Unit 3 Written Report

1

Edexcel AS Chemistry

Unit 3B: Chemistry

Laboratory Skills I Alternative

Unit 1 Practical

Methods and Procedures


A. 25cm3 of 1 mol dm–3 H2SO4

n H2SO4 = MV

1000

B. 1 mol H2SO4 reacted with 1 mol NiCO3

∴ answer from A = n NiCO3 Mass of NiCO3 = n NiCO3 × Mr NiCO3; weight it with slight excess

~3.00 g

Practical 1.1

0.00 2.00

2


C. Add little by little, stir continuously until the solid

dissolves and stop heating for a while. Then continue the same steps until all solid finishes.

D. to G. Hot mixture

Practical 1.1

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Practical 1.1 Let the mass of the crystal obtained is 4.55 g. A green shiny crystal obtained. 1) The theoretical yield of hydrated nickel sulfate is

0.025 × [58.7 + 32.1 + 4(16) + 7(18)] = 7.02 g

2) The percentage yield of the crystal obtained from this experiment is 4.55

7.02 × 100% = 64.8 %

3) It was found that the actual yield is much lower than expected due to some reason: a. Some product left behind in the apparatus when

transferring the product from one apparatus to another. b. Some reactant may be loss during the experiment.

3


A. 25cm3 of 2 mol dm–3 H2SO4

n(H2SO4) = MV

1000 1:1 reaction; n(H2SO4) = n(Fe)

B. m(Fe) = n(Fe) × Ar Fe

Do the same step until all

solid is added

Practical 1.2 (1)

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C. D. Addition of 10% of acid 5cm3 of 2 mol dm–3 H2SO4

Practical 1.2 (1)

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A. 25cm3 of ammonia solution 2 mol dm–3 H2SO4

𝑀𝑎𝑉𝑎

𝑎 =

𝑀𝑏𝑉𝑏

𝑏 → V(NH3) =

𝑀𝑎𝑉𝑎

𝑎 ×

𝑏

𝑀𝑏

B. & C. Gently heat in the fume hood until it became

concentrated

Practical 1.2 (2)

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A. B. & C. Solution from (1) Solution from (2) Allow to cool

D. to F.

Practical 1.2 (3)

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Practical 1.2 Result and observation: A light green shiny crystal obtained.

Let the mass of the crystal obtained is ________ g.

When both the crystal and iron(II) sulphate were exposed to air overnight, it was found that the crystal does not change colour but iron(II) sulphate turns colour from light green to brown. Discussion 4) The reaction between ammonia solution and sulphuric acid is

neutralization. This is due to the fact that the oxidation number of all element remain unchanged. Whereas the reaction between iron sulphuric acid is redox. This is because there is a change of oxidation number where iron changing from 0 to +2 (oxidation) and hydrogen changing from +1 to 0 (reduction).

5) The ions present in the double salts are ammonium ion (NH4+),

iron(II) ion (Fe2+), and sulphate ion (SO42–)


Practical 1.2 Discussion

6) Theoretical yield of the crystals which could be made is as follows:

Mr (NH4)2SO4•FeSO4•6H2O = 392

1 mole (NH4)2SO4 requires to react with 1 mole FeSO4

∴n(NH4)2SO4 = 0.05 mol

Theoretical yield = 0.05 × 392 = 19.59 g

7) Percentage yield = 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 𝑦𝑖𝑒𝑙𝑑

19.6×100%

8) The pure iron(II) sulphate turns colour from light green to brown because Fe2+ ion was oxidized to Fe3+ ion when exposed to air.

6


Practical 1.3

Solubility and Precipitation Reactions Insoluble Dispose to sink

NaOH Soluble Result and Observation:

Name Formula Appearance Solubility Colour of Solution

Addition of NaOH

Copper(II) chloride CuCl2 Soluble

Iron(II) sulphate FeSO4 Soluble

Magnesium

carbonate MgCO3 Insoluble

Iron(III) nitrate Fe(NO3)3 Soluble

Iron(III) oxide Fe2O3 Insoluble

Potassium

carbonate K2CO3 Soluble No change


Practical 1.3

Solubility and Precipitation Reactions Discussion:

1) There were some of the metal compounds are coloured (copper and iron compounds). Metals that give colour to the compounds are the transition metals

2) The formulae for copper(II) chloride, iron(II) sulphate and iron(III) nitrate all contain water which indicate that they are hydrated compounds. These compounds might reasonably by expected to be soluble in water.

