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ESCO-NIGERIA TECHNICAL &
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ONAL EDUC
CATION REVITALIS
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ECT-PHASE
E II
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YEAR II SECO
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Version
1: Decemberr 2008
TABLE OF CONTENT
WEEK 1 CALIBRATION OF A PIPETTE………………………………………………………….3
WEEK 2 USE OF LINEAR REGRESSION……………………………………………………… ...6
WEEK 3 THIN- LAYER CHROMATOGRAPHIC
SEPARATION OF AMINO ACIDS…………………………………………………………………7
WEEK 4 PRINCIPLE OF PARTITION: EXTRACTION
OF A MATERIAL FROM ONE PHASE INTO A SECOND PHASE …………………………......9
WEEK 5 EXPERIMENT: IDENTIFICATION OF
COLOURLESS MATERIALS IN THIN LAYER CHROMATOGRAPHY ………………………11
WEEK 6 COMPLEX ELUTION OF IRON AND COPPER USING
A CATION-EXCHANGE RESIN…………………………………………………………………..12
WEEK 7 PH DEPENDENCE OF ELECTRO PHORESIS OF AMINO ACIDS……………………14
WEEK 8 QUANTITATIVE AND QUANTITATIVE ANALYSIS OF FRUIT JUICES FOR
VITAMIN C USING HIGH PEAFORMANCE LIQUID CHROMATOGRAPHY………………..16
WEEK 9 MEANING OF MEAN DEVIATION, STANDARD DEVIATION,
ABSOLUTE ERROR, AND RELATIVE ERROR…………………………………………………18
WEEK 10 ABSOLUTE AND RELATIVE UNCERTAINTY……………………………………..21
WEEK 11 TO INTRODUCE THE BASIC STATISTICAL CONCEPTS REQUIRED TO
CHARACTERIZED ACCURACY, PRECISION AND UNCERTAINTY………………………..25
WEEK 12 STANDARDISATION OF HYDROCHLORIC ACID WITH SODIUM
TRIOXOCARBONATE (IV) STANDARD SOLUTION………………………………………….27
WEEK 13ANALYSIS OF ASPIRIN………………………………………………………………..28
WEEK 14 EXPERIMENT 5 GRAVIMETRIC DETERMINATION OF
NICKEL IN A NICHROME ALLOY………………………………………………………………30
WEEK 15 GRAVIMETRIC DETERMINATION OF SO3 IN A SOLUBLE
SULFATE……………………………………………………………………………………………32
2
WEEK 1 CALIBRATION OF A PIPETTE
1.1 Principle
Volumetric glassware is calibrated by measuring the mass of a liquid (usually distilled
or deionized water) of known density and temperature that is contained in (or delivered
by) the volumetric ware. In carrying out a calibration, a buoyancy correction must be
made, since the density of the masses. Glassware are usually generally calibrated to five
significant figures, the maximum precision likely to be attained in filling or delivering
solutions.
Example: A 10mL pipette may be calibrated to 9.9997mL which is as accurate as
10.003mL, i.e. the last figure is 1 part in 10,000).
Figure 1.1 (a) Variable-volume automatic pipette 100-1000 μL .
3
(b) Handheld, battery-operated, computer-controlled motorized pipette.
Figure 1.2 Typical pipettes: (a) Volumetric pipette (b) Mohr pipette (c) Serological pipette (d)
Eppendorf micro pipette (e) Ostwald-Folin pipette (f) Lambda pipette.
1.2 Procedure
Determine the empty mass of the dry receiver Stoppered Erlenmeyer flask or a
weighing bottle). Fill the pipette with distilled water (whose temperature you have
recorded). Transfer a portion of the temperature- equilibrated water to the receiver with
the pipette, weigh the receiver and its contents (again, to the nearest milligram), and
calculate the mass of water delivered from the difference in these masses. With the aid
of the table in appendix 1,calculate the volume delivered. Repeat the calibration several
times;
4
1.3 Calculations
Calculate the mean volume delivered and its standard deviation.
5
WEEK 2 USE OF LINEAR REGRESSION
2.1 Principle
The analyst is frequently confronted with plotting data that fall on a straight line, as in
an analytical curve, graphing, curve fitting is critically important in obtaining accurate
analytical data. It is the calibration graph that is used to calculate the unknown
concentration. Straight-line predictability and consistency will determine the accuracy
of the unknown concentration.
Graphing is often done intuitively, that is by simply “eyeballing” the best straight line.
By placing a ruler through the points which invariably have some scatter. A better
approach is to apply statistic to define the most probable straight line fit of the data.
If a straight-line relationship is assumed, the data fit the equation below known as the
regression model:
y= mx+b
Where:
y is the dependent variable
x is the independent variable
m is the slope, of the curve
b is the intercept on the ordinate (y-axis)\
y is usually the measured variable plotted as a function of changing x. In a
spectrophotometric calibration curve, y would represent the measured absorbances and
x would be the concentrations of the standards.
2.2 Procedure:
Prepare standards of 0, 10, 20, 30 and 40 ppm Na by diluting 0, 5, 10, 15 and 20 mL of
the stock Nacl solution to 50ml. (It may be better with some instruments to prepare
solutions five times more dilute than this by adding
0, 1,2,3,4, and 5mL of solution to the flasks. This may result in a more linear calibration
curve. Follow your instructor’s directions).
Run the standards and record their absorbances. Plot a graph of absorbance against
concentration
6
WEEK 3 THIN- LAYER CHROMATOGRAPHIC SEPARATION OF AMINO
ACIDS
3.1 Principle
The amino acids are separated on a TLC sheet e.g. silical gel, using a choice of two
developing solvents. The locating reagent is ninhydrin.
