Computing Uncertainties and errors in Laboratory Data and Result
- Accuracy is the closeness of agreement between a measured value and the true value.
- Precision is the closeness of agreement between independent measurements of a quantity under the same conditions.
- Uncertainty is the component of a reported value that characterizes the range of values within which the true value is asserted to lie.
- Error is the difference between a measurement and the true value of the object being measured. Error does not include mistakes.
We will consider the error and uncertainty in experimental measurements and calculated results. You wil be asked to show the measurements and errors in your labs/exams.
These are some important tips you will need.
Every measurement has an uncertainty associated with it, unless it is an exact, counted integer,
The two following examples exemplify these situations:
|MEASURED VALUE||COUNTED INTEGER|
|21.70 cm3, 37.0 cm3, 5.35 cm3
for all measurements, the last digit is estimated
|quantity: 4 ELEPHANTS
Counting items does not contain errors
- The numerical value “plus or minus” (±) the uncertainty value tells you the range of the result.
- For example a result reported as 5.35 ± 0.05 means that the experimenter has some degree of confidence that the true value falls in between 5.30 and 5.40
- Every calculated result also has an uncertainty, related to the uncertainty in the measured data used to calculate it.
- This uncertainty should be reported using the appropriate number of significant figures.
To consider error and uncertainty in more detail, we begin with definitions of accuracy and precision.
Accuracy and Precision WATCH THIS VIDEO
- The accuracy of a set of measurements is how close the measurement is to the true value of the quantity.
- The precision of a set of measurements is how close the measurements are to each other. (if an instrument is not properly calibrated, you may have high precision but not accuracy)
The relationship of accuracy and precision may be illustrated by using the following example:
You can see that good precision does not necessarily imply good accuracy. In #2, we have a systematic error.
Types of Error
The error of an observation is the difference between the observation and the actual or true value of the quantity observed.
Errors are often classified into two types:
Random errors vary in a completely non-reproducible way from measurement to measurement. but can be treated statistically, and so relate the precision of a calculated result to the precision with of the measurements taken.
Systematic errors may be caused by problems in either the equipment, the observer, or the use of the equipment. Systematic errors can result in high precision, but poor accuracy. They are difficult to discover. Examples:
♦ A student may overshoot the endpoint of a titration over and over again.
♦ A balance may always read 0.001 g more because it was zeroed incorrectly. .
Take, for example, the simple task of measuring the volume in the following picture:
You would probably , align your eyes to be able to see a imaginary horizontal line just underneath the meniscus.
Systematic errors come from several sources:
(1) Instrumental error
- (How “well calibrated” is the burette?
- How thin and how closely spaced are the burette’s marks?)
(2) Uncertainties in the thing being measured
- How smooth are the limits in the object measured?
- Is the liquid subject to temperature and humidity changes?)
A third source of error exists, related to how any measuring device is used.
(3-A) Parallax error
When using any measuring device you should look at the scale in a right angle. If this is not done you will produce parallax error.
this error applies not only to volumes but distances and masses as well.
(3-B) Using the equipment correctly
When you perform a titration, you fill the burette to the top mark and record 0.00 mL as your starting volume.
This is INCORRECT.
The correct way of using it, is:
- Add enough liquid so that the burette is nearly full,
- Then simply read the starting value to whatever precision the burette allows and record that value.
- It will be subtracted from your final burette reading to yield the most unbiased measurement of the delivered volume.
Absolute Error and % Error:
In your Cambridge Examination you will be asked to express the error in your measurements as absolute or %.
Although the detail in your Chemistry examination is much less than the required in Physics, you cannot express the sources of error as “I did not rinse the burette properly” this is not considered error but incompetence. Be sure not to mention HUMAN ERROR in the exam.
Consider weighing 1g of solid. If you use a two decimal place balance, the mass recorded will be to the nearest 0.01g.
We should express this measurement as (1.00 ± 0.01) g, where 0.01 is the absolute error.
In this example, the % error will be:
0.01 g x 100 = 1%
Error Propagation and Precision in Calculations
Most of the time, you will not asked to calculate the ERROR PROPAGATION in your exam, but you need to express the results with the correct amount of significant figures:
Lets work it out through an example:
We need to calculate the density of an object. You measure the mass using a 2 decimal place balance. You read 30.05 g, but you should record this as:
mass: 30.05 g ± 0.05 g – 4 SIGNIFICANT FIGURES % error: 0.166%
Then you use a 50 mL graduated cylinder and the volume measured is 25.5 mL, you should record this as:
volume: 25.5 mL ± 0.5 mL – 3 SIGNIFICANT FIGURES % error: 1.96%
Calculating the density:
DENSITY = MASS/VOLUME = 30.05 g / 25.5 mL = 1.17843 g/mL => this answer is WRONG!
The correct answer should show only 3 SIG FIG. so the correct answer will be
DENSITY: 1.18 g/mL
Error calculation in multiplications WILL NOT BE INCLUDED in this level.
Error in Titrations
One example of error propagation would be calculate the maximum error obtained when measuring the burette or pipette for a titration.
Maximum error is usually marked on the glassware.
- A 25 cm3 pipette has a maximum error of 0.02 or 0.03 cm3 in each measurement.
- A 50 cm3 burette has a maximum error of 0.05 cm3 in each measurement.
Calculation of percentage error
Percentage error = (maximum error ÷ quantity measured ) x 100%
Error in burettes
A burette is graduated in divisions every 0.1 cm3.
Using the half-division rule, the estimation is 0.05 cm3.
Burette is recorded to two decimal places with the last figure either ‘0’ or ‘5’.
The maximum error in each measurement = 0.05 cm3.
The overall maximum error in any volume measured always comes from two measurements, so
the overall maximum error = 2 x 0.05 cm3 = 0.1 cm3.