3) All compounds provided in this experiment obeyed the solubility rules.

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Practical 1.3

Further precipitation reactions Result and Observation:

Combination of solution

Observation Ionic equation

NaCl + Pb(NO3)2

NaCl + CuSO4

NaOH + Pb(NO3)2

NaOH + CuSO4

Pb(NO3)2 + Na2SO4

Pb(NO3)2 + CuSO4

Pb(NO3)2 + Na2CO3

CuSO4 + Na2SO4

CuSO4 + Na2CO3


Practical 1.3

Redox Reactions Result and Observation:

Solution Ionic equations Colour of Solution

Ionic equation

after adding NaOH Before After

H2O2 2Fe2+ → 2Fe3+ + 2e- 2H+ + H2O2 + 2e- → 2H2O 2Fe2+ + 2H+ + H2O2 → 2Fe3+ + 2H2O

Green solution

Brown solution

Fe3+(aq)+ 3OH–(aq) → Fe(OH)3(s) (Brown ppt)

KMnO4

5Fe2+ → 5Fe3+ + 5e- 8H+ + MnO4

- + 5e- → Mn2+ + 4H2O 5Fe2+ + 8H+ + MnO4

- → 5Fe3+ + Mn2+ + 4H2O

Green solution

Brown solution


Warm conc.

HNO3

6Fe2+ → 6Fe3+ + 6e- 8H+ + 2NO3

- + 6e- → 2NO + 4H2O 6Fe2+ + 8H+ + 2NO3

- → 6Fe 3+ + 4H2O + 2 NO

Green solution

Brown solution


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Practical 1.3

Reduction Result and Observation:

Substance Ionic Eqation

Colour of Solution Addition of NaOH

Ionic equation after adding

NaOH

Before After

Zinc, warm

2Fe3+(aq) + 2e- → 2Fe2+ (aq) Zn(s) → Zn2+(aq) + 2e- 2Fe3+(aq)+Zn(s)→ 2Fe2+(aq) + Zn2+(aq)

Brown solution

Green solution

Fe2+(aq)+ 2OH–(aq) → Fe(OH)2(s) (green ppt)

NaSO3, warm

2Fe3+ + 2e- → 2Fe2+ H2O + SO3

2– → SO32– +2H+ + 2e-

2Fe3+ + H2O + SO32–→ 2Fe2+

+ SO32– +2H+

Brown solution

Green solution

Fe2+(aq)+ 2OH–(aq) → Fe(OH)2(s) (green ppt)


Practical 1.4 The reaction between:

1) copper(II) sulphate and zinc

m(Zn) = 0.01 × Ar Zn

2) Citric acid and sodium hydrogen carbonate

m(NaHCO3) = 0.01 × Mr NaHCO3

Thermometer

Beaker Reaction mixture

Foam polysterene

cup and lid

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Practical 1.4 Results:

Experiment 1

(Limiting reagent: CuSO4) 2

(Limiting reagent: C6H8O7)

Moles used MV

1000

MV

1000

Initial temperature

(oC)

Final temperature

(oC)

Temperature change, ∆T (oC)


Practical 1.4 Calculation:

Experiment 1

(Limiting reagent: CuSO4) 2

(Limiting reagent: C6H8O7)

Energy transfer, q

m(solution) × 4.18 × ∆T m(solution) × 4.18 × ∆T

Enthalpy change ∆H1 = – 𝑞

𝑛(CuSO4) ∆H2 = –

𝑞

𝑛(C6H8O7)

Thermochemical equation

Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s) ∆H1 = _____

C6H8O7(aq) + 3NaHCO3(s) → C3H5O7Na3(aq) + 3CO2(g) +

3H2O(l) ∆H2 = _____

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Practical 1.4 Discussion:

4) The reaction between copper(II) sulphate and zinc is _______ whereas the reaction between citric acid and sodium hydrogencarbonate is ________.