3.2 Solution and chemicals required
Provided:
(a)
developing solvent: No I butyl alcoholacetic acid-water (80:20:20v/v); No.2
propylalcohol- water (7:3v/v).
(b)
Locating reagent: Ninhydrin solution 0.3% ninhydrin (1,2,3-triketohydrindene
Eastman No. 2495) in butyl alcohol containing 3% glacial acetic acid
3.3 Chromatographic equipment.
Fisher scientific TLC kit A or equivalent TLC sheets sprayer (for application of
reagent), developing apparatus (e.g. fisher scientific TLC kit)
3.4 Procedure
Obtain an unknown mixture from your instructor. The mixtures to be separated will
contain approximately 1mg of each amino acid per milliliter 0.5makololic hydrochloric
acid solution (see discussion below). Spot approximately 1 μL of the sample solution on
the chromagram sheet about 2cm from the lower edge. (activation of the sheet is not
necessary for this seperation). The unknown should contain acids whose Rf values are
sufficiently different with the solvent system used so that they will be readily separated.
Your instructor will advise you which standard amino acid solution to run with your
unknown so that you con distinguish between two amino acids that might have an Rf
value close to one in your mixture. Allow is to 20 min drying to ensure complete
evapouration of the hydrochloric acid.
Develop the chromagram sheet in the solvent of choice for a distance of 10cm or for
approximately 90min (see the list of Rf values in the table and follow your instructors
direction for the solvent of choice for your unknown mixture.) Dry the developed
chromagram sheet and spray with the ninhy- drin solution. Heat gently for several
minutes until separated zones appear clearly visible.
7
3.5 Results and determination
Lists the approximate Rf values obtained when the two solvent systems are applied to
the seperation of 13 different amino acids. From this table and the standard acids you
run determine what amino acid are present in your unknown.
8
WEEK 4 PRINCIPLE OF PARTITION: EXTRACTIONOF A MATERIAL FROM
ONE PHASE INTO A SECOND PHASE
4.1 Principles
If a solute is added to a mixture of two immiscible solvents in both of which it is
soluble, it will distribute itself in the two solvents. For example, when iodine is added a
beaker containing water and 1,1,1-trichloroethane,iodine will dissolve in both layers
.After stirring the mixture several times ,the amount of iodine in the two layers become
constant. i.e obeying the partition law .
Mathematically, the partition law can be represented by the following equation.
KD=
concentration of solute in organic solvent
concentration of solute in aqueous solvent
KD = [s]org
[s]aq
KD = partition is a method of separating a substance from mixture, using the principle
of partition equilibrium of a solute between two immiscible solvents.
4.2 Procedure:
Dissolve a given organic compound X in water. Add a suitable organic solvents and
shake the separatory funnel .After shaking the apparatus, the mixture separates into two
layers of liquids collect the two different layers separately. Distill off the organic
solvent to obtain the purified organic product left behind
9
Figure 4.1 A separatory funnel
Figure 4.2 Proper way of
shaking and
venting a separatory funnel
4.3 Calculations:
If the organic compound X had a partition coefficient of 30 in the two solvents ;
suppose 3.1g X is dissolved in 50cm3 water and 50cm3 of the organic solvents .How
much X is extracted by the organic solvents , (3.1-9)g would be the amount of x left in
water.
10
WEEK 5 EXPERIMENT: IDENTIFICATION OF COLOURLESS MATERIALS IN
THIN LAYER CHROMATOGRAPHY
5.1 Principle:
Thin layer chromatography (TLC) is a planer form of chromatography useful for widescale qualitative analysis screening and can also be used for quantitative analysis. The
stationary phase is a thin layer of finely divided adsorbent supported on a glass or
aluminium plate, or plastic strip.
5.2 Solutions and chemicals required
5.2.1Locating reagent: Iodine vapour
5.2.2 Chromatographic equipment: TLC plate incorporated with a fluorescent dye in the
powdered adsorbent.
5.3 Procedure:
Make a spot of the colourless solution onto the plate with a micropipette, about 2cm
from the lower edge.
Develop the plate (chromatogram) by placing the bottom of the plate or strip (but
not the sample spot) in a suitable solvent. The solvent is drawn up to the plate by
capillary action and the sample components more up the plate at different rates,
depending on their solubility and their degree of retention by the stationary phase.
Visualize the colourless or non fluorescent spots by exposes the developed plate
to iodine vapour. The iodine vapour interacts with the sample components, either
chemically or by solubility, to produce a colour when the plates are held ultraviolet
light, dark spots will appear where sample spots occur due to quenching of the plate
fluorescence.
5.4 Determinations
Using a table of standard/approximate Rf values, determine what solute substances are
present in the unknown colourless sample.
Rf =
dis tan ce solute moves
dis tan ce solvent front moves
11
WEEK 6 COMPLEX ELUTION OF IRON AND COPPER USING A CATIONEXCHANGE RESIN
6.1 Objective
To separate a mixture of copper (II) and iron (III) on a cation-exchange resin by elution using the
technique of phosphate complexation.
6.2 Introduction
The separation involves the stepwise elution of iron (III) by phosphoric acid and then of copper (II) by
hydrochloric acid, from a strongly acidic cation-exchange resin such as ZeoKarb 225, Amberlite IR120 or Dowex 50. If hydrochloric acid alone is used the other of removal of the ions from the column
is reversed, and a poorer separation is obtained.
The phosphoric acid eluant modifies the activities of the ions, possibly by complex formation, giving
improved resolution. The order of elution is also reversed with a phosphate eluant, iron having the
smaller retention volume.