In a titration, a burette will typically deliver around 25 cm3 so the percentage error is small.
In your practical exam, if they ask you to calculate the percent error in the titration, you should use the average of the volume you calculated in the titration as the volume measured.
Example, you perform the titration, and you calculate that the average volume is 24.50 cm3
Percentage error = (2 × 0.05 cm3 ÷24.50 cm3) x 100% = 0.41% (no units)
The percentage error becomes more significant when burette is used to deliver small volume
For delivery of 5.50 cm3,
Percentage error = (2 × 0.05 cm3 ÷ 5.50 cm3)× 100% = 1.82% (no units)
Error in 25 cm3 Pipettes
The error for a 25 cm3 pipette, is written on the apparatus. Most of the time it will be around 0.02mL (cm3), so the error for the pipette will be
absolute error: 0.02 cm3
Percentage error = ( 0.02 cm3 ÷ 25.00 cm3)× 100% = 0.08%
Since the error when measuring with the burette is much bigger, it will be the one affecting the results the most.
How Many Digits Should We Show in our Measurements?
Significant digits or significant figures are numbers that are important in a measurement or calculation. Every time you perform a measurement, you need to estimate the number of significant digits to show, and that depends on the instrument that is used for that purpose.
A very useful method is to calculate the amount measured between two consecutive lines of the instrument and then divide that amount by two. The answer will give you an idea on how many decimals you must show in your measurements and how much would be the error involved.
- When using a 50 cm3 burette:
The arrows show two consecutive lines in each burette.
A – 18.4 cm3 – 18.3 cm3 = 0.1 cm3 → 0.1 cm3 / 2 = 0.05 cm3 → measurement: 18.75 cm3
B – 0.2 cm3 – 0.1 cm3 = 0.1 cm3 → 0.1 cm3 / 2 = 0.05 cm3 → measurement: 9.25 cm3
C – 23.4 cm3 – 23.3 cm3 = 1 cm3 → 0.1 cm3 / 2 = 0.05 cm3 → measurement: 23.00 cm3
- The scale in the burette is upside down.
- All measurements have 2 decimals and the last digit will be 0 or 5.
- All measurements will have 3 or 4 significant digits.
- The last digit will always be 0 or 5 depending on the position of the lower meniscus.
- When using a graduated cylinder, the accuracy depends on the graduated scale that it contains.
The arrows show two consecutive lines in each graduated cylinder.
A – 23.4 cm3 – 23.2 cm3 = 0.2 cm3 è 0.2 cm3 / 2 = 0.1 cm3 è measurement: 24.0 cm3
B – 9.7 cm3 – 9.6 cm3 = 0.1 cm3 è 0.1 cm3 / 2 = 0.05 cm3 è measurement: 9.25 cm3
C – 79 cm3 – 78 cm3 = 1 cm3 è 1 cm3 / 2 = 0.5 cm3 è measurement: 75.5 cm3
- All measurements given have 3 significant digits.
- Not all measurements have the same number of decimals.
How Many Digits Should We Show in our calculations?
Many times, your calculator will display an answer containing more digits than the number of significant figures you obtained in the measurements.
You calculate through a titration that 20.35 mL of NaOH solution neutralize 25.0 mL of a 0.110 mol.dm-3 solution of HCl, you should perform the following calculations:
- Find the number of moles of Hydrochloric acid used è0 / 1000 x 0.110 = 0.00275
- Find the concentration of sodium hydroxide solution. è00275 x 20.35 / 1000 = 0.1351351
- Burette reading for HCl volume used = 20.35 cm3- (4 sig dig)
- Volume of NaOH = 25.0 cm3- (3 sig dig)
- Concentration of NaOH = 0.110 mol/dm3- (3 sig dig)
You may notice that your calculator shows 0.1351351, but since the volume of NaOH was given with 3 significant digits, and the concentration of HCl is also given with 3 significant digits, the correct value is 0.135 mol dm-3.
How many decimals? How many Significant digits?
Rules for calculations vary based on the calculation performed.
|Addition and Subtraction||Multiplication and Division|
| 1.4767g + 3.2g + 3.507 g = 8.1787 g
(4) (1) (3 ) = (1) (lowest)
( ) denotes the number of decimals
| 1.4767 cm x 3.20 cm x 3.507 cm = 16.5485 cm3
(5) (3) (4 ) = (3) (lowest)
( ) denotes the number of significant digits
|Calculator shows: 8.1787||Calculator shows: 16.5485|
|Correct answer: 8.2 g 1 decimal||Correct answer: 16.5 cm3 3 sig dig|
How to Round Off
When performing calculations, you need to know which digits are important (significant) and which are not. The significant figures are kept and the insignificant figures are dropped. You need to know what to do with the last significant digit in our number, and that depends on the FIRST NON SIGNIFICANT DIGIT.
If the number AFTER the last significant digit is 5 or more, round up
25.0 x 0.110 / 20.00 = 0.1375 → 0.138 (3 sig fig)
25.0 x 0.200 / 11.50 = 0.43478 → 0.435 (3 sig fig)
If the number AFTER the last significant digit is lower than 5, you just drop the numbers.
25.0 x 0.110 / 21.75 = 0.12644 → 0.126 (3 sig fig)
25.0 x 0.010 / 12.25 = 0.02041 → 0.0204 (3 sig dig) → 0.020 (2 sig fig)
(you should show 0.020 since the concentration of the acid is given with 2 sig dig)
Final Clarification: In your Cambridge Examination, you will be asked for the correct amount of sig fig in your answer. If nothing is stated, Most of the time, you should use 3 or 4 significant figures. Look for clues in the numbers given.
here you have a word document with the instructions given above