5) As excess of solid is used in each experiment is to ensure that all solution is completely reacted.

6) The things that we should ignore when doing the calculation include the mass of solid reactant added and the specific heat capacity of the solution.



7) A large container was used for the second experiment is because the carbon dioxide gas is formed which may produces bubbles or effervescences.

8) The main sources of error in this experiment include heat losses to the surrounding (environment) such as air, glass, calorimeter or even the products of the reactions, the concentration of the solutions are assumed to be correct, and slight inaccuracies in weighing the solid reactants should make little difference, as long as there is still excess.

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9) The percentage error for each enthalpy change are as follows:

Experiment 1:

% error = 2×0.1

∆𝑇1 × 100%

Experiment 2:

% error = 2×0.1

∆𝑇2 × 100%



Alcohol Propan-1-ol Butan-1-ol Octan-1-ol

Initial mass of spirit lamp + alcohol + cap (g)

Final mass of spirit lamp + alcohol + cap (g)

Mass of alcohol burned, m (g)

Initial temperature of water (oC)

Final temperature of water (oC)

12


Practical 1.5 Calculation:

1) a) n(propan-1-ol) = 𝑚(propan−1−ol)

𝑀𝑟(propan−1−ol)

b) Energy produce, q = answer from (1) × (–2021kJmol–1) c) Energy produce, q = heat capacity of apparatus, C ×

temperature rise

∴ C = 𝑎𝑛𝑠𝑤𝑒𝑟 𝑓𝑟𝑜𝑚 (2)

∆𝑇

2) d) Energy produce for butan-1-ol, q1 = answer from (3) × ∆T

Energy produce for octan-1-ol, q2 = answer from (3) × ∆T

e) n(butan-1-ol) = 𝑚(butan−1−ol)

𝑀𝑟(butan−1−ol) n(octan-1-ol) =

𝑚(octan−1−ol)

𝑀𝑟(octan−1−ol)

∆Hc butan-1-ol = – 𝑞1

𝑛(butan−1−ol) ∆Hc octan-1-ol = –

𝑞2

𝑛(octan−1−ol)


Practical 1.5 Discussion: 1) Complete combustion may not have taken place where

carbon of carbon monoxide might have been formed instead.

2) The main sources is the heat losses to the surrounding and the precautions which should be taken is to use a ‘chimney’ or other methods of draught proofing.

3) The heat loss and incomplete combustion both will tend to give results that are less negative than the true value.

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Practical 1.5 Discussion: 4) It was found that the enthalpy change of combustion

increase with the number of carbon atom. The difference is about 655 kJ for each extra carbon atom.

5) This is because more energy is needed to break bonds in the reactants as the number of carbon atoms increases. However, proportionally more energy is released when the new bonds form in the products. So overall reactions becomes more exothermic.



Experiment 1

(Solid reagent: Ca) 2

(Solid reagent: CaCO3)

Moles used m

𝐴𝑟 Ca

m

𝑀𝑟 CaCO3

Initial temperature

(oC)

Maximum temperature

(oC)

Temperature change, ∆T (oC)

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1) to 4)

The enthalpy change for both reactions are as follows:

Experiment 1 2

Energy transfer, q

m(solution) × 4.18 × ∆T m(solution) × 4.18 × ∆T

Enthalpy change ∆H1 = – 𝑞

𝑛(Ca) ∆H2 = –

𝑞

𝑛(CaCO3)

Thermochemical equation

Ca(a) + 2HCl(aq) → CaCl2(aq) + H2(g) ∆H1= _____

CaCO3(s) + 2HCl(s) → CaCl2(aq) + CO2(g) + H2O(l)

∆H2 = _____


Practical 1.6 Discussion: 5) The Hess’s cycle for the formation of calcium

carbonate is shown below:

Ca(s) + 3

2O2(g) + C(graphite) CaCO3(s)

2HCl(aq) 2HCl(aq)

CaCl2(aq) + H2O(g) + CO2(g) + H2(g) 6) From the Hess’s cycle above, the ∆Hf

[CaCO3(s)] is calculated as follows: ∆Hf [CaCO3(s)] + ∆H2 = ∆H1 + ∆Hf [CO2(g)] + ∆Hf [H2O (g)]

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Practical 1.6 Evaluation: 1) The two assumptions that must be made in this

experiment are that the specific heat capacity of HCl is same as that of water and that there is no heat loss to the surroundings.