6.3 Materials and equipment
Analar copper sulphate pentahydrate and ferric nitrate;
ZeoKarb 225;
hydrochloric acid (0.5M);
phosphoric acid (0.2M);
potassium cyanoferrate indicator solution;
two glass columns
6.4 Method
Make up two columns about 1.5cm in diameter and 10-15cm long using ZeoKarb 225 (in the sodium
form) mesh size 52-100 (8% DVB). Slurry the resin with water, allow to settle, and decant the
supernatant. Repeat until the washings are clear, then pack the column, and wash with water until the
eluate is no longer acid. The column is then in the hydrogen form.
12
Figure 6.1 Preparation of a stationary phase
Load each column with a solution containing about 0.5g each of copper sulphate and ferric nitrate.
Wash the column with about 100ml of water to remove the acid liberated in the exchange process.
Elute one column 0.5M HCL at a rate of ≈ 100ml h-1. Collect the column effluent in 5ml fractions.
Estimate in a semi – quantitative manner the amount of copper and iron present in each fraction by
transferring to a test tube and comparing with a solution of known concentration.
Elute the second column with 0.2M H3PO4 at ≈ 100ml h-1, Collect 2ml fractions, and, since the
phosphate complex is colourless, test each fraction with potassium cyanoferrate. When no more iron
can be detected change the eluting agent to hydrochloric acid (equal volumes of concentrated acid and
water); the copper is rapidly removed from the column. Collect fractions as before and test for copper
with the same reagent (K4[Fe(CN)6]).
Note (a) the volume eluted before iron first appears; (b) the volume which contains iron; (c) the
volume collected after all the iron has been eluted, and before copper first appears: and (d) the volume
which contains copper.
6.5 Remarks
Average values are: (a) 15ml (b) 100ml (c) 20ml and (d) 10ml.
Metals which readily form complexes with the eluent will tend to be eluted first.
In the above experiment iron complexes strongly with phosphate whilst copper hardly complexes at
all. Hence the iron is readily eluted
13
WEEK 7 PH DEPENDENCE OF ELECTROPHORESIS OF AMINO ACIDS
7.1 Principle:
Electrophoretic methods are used to separate substances based on their charge – to –
mass ratios, using the effect of an electric field on the charges of these substances.
These techniques are widely used for charged colloidal particles or macro-molecular
ions such as those of proteins, nucleic acids and polysaccharides. There are several
types of electrophoresis, zone electrophoresis being one of the most common.
In zone electrophoresis, proteins are supported on a solid so that, in addition to
the electric migration forces, convertional chromatographic forces may enter into the
separation efficiency. There are different types of supports, which include starch gels,
polyacrylaride gels, polyurethane foam and paper.
The migration rate of each substance depends on the applied voltage and on the
PH
of the buffer employed. The applied voltage is expressed in volts per centimeter. It is
up to 500V in low-voltage electrophoresis and can be several thousand volts in highvoltage electrophoresis. The latter is used for is used for high-speed separation of lowmolecular-weight substances.
Zone electrophoresis is used largely for the separation of amino acids which
depend on pH. At a certain pH, the net charge of an amino acid is zero and it exists as a
Zwitterion that exhibits no electophoretic mobility.
A relatively new separation technique that is capable of separating minute
quantities of substances in relatively short time with high resolution is capillary
electrophoresis.
7.2 Procedure:
Set up a capillary electrophoretic system as that shown in fig 7.1. The seperation
medium is a fused silica capillary tube containing an appropriate electrolyte. Introduce a
small volume of sample into one end of the capillary. Insert each end of the capillary
into an electrolyte buffer solution with voltages from 1000 to 30,000V.
Place a focused beam through the capillary determine the amino acid present.
14
Detector
Capillary
outlet
Capillary
inlet
Electrolyte
Buffer
Or sample
Electrolyte
buffer
Reservoir
Reservoir
D.C High-voltage
Supply
Pt electrode
(- anode)
Pt electrode
(+anode)
Figure 7.1 Capillary electrophoresis system.
15
WEEK 8 QUANTITATIVE AND QUANTITATIVE ANALYSIS OF FRUIT JUICES
FOR VITAMIN C USING HIGH PEAFORMANCE LIQUID CHROMATOGRAPHY
8.1 Principle
Samples of several juices are chromatographer directed on a strong – base anion
exchange column using a uv detector at 254nm .By comparison of retention times with
that of a vitaminC standard (l-ascorbic acid), the presence or absence of vitamin C is
ascertained .Peak area measurements are used for quantitative determine of the
concentration of vitamin C in those juices in which it is found.
8.2 Solutions and reagents required
8.2.1 Provided
Mobile – phase solution (1.36g KH2PO4 /LD distilled
deionized water ), L-ascorbic
acid.
8.2.2 To prepare
Vitamin C standard weigh 50mg of L-ascorbic acid to the nearest 0,1mg, dissolve in a
50ml volumetric flask, and dilute to volume with distilled deionized water .Prepare
fresh the say of use, It gives you s 0.1% solution ,prepare serial dilutions to obtain
0.05% and 0.02% standards.
8.3 Procedure
8.3.1 Calibration
your instrument should contain column of a strong-base anion exchange resin. At a
pressure of 1000psi, and a flow rate of about 0.5ml/min.
Using a 10 or 25ul syringe, inject a 10ul aliquot of the 0.02% vitamin C standard and
record the chromatogram. Repeat for the 0.05% and 0.1% standards. Measure the
retention times and the areas. The peak areas may be obtained from the height times
width at half height for each. Alternatively, the peaks may be cutout and weighted.
On the instrument may print out the integrated areas. Plot a calibration curve of the
peak area against concentration.
