ANALYTICAL CHEMISTRY (Errors in Chemical Analysis)

1. ANALYTICAL
CHEMISTRY
(Errors in Chemical
Analysis)

2. Mean
value
A mean value is obtained by dividing the sum of a
set of replicate measurements by the number of
individual results in the set.
For example, if a titration is repeated four times and
the titre values are 10.1, 9.9, 10.0 and 10.2ml
Mean = 10.1 + 9.9 + 10 + 10.2
4
= 40.2
4
= 10.05
This mean value is also called arithmetic mean or
average.

3. Mean
value
A mean value is obtained by dividing the sum of a
set of replicate measurements by the number of
individual results in the set.
For example, if a titration is repeated four times and
the titre values are 10.1, 9.9, 10.0 and 10.2ml
Mean = 10.1 + 9.9 + 10 + 10.2
4
= 40.2
4
= 10.05
This mean value is also called arithmetic mean or
average.

4. The
median
This is a value about which all other values in
a set are equally distributed. Half of the
values are greater and the other half smaller
numerically, compared to the median.
For example: If we have a set of values like
1.1, 1.2, 1.3, 1.4 and 1.5 the median value is
1.3.
When a set of data has an even number of
values, then the median is the average of the
middle pair.

5. The
median
This is a value about which all other values in
a set are equally distributed. Half of the
values are greater and the other half smaller
numerically, compared to the median.
For example: If we have a set of values like
1.1, 1.2, 1.3, 1.4 and 1.5 the median value is
1.3.
When a set of data has an even number of
values, then the median is the average of the
middle pair.

9. Precision
• Precision is defined as the agreement
between the numerical value of two or more
measurements of the same object that have
been made in an identical manner. Thus, a
value is said to be precise, when there is
agreement between a set of results for the
same quantity.
• Good precision does not guarantee accuracy

10. Methods of expressing
precision
_
• Precision can be expressed in an absolute
method. In the absolute way the
deviation from the mean
│xi ‒X│ expresses precision without
considering sign
S.No
Sample of an
organic compound
% of carbon
Deviation from mean
_
│xi ‒X │
1 X1 38.42 0.20
2 X2 38.02 0.20
3 X3 38.22 0.00
_
X = 38.22
0.40 = 0.133
3 (Average deviation)

11. Methods of expressing
precision
_
• Precision can be expressed in an absolute
method. In the absolute way the
deviation from the mean
│xi ‒X│ expresses precision without
considering sign
S.No
Sample of an
organic compound
% of carbon
Deviation from mean
_
│xi ‒X │
1 X1 38.42 0.20
2 X2 38.02 0.20
3 X3 38.22 0.00
_
X = 38.22
0.40 = 0.133
3 (Average deviation)

12. Methods of expressing
precision
_
• Precision can be expressed in an absolute
method. In the absolute way the
deviation from the mean
│xi ‒X│ expresses precision without
considering sign
S.No
Sample of an
organic compound
% of carbon
Deviation from mean
_
│xi ‒X │
1 X1 38.42 0.20
2 X2 38.02 0.20
3 X3 38.22 0.00
_
X = 38.22
0.40 = 0.133
3 (Average deviation)

13. Methods of expressing
precision
_
• Precision can be expressed in an absolute
method. In the absolute way the
deviation from the mean
│xi ‒X│ expresses precision without
considering sign
S.No
Sample of an
organic compound
% of carbon
Deviation from mean
_
│xi ‒X │
1 X1 38.42 0.20
2 X2 38.02 0.20
3 X3 38.22 0.00
_
X = 38.22
0.40 = 0.133
3 (Average deviation)

14. Accura
cy
• Accuracyrepresents
the
nearness to
a
measurement to its expected value.
Any difference between the measured
value and the expected value is
expressed as error.
• For example: The dissociation constant
for acetic acid is 1.75×10‒5 at 25 °C. In
an experiment, if a student arrives at
exactly this value, his value is said to
be accurate.

