Chapter 6:
Random Errors in Chemical
Analysis
TMHsiung@2014 1/48
Contents in Chapter 06
1. The Nature of Random Errors
1) Random Errors and Their Mathematical Equations
2) Central Limit Theorem
2. Gausian (Normal) Distribution
1) Define Gausian Distribution
2) Z Value Transformation
3) Z value Applications
4) Constructing Gaussian Curve From Experimental
Data
3. Standard Deviation of Calculated Results
4. Significant Figures
TMHsiung@2014 2/48
1. The Nature of Random Errors
1) Random Errors and Their Mathematical Equations
 Random (Indeterminate, Statistic) error:
Errors affecting the precision (uncertainty), caused
by uncontrollable variables. Positive and negative ()
fluctuation occur with approximate equal frequency.
 Population:
The set of infinite objects in the system being
investigated.
 Sample:
The finite members of a population that we actually
collect and analyze.
TMHsiung@2014 3/48
 Spread (range, w): The difference between extreme
values in a set of d
 Mean (Average)
 xi
mean  i
symbol :  for population , x for sample
n
 Standard deviation
N
- For population:  
 ( xi   ) 2
i 1
N
N
- For sample:
●
s
If n  20, s  σ, x  μ
2
(
x

x
)
 i
i 1
N 1
TMHsiung@2014 4/48
 Variance = s2
*variance
are additive
 Relative standard deviation (RSD) by 100%
(also called coefficient of variation, CV):
s
RSD in percent  100%
x
 Pooled standard deviation:
For several subsets of data, estimating standard
deviation by pooling (combining) the data.
S pooled 
N1
N2
N3
i 1
j 1
k 1
2
2
2
(
x

x
)

(
x

x
)

(
x

x
)
 i 1  j 2  k 3 ...
N1  N 2  N 3  ...  N t
TMHsiung@2014 5/48
 Degree of freedom (dof)
In statistics, the number of independent observations
on which a result is based.
- For standard deviation:
dof = n-1,
n is munber of measurement.
- For pooled standard deviation:
dof = N – M,
N: Total number of measurements
M: Number of the subset.
TMHsiung@2014 6/48
2) Central Limit Theorem
i) Definition
 For the probability distribution plot for the frequency
of individual values. The measurements subject to
indeterminate errors arise a normal (Gaussian)
distribution.
 The sources of individual error must be independent.
 The individual error must have similar magnitude (no
one source of error dominates the final distribution).
TMHsiung@2014 7/48
ii) Simulated Central Limit Theorem
(Frequency of Combinations of four equal-sized uncertainty, u)
Combination of uncertainty
Magnitude of
Combination
Frequency of
combinations
Relative
frequency
+U1+U2+U3+U4
+4U
1
1/16=0.0625
+2U
4
4/16=0.250
0
6
6/16=0.375
-2U
4
4/16=0.250
-4U
1
1/16=0.0625
-U1+U2+U3+U4
+U1-U2+U3+U4
+U1+U2-U3+U4
+U1+U2+U3-U4
-U1-U2+U3+U4
+U1+U2-U3-U4
+U1-U2+U3-U4
-U1+U2-U3+U4
-U1+U2+U3-U4
+U1-U2-U3+U4
+U1-U2-U3-U4
-U1+U2-U3-U4
-U1-U2+U3-U4
-U1-U2-U3+U4
-U1-U2 -U3 -U4
TMHsiung@2014 8/48
 4 random uncertainties
TMHsiung@2014 9/48
 10 random
uncertainties
 Infinite number of
random uncertainties
- A Gaussian (normal)
distribution curve.
- Symmetrical about the
mean.
TMHsiung@2014 10/48
2. Gausian (Normal) Distribution
1) Define Gausian Distribution
A “bell-shaped” probability distribution curve for
measurements showing the effect of random error,
which encountered continuous distribution.
The equation for Gaussian distribution:
x : individual value
y  f(x) 
1 e
 2

 
1
 2
2
(x


)

2 2
e
y : relative frequency (0 ~ 1)
 : true value
 : standard deviation
 ( x   ) 2 / 2 2
dx  1
TMHsiung@2014 11/48
IF: Same Mean, Different SD
1
y
e
 2