2) The reliability of this experiment could be improved by repeating the experiment and averaging them.

3) Error could be minimised by using a lid to minimise heat loss and by a cooling curve method.

4) The precision of the experiment could be increased by using apparatus which has a low in precision error such as pipette or thermometer that measures to ±0.1oC.


Practical 1.7 Result and Observation: Experiment Observation Inferences

1 Combustion

Combustion of cyclohexane gives a yellow (luminous) and sooty flame. The poly(ethene) softens and slowly becomes a liquid on gentle heating.

Cyclohexane has a high C:H ratio

2 Oxidation

No changes observed but slightly brown solution observe when react with poly(ethene)

Alkane does not react with the oxidising agent.

3 Action of bromine

No changes observed but bromine water colour fades when react with poly(ethene)

Alkane does not react with the bromine water at room condition

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4 Action of

bromine in sunlight

The colour of bromine water fades and white fume evolve when the test tube was irradiate under sunlight but no changes observe when test tube leave in a dark place

Alkane reacts with bromine through free radical substitution and the white fume is the hydrogen bromide.

5 Action of sulphuric

acid

No changes observed Alkane does not react with the sulphuric acid.

6 Action of

alkali No changes observed

Alkane does not react with the alkali.



7 Catalytic cracking

a) Flammability test The gas obtained is flammable.

The gas obtained is hydrocarbon.

b) Bromine test The bromine colour fades then colourless.

Alkene is present.

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1 Combustion

Combustion of cyclohexene and limonene gives a yellow (luminous) and slightly sooty flame.

Cyclohexene has a low C:H ratio.

2 Oxidation

The colour of potassium manganate(VII) decolourised. Two layers observed.

Alkene reacts with the oxidising agent due to the presents of C=C.

3 Action of bromine

The colour of bromine water decolourised. Two layers observed.

Alkene reacts with the bromine water due to the presents of C=C.



4 Action of

sulphuric acid

The solution turns dark and the bottom of the test tube is hot.

Alkene react with the sulphuric acid.

5 Polymerisation

CH3 n CH2=C CO2CH3 CH3 ~ C–CH2 ~ CO2CH3 n

The product have the same empirical formula as the starting material.

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Unit 3B: Chemistry

Laboratory Skills I Alternative

Unit 2 Practicals


1) The liquid is deflected towards the plastic rod.

2) Alcohols and ketones show deflection while hydrocarbons show no deflection.

3) This is because polar molecules posses a permanent dipole moment and the molecules tends to align its dipole moment along the field of the nearby charged object. Hydrocarbons do not posses a permanent dipole moment may be observed to be very slightly deflected, as a result of the induced dipole. However, hydrocarbons that show no deflection indicate that the structure is symmetry and the dipole moment are cancel out.

Practical 2.1 Finding the effect of electrostatic force on

jets of liquid

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Water Ethanol Cyclohexane

Experiment 1

Insoluble Slightly soluble;

solution is slightly brown

Soluble; purple solution obtained

Experiment 2

Soluble Slightly soluble Insoluble

Experiment 3

Soluble; dense white precipitate obtained

Slightly soluble; slight white

precipitate observed

Insoluble; no precipitate form

Practical 2.2 Solubility of simple molecules in different

solvents

1) Iodine and sugar both have covalent bond whereas calcium chloride is ionic.

2) Iodine molecules are held together by London forces. Sugar has hydrogen bonding and calcium chloride does not have intermolecular force.


