amount of Acid? What will be minimum amount (Moles) of HCl
required?
▫ What will be the volume of CO2 produced if
• the concentration of Ca2+ is 0.15 mol./litre?
• The concentration of Cl1- is 15mol./litre?
CONCENTRATIONS OF SOLUTIONS; Mole Fraction.
• Mole Fraction (xi)/Mole percent
ammount of component i in moles
xi 
Total ammount of all component in moles
CONCENTRATIONS OF SOLUTIONS; Molality
Molality (m)
• Used when the experiment requires high precision.
• If the temperature of a solution is changed, the amount of a
solute in moles/weight remains constant.
• However the volume of the solution changes.
• Therefore there is a slight change of molarity of the
solution.
Amount of solute in moles 
molality (m) 
mass of Solvent in kg 
CONCENTRATIONS OF SOLUTIONS; Normality
• Normality is the number of gram equivalents of a substance
dissolved in one litre of a solution.
Number of gram  equivalents
Normality 
Number of litres
CONCENTRATIONS OF SOLUTIONS
• Concentration of very dilute solutions.
• This is expressed in ppm, ppb, ppt etc.
CONCENTRATIONS OF SOLUTIONS
• Examples 1. A typical shot of hard liquor drink is 40% by
mass of C2H5OH. Assuming the rest is water, how much
alcohol would we consume if the drink has a mass of 30g?
Calculate the molality, molarity and the mole fraction of the
solution.
Solution
• % by mass of C2H3OH = 40%, mass of drink = 30g
% compositionof etOH  Mass of Solution
Mass of EtOH 
100%
40%

 30 g of So ln  12 g etOH
100%
CONCENTRATIONS OF SOLUTIONS
• Molality of ethanol
• First of all calculate the mass of the solvent in kg = total
mass – mass of Ethanol
• 30g of solution – 12g of ethanol = 18g = 0.018kg.
• Then calculate the number of moles of Ethanol nethanol
nethanol
mass of dissolved ethanol 12 g


0.26mol.
molar mass of ethanol
46.1g
Mole Fraction of Ethanol From above massof water = 18g = 1
mole of H2O.
ammount of component i in moles
0.26mol EtOH
xi 

 0.21
Total ammount of all component in moles 0.26mol EtOH  1molH 2 O
CONCENTRATIONS OF SOLUTIONS
• Example 2. A 2.00 M aqueous NaCl solution has a density of
1.08 g/mL. What is the molality of the solution?
Solution
• First of all calculate the mass of the solution in using the
density given
1.08 g
Mass  desity Volume 
 1000mL 1080 g
1mL
Then calculate the
mass of NaCl.
Mass of solute  Molarity ( M )  Volume of solution  Molar mass of solute
 2M  1L  58.449g/mo le  117 gNaCl.
Then calculate the mass of solvent in kg
Mass of solvent =total mass – mass of a solute = 1080g117g = 963g = 0.963kg of H2O
CONCENTRATIONS OF SOLUTIONS
• Finally calculate molality
nsolute
2mol NaCl
m

 2.08 molal NaCl
msolvent in kg 0.963kg H 2 O
CONCENTRATIONS OF SOLUTIONS; dilution
• Dilution is the process of preparing a more dilute
solution by adding solvent to a more concentrated one.
• Is the process of reducing the concentration of a solute
in solution, usually simply by adding more solvent.
• Addition of solvent does not change the amount of
solute in a solution but does change the solution
concentration.
• the volume times molarity (ViCi) before dilution is equal
to the volume times molarity (VfCf) after dilution.
• This gives the following general equation; ViCi= VfCf.
• The concentrations can be in molarities, molalities,
percentages etc.
• We must use the same concentration units thus the
general equation becomes ViCi= VfCf.
Same number of
solute molecules after
dilution
Added solvent
volume, therefore
less concentrated
Concentrated
solution
Dilute
solution
CONCENTRATIONS OF SOLUTIONS; dilution
• Example: How many milliliters of 18.0 M H2SO4 are
required to prepare 1.00 L of a 0.900 M solution of
H2SO4?
