Pre-DP 3

Content

● 1. Safety, equipment

● 2. Error, uncertainty

● Lab: Measurements and uncertainty

● Lab: Preparation of a solution

● 3. Stoichiometry, Limiting reactant, percentage yield

● Calculations

● 4. Writing a lab report

● Lab: Number of water molecules in a salt

● 5. Acids and bases

● 6. Acid-base titration

● Lab: Acid-base titration (Standardization of a solution)

Assessment● Based on the IB assessment model:

● Internal assessment i.e. practical work: 20 %● External assessment i.e. written exam: 80 %

● DEADLINE Lab report: - first draft 6.5.2014

- Final version 15.5.2014

● Revision 23.5 at 12.15 pm● EXAM 27.5 at 9.00-11.45● Returning exams 28.5 at 12.45

1.1 Safety

The European chemical hazard symbols

● Dangerous to the environment

● Oxidising● Longer term health hazards such as carcinogenicity andrespiratory sensitisation

● Toxic● Caution - used for less serious

health hazards like skin irritation● Gas under pressure

● Explosive● Corrosive

● Flammable

1.2 Equipment

https://www.youtube.com/watch?v=zcavPAFiG14

The understanding of science

● There is no single scientific method, but scientists have a common terminology and a common reasoning process.

● All reasons are based on evidence and argument.

● An idea is scientific if - it focuses on the natural world - has natural explanations- is testable

http://undsci.berkeley.edu/article/scienceflowchart

2. Errors and uncertainties in measurements

● In the laboratory, amounts are normally measured using either mass or volume.

2.1 Solids

● A solid is usually weighed to obtain its mass.

● The SI unit of mass is kilograms (kg), but chemists usually use grams (g).

1000 g = 1 kg

● The precision of the balance is indicated by the number of decimal places.

● 5,0 g of a substance would be recorded as 5,000 g on a balance weighing to ±0,001g.

2.2 Liquids

● Liquids may be weighed, or their volume may be measured.

● The mass can be obtained from the volume using:

density = mass

volume

● Density is measured in g cm-3 or kg dm-3

● Different apparatus is used for measuring volume depending on how precisely the volume is required.

d = m V

2.3 Absolute uncertainty

● An uncertainty range applies to any equipment used and occurs due to the limitations of the apparatus itself and the taking of readings.

● Analogue instruments (thermometer, measuring cylinder, burette etc.): The uncertainty is ± half the smallest division.

● Digital instruments (balance): The uncertainty is ± the smallest scale division.

2.4 Percentage uncertainty

● If the volume used from the burette is 20,0 cm3, the absolute uncertainty is ± 0,1 cm3 and the percentage uncertainty is:

0,1 x 100 = 0,5 %

20,0

2.5 Random errors

● Are a result of the uncertainties of the apparatus used and can not be avoided.

● Chance alone determines if these errors make the measured value smaller or larger than the true value, both are equally probably.

● Random errors are predictable and the degree of error can be calculated.

● eg. reading a scale

2.6 Systematic errors

● The systematic error makes the measured value smaller or larger than the true value, but not both (causes a bias in only one direction).

● eg. reaction time, incorrect calibration of equipment, standard solution etc

2.7 Precision

● how close several trials are to each other

2.8 Accuracy

● how close the result is to literature reference value

2.9 Percentage error

● A measure of how close the value obtained in the experiment is to the literature or accepted value.

percentage error = accepted value - experimental value · 100%accepted value

2.10 Total uncertainty● The overall percentage uncertainty is obtained by summing all

the individual uncertainties.

● When adding or subtracting uncertain values, add the absolute uncertainties:

initial temperature = 34.50ºC (±0.05ºC)

final temperature = 45.21ºC (±0.05ºC)

change in temperature: 45.21 – 34.50ºC = 10.71ºC

(±0.05ºC + 0.05) = ±0.1ºC

● Should be reported as 10.7ºC ± 0.1ºC

● When multiplying or dividing, add the percentage uncertainties:

mass = 9.24 g ± 0.005 g

volume = 14.1 cm3 ±0.05

density = m = 9.24 g = 0.655 g cm-3

V 14.1 cm3

● Convert the absolute uncertainties to percentage uncertainties

● Add the percentage uncertainties

● Convert the total uncertainty back to an absolute uncertainty

2.11 Significant figures

● When a measurement is taken there will be a random uncertainty in the reading. The significant figures must therefore be used in consensus with the measuring equipment.

● Zero can cause problems when determining the number of significant figures.

● Zero only becomes significant when it comes after a non-zero digit (1,2,3,4,5,6,7,8,9).

0.0001234 1.0234 1.2340

zero not a significant figure zero is a significant figure

values quoted to 4 sig. figs. values quoted to 5 sig. Figs.

● Zeros after a non-zero digit but before the decimal point may or may not be significant depending on how the measurement was made, e.g 123000

● This problem can be neatly overcome by using scientific notation.

e.g. 1.23000 x 106 quoted to six significant figures or 1.23 x 106 quoted to three significant figures.

● When adding or subtracting: The result should be expressed based on the measurement with the smallest number of decimal places.

e.g. 7.10 g + 3.10 g = 10.20 g 3 sig. figs. 3 sig. figs. 4 sig. figs

● When multiplying or dividing: The result should be expressed based on the measurement with the smallest number of significant figures.

● Ex. The mass of a beaker and a piece of aluminium metal is 35.4200 g. The mass of the empty beaker is 28.9200 g. If the aluminium displaces 2.41 cm3 of water, calculate the density of aluminium (in g cm-3)

ρ = m V

● The IB does not tend to penalise in exams if the number of significant figures in an answer differs by one from the correct number (unless the question specifically asks for them).

