Transcript analytical chem-lect 4

ERT 207
Analytical Chemistry
ERT 207 Analytical Chemistry
Lecture 4
23rd July 2007
ERT 207 Analytical Chemistry
Titles to be covered:
• 2.3 The effect of errors towards data
analysis
• 2.4 Significant Figures
• 2.5 Standard Deviation
• 2.6 Propagation of error
ERT 207 Analytical Chemistry
2.3 THE EFFECT OF ERRORS
TOWARDS DATA ANALYSIS
2.3.1 WAYS OF DESCRIBING
ACCURACY
2.3.2 WAYS OF DESCRIBING
PRECISION
ERT 207 Analytical Chemistry
2.3.1 WAYS OF DESCRIBING
ACCURACY
•
•
•
•
•
•
Recall: Accuracy is expressed as absolute error
or relative error.
Errors will show the closeness of measurements to
the accepted or correct values.
Absolute errors: E = O – A
E = absolute error
O = observed error
A = accepted value
ERT 207 Analytical Chemistry
• Absolute errors is the difference between true
value and measured value.
• For example, if 2.62 g sample analyzed to be 2.52
g:
• Therefore, the absolute error = 2.52 g -2.62 g
•
= -0.10 g.
• The positive and negative sign is assigned to show
whether the errors are high or low.
ERT 207 Analytical Chemistry
• If the measured value is average of several
measurements, the error is called the mean
error.
• Q1: How to calculate mean error?
• A1: By taking the average difference wrt
sign, of the individual test results from the
true value.
ERT 207 Analytical Chemistry
•Q1: What is
relative error?
Relative errors
ERT 207 Analytical Chemistry
• A1: Relative error is the absolute or
mean error expressed as a percentage
of the true value.
ERT 207 Analytical Chemistry
•
•
•
•
•
•
Consider the previous case,
Measured value = 2.52 g
True value = 2.62 g
The relative error = -0.10 X 100%
2.62
= -3.8%
ERT 207 Analytical Chemistry
• Now, Q2: What is relative accuracy?
• A2: Relative accuracy is the measured
value or mean expressed as a
percentage of the true value.
ERT 207 Analytical Chemistry
• Task: Calculate the relative accuracy of
the above case.
• Solution:
• Relative accuracy = 2.52 X 100%
•
2.62
•
= 96.2 %
ERT 207 Analytical Chemistry
• Example (3.6 from Gary D. Christian):
• The result of an analysis are 36.97 g, compared
with the accepted value of 37.06 g. What is the
relative error in parts per thousand?
• Solution:
• Absolute error = 36.97 g – 37.06 g
•
= -0.09 g
• Relative error = -0.09 g X 100%o
•
37.06 g
•
= -2.4 ppt
• 100%o indicates parts per thousand.
ERT 207 Analytical Chemistry
2.3.2 WAYS OF DESCRIBING PRECISION
Recall: Precision can be
expressed as standard deviation,
deviation from the mean,
deviation from the median, and
range or relative precision.
ERT 207 Analytical Chemistry
Example 1: The analysis of chloride ion on
samples A, B and C gives the following
result
Sample %Cl-
Deviation from
mean
Deviation
from
median
A
B
C
0.10
0.09
0.01
0.11
0.08
0.00
đ = 0.07
(d bar)
0.06
24.39
24.20
24.28
= 24.29
ERT 207 Analytical Chemistry
• Deviation from mean
• Deviation from mean is the difference between the
values measured and the mean.
• For example, deviation from mean for sample A=
0.10%
• Q1: How to get X ?
• A1: = 24.39 + 24.20 + 24.28
•
3
•
= 24.29
ERT 207 Analytical Chemistry
• Deviation from median
• Deviation from median is the difference
between the values measured and the
median.
• Example, deviation from the median for
sample A = 0.11%.
• Now, identify median = 24.28
• Therefore, deviation from median for
sample A = 24.39- 24.28 = 0.11%
ERT 207 Analytical Chemistry
• Range is the difference between the
highest and the lowest values.
• For example, range = 24.39- 24.20 =
0.19 %.
ERT 207 Analytical Chemistry
• Sample standard deviation
For N (number of measurement) < 30
ERT 207 Analytical Chemistry
For N > 30
ERT 207 Analytical Chemistry
2.4 SIGNIFICANT FIGURES
Q1 : What is
significant figures?
ERT 207 Analytical Chemistry
• A1: The number of digits necessary to
express the results of a measurement
consistent with the measured precision.
ERT 207 Analytical Chemistry
• A2: Digits that are known to be certain plus
one digit that is uncertain.
• -The zero value is significant if it is part of
the numbers.
• -The zero values is not significant figure
when it is used to show magnitude or to
locate decimal points.