17. Accuracy versus precision
1.Accuracy is the closeness of a measurement
to the true (or accepted) value (μ or xt).
Accuracy is expressed by the absolute error
or the relative error:

18. Absolute
error
• The term accuracy is denoted in terms of
absolute error E, E is the difference
between the observed value (Xi) and the
expected value (Xt).
E = │
Xi – Xt│
• If a student obtains a value of 1.69×10‒5 for
the dissociation constant of acetic acid at
25°C, the absolute error in this
determination is
E = │1.69 ×10‒5 –1.75×10‒5 │
= │0.06 ×10‒5 │

19. Absolute
error
• The term accuracy is denoted in terms of
absolute error E, E is the difference
between the observed value (Xi) and the
expected value (Xt).
E = │
Xi – Xt│
• If a student obtains a value of 1.69×10‒5 for
the dissociation constant of acetic acid at
25°C, the absolute error in this
determination is
E = │1.69 ×10‒5 –1.75×10‒5 │
= │0.06 ×10‒5 │

20. Relative
error
• Sometimes the term relative error is used to
express the uncertainty in data. The relative
error denotes the percentage of error
compared to the expected value. For the
dissociation constant value reported.
Relative error = 0.06 ×10‒5 × 100
1.75×10‒5
= 3.4%

21. Relative
error
• Sometimes the term relative error is used to
express the uncertainty in data. The relative
error denotes the percentage of error
compared to the expected value. For the
dissociation constant value reported.
Relative error = 0.06 ×10‒5 × 100
1.75×10‒5
= 3.4%

22. Proble
m:
• The actual length of a field is 500 feet.
instrument shows the length to
be 508
A
measuri
ng feet.
Find:
a.) the absolute error in the measured length of the
field. b.) the relative error in the measured length of
the field.
Solution:
• (a)The absolute error in the length of the field is 8 feet.
E =│ Xi – Xt│ =508-500 = 8 feet.
• b.) The relative error in the length of the field is
Relative error = 8 × 100
500
= 1.6%

23. Proble
m:
• The actual length of a field is 500 feet.
instrument shows the length to
be 508
A
measuri
ng feet.
Find:
a.) the absolute error in the measured length of the
field. b.) the relative error in the measured length of
the field.
Solution:
• (a)The absolute error in the length of the field is 8 feet.
E =│ Xi – Xt│ =508-500 = 8 feet.
• b.) The relative error in the length of the field is
Relative error = 8 × 100
500
= 1.6%

24. Proble
m:
• The actual length of a field is 500 feet.
instrument shows the length to
be 508
A
measuri
ng feet.
Find:
a.) the absolute error in the measured length of the
field. b.) the relative error in the measured length of
the field.
Solution:
• (a)The absolute error in the length of the field is 8 feet.
E =│ Xi – Xt│ =508-500 = 8 feet.
• b.) The relative error in the length of the field is
Relative error = 8 × 100
500
= 1.6%

25. Error
s
two main types
• Determinate errors
• Indeterminate errors

26. Determinate errors:
These errors are determinable and are
avoided if care is taken. Also known as
Systemic errors.
•predictable
•correctable (using a ref.)
•in the same direction ( + or -)

27. Sources of Systematic (Determinate) Errors
•Instrumental errors
•Operative / Personal errors
•Method errors

28. Instrumental
error
• Instrumental errors are introduced due to the
use of defective instruments.
• For example an error in volumetric analysis
will be introduced, when a 20ml pipette,
which actually measures 20.1ml, is used.
• Sometimes an instrument error may arise
from the environmental factors on the
instrument.
• For example a pipette calibrated at 20°C, if
used at 30°C will introduce error in volume.
• Instrumental errors may largely be eliminated
by

29. Operative
errors
• These errors are also called personal errors
and are introduced because of variation of
personal judgements.
• For example due to colour blindness a
person may arrive at wrong results in a
volumetric or
colorimetric analysis.
• Using incorrect mathematical
equations
an
d
caus
e
committingarithmetic mistakeswill
also operative errors.

