Topic 16Unit 53 - STMGSSintranet.stmgss.edu.hk/STUDENT/SUBJECTS/Chemistry/CHEM_TB_AN… · Topic 16Unit 53 Suggested answers to in-text activities and unit-end exercises In-text activities

1

To

pic

16

U

nit 5

3

Unit 53Topic 16Suggested answers to in-text activities and unit-end exercises

In-text activities

Checkpoint (page 11)

1 Pair of solutions (i) Precipitate produced? (ii) Equations

(a) CaCl2(aq) + Na2SO4(aq) a precipitate produced CaCl2(aq) + Na2SO4(aq) CaSO4(s) + 2NaCl(aq)Ca2+(aq) + SO4

2–(aq) CaSO4(aq)

(b) ZnSO4(aq) + NaNO3(aq) no precipitate produced —

(c) Pb(NO3)2(aq) + MgCl2(aq) a precipitate formed Pb(NO3)2(aq) + MgCl2(aq) PbCl2(s) + Mg(NO3)2(aq)Pb2+(aq) + 2Cl–(aq) PbCl2(s)

2 a) Potassium ion

b) X contains carbonate ions.

The carbonate ions in the aqueous solution of X form a precipitate (copper(II) carbonate) when mixed with copper(II) sulphate solution.

There is no reaction between aqueous solutions of potassium chloride / potassium nitrate and copper(II) sulphate.


1 a) Add dilute aqueous solution of sodium hydroxide / dilute aqueous ammonia to each solution.

The aqueous solution of magnesium sulphate gives a white precipitate.

Mg2+(aq) + 2OH–(aq) Mg(OH)2(s)

The aqueous solution of zinc sulphate gives a white precipitate that is soluble in excess alkali to form a colourless solution.

Zn2+(aq) + 2OH–(aq) Zn(OH)2(s)

Zn(OH)2(s) + 2OH–(aq) [Zn(OH)4]2–(aq) / Zn(OH)2(s) + 4NH3(aq) [Zn(NH3)4]

2+(aq) + 2OH–(aq)

b) Warm each solid with dilute aqueous solution of sodium hydroxide.

Ammonium chloride gives a gas that turns moist red litmus paper blue (ammonia).

NH4Cl(s) + NaOH(aq) NaCl(aq) + H2O(l) + NH3(g)

There is no observable change for potassium chloride.

c) Add dilute aqueous solution of sodium hydroxide / dilute aqueous ammonia to each solution.

Aqueous solution of copper(II) nitrate gives a pale blue precipitate.

Cu2+(aq) + 2OH–(aq) Cu(OH)2(s)

2

To

pic

16

U

nit 5

3

Aqueous solution of lead(II) nitrate gives a white precipitate.

Pb2+(aq) + 2OH–(aq) Pb(OH)2(s)

OR

Add dilute hydrochloric acid to each solution.

Aqueous solution of lead(II) nitrate gives a white precipitate (lead(II) chloride).

Pb2+(aq) + 2Cl–(aq) PbCl2(s)

There is no observable change for the aqueous solution of copper(II) nitrate.

2 a) i) Silver chloride

ii) Ag+(aq) + Cl–(aq) AgCl(s)

b) Zinc ion

Zinc ion forms with NH3(aq) a precipitate that is soluble in excess NH3(aq) to give a solution.

Iron(II) ion forms with NH3(aq) a precipitate that does not dissolve in excess NH3(aq).

Hence filtrate C contains zinc ions while precipitate D is iron(II) hydroxide.

c) Pale green

Fe2+(aq) ions are pale green in colour while the other ions in solution X are colourless.

Problem Solving (page 18)

Scheme 1

Distinguish lead(II) chloride and sugar from the other solids

Add water to each solid.

Only lead(II) chloride is insoluble.

Test the electrical conductivity of the four solutions obtained.

Only the aqueous solution of sugar does not conduct electricity.

Distinguish between ammonium chloride, magnesium chloride and sodium chloride

Add dilute aqueous solution of sodium hydroxide to the aqueous solution of each solid.

Only the aqueous solution of magnesium chloride gives a white precipitate (magnesium hydroxide).

