Error Analysis
Significant Figures in Calculations
Every lab report must have an error analysis. For many experiments, significant
figure rules are sufficient. For a brush up on significant figure rules, see your General
Chemistry or Analytical text. Remember to carry at least one extra significant figure to
avoid round off error through intermediate calculations. Non-significant figures are
written smaller or like "subscripts" to avoid confusion. Insignificant figures can also be
underlined. For example, 0.1234 0.001 or 0.002 would be written 0.1234 or 0.1234. On
the other hand, 0.1234 0.003 would be written 0.123 or 0.123. Keeping track of
significant figures in long calculations is easy. Just underline the insignificant digit in
your Excel spreadsheets for all the intermediate and final results using a pen or pencil.
To find the uncertainties and approximate number of significant figures when using
volumetric glassware use Table 1.
Table 1. Capacity Tolerances for Class A Volumetric Glassware.
Pipets
1 ml.
2
5
10
15
20
25
±0.006
0.006
0.01
0.02
0.03
0.03
0.03
Sig. Figs
3
3
3
4
4
4
4
Flasks
10 ml.
25
50
100
200
250
1000
±0.02
0.03
0.05
0.08
0.10
0.12
0.30
Sig. Figs
4
3
3
4
4
4
4
A 10-ml pipet is listed as 10.00 0.02, which is close enough to 4 significant figures,
10.00 ml. But a 1-ml pipet is listed as 1.000 0.006, which is really only 3 significant
figures, 1.00 ml.
The significant figure rules are:
SF Rule 1: In multiplication and division the number of significant figures in the result
is the same as the smallest number of significant figures in the data.
SF Rule 2: In addition and subtraction the number of decimal places in the result is the
same as the smallest number of decimal places in the data.
SF Rule 3: The number of significant figures in the mantissa of log x is the same as the
number of significant figures in x. Use the same rule for ln x. (In log 4.23x10-3 = -2.374,
the mantissa is the .374 part.)
SF Rule 4: The number of significant figures in 10x is the number of significant figures
in the mantissa of x. Use the same rule for ex.
Example 1: Concentration Calculations: A solution is made by transferring 1 ml of a
0.12453 M solution, using a volumetric pipet, into a 200-ml volumetric flask. Calculate
the final concentration.
…Solution: The 1-mL volumetric pipet has 3 significant figures; all the other values have
4. The calculations all involve multiplication and division, so the final answer should be
expressed with 3 significant figures.
1.00 x 0.12453 M / 200.0 = 0.0006226 M = 6.22x10-4 M
Example 2: Logs: The equilibrium constant for a reaction at two different temperatures
is 0.0322 at 298.2 and 0.473 at 353.2 K. Calculate ln(k2/k1).
…Solution: Both k’s have 2 significant figures, so k2/k1 should also have 2 significant
figures: k2/k1 = 13.89. Then using SF Rule 3 shows that ln k2/k1 should have 2 significant
figures in the “mantissa:”
ln k2/k1 = ln 13.89 = 2.631
Example 3: Antilogs: The rate of a reaction depends on temperature as
ln k = ln A –Ea/RT. Using curve fitting it was found that ln A = 9.874. Calculate A.
…Solution: The result is e9.874 = 1.9427x104. The mantissa has only 2 significant figures,
so the result should only have two significant figures (SF Rule 4): 1.94x104, or just
1.9x104.
Example 4: Antilogs: The rate of a reaction depends on temperature as
ln k = ln A –Ea/RT. Using curve fitting it was found that ln A = 9.874 and Ea= 28.26
kJ/mol. Calculate k at T = 298.15 K.
…Solution: The result for ln k has 2 significant figures: both ln A and Ea have 3
significant figures (SF Rule 1), but the difference has only one past the decimal point:
limiting term on subtraction
ln k = 9.874 – 28.26x103/8.314/298.15 = 9.874 – 11.40 = -1.5266
The mantissa has only 1 significant figure. Then solving for k and using SF Rule 4 gives
k= e-1.5266 = 0.217 or finally just 0.2.
2
Propagation of Errors
Significant figure rules are sufficient when you don't have god estimates for the
measurement errors. If you do have good estimates for the measurement errors then a
more careful error analysis based on propagation of error rules is appropriate. Least
squares curve fitting provides very good estimates for uncertainties. For error analysis
with the slope or intercept from least squares curve fitting, a little more care is justified
than is provided by significant figure rules. Use propagation of error rules to find the error
in final results derived from curve fitting.
The propagation of error rules are listed below. The variance of x, s(x)2, is the square
of the standard deviation.
Rule 1: Variances add on addition or subtraction. s(z)2 = s(x)2+s(y)2
s(z)2 s(x)2 s(y)2
Rule 2: Relative variances add on multiplication or division. z2 = x2 + y2
s(x)2
2
Rule 3: The variance in ln(x) is equal to the relative variance in x. s(z) = x2 .
The variance in log(x) is the (relative variance in x)/(2.303)2.
s(z)2
Rule 4: The relative variance in ex is equal to the variance in x. z2 = s(x)2.
The relative variance in 10x is equal to the (variance in x)(2.303)2.