8.3.2 Unknown
You will be provided with three or more unknown juices. Inject 10 μL portions of
each and record the chromatograms. Several peaks may be obtained. Run at least
16
two chromatograms on each. From a comparison of retention time, identify the juice
that contain vitamin C. measure the areas of the vitamin C peaks, and from the
calibration curve, determine and report the concentration in the unknowns. Report
the acreage concentration and the standard deviation. The vitamin C peak may be
partially overlapped by another. In this case, extrapolate to the baseline and measure
the area from the baseline.
Figure 8.1 Equipment for high performance liquid chromatography (HPLC)
17
WEEK 9 MEANING OF MEAN DEVIATION, STANDARD DEVIATION,
ABSOLUTE ERROR, AND RELATIVE ERROR.
9.1 Principle
There are various ways and units in which the accuracy of measurement can be expressed, an accepted
true value for comparison is being assumed.
9.2 Ways of expressing accuracy
9.2.1 Absolute error
The difference between the true value and the measured value, with regard to the sign is know as the
absolute error; and it is reported in the same units as the measurement
For example,
If a 5.15g sample of a material is analyzed to be 5.02g, the absolute error is -0.13g.
If the measured value; 5.02g is the average of several measurements, the error is called mean
error.
For example,
If in a series of measurements, the following results were obtained; in grams, 5.05, 5.03, 5.02,
5.02, 5.02, 5.01, 5.01, 5.01, 5.01 and 5.02.
The average (mean) of the results obtained
x1
∑N
= x = 5.02
Thus the absolute error, - 0.13g is also the mean error.
9.3 Relative error
This is the absolute or mean error expressed as a percentage of the true value.
Thus the above analysis has a relative error of (- 0.13/5.15) x 100% = -2.52%
9.4 Relative accuracy
This is the measured value or mean expressed as a percentage of the true value (5.02/5.15) x 100% =
97.4757
≈ 97.5%
It should be emphasized here that neither number is known to be “true” and the relative error or
accuracy is based on the mean of two sets of measurements. The relative error can be expressed in
units other than percentages. In very accurate work, we usually deal with relative error of less than
1%, and it is convenient to use to a smaller unit. A 1% error is equivalent to 1 part in 1000. it is also
equivalent to 10 parts in 1,000. This letter unity is commonly used for expressing small uncertainties.
That is, the uncertainty is expressed in parts per thousand, written as ppt.
Example:
18
The results of an analysis are 36.97%g compared with the accepted value of 37.06g. what is the
relative error in parts per thousand?
Solution
Absolute Error = 36.97g – 37.06g
= - 0.09g.
Relative Error = - 0.09 x 1000 0 00
37.6
= - 2.4 ppt.
9.5 Ways of expressing precision
Each set of analytical results should be accompanied by an indication of the precision. Various ways
of indicating precision are acceptable:
9.5.1 Standard deviation
The standard deviation S, of a finite set of experimental data is theoretically given by
S=
∑ (x − x )
2
1
N −1
(generally N<30)
Where
x1 = measurements
x = mean of limited number of measurements
N = number of measurements.
N – 1 = number of degrees of freedom
While the standard deviation σ of an infinite set of experimental data is theoretically given by
σ=
∑ (x − μ )
2
1
N
Where
∞
µ = mean of the infinite number of measurement
N →∞
N = number of infinite measurements.\
The value of S is only an estimate of σ and will more nearly approach σ as the number of
measurements increases. Since we deal with small numbers of measurements in an analysis, the
precision is necessarily represented by S.
Example:
Calculate the mean and the standard deviation of the following set of analytical results:
15.67, 15.69 and 16.03g
19
Solution
x1
x1 − x
( x 1 − x )2
15.67
0.13
0.0169
15.69
0.11
0.0121
16.03
0.23
0.0529
∑
47.39
∑ 0.47
x=
∑x
=
S=
1
N
∑ 0.0819
47.39
3
=15.80
0.0819
= 0.20g
3 −1
The standard d
20
WEEK 10 ABSOLUTE AND RELATIVE UNCERTAINTY
10.1 Absolute uncertainty
Is an expression of the margin of uncertainty associated with a measurement. If the estimated
uncertainty a perfectly calibrated burette is + 0.02ml, we call the quantity + 0.02ml the absolute
uncertainty associated with the reading.
10.2 Relative Uncertainty
Is an expression comparing the size of the absolute uncertainty to the size of its associated
measurement. The relative uncertainty of a burette reading of 12.35 + 0.02mL is
Relative uncertainty = absolute uncertainty
Magnitude of measurement
= 0.02mL = 0.002
12.35mL
(10-1)
The percent relative uncertainty is simply
Percent relative uncertainty = 100 x Relative uncertainty
(10-2)
In the above example the percent relative uncertainty is 0.2%.
A constant absolute uncertainty leads to smaller relative uncertainty as the, magnitude of the
measurement increases. If the uncertainty in reading a burette is constant at + 0.02mL, the relative
uncertainty is 0.2% for a volume of 10mL and 0.1% for a volume 20mL
10.3 Propagation of uncertainty
It is usually possible to estimate or measure the random error associated with a particular
measurement, such as the length of an object or the temperature of a solution. The uncertainty should
be base on your estimate of how well you can read and instrument or on your experience with a
particular method. When possible, uncertainty is usually expressed as standard deviation of a series of
replicate measurement. The discussion that follows applies only to random error; it is assumed that
any systematic error as being detected and corrected.
In most experiments, it is necessary to perform arithmetic operation on several numbers each
of which as it associated error. The most likely uncertainty in the result is not the some of individual
errors, the some of these are likely to be positive and some negative. We expect a certain amount of
cancellation of errors.