30. Detection of Systematic Instrument and
Personal Errors
• analysis of standard samples
• independent analysis
• blank determinations
• variation in sample size

31. Method errors
• These errors are caused by adopting
defective experimental methods.
• For example in volumetric analysis the use of
an improper indicator leading to wrong
results is an example for methodic error.
• Proper understanding of the theoretical
background of the experiments is a
necessity for avoiding methodic errors.

32. Method Errors
•Instability of the reagent
•Slowness of some reactions
•Loss of solution by evaporation
•Interferences (pH measurements at
high/low pHs)
•Contaminants

33. Indeterminate
errors
• These errors are also called accidental
errors. Indeterminate errors arise from
uncertainties in a measurement that are
unknown and which cannot be controlled
by the experimentalist.
• For example: When pipetting out a liquid,
the speed of draining, the angle of
holding the pipette, the portion at which
the pipette is held, etc, would introduce
indeterminate error in the volume of the
liquid pipette out.

34. Random (Indeterminate)
Errors
•Affect precision but not
accuracy
•Follows a Gaussian or normal
distribution
•Most values fall close to the
mean, with values farther away
becoming less likely. The width
of the distribution tells us
something about the precision of
our measurement.

35. Random Error
• always present
• unpredictable
• non-correctable (equal probability of being + or -)
• can be reduced by averaging multiple measurements
• can be treated mathematically (with statistical methods)

36. Gross Error—(Human) silly mistakes:
•occur only occasionally
•often large (+ or -)
•undetected mistakes during the experiment
•can be verify by “Q-test”
Examples:
0.1000 recorded as 0.0100
1.00 g as 1.00 mg
Wrong connection of electrode wires

38. Normal error
Curve
• Thenormalerror curvewasfirst studied
by Carl FriedrichGaussas a curve for
the
the
distributionof errors. He found that
distribution of errors could be
closely
approximated by a curve
called the normal
curve of errors.

39. Gaussian Distribution
Two parameters define a Gaussian distribution for a population, the
population mean, μ, and the population standard deviation, σ.

40. General properties of a normal error curve
(a normalized Gaussian distribution of errors)
a.The mean (or average) is the central point of maximum
frequency (i.e., the top of the bell curve).
b.The curve is symmetric on both sides of the mean (i.e.,
50% per side).
c.There is an exponential decrease in resulting frequency
as you move away from the mean.
d.If time and expense permit, you need to perform more
than 20 replicates when possible to be sure that the
sample mean and standard deviation are sufficiently
close to the population mean and standard deviation.

41. Normal error
Curve
• This normal distribution curve is a useful
one to measure the extent of
indeterminate error. It is given by
 is the standard deviation
x = value of the continuous random
variable. µ = mean of the normal
random variable
π =constant = 3.14

42. Normal error
Curve
• In normal error curve, the frequency is
plotted against mean deviation.
• When the frequency is maximum the error is
nil.
• When the frequency decreases,
the magnitude of the error increases

43. Normal error
Curve
• When  is very large, the curve
obtained is bell shaped. When  is
very small, then a sharp curve is
obtained.
• When frequency
increases, the 
decrease → sharp curve → nil
error.
• When frequency decreases the 
increase → bell shaped curve →
increases
wil
l
wil
l
erro
r

44. Normal error
Curve
• The normal distributions are extremely
important in statistics and are often
used in science for real valued random
variables whose distributions are not
known.

45. Significant
figure
• Data have to be reported with care keeping in
mind reliability about the number of figures
used.
• For example, when reporting a value as many
as six decimal numbers can be obtained,
when one uses a calculator.
• However, reporting all these decimal numbers
is meaningless because, as is generally true,
there may be uncertainty about the first
decimal itself.
• Therefore, experimental data should be
rounded off.