Warm the remaining two solution mixtures.

Only the aqueous solution of ammonium chloride gives a gas that turns moist red litmus paper blue (ammonia).

The remaining solid is sodium chloride.

Scheme 2

Distinguish ammonium chloride and sugar from the other solids

Heat each solid strongly.

3

To

pic

16

U

nit 5

3

Ammonium chloride gives a white sublimate.

Only sugar chars.

Distinguish between lead(II) chloride, magnesium chloride and sodium chloride

Add water to each solid.

Only lead(II) chloride is insoluble.

Add dilute aqueous solution of sodium hydroxide / dilute aqueous ammonia to each aqueous solution obtained.

Only the aqueous solution of magnesium chloride gives a white precipitate (magnesium hydroxide).

The remaining solid is sodium chloride.


We need to look at the properties of the cations:

• Ag+(aq) ions form an insoluble chloride.

• Ca2+(aq) ions give a white precipitate with NaOH(aq) but no precipitate with NH3(aq).

• Mg2+(aq) ions give a white precipitate when mixed with NaOH(aq) and NH3(aq) separately.

• K+(aq) ions give no precipitate with NaOH(aq) and NH3(aq).

The following flow diagram summarizes the separation scheme for the cations.

precipitate of Ca(OH)2(s)

Step 3 — add HCl(aq) followed by NaOH(aq)

solution containing Ag+(aq),Ca2+(aq), K+(aq), Mg2+(aq)

precipitate of AgCl(s)

precipitate of Mg(OH)2(s)

Step 2 — add NH3(aq)

Step 1 — add HCl(aq)

solution containing Ca2+(aq), K+(aq), Mg2+(aq)

solution containing Ca2+(aq), K+(aq)

solution containing K+(aq)

4

To

pic

16

U

nit 5

3


1 a) Any one of the following:

• Add barium chloride solution / barium nitrate solution to the acids.

Dilute sulphuric acid gives a white precipitate while dilute hydrochloric acid gives no precipitate.

• Add dilute nitric acid followed by silver nitrate solution to the acids.

Dilute hydrochloric acid gives a white precipitate while dilute sulphuric acid gives no precipitate.

• Add calcium carbonate to excess acids.

Calcium carbonate disappears after reaction when added to dilute hydrochloric acid.

Calcium carbonate does not react completely with dilute sulphuric acid due to the formation of insoluble calcium sulphate on its surface.

b) Add concentrated sulphuric acid to each chemical.

Table salt gives steamy fumes.

White sugar chars.

c) Add aqueous chlorine to each aqueous solution.

The aqueous solution of potassium bromide turns yellow-brown.

There is no observable change for the aqueous solution of potassium chloride.

2 a) i) Chloride ion


b) i) Zinc ion

ii) The white precipitate formed initially was zinc hydroxide.

Zinc hydroxide was soluble in excess aqueous ammonia due to the formation of a soluble complex salt.

Zn2+(aq) + 2OH–(aq) Zn(OH)2(s)

Zn(OH)2(s) + 4NH3(aq) [Zn(NH3)4]2+(aq) + 2OH–(aq)

5

To

pic

16

U

nit 5

3


a) Test Procedure Observations Deduction

1 Describe the appearance of Y. Y is a white crystalline solid.

Y does not contain coloured transition metal ions.

2 Carry out a flame test on Y. A lilac flame appears. Y contains potassium ions.

3 Prepare an aqueous solution of Y. A colourless solution forms. —

a) Add dilute nitric acid followed by aqueous solution of silver nitrate.

Divide the reaction mixture into two separate portions.

A creamy white precipitate forms.

Y contains halide ions.

b) To the first portion, add excess dilute aqueous ammonia.

The precipitate does not dissolve.

Y does not contain chloride ions.

c) To the second portion, add excess concentrated aqueous ammonia.

The precipitate dissolves, giving a colourless solution.

Y contains bromide ions.

b) Y contains potassium ions because Y gives a lilac flame in the flame test.

Y contains halide ions because the aqueous solution of Y gives a creamy white precipitate with acidified aqueous solution of silver nitrate.

