Presentation is loading. Please wait.

Presentation is loading. Please wait.

Errors in Chemistry Experiments Rohana Sirimanna.

Similar presentations

Presentation on theme: "Errors in Chemistry Experiments Rohana Sirimanna."— Presentation transcript:

1 Errors in Chemistry Experiments Rohana Sirimanna

2 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

3 Pipette

4 Pipette

5 Burette We have checked with our glassware suppliers and have decided the error will be ± 0.05 mL

6 Volumetric Flasks 250.0 mL ± 0.15 mL100.0 mL ± 0.1 mL

7 Other Equipment Thermometer: Glass ± 0.25 o CDigital ± 0.1 o C Measuring Cylinders: –100 mL ± 1 mL, – 10 mL ± 0.2 mL, –25 mL ± 0.5 mL


9 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

10 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.)

11 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

12 Using data in formulae (Data Analysis) Most commonly this is done using n = m/M r c = n/V Δ H = msΔT or mole ratios.

13 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.

14 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

15 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.

16 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.

17 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. Eg0.04 has 1 significant figure. 0.00025 has 2 significant figures.

18 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.

19 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. Eg0.50 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.

20 If it is 3.50 x 10 4 there are 3 significant figures. If it is 3.5 x 10 4 there are 2 significant figures. If it is 3.500 x 10 4 there are 4 significant figures. If it is 3.5000 x 10 4 there are 5 significant figures.

21 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.

22 Worked examples Pure washing soda crystals have the formula Na 2 CO 3. 10H 2 O. 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: 1.A crucible and lid were weighed. Some crystals were placed in the crucible and weighed with the lid. 2.The crucible was heated, gently at first, and then more strongly, with the lid being left slightly ajar. 3.The crucible, lid and residue were allowed to cool and were then re-weighed. 4.The heating, cooling and weighing were then repeated.

23 Calculation The results were:(mass in g ± 0.01g) Mass of crucible + lid 19.41 Mass of crucible + lid + crystals 24.76 Mass of crucible + lid + crystals after 1st heating 22.06 Mass of crucible + lid + crystals after 2nd heating 22.06

24 Calculation


26 2Titration between 25.0 mL of 0.450 mol dm -3 NaOH and H 2 SO 4 gave the following titres:




30 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.

31 Notice how SMALL this error is and therefore how accurate your school laboratory equipment is. Remember this when you do your EVALUATION.


Download ppt "Errors in Chemistry Experiments Rohana Sirimanna."

Similar presentations

Ads by Google