21
10.4 Addition and subtraction
Suppose you wish to perform the following arithmetic, in which the experimental uncertainties are
given in parenthesis:
1.76 ( ± 0.03 ) ← e1
+1.89 ( ± 0.02 ) ← e2
-0.59 ( ± 0.02 ) ← e3
3.06 (+e4)
(10-3)
The arithmetic answer is 3.06, but what is the uncertainty associated with this result?
We start by calling the three uncertainties e1, e2 and e3. for addition and subtraction, the uncertainty in
the answer is obtained by manipulating the absolute uncertainties of the individual terms:
e4 =
e12 + e 22 + e 32
(10-4)
For the sum in Equation 10-3 we can write.
e4 =
(0.03)2 + (0.02)2 + (0.02)2
= 0.041
(10-5)
The absolute uncertainty associated with the sum is + 0.04, and we can write the answer as 3.06 +
0.04. Although there is only one significant figure in the uncertainty, we wrote it initially as 0.041,
with the first insignificant subscripted. The reason for retaining one or more insignificant figures is to
avoid introducing round-off errors into later calculations through the number 0.041. The insignificant
figure was subscripted to remind us where the last significant figure should be at the conclusion of the
calculations.
If we wish to express the percent relative uncertainty in the sum of Equation 10-3, we may
write
Percent relative uncertainty =
0.041
x 100 = 1.3%
3.06
(10-6)
The uncertainty, 0.041, is 1.3% of the result, 3.06. The subscript 3 in 1.3% is not significant. It would
now be sensible to drop the insignificant figures and express the final result as
3.06 ( +0.04) (absolute uncertainty)
or
3.06 ( +1%)
(relative uncertainty)
22
10.5 Multiplication and division
For multiplication and division we first convert all uncertainties to percent relative uncertainties (or
relative uncertainties). Then we calculate the error of the product or quotient as follows:
0
0
e4 =
(%e1 )2 + (%e 2 )2 + (%e 3 )2
(10-7)
For example, consider the following operations:
1.76 (± 0.03) × 1.89 (± 0.02)
= 5.6 4 ± ?
0.59 (± 0.02)
(10-8)
First convert all the absolute uncertainties to percent relative uncertainties:
1.76 (± 1.7 % ) × 1.89 (± 1.1 % )
= 5.6 4 ± ?
0.59 (± 3.4 % )
(10-9)
Then find the relative uncertainties of the answer using Equation 10-7:
0
0
e4 =
(1.7 )2 + (1.1 ) + (3.4 )2
= 4.0%
(10-10)
The answer is 5.64 (± 4.0 % ).
To convert the relative uncertainty to absolute uncertainty, find 4.0% of the answer:
4.0% x 5.64 = 0.040 x 5.64 = 0.23
(10-11)
The answer is 5.64 (+0.23). Finally, we drop all the figures that are not significant. The result may be
expressed as
5.6 (+0.2)
(absolute uncertainty)
5.6 (+4%)
(relative uncertainty)
There are only two significant figures because we are limited by the denominator, 0.59, in the original
problem.
10.6 Mixed Operations
As a final example, consider the following mixed operations:
[1.7 (± 0.03) - 0.59 (± 0.02) ] = 0.619 ± ?
0
1.89 (± 0.02)
(10-12)
First work out the difference in brackets, using absolute uncertainties:
1.76 (+0.03) – 0.59 (+0.02) = 1.17 + 0.036
Since
(0.03)2 + (0.02)2
(10-13)
= 0.036
23
1.17(±0.03 6 ) 1.17 (± 3.1 % )
=
= 0.619 0 (± 3.3 % )
1.89(± 0.02) 1.89(± 1.1 % )
Since
(3.1 )2 + (1.1 )2
(10-14)
= 33.
The relative uncertainty in the result is 3.3%. The absolute uncertainty is 0,033 × 0.6190 = 0.020. T he
final answer could be written as
0.619 ( ± 0.02 0 )
(absolute uncertainty)
Or
0.619 (3.3%)
(relative uncertainty)
Since the uncertainty spans the last two places of the result, it would also be reasonable to write the
result as
0.62 (+0.02)
Or
0.62 (+3%)
24
WEEK 11 TO INTRODUCE THE BASIC STATISTICAL CONCEPTS REQUIRED
TO CHARACTERIZED ACCURACY, PRECISION AND UNCERTAINTY.
1. Samples of a certified reference material (CRM) were used to assess the accuracy and
precision of a method for the determination of a pesticide. Six individual applications of the
method provided the following results:
2. 3.21, 3.30, 3.35, 3.28, 3.40 and 3.25 μg / kg. The CRM was certified to contain an amount of
pesticide of 3.40 and ± 0.02 μg / kg. Calculate and discuss the parameters that define
accuracy, precision and uncertainty.
11.1 Precision
¾ The data set is statistically processed, whether by hand or using a pocket calculator.
First, the mean of the six results is calculated to be x =
∑ x / n. then,
i
d = x i − X , and
d2 are calculated as follows:
d2
xi
d
3.21
0.09
0.0081
3.30
0
0
3.35
0.05
0.0025
3.28
0.02
0.0004
3.40
0.10
0.010
3.25
0.05
0.0025
∑d
2
= 0.0235
Using the calculator to perform the statistical calculations provide the following parameter values:
x = 3.29833 ≈ 3.30
s = 0.06853 ≈ 0.07
n=6
¾ The limits of confidence at the 95% confidence level (p = 0.05) are calculated from
(
)
x ± ts/ n , where t is student’s parameter. With U = 5-1 degree of freedom and p =
0.05, t = 2.57
3.30 ± 2.57
0.07
n
= 3.30 ± 0.070
{
3.23
3.37
μg / kg
¾ The mean characterizes the method.
