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QUANTITATIVE ANALYTICAL
CHEMISTRY
1) Volumetric
(titrimetric)
analysis
2)
Gravimetric
analysis
3) instrumental
analysis
Acid-Base
Electrochemistry
preciptimetry
Spectrophotometry
Complexmetry
Spectrofluorometry
Redox
Flame photometric
methods
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Statistics
Redox theory
Redox application
Complexometry
‫محمد عبد الجليل‬:‫د‬
‫حريه‬.‫د‬.‫أ‬
‫حسن عزقل‬.‫د‬.‫أ‬
‫سميحه‬.‫د‬.‫أ‬
STATISTICS
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Error: deviation from the absolute value .
Absolute error (E) : the difference between an
observed or measured value (O) and the true value
(T) , with regard to the sign and it is reported in
the same units as the measurement.
E=O–T
Mean error: the difference between the average
of several measurements and the true value (T).
Relative Error: absolute or mean error (E)
expressed as a percentage (%) of the true value
(T)
Example,1:
 If a 2.62 g sample of material analyzed to be
2.52 g.
 The absolute error (E) = 2.52 – 2.62 = -0.10 g.
 Example,2:
 In the titration of 10 ml of 0.1 N NaOH with
laboratory prepared 0.1 N HCl, the true value
is 9.9 ml, and we have:
 10.1, 9.9, 9.8, 10.2, 10.1 observed values,
 So, Mean = summation of observed values / their
number
= (10.1 + 9.9 + 9.8 + 10.2 + 10.1) / 5 = 10.02 ml.
 And mean error = 10.02 – 9.9 = 0.12 ml.
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In example,1:
Relative Error = (-0.10/2.62) x 100% = - 3.8
%
 In example,2:
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Relative Mean Error = ( 0.12/9.9) x 100% = 1.21%
Types of Errors:
 (A) Determinate or systemic (constant)
errors:
 can be determined, (can be avoided)
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(B) Indeterminate (random, accidental
or chance) errors:
cannot be determined or corrected.
Accuracy:agreement of a measurement with the
true value.
 Determination of accuracy:
Absolute method
 Accuracy is determined from the relative error;
 In example,1: Relative Error = (-0.10/2.62) x 100%
= -3.8 %
 And accuracy = 100.0 – 3.8 = 96.2 %.
 In example,2: Relative Mean Error = ( 0.12/9.9) x
100% = 1.21%
 And accuracy = 100.00 – 1.21 = 98.79 %.
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Example 3:
In practical exam of volumetric analysis,
three students get the following results:
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Precision :The agreement between several
measurements of the same substance.
Mean (X):It is the arithmetic average of all
measured values.
The range (w): the"spread":
It is the difference between the highest
measurement and the lowest one.
 The median: It is the measurement in the
middle of the arranged measurements where
the numbers of higher and lower
measurements are equal.
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standard deviation (s):
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Variance =The square of the standard
deviation = S2
Relative standard deviation (RSD)
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Coefficient of variation (C.V.) :
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Example:
 Analysis of a sample of iron ore gave the
following % values:
7.08, 7.21, 7.12, 7.09, 7.16, 7.14, 7.07,
7.14, 7.18, 7.11.
 Calculate the mean, standard deviation, the
variance and coefficient of variation;
 Find also the median and the range for these
data.
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0.0182
Standard deviation 
 0.0020  0.045
10  1
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Variance (S2) = 0.002
(b) The arranged data are:
7.07, 7.08, 7.09, 7.11, 7.12, 7,14, 7,14,
7.16, 7.18, 7.21
The median is : (7.12 + 7.14) / 2 = 7.13
The range is : 7.21 – 7.07 = 0.14
Rejection of a result (The Q test):
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The Q test is used to determine if an “outlier”
is due to a determinate error or due to
indeterminate error.
If it is due to a determinate error, it should
be rejected.
If it is not due to a determinate error, then it
falls within the expected random error and
should be retained.
The ratio Q is calculated by arranging the
data in decreasing order of numbers.
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The difference (a) between the suspect
number (the outlier) and its nearest
neighbour number is divided by:
the range (w), which is the difference
between the highest number and the lowest
number,
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The ratio is compared with the tabulated
values of Q (see the Table).
If Q measured is equal or greater than the
tabulated value, the suspected observation
can be rejected.
If it is smaller than the tabulated value, the
suspected observation is retained
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Example:
The following set of chloride analysis on
separate aliquots of serum were reported;
103, 106, 107 and 114 meq/L. one value
appears suspect.
Determine if it may be rejected or not.
Answer:
 The suspected result is 114 meq/L.
 It differs from the nearest neighbor by (a) : 114 –
107 = 7 meq/L.
 The range (w) is : 114 – 103 = 11 meq/L.
 Therefore, Q = a/w = 7/11 = 0.64
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The tabulated Q value (4 observations, 95%
confidence level) is: 0.829
Since the calculated Q value is less than the
tabulated Q value,
the suspected no. (114 meq/L) retained.
Significant figures
 ‘digit’ = 0, 1, 2, ………..8,9
 A significant figure = is a digit which denotes the amount
of quantity in the place in which it stands.
 The digit 0 is a significant figure except when it is the first
figure in a number.
 In 1.2680 g and 1.0062 g
5
 the zero is significant,
 but in the quantity 0.0025 kg
2
 the zero is not significant, because 0.0025 kg = 2.5 g.
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1g
means that it is between 0.9
and 1.1 g
1.0 g
means that it is between 0.99 and 1.01 g
1.00 g means that it is between 0.999 and 1.001g
Take 10.0 ml of Zn2+ sample, add 10 ml of NH3-buffer
Weigh 1.000 g of powdered drug sample, add 2 g of
hexamine reagent …..
1 kg of tomato xxxxxxx 1.000 kg of gold !!!
volume which is known to be between 20.5 ml and 20.7 ml
should be written as 20.6 ml; but not as 20.60.
20.60 ml indicates that the value lies between 20.59 ml
and 20.61 ml.
Rejection quotient, Q,
at different confidence levels
No. of
Observations
3
4
5
6
7
8
9
10
15
20
25
30
Q90
0.941
0.765
0.642
0.560
0.507
0.468
0.437
0.412
0.338
0.300
0.277
0.260
Confidence level
Q99
Q95
0.994
0.970
0.926
0.829
0.821
0.710
0.740
0.625
0.680
0.568
0.634
0.526
0.598
0.493
0.568
0.466
0.475
0.384
0.425
0.342
0.393
0.317
0.372
0.298