46. Significant
Figures
Rules for Counting Significant Figures
2. Zeros
a. Leading zeros - never count
0.0025 2 significant figures
b. Captive zeros - always count
1.008 4 significant figures
c. Trailing zeros - count only if the number is written
with a decimal point
100 1 significant figure
100. 3 significant figures
120.0 4 significant figures

47. Significant
figure
• A zero is not a significant figure, when used
to locate a decimal. However, it is significant
when it occurs at the end.
• For example 0.00405 has three significant
figures, the two zeros before the 4 being
used to imply only the magnitude, but
0.04050 has four significant figures, the zero
beyond the 5 being significant.

48. Significant
figure
• The number of significant figures in a given
number is found by counting the number
figures from the left to right in the number
beginning with the first non-zero digit and
continuing until reaching the digit that
contains the uncertainty. Each of the
following has three significant figures.
646 0.317 9.22 0.00149 20.2

49. • How many significant figures
are in: 1. 12.548
2. 0.00335
3. 504.70
4. 4000
5. 0.10200

50. • How many significant figures
are in: 1. 12.548 - 5
2. 0.00335
3. 504.70
4. 4000
5. 0.10200

51. • How many significant figures
are in: 1. 12.548 - 5
2. 0.00335 - 3
3. 504.70
4. 4000
5. 0.10200

52. • How many significant figures
are in: 1. 12.548 - 5
2. 0.00335 - 3
3. 504.70 - 5
4. 4000
5. 0.10200

53. • How many significant figures
are in: 1. 12.548 - 5
2. 0.00335 - 3
3. 504.70 - 5
4. 4000 - 1
5. 0.10200 - 5

54. Correct answer: 10004.5

55. Correct answer: 10004.5

56. Correct answer: 10004.5

57. Write the sum of 1.586 + 2.31 with the
correct number of significant figures.
1.586 + 2.31 = 3.896
3.896 = 3.90
Write the difference of 0.954 - 0.3109 with
the correct number of significant figures.
0.954 - 0.3109 = 0.6431
0.6431 = 0.643

58. Write the sum of 1.586 + 2.31 with the
correct number of significant figures.
1.586 + 2.31 = 3.896
3.896 = 3.90
Write the difference of 0.954 - 0.3109 with
the correct number of significant figures.
0.954 - 0.3109 = 0.6431
0.6431 = 0.643

59. Write the sum of 1.586 + 2.31 with the
correct number of significant figures.
1.586 + 2.31 = 3.896
3.896 = 3.90
Write the difference of 0.954 - 0.3109 with
the correct number of significant figures.
0.954 - 0.3109 = 0.6431
0.6431 = 0.643

65. Write the product of 2.10 × 0.5896 with
the correct number of significant figures
2.10 × 0.5896 = 1.23816
1.23816 = 1.24
Write the quotient of 16.15 / 2.7 with the
correct number of significant figures.
16.15 / 2.7 = 5.98148148
5.98148148 = 6.0

66. Write the product of 2.10 × 0.5896 with
the correct number of significant figures
2.10 × 0.5896 = 1.23816
1.23816 = 1.24
Write the quotient of 16.15 / 2.7 with the
correct number of significant figures.
16.15 / 2.7 = 5.98148148
5.98148148 = 6.0

67. Write the product of 2.10 × 0.5896 with
the correct number of significant figures
2.10 × 0.5896 = 1.23816
1.23816 = 1.24
Write the quotient of 16.15 / 2.7 with the
correct number of significant figures.
16.15 / 2.7 = 5.98148148
5.98148148 = 6.0

68. Rules for Rounding Off Numbers
•When the number to be dropped is less
than 5 the preceding number is not
changed.
•When the number to be dropped is 5 or
larger, the preceding number is
increased by one unit.
•Round the following number to 3 sig
figs: 3.34966 x 104
=3.35 x 104

69. Rules for Rounding Off Numbers
•When the number to be dropped is less
than 5 the preceding number is not
changed.
•When the number to be dropped is 5 or
larger, the preceding number is
increased by one unit.
•Round the following number to 3 sig
figs: 3.34966 x 104
=3.35 x 104

70. Thank you and have a good
day!