Solution
• Vf =1.00 L, Cf=0.900 M and Ci=18.0M
• Therefore, the relationship ViCi=VfCf can be used,with
subscript i for the initial acid solution and subscript f for
the dilute solution.
• We solve ViCi=VfCf for Vi
Vi 
Vf C f
Ci
1.00 L  0.900M

 0.050 L  50.0mL
18.0M
USING SOLUTIONS IN CHEMICAL REACTIONS
• If we plan to carry out a reaction in a solution, we must calculate the
amounts of solutions that we need.
• If we know the molarity of a solution, we can calculate the amount of
solute contained in a specified volume of that solution.
• Example: Calculate the volume in liters and in milliliters of a 0.324
M solution of sulfuric acid required to react completely with 2.792
grams of Na2CO3 according to the equation,
H2SO4+Na2CO3→Na2SO4+CO2+H2O
• Solution: From the balanced equation the mole ratio for all species
is one to one
• We convert g Na2CO3 to mol Na2CO3 to mol H2SO4 to L H2SO4 soln
? L . H 2 SO4  2.792 g . Na2CO3 
1.00mol.Na2CO3 1.00mol.H 2 SO4
1.00 L.H 2 SO4


106.0 g .Na2CO3 1.00mol.Na2CO3 0.324 mol.H 2 SO4
 0.0818 L.H 2 SO4 So ln  81.3mL.H 2 SO4 so ln
134
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Gravimetric methods
135
Gravimetry or gravimetric methods
03:08
• Quantitative methods that involve measurement of
mass of a pure compound to which the analyte is
chemically related
Types:
• volatilization methods
• precipitation methods
• electro-analytical methods
• extraction and chromatographic methods
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03:08
•1.1
Involves
collecting andmethod
weighing of a volatilized
Volatilization
product OR Determining mass of product from loss
in mass of a sample, following volatilization of an
analyte or its decomposition product.
• e.g. A sample containing a bicarbonate may be
treated with sulphuric acid to evolve CO2 ie
NaHCO3 (aq) + H2SO4 (aq)  CO2(g) +
H2O(l)+NaHSO4(aq)
• The liberated CO2 be dried to remove H2O and
finally retained in weighed tube that contains an
adsorbent [NaOH (to trap CO2)].
Volatization method
137
03:08
• The difference in mass of the tube before and after
adsorption can be used to calculate the amount of
carbonate in the sample.
138
1.2 Precipitation methods
03:08
• The analyte is converted to a sparingly soluble
precipitate filtered, washed free of impurities and
converted to a product of known composition by
suitable heat treatment.
• The product is finally weighed to determine its mass:
e.g. A sample known to contain Cl¯ as analyte is
dissolved in water and the Cl¯ is converted to a
precipitate of AgCl by adding excess AgNO3.
139
03:08
• Ag+ (aq) + Cl(aq)  AgCl(s)
• The precipitate is washed with water in glass
crucible, and dried in an oven. It is cooled and
weighed to obtain the weight of AgCl.
• Thus chloride content in a sample can be
established.
140
Gravimetry or gravimetric methods
03:08
1.2.1
Advantages of gravimetric analysis
• Accurate and precise when using modern analytical
balances
• Determination can be carried out using relatively
inexpensive apparatus (except for muffle furnace and
platinum crucibles)
• Possible sources of errors are readily checked. Eg.
Completeness of precipitation can be readily tested.
1.3 Gravimetric Calculations
141
03:08
• Results of gravimetric analysis are computed from mass of
product of known composition and mass of sample
mass of x  100%
% of x 
mass of sample
• Example: 0.3516 g of sample of commercial phosphate
detergent was ignited to destroy all the organic matter. The
residue was then dissolved in hot HCl which converted the P to
H3PO4. The phosphate was precipitated as MgNH4PO4.6H2O by
addition of Mg2+ followed by ammonia. After being filtered and
washed, the precipitate was converted to Mg2P2O7 (222.57
g/mol) by ignition at 1000oC. This residue weighed 0.2161 g.
Calculate the percentage P (30.974 g/mol.) in the sample.
142
1.3 Gravimetric Calculation
03:08
• Solution
mass of P
%P 
100%;
mass of sample
but mass of
P  0.2161gMg 2 P2O7 
2mol.P  30.974 gP / mol.