Practical 2.2 Solubility of simple molecules in different

solvents

3) Iodine dissolves in cyclohexane due to the ability for the molecules to interact with hexane molecules through van der Waals forces. It does not dissolve in water or very well in ethanol because the molecules are both polar and therefore cannot form hydrogen bonds to the iodine molecules. Sugar is the opposite and dissolves only in polar solvents like water and ethanol (to some extent). Calcium chloride, being ionic, dissolves in polar solvents because the water molecules interact with the ionic lattice, therefore breaking the lattice and dispersing the ions.

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Procedures: Little amount of carbonate compound are added into a test tube and heated. The gas emitted was allow to bubble through a delivery tube into another test tube containing a small quantity of limewater

Practical 2.3 Thermal decomposition of group 2 nitrates

and carbonates

Effect Ionic radius of cation/nm

MgCO3 Limewater turns cloudy 0.072

CaCO3 No change 0.100

SrCO3 No change 0.113

BaCO3 No change 0.136


Procedures: weigh certain amount of nitrate compound with crucibles. The compound are strongly heated in crucibles for few minutes and allow it to cool then re-weighing the crucible.


and carbonates

Effect Ionic radius of cation/nm

Mg(NO3)2 Brown fumes observe 0.072

Ca(NO3)2 No change 0.100

Sr(NO3)2 No change 0.113

Ba(NO3)2 No change 0.136

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Interpretation: As going down the group, the size of the cations increases, the charge density decrease. This makes the cation decrease in polarizing power and therefore less able to polarize the neighbouring anion. Hence, within the anion, less able to weaken the bonds, making the anion having less easily to be decomposed.

MgCO3 (s) → MgO (s) + CO2 (g)

2 Mg(NO3)2 (s) → 2 MgO (s) + O2 (g) + 4 NO2 (g)


and carbonates


1) The coloured flame produced due to electronic excitations of electrons in the metal to a higher electronic energy levels when electrons absorb energy and then return to the lower energy levels with the emission of light energy in the visible region that corresponds to a particular colour.

2) Fireworks.

Practical 2.4 Flame tests on compounds of group 1 and 2

Compound Observation Compound Observation

Li+ Deep red Ca2+ Brick red

Na+ Yellow Sr2+ Blood red

K+ Lilac Ba2+ Apple green

Rb+ Bluish red

Cs+ Blue

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Observation: T = 24 – 25oC; metyl orange: yellow → reddish orange

bromophenol: blue → yellow

1) Ca(OH)2 (aq) + 2 HCl (aq) → CaCl2 (aq) + 2 H2O (l)

2) 2 mol HCl reacted with 1 mol Ca(OH)2

∴ 1 mol HCl reacted with 0.5 mol Ca(OH)2

3) Number of moles HCl = 0.050×𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑜𝑙𝑢𝑚𝑒

1000

4) Number of moles in Ca(OH)2 in 10.0 cm3

= 0.5 × answer from (3) = x

∴ In 1000.0 cm3, number of moles Ca(OH)2

= 1000

10 × x

5) Molecular mass of Ca(OH)2 = 40 + 2(16) + 2(1) = 74

Mass of Ca(OH)2 in 1000.0 cm3 = answer from (4) × 74

6) answer from (5) g dm–3 at 24 oC

7) Percentage error = 0.1×2

𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 × 100%

Practical 2.5 Simple acid-base titration


Practical 2.6 Oxidation of metal and non-metallic

elements and ions by halogens

Chlorine Bromine Iodine

Experiment 1

Solution turns orange/brown

Solution turns orange/brown

No change

Experiment 2

The colour turns from pale yellow to

colourless

The colour turns from orange to

colourless

The colour turns from brown to

colourless

Experiment 3

Reaction with KI: solution turns from colourless to brown; Addition of hexane: Top layer: purple, bottom layer: pale

yellow to colourless

– –

Experiment 4

– –

Addition of starch: Black; addition of

sodium thiosulfate: colourless

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elements and ions by halogens 1) Experiment 1

Ox. half equation: 2 Fe2+ (aq) → 2 Fe3+ (aq) + 2 e–

Red. Half equation: Cl2 (aq) + 2 e– → 2 Cl– (aq)

Br2 (aq) + 2 e– → 2 Br– (aq)