( x )2
2 2
982_Ch04
_Treat_Ra
ndDistr_D
ata.xls
TMHsiung@2014 12/48
IF: Different Mean, Same SD
982_Ch04
_Treat_Ra
ndDistr_D
ata.xls
TMHsiung@2014 13/48
IF: Different Mean, Different SD
982_Ch04
_Treat_Ra
ndDistr_D
ata.xls
TMHsiung@2014 14/48
2) Z value Transformation
 For x-axis: transform x to z by:
z
x

 For y-axis: transform to f(z) by :
1
y  f ( z) 
2
 1
 
2
z2

e 2
e
z2 / 2
dz  1
TMHsiung@2014 15/48
Now, the Gaussian Curve Always Consistent
982_Ch04
_Treat_Ra
ndDistr_D
ata.xls
z
x

1
y
e
2
z2

2
TMHsiung@2014 16/48
教戰守則
1. 經過 z transform, Gaussian curve 長得一模一樣，
此時 x-axis 的單位為σ。
2. 看到標示為 x，單位為 σ 時，代表已經經過 z
transform。
3. Relative frequency (y) at mean (x), 約為 0.4。
4. The entire area of the Gaussian curve is 1.
TMHsiung@2014 17/48
3) Z value Applications
Area (probability, percentage) in defined z interval
Area within  1σ
Area within  2σ
Area within  3σ
Area within  
1
1
1
e
 z 2 /2
dz  0.6826
2π
 z 2 /2
2 1
e
dz  0.9546
 2
2π
 z 2 /2
3 1
e
dz  0.9973
 3
2π
 z 2 /2
 1
e
dz  1
 
2π
TMHsiung@2014 18/48
4) Constructing Gaussian Curve From Experimental
Data
i) General procedure
Step 1: Raw data collection
Step 2: Arrange the data in order from lowest to
highest
Step 3: Condense the data by grouping them into cells
Step 4: Pictorial representation of the frequency
distributions
Step 5: Estimating the σ from s
Step 6: Plotting relative frequency versus x or z
TMHsiung@2014 19/48
Example: Replicate Data for the Calibration of a 10 mL Pipet
982_Ch04
_Treat_Ra
ndDistr_D
ata.xls
TMHsiung@2014 20/48
Cont’d
982_Ch04
_Treat_Ra
ndDistr_D
ata.xls
TMHsiung@2014 21/48
982_Ch04
_Treat_Ra
ndDistr_D
ata.xls
TMHsiung@2014 22/48
Relative frequency versus x
Number in range
Percentage in range
(x  μ) 2
(total pipet)(vol ume per bar)  2σ 2
y
e
s 2π

(50)(0.003 )
(0.006) 2π

y
(volume per bar)
s 2π
(x  9.982 ) 2
e 2(0.006)
2

(0.003)
(0.006) 2π
X-axis 刻度為體積之間隔


(x  μ) 2
e
962_Ch04_Trea
t_RandDistr_D
ata.xls
2σ 2
(x  9.982 ) 2
e 2(0.006)
2
TMHsiung@2014 23/48
Matching Histogram with Gaussian curve
TMHsiung@2014 24/48
Z Transformed Gaussian Curve
982_Ch04
_Treat_Ra
ndDistr_D
ata.xls
TMHsiung@2014 25/48
4. Standard Deviation of Calculated Results
(for Random Errors)
TMHsiung@2014 26/48
1) Sum or Difference
Example 1: The calibration result of class A 10 mL pipet showed that
the marker reading is 9.9920.006 mL. When it is used to deliver
two successive volumes. What is the absolute and percent relative
uncertainties for the total delivered volume.
Solution:
Total volume = 9.992 mL + 9.992 mL = 19.984 mL
sy  (0.006)2  (0.006)2  0.00848
Ans:
19.984(0.008) mL
TMHsiung@2014 27/48
Example 2: For a titration experiment, the initial reading is
0.05(0.02) mL and final reading is 17.88 (0.02) mL. What is the
volume delivered?
Solution:
Delivered volume = 17.88 mL – 0.05 mL = 17.83 mL
sy  (0.02)2  (0.02)2  0.028
Ans:
17.83(0.03) mL
TMHsiung@2014 28/48
2) Product or Quotient
Example 1: The quantity of charge Q = I x t. When a current of
0.150.01 A passes through the circuit for 1201 s, what is the total
charge?
Solution:
Total charge 0.15 A x 120 s = 18 C
sy
0.01 2
1 2
) (
)  0.067
y
0.15
120
sy  18  0.067  1.2
Ans:
 (
18(1) C
TMHsiung@2014 29/48
3) Mixed operations
TMHsiung@2014 30/48
4) Exponents and logarithms
Example: The pH of a solution is 3.720.03, what is the [H3O+] of
this solution?
sy
x
y  10
 2.3026 s x
Solution: [H3O+] = 10–pH
y
y = 10–3.72 =1.91x10–4
sy
y