1mol.Mg 2 P2O7  222.57 gMg 2 P2O7 / mol
Therefore;
0.2161gMg 2 P2 O7 
%P 
2mol.P  30.974 gP / mol.
1mol.Mg 2 P2 O7  222.57 gMg 2 P2 O7 / mol
 100%  17.11%
0.3516 g sample
143
Example 2
Magnetite is a mineral with the formula Fe2O4. A
1.1324g sample of magnetite ore was dissolved in
concentrated HCl to give a solution that contained a
mixture of Fe3+ and Fe2+. Nitric acid was added and
solution boiled for a few minutes to convert all Fe2+ to
Fe3+. The Fe3+ was then precipitated as Fe2O3 x H2O
by the addition of NH3. After filtration, the residue
was ignited at high temperature to give 0.5394g of
pure Fe2O3 (159.69g/mol). Calculate (a) percent Fe
(55.847 g/mol), (b) percent Fe3O4 (231.54 f/mol).
03:08
144
1.3 Gravimetric Calculation
03:08
Solution,
(a)
% Fe 
mass of Fe
100%;
mass of sample
1molFe2O3
2molFe
55.847 gFe
but mass of Fe  gFe2O3 


159.69 gFe2O3 1molFe2O3
1molFe
2mol.Fe  55.847 gFe
 0.5394 gFe2O3 
159.69 gFe2O3
2mol.Fe  55.847 gFe
0.5394 gFe2O3 
159.69 gFe2O3
Percent Fe 
 100%  33.32%
1.1324 g magnetite ore
145
1.3 Gravimetric Calculation
03:08
(b) In this calculation, we assume that 3moles of Fe2O3 are
formed from 2 moles of Fe3O4.
i.e. 2Fe3O4+0.5O2→ 3Fe2O3., Then;
1molFe2O3
2molFe3O4 231.54 gFe3O4
mass of Fe3O4  gFe2O3 


159.69 gFe2O3 3molFe2O3
1molFe3O4
2mol.Fe3O4  55.847 gFe3O4
 0.5394 gFe2O3 
3molFe2O3 159.69 gFe2O3
Therefore;
2mol.Fe3 O4  55.847 gFe3O4
3molFe2 O3  159.69 gFe2 O3
 100%  46.04%
1.1324 gmagnetite ore
0.5394 gFe2 O3 
Percent Fe3 O4 
2.0.Precipitates and precipitating reagents
146
03:08
• An ideal precipitating agent used in gravimetric analysis
should react either selectively or specifically with the analyte
of interest.
• For example, dimethylglyoxime is a specific reagent for Ni2+
under alkaline conditions; AgNO3 is a selective reagent which
forms precipitates with Cl¯, l¯ Br¯, SCN¯.
• In addition the product formed must have the following
properties:
(i) Readily filtered and washed free of contaminants
(ii) Sufficiently low solubility to minimize loses during
filtration and washing
(iii) Inert to constituents of the atmosphere
(iv) Known composition after filtration or ignition.
2.0 Precipitates and precipitating
reagents
147
03:08
2.1 Particle size and filterability of precipitates
• For gravimetric analysis, large particles are more desirable
than the finely divided precipitates because they are easy to
filter and wash free of contaminations.
• Particles formed in liquids vary enormously in size from
colloidal suspensions to crystalline suspension.
• Colloid is a solid consisting of particles with diameter less
than 10-4 cm.
148
• Thus colloidal suspension consists of tiny
particles which are invisible to human eye.
• Such solutions are difficult to filter and particles
remain permanently suspended in solution.
• However, colloidal particles can be made to stick
together to give filterable mass.
03:08
149
2.0 Precipitates and precipitating reagents
03:08
• Crystalline suspension is made up of particles which are
at least 0.1 mm;
• Crystalline suspension tend to settle in the solution and
can be readily filtered.
• Particles size (Crystalline or colloidal) is influenced by
experimental variables: temperature, precipitate solubility,
reactant reactions and rate at which reactants are mixed.
• The net effect of these variables can be quantitatively
accounted by a single property of a system known as
relative supersaturation.