¾ The absolute uncertainty is ± U = 0.072 μg / kg. It can also be calculated from the
expression UR = R.SR; since R = 2 (p = 0.05) ⇒ ± U R = 2 × 0.07 = 0.14 μg/kg. The
divergence arises from the fact that n is small (distant from the proposed Gaussian
distribution).
25
¾ The relative uncertainty will be ±
U
× 100 = 2.18 0 0.
3.30
11.2 Accuracy
¾
x̂ ' , the value held as true and the result of operating under near-excellence
conditions, will be given by
x̂ ' = 3.45 ± 0.02 μg/kg
¾ One can refer to accuracy since an appropriate reference, x, is available
¾ It is accuracy of a method since one intends to obtain the mean x of 6
results and, specifically, to quantify the bias (n<30).
¾ The absolute systematic error will be x − x̂' = 3.30 - 3.45 = - 0.15 μg/kg.
¾ The relative systematic error will be ±
x.x̂'
x̂'
= - 4.35 o o.
¾ The bias of the method is thus negative (results are underestimated).
¾ Other considerations
¾ The precision of the method used is lower than that of the CRM since
the uncertainty in the former is greater:
U x > U x̂' 0.07 > 0.02
x
bias
320
330
340
x̂'
¾ If U x̂ is assumed to have obtained at the same probability level, then
the values of x̂ ' and x can be compared with their ranges.
¾ The method used is scarcely accurate because the ranges ± U x and ±
U x̂' do not coincide. Therefore, the relative error in the negative bias (-
4.35%) is misleadingly low.
26
WEEK 12 STANDARDISATION OF HYDROCHLORIC ACID WITH SODIUM
TRIOXOCARBONATE (IV) STANDARD SOLUTION.
12.1 Principle
In standardization, the unknown concentration of a given solution can be determined by
titrating it with another solution of known concentration.
12.2 Requirement
Hydrochloric acid (HCl), standard sodium trioxocarbonate IV solution, 50cm3 burette, 25cm3
pipette, methyl orange indicator.
12.3 Procedure
Fill the burette with hydrochloric acid and adjust its level to the zero mark. Pipette 25cm3 of
the sodium trioxocarbonate IV solution into a conical flask and add two drops of methyl orange.
Titrate the acid against the alkali, swirling the flask all the time to mix the solutions. Keep titrating till
the end point is reached. (i.e. when the indicator turns orange) Note the volume of acid used to
neutralize the alkali. Carry out two more accurate titrations. When approaching the end point, add in
the acid drop by drop until one drop of the acid changes the colour of the indicator permanently.
Record the burette readings.
12.4 Equation
2 HCl + Na 2 CO3 → 2 NaCl + H 2 O + CO2
12.5 Calculations
i.
Determine the average volume of acid used.
ii.
Given that the concentration of sodium carbonate is 0.1moldm-3 determine the concentration of
hydrochloric acid using the formula:
acid
M AV A
= mole ratio of
base
M BVB
Where
MA, VA are the concentration and volume of the acid and
MB, VB are the concentration and volume of the base.
27
WEEK 13ANALYSIS OF ASPIRIN
13.1 Background
Aspirin is a pain reliever and reduces fever-acts as antiphretic. It is used for different ailments such
as aches. Fever due to cold. tension. rheumatism and arthritis. Aspirin tablets consist of pure
aspirin and a binder. The amount of pure aspirin should be specified on the label. However
sometimes there is a need to check it. As aspirin has an acidic property the titration of aspirin
tablets with standard NaOH solution provides a means of determining the number of milligrams of
pure aspirin per tablet. Aspirin is acetylsalicylic acid having the following structural formula:
COOH
Acetylsalicylic acid (aspirin)
OCOCH3
The molar mass of aspirin is 180gmol-1
13.2 Materials and reagents
a)
one burette
b)
standard NaOH solution-about 100cm3 (record the concentration)
c)
two tablets of aspirin
13.3 Procedure
1.
Rinse a burette twice with a few cm3 of standard NaOH solution, and then fill it above the zero
mark with the same solution and drain to the mark, making sure the burette tip is full.
2.
Dissolve one tablet of aspirin in 50cm3 of distilled water in a 250cm3 conical flask.
3.
Heat the solution to about 900C to help the aspirin to dissolve, as it is only slightly soluble in
water. The solution does not become completely clear as the binder is insoluble. But it does not
affect your determination.
4.
Cool the solution add three drops of phenolphthalein indicator and titrate with standard NaOH
solution. Swirling the. Flask to ensure good mixing until the first appearance of a permanent
pink colour. After standing for several minutes, the pink colour may disappear due to a
secondary reaction.
5.
Repeat the determination once more using the second tablet of aspirin.
6.
Calculate the weight of aspirin in mg per tablet using the following equations: weight of aspirin
=MNaOH X VNaOH X molar mass of aspirin
13.4 Questions
1.
Vinegar contains acetic acid. CH3COOH. Titration of 5.000g vinegar with 0.100M NaOH
requires 33.0cm3 to reach the equivalence point
28
a.
what is the weight percentage of CH3COOH in vinegar?
b.
If the vinegar has a density of 1.005g cm3 what is the molarity of CH3COOH in vinegar?
2.