Rule 5: In calculations with only one error term, you can work with standard deviation
instead of variance.
Rule 6: The variance of an average of N numbers, each with variance s2, is s2/N.
(The standard deviation in the average improves as 1/ N ).
Example 5: Subtraction: If Z = A – B with A = 163.455±0.002 and B = 1.34±0.02
calculate Z. Find the uncertainty in the result.
...Solution: Work with absolute variances (Rule 1) and note that variances always add
(i.e. the error always builds up):
variance in A = (0.002)2
+ variance in B = (0.02)2
Variance in result (0.02)2
Standard deviation = 0.02
Result (163.455±0.002) – (1.34±0.02) = 162.12±0.02
3
Example 6: Multiplication: The result of an experiment is given by (slope x Cp). Let
slope = 123.2±2.4 and Cp=4.184±0.031. Find the uncertainty in the result.
...Solution: The relative variance of the result is the sum of the relative variances of the
data (Rule 2).
2
Relative variance in slope = (2.4/123.2) = (0.019)2 =3.6x10-4
+ Relative variance in Cp = (0.031/4.184)2= (0.0074)2 =5.5x10-5
Relative variance in product
=4.2x10-4
-4
Relative st.dev in product= √4.2x10 = 0.020 or 2.0%
Result =( 123.2±2.4 )x( 4.184±0.031)= 515.47 ± 2.0% = 515. ± 10.
Example 7: Logs: The equilibrium constant for a reaction at two different temperatures
is K1 = 0.0322±0.0007 at 298.2 and K2 =0.473±0.006 at 353.2 K. Calculate ln(K2/K1) and
find the uncertainty in the result.
…Solution: Just like Example 6 the relative variance in k2/k1 is the sum of the relative
variances:
Relative variance in K2 = (0.0007/0.0322)2= (0.022)2
= 4.7x10-4
2
2
+ Relative variance in K1 = (0.006/0.473) = (0.013)
= 1.6x10-4
Relative variance in K2/K1
= 6.3x10-4
Then using Rule 3 shows that absolute variance in ln K2/K1 is the relative variance in
K2/K1.
Variance in ln K2/K1 = Relative variance in K2/K1 = 6.3x10-4
Standard deviation in result = √6.3x10-4 = 0.025
ln K2/K1 = ln 14.689 = 2.687± 0.025 = 2.69±0.03
See example 2 for the significant figure version of this error analysis.
Example 8: Antilogs: The rate of a reaction depends on temperature as
ln k = ln A –Ea/RT. Using curve fitting it was found that ln A = 9.874±0.0041. Calculate
A, and find the uncertainty.
…Solution: The result is e9.874 = 1.9427x104. The relative variance of the result is the
absolute variance of A (Rule 4):
Relative variance of e9.874 = variance of A = (0.0041)2 = 1.68x10-5
Relative st. dev in result = √1.68x10-5 = 4.1x10-3 or 0.41%
Result = 1.9427x104± 0.41% = 1.943x104 ± 0.080x104.
See example 3 for the significant figure version of this error analysis.
4
Example 9: Multiplication with 'Certain' Numbers: The result of a calculation is
(slope x R), where R is the gas constant in J K-1 mol-1. Let slope = 1.23±0.02. Find the
uncertainty in the result.
...Solution: Since R is known to several more significant figures than the slope, the
uncertainty in R will add very little to the error in the final result. Therefore, R is 'certain'
for this calculation. Rule 5 applies since only the slope is in error. Therefore, the error in
the final result is then just the standard deviation in the slope multiplied by R:
Result = slope x R = 1.23 ±0.02 x 8.3144126 = 10.23 ± 0.17
Example 10: The Inverse of the Slope or Intercept: The result of an experiment is the
inverse of the intercept from a graph, 1/b. Let b=0.523 ± 0.043. Find the uncertainty in
the result.
...Solution: The relative variance in the result is equal to the relative variance in the
intercept (Rule 2). We can also work directly in terms of standard deviation (Rule 5):
Relative st.dev in b = 0.043/0.523 = 0.082 or 8.2%
Relative st. dev of Result = Relative st. dev in b = 8.2%
Result = 1/b = 1/0.523 = 1.912 ± 8.2% = 1.91 ± 0.16
Example 11: Division and Subtraction: The term (1/T2- 1/T1) is a very common factor
in many equations. For T1 = 298.2 ±0.2 K and T2= 353.2 ±0.2 K calculate
(1/T2- 1/T1). Find the uncertainty in the result.