sy
1.91 10
4
 2.3026  0.03  0.070
s y  y  (0.070 )  (1.91 104 )(0.070 )  1.3 105
y(sy) = 1.91 (0.13)x10–4
Ans:
1.9 (0.1)x10–4 M
TMHsiung@2014 31/48
4. Significant Figures
1) Statement of significant figures
 The digits in a measured quantity, including all digits
known exactly and one digit (the last figure) whose
quantity is uncertain.
 The more significant digits obtained, the better the
precision of a measurement.
 The concept of significant figures applies only to
measurements.
 The Exact values (e.g., 1 km = 1000 m) have an
unlimited number of significant figures.
TMHsiung@2014 32/48
2) Recording with significant figures
0
1
2
Recording: 1.4 or 1.5 or 1.6
0
1
3
2 significant figure,
1 certain, 1 uncertain
2
Recording: 1.51 or 1.52 or 1.53
3
3 significant figure,
2 certain, 1 uncertain
TMHsiung@2014 33/48
3) Rules for Zeros in Significant Figures
 Zeros between two other significant digits are
significant
e.g. 10023
no. of sig. fig.: 5
 A zero preceding a decimal point is not significant
e.g., 0.10023
no. of sig. fig.: 5
 Zeros preceding the first nonzero digit are not
significant
e.g. 0.0010023
no. of sig. fig.: 5
 Zeros at the end of a number are significant if they are
to the right of the decimal point
e.g. 0.1002300
no. of sig. fig.: 7
TMHsiung@2014 34/48
 Zeros at the end of a number may or may not be
significant if the number is written without a decimal
point
e.g. 92500
no. of sig. fig.: N/A
Scientific notation is required:
e.g. 9.25x104
no. of sig. fig.: 3
e.g. 9.250x104
no. of sig. fig.: 4
e.g. 9.2500x104
no. of sig. fig.: 5
TMHsiung@2014 35/48
2. Significant Figures in Arithmetic
1) Rules for rounding off numbers
i) If the digit immediately to the right of the last sig. fig.
is more than 5, round up.
ii) If the digit immediately to the right of the last sig. fig.
is less than 5, round down.
iii) If the digit immediately to the right of the last sig. fig.
is 5 followed by nonzero digits, round up.
iv) If the digit immediately to the right of the last sig. fig.
is 5
 round up if the last sig. fig. is odd.
 round down if the last sig. fig. is even.
v) If the resulting number has ambiguous zeroes, it should
be recorded in scientific notation to avoid ambiguity.
TMHsiung@2014 36/48
Examples:
35.76 in 3 sig. fig. is
35.8
35.74 in 3 sig. fig. is
35.7
24.258 in 3 sig. fig. is
24.3
24.35 in 3 sig. fig. is
24.4 (rounding to even digit)
24.25 in 3 sig. fig. is
24.2 (rounding to even digit)
13,052 in 3 sig. fig. is
1.31 x 104
TMHsiung@2014 37/48
2) Addition and Subtraction:
 The reported results should have the same number of
decimal places as the number with the fewest decimal
places
Example 1:
49.146
+ 72.13
–
9.1
112.176
Ans: 112.2 m
Example 2: MW of KrF2
m
m
m
m
18.9984032 (F)
+ 18.9984032 (F)
+ 83.798
(Kr)
121.7948064
Ans: 121.795 g/mol
TMHsiung@2014 38/48
Note:
To avoid accumulating “round-off ” errors in calculations:
1) Go through all calculation by calculator, then
rounding on the final answer
OR