The approximate concentration of hydrochloric acid. HCl in the stomach (stomach acid) is
0.17M. Calculate the mass of the following antacids required to naturalize 50cm3 of this acid.
a)
bicarbonate of soda. NaHCO3
b)
Aluminum hydroxide Al (OH)3
29
WEEK 14 EXPERIMENT 5 GRAVIMETRIC DETERMINATION OF NICKEL IN A
NICHROME ALLOY
14.1 Principle
Nickel forms a red chelate with diethylglyoxime (DMG), which is quite suitable for gravimetric
analysis. Precipitation of the chelate is complete in an acetic acid-acetate buffer or in ammoniacal
solution. Acetate buffer is generally used when Zn, Fe or Mn is present in the alloy. The sample given
to you is a nichrome alloy that has Ni (approximately 60%), Cr, and Fe as the major constituents.
Interference form Cr and Fe is removed by complexation with tartrate or citrate ions. Precipitation is
then carried out in an ammoniacal solution. The Ni content is calculated from the weight of the
precipitate
14.2 Equation
Ni2++2HDMG → Ni (DMG)2+2H+
14.3 Solutions and chemicals provided
HCl (1:1), HNO3 (1:1), NH3 solution (conc. and 1:1), 1% DMG in ethanol, 20% citric acid solution,
30% ethanol.
14.4 Thing to do before the experiment
1.
Obtain the alloy sample from your instructor.
2.
Prepare three sintered-glass filter crucibles. Dry to a constant weight of ± 0.3 to 0.4mg.
14.5 Procedure
14.5.1 Dissolving the sample
Weigh accurately triplicate 0.10-to 0.12-g samples of the alloy. Place in separate numbered 400-ml
beakers and add 25mL each of HCl (1:1) and HNO3 (1:1). Heat moderately (under the hood, please)
until the brown fumes are driven off-40 to 50min. keep the beakers covered partially with watch
glasses. (You may notice some undissolved carbon particles in the residue. If this occurs, then, using
quantitative technique, filter the contents of each beaker though what man filter paper No. 40 into
separate 600- mL beakers. Wash the 400-mL beaker several times with small amounts of water and
transfer the washings to the 600mL beaker through the filter. Wash the filter several times with small
amounts of water. Remove the filter paper and wash the funnel and stem with small amounts of water
before removing from the receiving beaker).
14.5.2 Precipitation
Add 4 to 6 mL citric acid solution and 100 to 150mL water to each beaker. Add conc. NH3 drop wise
with stirring until the solution is slightly basic (smell of ammonia in the solution). This should take 2
to 3mL. Test to see that the pH is about 8 by touching the wet stirring rod to pH paper. Be careful that
no drops adhere to the rod keep solution loss negligible. If a precipitate is formed, insufficient citric
acid has been added. Dissolve the precipitate by adding drop wise HCl (1:1), about 3mL) until the
30
solution is slightly acidic (no ammonia odor in solution). Heat the beakers on the hot plate to about 70
to 800C. Do not boil. Remove from the hot plate and add about 50mL 1% DMG solution drop wise
with stirring. If a red precipitate forms, add HCl (1:1) drop wise until it dissolves. The precipitation is
begun. Add NH3 (1:1) drop wise with stirring until the solution strongly smells of ammonia; this may
take 3 to 4mL. A red precipitate is formed. It is important that the solution be distinctly alkaline at this
point. Since the nose may retain the odor of the ammonia reagent, take care that the odour comes from
the beaker. Test to be sure the pH is greater than 8.5 by touching the wet stirring rod to pH paper to
test the pH. This is best done after the bulk of the precipitate is formed and allowed to settle. Cover the
beakers with watch glasses and set aside for at least 2h (preferably overnight).
14.5.3 Filtering and washing the precipitate
Filter the precipitate through previously weighed glass crucibles using gentle suction. After
transferring the bulk of the precipitate, wash the precipitate with lukewarm distilled water and transfer
quantitatively into the crucible. If traces of the precipitate (which are difficult to transfer with a rubber
policeman) stick to the precipitate (which are difficult to transfer with a rubber policeman) stick to the
sides of the beaker, dissolve in a few drops of hot HCl (1:1) and reprecipitate by adding a few drops of
1% DMG and NH3 (1:1). The precipitate now obtained will not stick and can be transferred to the
crucibles. Wash the precipitate in the crucible at least three to four time with warm water. Finally,
wash the precipitate once with 30% alcohol, which dissolves any excess DMG from the precipitate.
14.5.4 Drying and weighing the precipitate
Place the crucibles in beakers (labeled with your name and covered with watch glasses). Dry in the
oven for 1-2h weigh accurately. Reheat for 1-h periods necessary to obtain weights constant to within
± 0.3 to 0.4mg.
14.6 Calculation
Calculation and report the percent nickel in your unknown for each portion analyzed report each
individual result, the mean, and the relative standard deviation.
31
WEEK 15 GRAVIMETRIC DETERMINATION OF SO3 IN A SOLUBLE SULFATE
15.1 Principle
Sulfate is precipitated as barium sulfate with barium chloride. After the filtering with filter paper, the
paper is charred off, and the precipitate is ignited to constant weight. The SO3 content is calculated
from the weight of BaSO4.
Equation
SO42- + Ba2- → BaSO4
15.2 Solution and chemicals required
1.
provided. Dil (0.1M) AgNO3, conc. HNO3, conc. HCl.
2.
To prepare. 0.25M BaCl2. Dissolve about 5.2 g BaCl2 (need not be dried) in 100mL distilled
water.
15.3 Thing to do before the experiment
1.
Obtain your unknown and dry. Check out a sample of a soluble sulfate form the instructor and
dry it in the oven at 110 to 120oC for at least 2 h. Allow it to cool in a desiccator for at least 0.5 h.
2.