…Solution: Don’t be put off by multi-step problems, just work one step at a time. First,
get the uncertainty in 1/T2 and 1/T1. Since both of these are divisions the relative variance
of 1/T is just the relative variance of T (Rule 2). Then convert to absolute variance to
calculate the error in (1/T2- 1/T1) using Rule 1:
Relative variance in 1/T2 = (0.2/353.2)2 = 3.2x10-7
Relative variance in 1/T1 = (0.2/298.2)2 = 4.5x10-7
Variance in 1/T2 = 3.2x10-7 (1/353.2)2 = 2.6x10-12
+ Variance in 1/T1 = 4.5x10-7 (1/298.2)2 = 5.1x10-12
Variance in (1/T2- 1/T1)
= 7.7x10-12
-12
Standard deviation in result = √7.7x10 = 2.8x10-6
Result = -5.22x10-4± 3x10-6
5
Example 12: A Multi-Step Problem An example of a more realistic problem is the
temperature dependence of the equilibrium constant. Let’s assume we wish to evaluate
rH, knowing K2, K1, R, T2, and T1 in the equation:
K2
1
1
rH
ln(K ) = – R ( T - T )
1
2
1
where K1 =0.0322±0.0007 at 298.2 and K2 =0.473±0.006 at 353.2 K with T = ±0.2 K
(The same data as Examples 7 and 11!). Find the uncertainty in the result.
…Solution: Solving for rH:
K2
8.314 ln(K )
1
∆rH = - 1
= 42.785 kJ/mol
1
(T2 - T1)
We already know the uncertainty in ln(K2/K1) from Example 7, 2.687± 0.025, and the
uncertainty in (1/T2- 1/T1) from Example 11, -5.22x10-4± 2.8x10-6. R is a certain number.
The relative variance in ∆rH is then just the sum of the relative variances:
Relative variance ∆rH = (0.025/2.687)2 + (2.8x10-6/ 5.222x10-4)2 = 1.19x10-4
Relative standard deviation of ∆rH = √1.19x10-4 = 0.011 or 1.1%
Standard deviation of ∆rH = (0.011)(42.785) = 0.47 kJ/mol
Result ∆rH = 42.78 ± 0.47 kJ/mol
As the calculations get longer, the error analysis can get to be rather tedious. However,
spreadsheets can come to the rescue. When you set up your spreadsheets just include the
error analysis. For this problem you might set up the following:
B
4
5
6
7
8
9
10
11
12
13
14
15
16
17
C
K
K2
K1
D
±
0.473
0.0322
E
relative
variance
0.006 1.609E-04
0.0007 4.726E-04
F
T2
T1
G
T (K)
±
353.2
298.2
H
I
relative
variance
0.2 3.206E-07
0.2 4.498E-07
rel variance variance
±
1/T2-1/T1= -5.222E-04 2.798E-05 7.629E-12 2.762E-06 K-1
ln K2/K1=
rH=
rH=
2.6871289 8.773E-05 6.335E-04 0.0251694
4.278E+04 1.157E-04 2.118E+05 460.22886 J/mol
42.785
±
0.460 kJ/mol
R=
8.314447 J/mol/K
The important cells are
Variance in (1/T2- 1/T1)
6
E10: =(I6/G6^2+I7/G7^2)
Variance in ln K2/K1
Relative variance ∆rH
E12: =E6+E7
D14: =D10+D12.
It is a good habit to use spreadsheets for all your lab calculations and to include the
error analysis, if the lab requires a full propagation of errors treatment (see below).
Summary:
Keep this tutorial handy, you will need it all year long. Every lab requires an error
analysis for the final results, even calculations-only lab reports.
All lab reports should have an error analysis
If the calculations don’t use least squares curve fitting or the curve fitting is used at
a very early stage of the calculation with many subsequent calculations, just use
significant figure rules (e.g. for all the calorimetry experiments just use significant
figure rules).
From least squares curve fitting use propagation of error rules
7
HOMEWORK
Name_______________________
1. 5 ml of a 0.2134 M HCl is diluted to 100 ml using volumetric glassware. Calculate the
pH of the final concentration. Use significant figure rules. Present the results using the
proper number of significant figures.
2. The binding constant for a guest-host complex is given by slope/intercept of a graph.
Using curve fitting it was found that the slope=0.2265±0.0379 and intercept=1.601x10-4±
1.99x10-5. Calculate the binding constant and the uncertainty using propagation of errors
rules. (This is real Colby research data, by the way.)
3. The number of ligands in a complex is determined by the inverse of the intercept of a
graph. The intercept is b=0.341 ± 0.023. Calculate the number of ligands and the
uncertainty of the result. Is it safe to assume that the answer is = 3.00 to within
experimental error?
4. The rate of a reaction depends on temperature as ln k = ln A –Ea/RT. Using curve
fitting it was found that ln A = 9.874±0.023 and Ea= 28.26±0.03 kJ/mol. Calculate k at T
= 298.15 K. Use propagation of errors.
8
5. Write a spreadsheet to fit the following data. The data and the final result are listed
below to use to check your spreadsheet. But you use the formulas with the explicit
sums—you don’t use the Excel linest() function for your assignment!
Excel Least Squares
x
y
Curve Fitting
fit y
-1.5
1
3.5
6
-1.35
2.67
4.68
6.74
-0.76
1.871
4.499
7.127
slope
slope ±
r2
F
regression SS
1.0512
0.136972
0.967159
58.89904
34.53192
0.8198
0.4914789
0.7656958
2
1.17258
intercept
intercept ±
s(y) ±
degrees-of-freedom
residual SS
Test Cha rt
8
7
6
5
4
Y (units?)
3
y
f it y
2
1
0
-2
0
2
-1
-2
-3
X (units ?)
9
4
6