Overall ionic equation:

2Fe2+(aq)+Cl2(aq)→2Fe3+(aq)+2Cl–(aq)

2Fe2+(aq)+Br2(aq)→2Fe3+(aq)+2Br–(aq)

Experiment 2

Ox. half equation:

2OH– (aq) + ½Cl2 (aq) → ClO– (aq) + H2O (l) + e–

Red. Half equation: ½Cl2 (aq) + e– → Cl– (aq)


2OH– (aq) + Cl2 (aq) → ClO– (aq) + Cl– (aq) + H2O (l)



elements and ions by halogens 1) Experiment 3

Ox. half equation: 2I– (aq) → I 2 (aq) + 2 e–

Red. Half equation: Cl2 (aq) + 2 e– → 2 Cl– (aq)

Overall ionic equation:2I–(aq)+Cl2(aq)→I2(aq)+2Cl–(aq)

Experiment 4

Ox. half equation: 2 S2O32– (aq) → S4O6

2– (aq) + 2 e–

Red. Half equation: I2 (aq) + 2 e– → 2 I– (aq)


2 S2O32– (aq) + I2 (aq) → S4O6

2– (aq) + 2 I– (aq)

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elements and ions by halogens 2) From the reaction,

2NaOH(aq)+Cl2(aq) → NaClO(aq)+NaCl(aq)+H2O(l)

0 +1 –1

The oxidation number of chlorine changes from 0 to +1 and –1 where chlorine act both as oxidizing and reducing agent simultaneously.

3) Oxidizing strength of the halogens decreases down group 7

4) The liberated iodine reacts with the starch turning it to black.


Practical 2.7 Disproportionation reactions with cold and

hot alkalis

1) 2OH– (aq) + Cl2 (aq) → ClO– (aq) + Cl– (aq) + H2O (l)

2) The chlorine undergoes two change: part of it changes from 0 to –1 and part from 0 to +1.

3) Chlorine changes in oxidation number from 0 to +5 and from 0 to –1.

4) 6KOH(aq) + 3I2(aq) → 5KI(aq) + KIO3 (aq) + 3H2O(l)

Halogen solution

Observation on adding alkali Equation

Chlorine The colour turns from pale

yellow to colourless 2NaOH(aq)+Cl2(aq) →

NaClO(aq)+NaCl(aq)+H2O(l)

Bromine The colour turns from orange

to colourless 2NaOH(aq)+Br2(aq) →

NaBrO(aq)+NaBr(aq)+H2O(l)

Iodine The colour turns from brown

to colourless 2NaOH(aq)+I2(aq) →

NaIO(aq)+NaI(aq)+H2O(l)

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Practical 2.8 Iodine/thiosulfate titration and the

determination of purity of potassium iodate(V) 1) A well-mixed, accurately made solutions prepared from

good quality reagents is required in order to have an excellent result.

2) The oxidation number of sulfur in sodium thiosulfate is +2 and in sodium tetrathionate is +2½.

Part 1

1) n Na2S2O3 = 0.010×𝑣

1000; v = average titre

2) 2 mol Na2S2O3 reacted with 1 mol I2

∴ answer from step 1 divided with 2

3) 3 mol I2 produced from 1 mol KIO3

∴ answer from step 2 divided with 3 4) Molecular mass of KIO3 = 40 + 127 + 3(16) = 214

Mass of KIO3 in 10 cm3 = answer from step 3 × 214

∴ Mass of KIO3 in 1000 cm3 = 1000

10 × Mass of KIO3 in 10 cm3

5) Percentage purity = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐾𝐼𝑂3 𝑖𝑛 1 𝑑𝑚3𝑓𝑟𝑜𝑚 4

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑟𝑢𝑑𝑒 𝐾𝐼𝑂3 × 100%


Practical 2.9 Reactions between halogens and halide

ions/some reactions of the halides 1) a) Observation:

Cl2 with KCl: no changes

Cl2 with KBr: the colour of the solution turns from colourless to yellow–orange

Cl2 with KI: the colour of the solution turns from colourless to brown

2KBr (aq) + Cl2(aq) → Br2 (aq) + 2KCl (aq)

2KI (aq) + Cl2(aq) → I2 (aq) + 2KCl (aq)

Addition of hydrocarbon helps in reaching a decision as the halogen dissolve well in hydrocarbons due to the formation of intermolecular interaction.