Prepare crucibles. Clean three porcelain crucibles and covers. Place them over Tirrill burners
and heat at the maximum temperature of the burner for 10 to 15 min; then place in the desiccator to
cool for at least 1 h. Weigh accurately each crucible with its cover. Between heating and weighing, the
crucibles and covers should be handled only with a pair of tongs.
3.
Prepare the 0.25M BaCl2 solution.
15.4 Procedure
1.
Preparation of the sample. Weigh accurately to four significant figures, using the direct
method, three sample of about 0.5 to 0.7 g each. Transfer to 400-mL beakers, dissolve in 200 to 250
mL distilled water, and add 0.5 mL concentrated hydrochloric acid to each.
2.
Precipitation. Assume the sample to be pure sodium sulfate and calculate the millimoles
barium chloride required to precipitate the sulfate in the largest sample. Heat the solutions nearly to
boiling on a wire gauze over a Tirrill burner and adjust the burner to keep the solution just below the
boiling point. Add slowly from a buret, drop by drop, 0.25 M barium chloride solution until 10% more
than the above calculated amount is added; stir vigorously throughout the addition. Let the precipitate
settle. Then test for complete precipitate by adding a few drops of barium chloride without stirring. If
additional precipitate forms. Add slowly, with stirring, 5 mL more barium chloride; let settle, and test
again. Repeat this operation until precipitation is complete. Leave the stirring rods in the beakers,
cover with watch glass and digest on the steam bath until the supernatant liquid is clear.(the initial
precipitate is fine particles. During digestion, the particles grow to filterable size.) This will require 30
to 60 min or longer. Add more distilled water if the volume falls below 200mL
32
3.
Filtration and washing of the precipitate. Prepare three 11-cm No. 42 what man filter paper or
equivalent for filtration; the paper should be well fitted to the funnel so that the stem of the funnel
remains filled with liquid, or the filtration will be very slow. See the discussion in chapter 2 for the
proper preparation of filters. Filter the solutions while hot; becareful not to fill the paper too full, as
the barium sulfate has a tendency to “creep” above the edge of the paper. Wash the precipitate into the
filter with hot distilled water, dream the adhering precipitate from the stirring rod and beaker with the
rubber policeman, and again rinse the contents of the beaker into the filter, examine the beaker very
carefully for particles of precipitate that may have escaped transfer. Wash the precipitate and the filter
paper with hot distilled water until no turbidity appears when a few milliliters of the washings
acidified with a few drops of nitric acid are tested for chloride with silver nitrate solution. Drying the
washing, rinse the precipitate down into the cone of the filter as much as possible. Examine the filtrate
for any precipitate that may have run through the filter.
4.
Ignition and weighing of the precipitate. Loosen the filter paper in the funnels and allow to
drain for a few minutes. Fold each filter into a package on closing the precipitate, with the triple
thickness of paper on top. Place in the weighed porcelain crucibles and gently press down into the
bottom impact the funnels for traces of precipitate; if any precipitate is found, wipe it off with a small
piece of moist ashless filter paper and add to the proper crucible. Place each crucible on a triangle on a
tripod or the ring of ring stand in an inclined position with the cover displaced slightly. Heat gently
with a small flame (Tirrill burner) until all the moisture has been driven off and the paper begins to
smoke and char. Adjust the burner so that the paper continues to char without catching fire. If the
paper inflames, cover the crucible to smother the fire, and lower the burner flame. When the paper has
completely carbonized and no smoke is given off, gradually raise the temperature enough to burn off
the carbon completely. A red glowing of the carbon as it burns is normal, but there should be no flame.
The precipitate should finally be white with no black particles. Allow to cool. Place the crucible in a
vertical position in the triangle, and moisten the precipitate with three or four drops of dilute (1:4)
sulfuric acid. Heat very gently until the acid has fumed off. (This treatment converts any precipitate
that has been reduce to barium sulfide by the hot carbon black to barium sulfate) then, cover the
crucibles and heat to dull redness in the full flame of the Tirrill burner for 15 min.
Allow the covered crucible to cool in the desiccator for at least 1h and then weigh them. Heat
again to redness for 10 to 15 min, cool in the desiccator, and weigh again. Repeat until two successive
weighings agree within 0.3 to 0.4 mg.
15.5 Calculation
Calculate and report the percent SO3 in your unknown for each portion analyzed report also the mean
and the standard deviation.
33
APPEDIX 1
Volume Occupied by 1.000g of water weighed in Air against stainless steel weights.
Volume, mL
corrected to
Temperature,T, oC
At T
20oC
10
1.0013
1.0016
11
1.0014
1.0016
12
1.0015
1.0016
13
1.0016
1.0018
14
1.0018
1.0019
15
1.0019
1.0020
16
1.0021
1.0022
17
1.0022
1.0023
18
1.0024
1.0025
19
1.0026
1.0026
20
1.0028
1.0028
21
1.0030
1.0030
22
1.0033
1.0032
23
1.0035
1.0034
24
1.0037
1.0036
25
1.0040
1.0037
26
1.0043
1.0041
27
1.0045
1.0043
28
1.0048
1.0046
29
1.0051
1.0048
30
1.0054
1.0052
Corrections for buoyancy (stainless steel weights) and change in container volume have been applied.
APPENDIX 2 critical values Q (P=0.05
APPENDIX 1 CRITICAL VALUES QF Q (P =0.05)\
Sample size
Critical value
4
0.831
5
0.717
34
6
0.621
7
0.570
8
0.524
9
0.492
10
0.464
Taken from E.P.King, Journal or American statistical Association, 1958, 4
8,531 by permission from the American Statistical Association,
eviation may be calculated also using the following equivalent equation:
S=
∑x
2
1
− (∑ x 1 )2 / N
N −1
35