2.12 Preparation of a solution

● Preparation of a solution by dissolving a known mass of a solid (solute) into a specific amount of water (solvent):

m = c · V · M

● Ex. Prepare 1 liter of 1.00 M NaCl solution.

1. Calculate the molar mass of NaCl.

2. Weigh 58.44g NaCl and place it in a 1 L volumetric flask.

3. Add a small volume of distilled, deionized water to dissolve the salt. Once the sodium chloride has dissolved completely (swirl the flask gently), add water to bring the volume up to the final 1 L line.

2.13 Dilution of a solution

● Adding more solvent to a solution results in a solution of lower concentration.

c1V

1 = c

2V

2

● Ex.How many millilieters of 5.5 M NaOH are needed to prepare 300 mL of 1.2 M NaOH?

5.5 M x V1 = 1.2 M x 0.3 L

V1 = 1.2 M x 0.3 L / 5.5 M

V1 = 0.065 L

V1 = 65 mL

● So, to prepare the 1.2 M NaOH solution, you pour 65 mL of 5.5 M NaOH into your volumetric flask and add water to get 300 mL final volume.

Primary standard

● When a substance of very high purity and large molar mass is dissolved in a known volume of solvent it creates a primary standard solution.

● The concentration of a primary standard solution can be determined accurately and it can therefore be used in acid-base titrations.

Lab: Preparation of solutions

● Show your calculations to the teacher before you start!

● 1. Prepare 100 ml of a NaOH (0,1M) solution.

● 2. Prepare 50 ml of an oxalic acid solution (0,1 M) that you save and then use as a primary standard in your titration lab (week 19).

3. Stoichiometry

https://www.youtube.com/watch?v=UL1jmJaUkaQ

3.1 Calculating yields

1. Write a balanced equation for the reaction.

2. Convert the mass (or volume) of reactants to amount in moles.

3. Using the mole ratio from the coefficients in the equation, calculate the amount of required product.

4. Convert the amount of the product to the appropriate units of mass (or volume).

1. Calculate the mass of calcium oxide that could be obtained by heating 2.5 grams of calcium carbonate, CaCO3.

CaCO3 (s) → CaO (s) + CO2 (g)

3.2 The limiting reactant and the reactant in excess

● If one of the reactant is present in excess, once the reaction is complete, some of that reactant will be left over.

● For example, the reaction between 4.8 grams of magnesium and 4.8 grams of sulfur:

1. Mg(s) + S(s) → MgS(s)

2. Amount of magnesium atoms:

Amount of sulphur atoms:

3. The coefficients indicate that one mole of magnesium atoms reacts with one mole of sulfur atoms to form one mole of MgS.

3.3 Theoretical and experimental yield

● The quantity of product that is calculated to be formed when all the limiting reactant reacts is called the theoretical yield.

● The mass (or volume) of a product actually obtained in a chemical reaction is called experimental yield.

● The experimental yield is always less than the theoretical yield because of:

– Incomplete reaction

– Side-products

– Loss of reactants or products

– Impurities present in the reactants

3.4 Percentage yield

● The efficiency of the procedure can be calculated:

4. Writing a lab report

● Lab: Number of water molecules in a salt

4.1 Ion-dipole-bond

● When a salt such as copper sulfate (CuSO

4) is dissolved in

water, it exists as the free ions Cu² ⁺ and SO₄²ˉ.

● In a solution, the partial charges of water molecules (called dipoles) are attracted to these charged ions, thus forming ion-dipole bonds.

(

http://www.youtube.com/watch?NR=1&v=EBfGcTAJF4o&feature=endscreen

4.2 Hydrous copper(II) sulfate

● If the water is let to evaporate, the salt crystallizes but some of the water molecules remain bonded to the ions and are built into the salt crystal pattern.

● The salt is therefore a hydrous salt and it contains crystallized water: CuSO

4 · x H

2O

● The crystallized water can be easily removed by heating the compound and anhydrous copper(II) sulfate is formed:

CuSO4 · x H

2O→ CuSO

4 + x H

2O

4.3 Writing a good lab report

● The new assessment model uses the following criteria to assess the work of both SL and HL students:

5.1 Acids and bases

● An acid is a molecule or ion that can donate a proton (H+) = proton donor

HCl (g) + H2O(l) H

3O+(aq) + Cl- (aq)

● A base is a molecule or ion that can accept a proton = proton acceptor

NH3 (g) + H

2O (l) NH

4

+ (aq) + OH- (aq)

https://www.youtube.com/watch?v=ANi709MYnWg

5.2 Neutralization

acid + base salt + water⇒

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

2. How many moles of H2O form when 25.0 mL of 0.100 M HNO3 solution is completely neutralized by NaOH?

3.

6. Acid-base titration

● A known volume of a solution (= analyte) is measured with a pipette and added to a flask.

● An acid-base indicator is added and a solution of known concentration (= standard solution or titrant) is added in small measured quantities from a buret.

● The titration is continued until the indicator changes color at the end-point.

http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/flashfiles/stoichiometry/a_b_phtitr.html

● At the end-point (= eqvivalence point) the two substances are present in stoichiometric (equal) quantities.

● Ex:

Sodium hydroxide reacts with hydrochloric acid according to the following equation:

NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)

Calculate the volume of 0,0500 mol dm-3 sodium hydroxide solution to react exactly with 25 cm3 of 0,20 mol dm-3 hydrochloric acid.