150
• supersaturated solution is an unstable solution which
contains more solute than it can hold at a particular
temperature.
• Eventually, supersaturation is relieved by precipitation
of the excess solute.
• According to Von Weiman, supersaturation plays an
important part in determining the particle size of the
precipitate.
03:08
QS
Von
Weiman
Equation
Re lative sup ersaturation 
• Where Q = concentration ofSspecies at any time; S =
equilibrium solubility.
2.0 Precipitates and precipitating
reagents
151
03:08
• Q-S will denote the super saturation at the moment
precipitation commences.
• Note: The expression applies only when Q is larger than S.
• When (Q-S)/S is large, a colloidal precipitate forms,
otherwise crystalline precipitant is Formed.
• Many salts with low solubility, S, eg. Oxides of Fe3+, Cr3+
form colloidal precipitates.
2.0 Precipitates and precipitating
reagents
152
03:08
2.1.1Mechanism of precipitation
• Precipitation occurs by two parallel mechanisms:
nucleation and particle growth.
• The size of the particles formed is determined by a faster
mechanism:
• If nucleation predominates, colloidal precipitates is
formed; if particle growth is faster, then crystalline
precipitate is formed
153
03:08
• Nucleation is a process where a minimum number
of atoms, ions or molecules come together to form
a stable solid.
• Nuclei may form on the surface of the solid
contaminant e.g. dust particles.
• Further precipitation is a competition between
additional nucleation and growth on existing
nuclei.
2.0 Precipitates and precipitating reagents
2.1.2
154
03:08
Experimental control of particles size
• Crystalline ppt formation depends on any
experimental
variables
which
minimize
supersaturation.
• They include;
(i)Elevated temperature (increase S of pt).
(ii)Dilute solution (Q is kept low),
(iii)Slow addition of precipitating reagent with good
stirring (minimize Q).
(iv)Control of pH can be used to obtain large particles
provided that solubility of the precipitate depends
upon pH.
2.0 Precipitates and precipitating reagents
155
03:08
2.2 Colloidal precipitates
• Colloidal precipitates cannot be analyzed effectively by
gravimetry, because the particles are difficult to filter.
• Heating, stirring, and/or adding electrolyte binds the
individual colloidal particles together to give
amorphous mass that settles out of the solution and is
filterable.
• A process of converting colloidal suspension into
filterable mass is known as coagulation or
agglomeration.
2.0 Precipitates and precipitating reagents
156
03:08
2.2.1 Coagulation of Colloids
• Colloidal suspensions are stable because they are made up of
similarly charged particles
• The charge originates from cations or anions which are bound on
the surface of the particles.
• This process is known as adsorption, a process by which ions are
retained on the surface of a solid.
• For Example, when Sodium chloride is first added to a solution
containing silver Nitrate, the colloidal particle of silver chloride
formed are positively charged as shown in the fig below.
• This charge is due to adsorption of some of the excess silver ion in
the medium.
• The charge on the particle becomes negative, however when
enough sodium chloride has been added to provide excess of
Chloride ion.
157
03:08
2.0 Precipitates and precipitating reagents
158
03:08
Coagulation may be brought about by:
(i)Short periods of heating accompanied by stirring;
since it decreases the number of adsorbed particles,
making it possible for particles to approach one
another. In addition, heating has an effect of
increasing the kinetic energy to overcome barrier
which keeps the particles apart.
159
03:08
(i) Increasing electrolyte concentration of a
solution; adding the right electrolyte increases
the concentration of counter ions around the
particle. Consequently, the volume of solution
containing the counter ions decreases.
2.0 Precipitates and precipitating reagents
160
03:08
2.2.2
Peptization
• Process by which a coagulated colloidal reverts to
its original dispersed state.
• Washing coagulated colloid results in leaching of
the electrolyte responsible for coagulation from
the internal liquid in contact with the solid
particles.
161
03:08
• Removing the electrolyte responsible for
coagulation increases the volume of layer
surrounding the particle.
• The repulsive forces which were responsible
for the original colloidal state is reestablished and particles detach themselves
from the coagulated mass.