2) Retaining one extra insignificant figure (a subscribe
digit) for intermediate results, then rounding on the
final answer
TMHsiung@2014 39/48
Example 3:
Write the answer with correct number of digits:
2.432x106 + 6.512x104 – 1.227x105 = ?
2.432
+ 0.06512
– 0.1227
2.37442
x 106
x 106
x 106
x 106
Ans: 2.374x106
TMHsiung@2014 40/48
3) Multiplication and Division:
 The reported results should have no more significant
figures than the factor with the fewest significant
figures
Example 1: 1.827 m × 0.762 m = ?
1.827 m x 0.762 m = 1.392174 m2 = 1.39 m2
Ans: 1.39 m2
Example 2:
(4.3179 x 1012)(3.6x10–19) = 1.554444x10–6 = 1.6x10–6
TMHsiung@2014 41/48
 Stoichiometric coefficients in a chemical formula,
and unit conversion factors etc, have an infinite
number of significant figures.
Example:
Results of four measurements: 36.4 g, 36.8 g, 36.0 g,
37.1 g. What is the average?
Solution:
(36.4+36.8+36.0+37.1)/4 = 36.6
Ans: 36.6 g
TMHsiung@2014 42/48
4) Logarithms and Antilogarithms
i) Mathematic terms
 For exponential expression, e.g. 4.2 x 10–2:
“4.2” is called the coefficient,
“–2” is called the exponent.
 For logarithm operation, e.g. log (4.2 x 10–2) = –1.38
“–1” is called the characteristic, the integer part
“38” is called the mantissa, the decimal part
TMHsiung@2014 43/48
ii) Logarithms operation
Example 1: log(56.7 x 106) = ?
Solution:
log(5.67 x 107) = 7 + log(5.67) = 7.754
3 sig. fig.
Example 2: log(0.002735) = ?
Solution:
log(0.002735) = – 2.5630
4 sig. fig.
Mantissa remain
equal number of
sig. fig. as the
original digit in
coefficient
Or
log(0.002735) = 2.735x10–3 = (– 3) + log(2.735)
= (–3) + (0.4370) = – 2.5630
TMHsiung@2014 44/48
Example 3: Percent transmittance (%T) is related to the
absorbance (A) by the equation: A = –log (%T/100). What
is correct digits of A if %T is 72.9.
Solution:
A = –log (%T/100) = –log (0.729) = – (– 0.137) = 0.137
Or
log (0.729) = – log(7.29x10–1) = – [(–1) + log (7.29)]
= – [(–1) + (0.863)] = 0.137
Example 4: The pH is defined as pH = –log [H3O+]. If the
[H3O+] is 3.8 x 10–2 M, what is the pH of the solution?
Solution:
pH = –log [H3O+] = –log (3.8 x 10–2) = –[(–2) + log(3.8)]
=– (–1.42) = 1.42
TMHsiung@2014 45/48
iii) Antilogarithms operation
Example 1: antilog(2.671) = ?
Solution:
antilog(2.671) = 102.671 = 100.671x102 = 4.69 x 102
Example 2: What is the correct digits of %T if the
absorbance is 0.931?
Solution:
Antilog (–0.931) = 10–0.931 = 0.117
Final digit has
%T = 0.117 x 100% = 11.7%
same number of
sig. fig. as
number of digits
in mantissa
TMHsiung@2014 46/48
Example 3: If the pH is 10.3, what is the [H3O+]?
Solution:
[H3O+] = 10–pH = 10–10.3 = 5 x 10–11
Or 10–10.3 = 100.7 x 10–11 = 5 x 10–11
Example 4: If the pH is 2.52, what is the [H3O+]?
Solution:
[H3O+] = 10–pH = 10–2.52 = 3.0 x 10–3
Or 10–2.52 = 100.48 x 10–3 = 3.0 x 10–3
TMHsiung@2014 47/48
Homework (Due 2014/3/6)
Skoog 9th edition, Chapter 06 Questions and Problems
6-1
6-4
6-5
6-9 (a) (c) (e)
6-11 (a) (c)
6-15
End of Chapter 06
TMHsiung@2014 48/48