• -The position of decimal points have no
relation with significant figures.
ERT 207 Analytical Chemistry
• The number of significant figures = the
number of digits necessary to express the
results of a measurement consistent with the
measured precision.
ERT 207 Analytical Chemistry
• Since there is uncertainty (imprecision) in
any measurement of at least ± 1 in the last
significant figure, the number of significant
figures includes all of the digits that are
known, plus the first uncertain one.
ERT 207 Analytical Chemistry
• Each digit denotes the actual quantity it
specifies. For example, in the number 237,
we have 2 hundreds, 3 tens, and 7 units.
ERT 207 Analytical Chemistry
• The digit 0 can be a significant part of a
measurement, or it can be used merely to
place the decimal point.
• The number of significant figures in a
measurement is independent of the
placement of the decimal point.
ERT 207 Analytical Chemistry
• Take the number 92,067. This number has
five significant figures, regardless of where
the decimal point is placed.
• For example. 92.067 μm, 9.2067 cm,
0.92067 dm and 0.092067 m all have the
same number of significant figures.
• They merely represent different ways
(units) of expressing one measurement.
ERT 207 Analytical Chemistry
• The zero between the decimal point and the 9 in
the last number is used only to place the decimal
point.
• There is no doubt whether any zero that follows a
decimal point is significant or is used to place the
decimal point.
• In the number 727.0, the zero is not used to locate
the decimal point but is a significant part of the
figure.
ERT 207 Analytical Chemistry
• Ambiguity can arise if a zero precedes a
decimal point. If it falls between two other
nonzero integers, then it will be significant.
Such was the case with 92,067.
• In the number 936,600, it is impossible to
determine whether one or both or neither of
the zeros is used merely to place the
decimal point or whether they are a part of
the measurement.
ERT 207 Analytical Chemistry
• It is best in cases like this to write only the
significant figures you are sure about and
then to locate the decimal point by
scientific notation.
• Thus, 9.3660 X 10 has five significant
figures, but 936,600 contains six digits,
one to place the decimal.
ERT 207 Analytical Chemistry
• Example:
• List the proper number of significant
figures in the following numbers and
indicate which zeros are significant.
0.216; 90.7; 800.0; 0.0670
ERT 207 Analytical Chemistry
• Solution
• 0.216 three significant figures
• 90.7 three significant figures; zero is
significant
• 800.0 four significant figures; all zeros are
significant
• 0.0670 three significant figures; only the
last zero is significant
ERT 207 Analytical Chemistry
Steps in Writing Significant Figures
• Make sure that there is no digit after the
first uncertainty figure.
• For example:32.5239
uncertainty figure
ERT 207 Analytical Chemistry
Rounding Off
• If the last digit to be removed is greater than
5, add one to the second last digit.
• Example, 22.486
22.49
• If the last digit to be removed is smaller
than 5, then the second last digit does not
change.
• Example, 31.392
31.39
ERT 207 Analytical Chemistry
• If the last digit is 5 and the second last digit
is an even number, thus the second last digit
does not change.
• Example, 73.285
73.28
• If the last digit is 5 and the second last digit
is an odd number, thus add one to the last
digit.
• Example, 63.275
63.28
ERT 207 Analytical Chemistry
Addition and Subtraction Operation
• The number of digits to the right of decimal
point in the operation of
addition/subtraction should remain.
• The answer to this operation has a value
with the least decimal point.
ERT 207 Analytical Chemistry
• Example : Give the answer for the
following operation to the maximum
number of significant figures.
ERT 207 Analytical Chemistry
43.7
4.941
+ 13.13
61.771
61.8
• answer is therefore 61.8 based on the key
number (43.7).
ERT 207 Analytical Chemistry
Multiplication and Division Operation
• The number of significant figures in this
operation should be the same as the number
with the least significant figure in the data.
• Example : Give the correct answer for the
following operation to the maximum
number of significant figures.
• 1.0923 x 2.07
ERT 207 Analytical Chemistry
• Solution:
• 1.0923 x 2.07 = 2.261061
2.26
• The correct answer is therefore 2.26 based
on the key number (2.07).
ERT 207 Analytical Chemistry
Exponential
• The exponential can be written as follows.
Example, 0.000250
2.50 x 10-4
ERT 207 Analytical Chemistry
•
•
•
•
2.5 STANDARD DEVIATION
The most important statictics.
Recall back: Sample standard deviation.
For N (number of measurement) < 30
ERT 207 Analytical Chemistry
For N > 30,
ERT 207 Analytical Chemistry
• Try these questions on standard deviations.
ERT 207 Analytical Chemistry
• Thank you
ERT 207 Analytical Chemistry