2.0 Precipitates and precipitating reagents
162
03:08
2.2.3
Treatment of colloidal ppt
• Colloids are best pppted from hot, stirred solution
containing enough electrolytes for coagulation.
• Colloidal ppt can be more easily filtered if digested
(process by which a ppt is heated for at least 1 h in
mother liquor).
2.0 Precipitates and precipitating reagents
163
03:08
2.3 Crystalline particles
• Crystalline particles are generally more easily filtered
and purified than are coagulated colloidal.
• Also the particles size and their filterability can be
improved.
2.0 Precipitates and precipitating reagents
164
03:08
2.4 Co precipitation
• Phenomenon, in which soluble compounds are
removed from solution by a precipitate.
• Note:
solution is not saturated with the
coprecipitated species.
4 types of coprecipitation
• Surface adsorption (equilibrium process)
• Mixed crystal formation (equilibrium process)
• Occlusion (kinetics of crystal growth)
• Mechanical entrapment (kinetics of crystal growth)
2.0 Precipitates and precipitating reagents
165
03:08
2.4.1
Surface adsorption
• Source of coprecipitation which occurs to a greater
extent with coagulated colloids than with crystalline
solids and is therefore a common source of
contamination with large specific surface area (i.e.
colloids).
• The net effect of surface adsorption is carrying down
of an otherwise soluble compound as surface
contaminant.
166
03:08
• E.g. coagulated AgCl formed in gravimetric
determination of Cl¯ is contaminated with
adsorbed Ag+, NO3¯.
• Consequently, silver nitrate, a soluble compound
is coprecipitated with AgCl.
2.0 Precipitates and precipitating reagents
03:08
167
Adsorbed impurities can be minimized by
(i)Digestion– water is expelled by this process to give a denser
mass of smaller specific surface area for adsorption.
(ii)Washing the coagulated colloid with a solution containing
volatile electrolyte e.g. volatile HCl be used to wash AgCl ppt.
(iii)Reprecipitation –
• Is a drastic but an effective way to minimize effects
adsorption.
• In reprecipitation the filtered solids are redisolved and
reprecipitated.
168
03:08
• It significantly reduces contamination originally
present in the ppt but substantially increase the
analysis time.
• E.g. Hydrous oxides of Fe3+ and Al3+ have
tendencies to adsorb the hydroxide Zn, Cd and can
be removed by reprecipitation.
2.0 Precipitates and precipitating reagents
03:08
169
2.4.2
Mixed crystal formation
• Coprecipitation where a contaminat ion replaces an
ion in the lattice of the crystal.
• E.g. Pb2+ or Sr2+ may replace Barium ions in the
BaSO4 crystals.
• For exchange to occur the ions must have similar
charge and size.
• Occurs in both colloid and crystalline ppts
Control
• Remove the contaminant ion before pptn
• Use an alternative ppting reagent which does not
give rise to mixed crystal.
170
2.0 Precipitates and precipitating
reagents
03:08
2.4.3
Occlusion and mechanical entrapment
• Occlusion – coprecipitation whereby a compound is
trapped within a pocket formed during rapid crystal
growth.
• The amount of occluded material is greatest in the part
of the crystal that forms first.
• Mechanical entrapment-Occurs when crystals lie
close together during growth, thereby trapping a portion
of the solution in tiny pocket.
171
03:08
• Both Occlusion and mechanical entrapment
are at a minimum when the rate of ppt formation
is low
• Digestion is often remarkably helpful in reducing
this type of coprecipitation.
• Rapid solution and reprecipitation, which occur at
elevated temperature of digestion open pockets
and frees impurities into solution.
2.0 Precipitates and precipitating reagents
172
03:08
2.4.4
Errors due to coprecipitation
• Coprecipitation can cause either positive or negative
errors.
2.0 Precipitates and precipitating reagents
03:08
173
2.5 precipitation from Homogeneous solution
• is a process in which a precipitate is formed by slow
generation a precipitating reagent homogeneously
throughout a solution.
• Local reagent excess do not occur because the
precipitating
agent
appear
gradually
and
homogeneously throughout the solution and reacts
immediately with the analyte.
174
• As the result, the relative supersaturation is kept
low during the entire precipitation.