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ions/some reactions of the halides 1) b) Observation: All solutions give out two layer

Cl2 with KCl: top layer: pale green

bottom layer: light green to colourless

Cl2 with KBr: top layer: orange

bottom layer: light yellow to colourless

Cl2 with KI: top layer: purple

bottom layer: light yellow to colourless

Halogen mixed well with hydrocarbon because both halogen and hydrocarbon having London force with similar strength which allows them to interact easily between the molecules.



ions/some reactions of the halides 2) A definite trend in reactivity is clearly observed in this

experiment because as the atomic radius of the halogen increases down the group, the tendency to accept electron decreases which also decreases the oxidizing power of the halogens.

2I– (aq) + Br2 (aq) → I2 (aq) + 2Br– (aq)

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ions/some reactions of the halides

The reactions between halogens and halide ions

Action on

Solution added

Water Potassium chloride

Potassium bromide

Potassium iodide

Chlorine followed

by hexane

With water: soluble forming

light green solution;

Addition of hexane: Top layer: pale

green, bottom layer: light

green to colourless

No changes

With Cl: solution turns

from colourless to yellow-

orange; Addition of

hexane: Top layer:

orange, bottom layer: pale yellow to colourless

With Cl: solution turns

from colourless to brown;

Addition of hexane:

Top layer: purple, bottom

layer: pale yellow to colourless





Action on

Solution added


Potassium bromide

Potassium iodide

Bromine followed

by hexane

With water: soluble and

forms orange to light brown

solution Addition of hexane: Top

layer: orange, bottom layer: pale yellow to

colourless

No change No change

With Br: solution turns

from colourless to brown;

Addition of hexane:

Top layer: purple, bottom

layer: pale yellow to colourless

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Action on

Solution added


Potassium bromide

Potassium iodide

Iodine followed

by hexane

With water: insoluble in

cold water but slightly soluble

in hot water and forms

brown solution Addition of hexane: Top

layer: purple, bottom layer: pale yellow to

colourless

No change No change No change




Some reactions of the halides

1) The silver halides

Solution

Action on

Silver nitrate Concentrated

ammonia solution Exposure to light

Potassium chloride

White precipitate obtained

The precipitate is soluble

Silver substance observed

Potassium bromide

Cream precipitate obtained

The precipitate is soluble

*Yellow substance observed

Potassium iodide

Yellow precipitate obtained

The precipitate is insoluble

No change observed

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2) The action of concentrated sulfuric acid on potassium salts

Action on

Potassium chloride Potassium bromide

Potassium iodide

Lead(II) ethaonoate

No change No change Turns from clear

to brown

Acidified potassium

dichromate(IV) No change

Turns from orange to green

No change

Ammonia solution

White fume observe

No change No change





3) The properties of the hydrogen halides

Action on

Hydrogen chloride Hydrogen bromide Hydrogen iodide

Solubility in Water

Soluble Slightly soluble Slightly soluble

Reaction with Ammonia gas

Dense white fume observed



Thermal stability

No change observed; the gas is

stable

Little amount of brown gas

observed; the gas is less stable

Vast amount of purple gas

observed; the gas is unstable

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Practical 2.10 Factors that influence the rate of chemical

reactions

1) MnO2 and PbO2

2) A catalyst works by providing an alternative reaction route of a lower activation energy


Practical 2.11 Effect of temperature, pressure and

concentration on equilibrium

1) Initially, brown gas is observed. When a hot water bath is used to heat up the test tube containing the gas, the intensity of brown colour is increased. When a ice bath is used to cool down the test tube, the intensity of brown colour decreasing until it reaches colourless.

2) When high pressure is applied, the intensity of brown colour decreases until it reaches colourless. But when lower pressure is applied, the intensity of brown colour increases.

