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Errors in Chemistry Experiments

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

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Errors in Chemistry Experiments

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  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. Pipettehttp://www.labsglassware.com

  4. Pipette

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

  6. Volumetric Flasks 250.0 mL ± 0.15 mL 100.0 mL ± 0.1 mL

  7. 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

  8. 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

  9. 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.)

  10. 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

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

  12. 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.

  13. 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

  14. 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.

  15. 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.

  16. 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 NOTsignificant. Eg 0.04 has 1 significant figure. 0.00025 has 2 significant figures.

  17. Zeroes 2. If the zeroes are between non-zero digits they AREsignificant. Eg 0.304 has 3 significant figures. 807 has 3 significant figures. 0.04092 has 4 significant figures.

  18. Zeroes • 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.

  19. 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.

  20. 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.

  21. Worked examples • 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.

  22. 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

  23. Calculation

  24. 2Titration between 25.0 mL of 0.450 mol dm-3 NaOH and H2SO4 gave the following titres:

  25. 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.

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

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