• Homogeneously formed precipitates, both
colloidal and crystalline, are better suited to
analysis than those formed by direct addition of
a precipitating agent.
03:08
• just below 100oC, Urea is used to homogeneous
generation of OH¯ as follows;
• (NH2)2CO+3H2O→CO2+2NH4++2OH¯.
2.0 Precipitates and precipitating
reagents
175
03:08
• For example, anhydrous oxides of iron and
aluminium are formed by adding a base in a
solution containing iron and aluminium.
• direct addition of a base (OH¯) results to
formation of gelatinous masses that are heavily
contaminated and difficult to filter.
• In contrast when same products are produced by
homogeneous generation of OH¯, they are dense
readily filtered and have considerable high purity.
3.0 Drying and ignition of the ppt
176
03:08
• After filtration, a gravimetric ppt is subjected to heat
until constant mass is obtained.
• Heating removes solvents and any volatile species
which may be carried down with ppt.
• In some cases the ppt are ignited to a solid compound
of known composition.
• E.g. CaC2O4 may be heated to 200oC to expel both
unbounded and bonded water and further heated to
900oC to convert it to CaO.
4.0 Application of gravimetric
methods
177
03:08
• Determination of most inorganic cations and anions as
well as neutral species e.g. H2O, SO2, CO2 and l2
• Determination of organic compounds – lactose,
salicylates nicotine, cholesterol, etc.
4.1 Inorganic precipitating agents
• These are inorganic reagents which form slightly
soluble salts or anhydrous oxides with the analyte.
• Most of them are not selective.
178
4.0 Application of gravimetric methods
03:08
4.2 Reducing precipitating reagents
• These are the reagents that convert an analyte to its
elemental form for weighing.
Table: Reducing agents employed in gravimetric methods.
Reducing agent
SO2
H2C2O4
H2
SnCl2
HCOOH
NaNO2
SnCl2
Electrolytic reduction
Analyte
Se, Au
Au
Re, Ir
Hg
Pt
Au
Hg
Co, Ni, Cu, Zn, Ag, In, Sn, Sb, Cd, Re, Bi
179
4.0 Application of gravimetric methods
4.3 Organic precipitating reagents
03:08
• Form intensely coloured chelating complexes with certain
species
• They are more soluble in organic solvents than in aqueous
solution 8-hydroxyquionoline.
8-Hydroxyquinoline
• It forms sparingly soluble compounds with most metal ions
except the alkali ions.
• The solubility of the different compounds depends strongly
on the acidity of the solution.
• Other metal ions that form sparingly soluble quinolates in a
near neutral solution (Co2+, Ni2+ etc.) should be absent.
• Ions that require a basic solution to be precipitated as
quinolates (e.g. Mg2+) will not disturb.
180
4.0 Application of gravimetric methods
03:08
• Aluminium ions are precipitated quantitatively by 8hydroxyquoinoline from a near neutral solution.
• Al3+ + 3C9H7NO + 3 H2O → Al(C9H6NO)3 + H3O+
181
03:08
OH
N
C
HC
HC
C
N
C
CH
C
CH
O
O
Al
C
N
N
O
8-Hydroxoquinoline
Aluminiumm quinolate
4.0 Application of gravimetric methods182
03:08
• Dimethylgyloxime DMG
• Only Nickel ions are precipitated quantitatively by
dimethylglyoxime from a weakly acidic or ammoniacal
solution.
• Ni2+ + 2 C4H8N2O2 + 2 H2O → Ni(C4H7N2O2)2 + 2 H3O+
183
03:08
O
H3C
H
N
O
H3C
C
H
O
N
N
C
Ni
C
H3C
C
N
O
H
H3C
N
H
Dimethylglyoxime
C
C
N
O
CH3
O
Nickel dimethylglyoximate
CH3
QUALITATIVE ANALYSIS
◦ Shows the chemical identity of the component/chemical
species/analytes in a sample.
◦ May cover confirming presence or absence of species
ANALYTICAL
CHEMISTRY
QUANTITATIVE ANALYSIS
◦
Establishes the amount (absolute, relative) of the
analyte contained in a sample.