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Practical 2.12 Reactions of alcohols

1) Solubility in water

The alcohols with shorter carbon chain are more soluble than the others due to their ability to form hydrogen bonds to water molecules. As the carbon chains is getting longer, the solubility decreases. This is because the carbon chain will interrupt (break) the hydrogen bonding and replace with London forces. Thus, the trend solubility of the alcohols is decreasing as the number of carbon chain increase.

Alcohols Observation

With water pH

Ethanol Soluble 7

Methanol Soluble 7

Propan–1–ol Slightly soluble, two layer observed 7

Butan–1–ol Slightly soluble; two layer observed 7

Pentan–1–ol Less soluble, two layer observed 7



2) Reaction with sodium

The sodium reacts vigorously with alcohols with shorter carbon chain. As the carbon chains is getting longer, the reaction is getting less reactive. The products produce are the alkoxide compounds. It is predicted that alcohols behave similarly with water however the reaction between alcohol and sodium is less vigorous than that between sodium and water. When the alcohols react with sodium, the O–H breaks. Thus, the trend reactivity of the alcohols is decreasing from methanol to pentan–1–ol as the number of carbon chain increase.


Ethanol Sodium dissolve vigorously in ethanol

Methanol Sodium dissolve vigorously in methanol

Propan–1–ol Sodium dissolve slightly vigorous in propan–1–ol

Butan–1–ol Sodium dissolve slightly vigorous in butan–1–ol

Pentan–1–ol Sodium dissolve less vigorous in pentan–1–ol

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3) Reaction with sodium

The acidified dichromate(VI) solution oxidized alcohols readily. However, the reaction of the alcohols becomes progressively slower with increasing length of carbon chain but reaction with methanol is also quite slow. The products obtained produce a mild fruity smell indicate that aldehydes are form.


Ethanol The solution turns colour from orange to green.

Methanol The solution turns colour from orange to green.

Propan–1–ol The solution turns colour from orange to green.

Butan–1–ol The solution turns colour from orange to green.

Pentan–1–ol The solution turns colour from orange to green.


Practical 2.13 Preparation of an organic liquid

(reflux and distillation) 1) The dehydration of propan–1–ol

The product obtained from this experiment has the ability to decolourized both of the reagent indicates that the product obtained having C=C. The name of the product is ___________.

Reagent Observation

Bromine water The colour of bromine water decolourized.

Acidified potassium manganate(VII)

The colour of the solution decolourized.

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Practical 2.13 Preparation of an organic liquid

(reflux and distillation) 2) The oxidation of propan–1–ol

• sour smell

• neutralises a relatively large volume of sodium carbonate solution.

• It does not give any change to the Fehling solution.

• indicate carboxylic acid obtained

• The name of the product is __________,

Reagent Observation

Sodium carbonate A mild effervescence observed

Fehling solution No change


Practical 2.14 Preparation of a halogenoalkane

from an alcohol 1) A halogenoalkane from a primary alcohol

• pungent smell

• Milky globules observed at the bottom of the conical flask

• C2H5OH (aq) + HBr (aq) → C2H5Br (aq) + H2O (aq)

• C–O bond is broken in ethanol

2) A halogenoalkane from a tertiary alcohol • ~16.5 g

• Reasons:

• The reactant used may not be pure

• The reaction may not be complete

• Product maybe left behind on the apparatus

• Product maybe evaporate

• Parallax factor

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Practical 2.15 Reactions of the

halogenoalkanes 1) Combustion

2) Reaction with aqueous alkali • The product obtained is alcohol

3) Comparison of halogenoalkanes • Iodo, bromo, chloro

• Tertiary, secondary, primary

Halogenoalkanes Observation

1-chlorobutane Readily burn with yellow flame

2-chloro-2-methylbutane Readily burn with yellow flame

1 – bromobutane Burn with yellow flame

1 – iodobutane Burn with yellow flame and emission of purple

fume


Practical 2.15 Reactions of the

halogenoalkanes 4) Reaction with alcoholic alkali

• The product obtained is alkene

• Elimination reaction take place

• Due to the presents of unchanged reactant which may be caused by strong heating

Documents

Unit 3 Written Report