The precipitate formed is silver bromide which does not dissolve in excess dilute aqueous ammonia but soluble in very concentrated aqueous ammonia.

Thus, Y contains bromide ions.


Experiment title: Investigating the chemical properties of hex-1-ene

Hazardous chemical, procedure or

equipment involvedNature of hazard Safety precautions

Source of information

Hex-1-ene flammable, irritant • wear safety glasses and protective gloves

• keep the reagent bottle tightly closed• ensure good ventilation in the

laboratory• dispose of hex-1-ene in a waste bottle

inside the fume cupboard

Material Safety Data Sheets (MSDS)

Continued on next page

6

To

pic

16

U

nit 5

3

Experiment title: Investigating the chemical properties of hex-1-ene

Hazardous chemical, procedure or

equipment involvedNature of hazard Safety precautions

Source of information

Aqueous bromine harmful • keep aqueous bromine inside the fume cupboard

• wear safety glasses and potective gloves

• carry out the experiment inside the fume cupboard

• wash affected areas with plenty of water if spilt on the skin or clothes; report the accident to the teacher

• dispose of aqueous bromine in a waste bottle inside the fume cupboard

Material Safety Data Sheets (MSDS)

Shaking the solution mixtures

spillage may occur if shaken too vigorously

• do not shake too vigorously previous instruction

Unit-end exercises (pages 49 – 57)

Answers for the HKCEE and HKALE questions are not provided.

1 a) Clean the end of a nichrome wire by dipping it into concentrated hydrochloric acid and then heat it in the Bunsen flame.

Dip the nichrome wire into concentrated hydrochloric acid and then into the solid under test.

Put the end of the wire in the Bunsen flame again. Observe the colour of the flame.

b) A calcium salt gives a brick-red flame.

A sodium salt gives a golden yellow flame.

2 Cation Action of NaOH(aq) Action of NH3(aq)

Aluminium ion, Al3+(aq) white precipitate, soluble in excess NaOH(aq) to give a colourless solution

white precipitate

Ammonium ion, NH4+(aq) colourless gas (NH3) given off on

warming—

Calcium ion, Ca2+(aq) white precipitate —

Copper(II) ion, Cu2+(aq) blue precipitate blue precipitate, soluble in excess NH3(aq) to give a deep blue solution

Iron(II) ion, Fe2+(aq) green precipitate, turning brown on standing

green precipitate, turning brown on standing

Iron(III) ion, Fe3+(aq) reddish brown precipitate reddish brown precipitate

Lead(II) ion, Pb2+(aq) white precipitate, soluble in excess NaOH(aq) to give a colourless solution

white precipitate

Zinc ion, Zn2+(aq) white precipitate, soluble in excess NaOH(aq) to give a colourless solution

white precipitate, soluble in excess NH3(aq) to give a colourless solution

7

To

pic

16

U

nit 5

3

3 Gas / vapour Test Observation

Ammonia Insert a piece of moist red litmus paper into the gas.

It turns moist red litmus paper blue.

Dip a glass rod in dilute hydrochloric acid and insert it into the gas.

Dense white fumes are formed.

Carbon dioxide Bubble the gas through limewater. The limewater turns milky.

Chlorine Insert a piece of moist blue litmus paper into the gas.

Chlorine turns the moist blue litmus paper red and then bleaches it.

Bromine Insert a piece of moist blue litmus paper into the vapour.

Bromine turns the moist blue litmus paper red and then bleaches it.

Iodine Insert a piece of moist starch paper into the vapour.

The moist starch paper turns blue-black.

Hydrogen Insert a burning splint into the gas. A ‘pop’ sound is heard.

Oxygen Insert a glowing splint into the gas. The glowing splint relights.

Hydrogen halides Dip a glass rod in aqueous ammonia and insert it into the gas.

Dense white fumes are formed.

Sulphur dioxide Insert a piece of filter paper soaked with acidified potassium dichromate solution into the gas.

The paper turns from orange to green.

Water vapour Insert a piece of blue cobalt(II) chloride paper into the vapour.

The blue cobalt(II) chloride turns pink.

