Errors in Chemistry
Experiments
Rohana Sirimanna
Reading errors (Data Collection)
• When reading off a scale the error is half
the last decimal place you can use or the
value given to you for this piece of
equipment.
• Electronic Balance
• If it reads to 0.001 g the
error / uncertainty is ± 0.001 g
Pipette
http://www.labsglassware.com
Pipette
Burette
We have checked with our glassware suppliers and have decided the error will be ± 0.05 mL
Volumetric Flasks
250.0 mL ± 0.15 mL
100.0 mL ± 0.1 mL
Other Equipment
• Thermometer:
• Glass ± 0.25 oC Digital ± 0.1 oC
• Measuring Cylinders:
– 100 mL ± 1 mL,
– 10 mL ± 0.2 mL,
– 25 mL ± 0.5 mL
Combining errors
(Data Analysis)
• When data is added or subtracted the
errors need to be added.
• Eg.
22.45 ± 0.01g - 20.32 ± 0.01 g
=
2.13 ± 0.02 g
When averaging data we will use
the half range rule for the errors.
• This means subtracting the lowest reading
from the highest reading and then dividing
this answer by two.
• (If this number is smaller than the reading
error, use the reading error.)
Average titration results
Average these titration results:
24.3 ± 0.1 mL,24.1 ± 0.1 mL, 24.5 ± 0.1 mL
• The average of 24.3, 24.1 and 24.5 is 24.3
•
The range will be 24.5 – 24.1 = 0.4.
Divide this by 2 to get 0.2
•
Therefore the average titre is
24.3 ± 0.2 mL
Using data in formulae
(Data Analysis)
• Most commonly this is done using
•
n = m/Mr
c = n/V
Δ H = msΔT
or
mole ratios.
The errors from above first need to be
converted into percentage errors.
2.13 ± 0.01 g
the percentage error is calculated by:
0.01 x 100
=
0.47%
2.13
The percentage errors for each piece of
data used in the arithmetic are added.
Then the percentage error is converted
back into a “real” error.
The errors from above first need to be
converted into percentage errors.
• An answer for a concentration of
0.26 mol dm-3 ± 2.3%
should be converted by:
2.3% of 0.26 is .023 x 0.26 = 0.00598
•
This should be written as:
0.26 ± 0.01 mol dm-3
Significant figures
• In IB you have to be aware
of using the correct number
of significant figures in your
calculations or you will lose
marks.
Significant figures
• Examples
•
•
•
•
3.65 has 3 significant figures.
52.9 has 3 significant figures.
374.85 has 5 significant figures.
0.5822 has 4 significant figures.
Zeroes
• We need to be careful about zeroes
because sometimes they are counted and
sometimes not.
• This will be apparent in the following
examples:
1.If the zeroes precede the first non-zero
digit they are NOT significant.
Eg 0.04 has 1 significant figure.
0.00025 has 2 significant figures.
Zeroes
2. If the zeroes are between non-zero digits they
ARE significant.
Eg
0.304 has 3 significant figures.
807 has 3 significant figures.
0.04092 has 4 significant figures.
Zeroes
3. If the zeroes follow the non-zero digits we can't
be sure if there is no decimal point. If there is a
decimal point we count the zeroes.
Eg 0.50 has 2 significant figures.
250.00 has 5 significant figures
21.60 has 4 significant figures.
35000 may have 2, 3, 4 or 5 significant
figures depending on the accuracy of the data.
Using scientific notation helps here.
Using scientific notation helps here.
If it is 3.50 x 104 there are 3 significant
figures.
If it is 3.5 x 104 there are 2 significant
figures.
If it is 3.500 x 104 there are 4 significant
figures.
If it is 3.5000 x 104 there are 5 significant
figures.
Writing the Final Answer
• The final answer must not have more
significant figures than the data with the
least number of significant figures.
• In this modern age of calculators you are
able to store many decimal places as the
calculation is done.
• You must do this and then convert your
answer to the appropriate number of
significant figures at the end.
Worked examples
•
•
•
1.
2.
3.
4.
Pure washing soda crystals have the formula
Na2CO3 . 10H2O.
A student was given some old crystals which were
white and flaky and was asked to find out their formula.
[It will not have 10 water molecules any more]
This is the procedure the student used:
“A crucible and lid were weighed. Some crystals
were placed in the crucible and weighed with the lid.
The crucible was heated, gently at first, and then more
strongly, with the lid being left slightly ajar.
The crucible, lid and residue were allowed to cool and
were then re-weighed.
The heating, cooling and weighing were then repeated.
Calculation
• The results were: (mass in g ± 0.01g)
Mass of crucible + lid
Mass of crucible + lid + crystals
Mass of crucible + lid + crystals after 1st heating
Mass of crucible + lid + crystals after 2nd heating
19.41
24.76
22.06
22.06
Calculation
2Titration between 25.0 mL of 0.450 mol dm-3
NaOH and H2SO4 gave the following titres:
• Therefore the concentration of the sulfuric acid is
0.235 ± 0.003 mol dm-3
• Answer has 3 s. f. because the data with the
lowest number of significant figures was the
pipette and the NaOH concentration which both
had 3. The burette volumes also had 3.
• The error should only affect the last decimal
place hence it is
± 0.003 mol dm-3 and
not
± 0.003008 mol dm-3.
• Notice how SMALL this error is
and therefore how accurate your
school laboratory equipment is.
• Remember this when you do your
EVALUATION.