4

Anion Action of HCl(aq)Action of

concentrated H2SO4(aq)

Action of HCl(aq), followed by BaCl2(aq)

Action of dilute HNO3(aq), followed

by AgNO3(aq)

Carbonate ion, CO3

2–

colourless gas (CO2) given off

vigorous reaction, colourless gas (CO2) given off

colourless gas (CO2) given off

—

Hypochlorite ion, OCl–

greenish yellow gas (Cl2) given off

— greenish yellow gas (Cl2) given off

—

Sulphite ion, SO3

2–

colourless gas (SO2) given off

vigorous reaction, colourless gas (SO2) given off

colourless gas (SO2) given off

—

Sulphate ion, SO4

2–

— — white precipitate insoluble in dilute HCl(aq)

—

Chloride ion, Cl– — steamy fumes (HCl) given off

— white precipitate soluble in NH3(aq)

Bromide ion, Br– — reddish brown vapour (HBr, Br2 and SO2) given off

— creamy precipitate soluble in concentrated NH3(aq)

Iodide ion, I– — purple vapour (I2, SO2 and H2S) given off

— yellow precipitate

8

To

pic

16

U

nit 5

3

5 a) i) Flammable and toxic

ii) Corrosive

iii) Oxidizing

iv) Corrosive

b) i) • As methanol is flammable, keep the reagent bottle tightly closed.

• There should be no naked flame in the laboratory during an experiment with methanol.

ii) • Wear safety glasses and protective gloves.

• Concentrated sulphuric acid is highly corosive. It gives severe burns to the skin. Handle with great care.

• NEVER add water to concentrated sulphuric acid.

• If you spill any acid on your skin or clothes, wash it off immediately with plenty of water. Then report the accident to your teacher.

• Wash containers of concentrated sulphuric acid with great care; do not add water to the acid directly.

• Dispose of the acid in a waste bottle inside the fume cupboard.

6 a) A pale blue precipitate forms initially.

The precipitate dissolves in excess dilute aqueous ammonia to form a deep blue solution..

b) Reddish brown vapour is formed.

The vapour condenses to a reddish brown liquid.

c) Purple vapour which condenses to a black solid is formed.

d) Anhydrous cobalt(II) chloride paper turns from blue to pink.

7 Pair of solutions

(i) Precipitate produced?

(ii) Colour of precipitate

(ii) Equation

(a) AgNO3(aq) + HCl(aq) a precipitate produced

white Ag+(aq) + Cl–(aq) AgCl(s)

(b) Ca(NO3)2(aq) + MgSO4(aq) a precipitate produced

white Ca2+(aq) + SO42–(aq) CaSO4(s)

(c) KCl(aq) + NH3(aq)no precipitate

produced— —

(d) ZnSO4(aq) + K2CO3(aq) a precipitate produced

white Zn2+(aq) + CO32–(aq) ZnCO3(s)

9

To

pic

16

U

nit 5

3

8 a) CuCO3 gives a black solid (CuO) upon heating.

CuCO3(s) CuO(s) + CO2(g)

b) FeSO4•7H2O gives a colourless vapour that turns blue cobalt(II) chloride paper pink (water vapour) upon heating.

FeSO4•7H2O(s) FeSO4(s) + 7H2O(g)

c) Na2S2O3 gives a colourless gas that turns acidified aqueous solution of potassium dichromate green (sulphur dioxide) upon reaction with dilute hydrochloric acid.

Na2S2O3(aq) + 2HCl(aq) 2NaCl(aq) + SO2(g) + H2O(l) + S(s)

d) CuCO3 gives a gas that turns limewater milky (carbon dioxide) upon reaction with dilute hydrochloric acid.

CuCO3(s) + 2HCl(aq) CuCl2(aq) + H2O(l) + CO2(g)

e) NaOCl gives a gas that bleaches moist blue litmus paper (chlorine) upon reaction with dilute hydrochloric acid.

NaOCl(s) + 2HCl(aq) NaCl(aq) + H2O(l) + Cl2(g)

f) KI gives a purple vapour (iodine) upon reaction with concentrated sulphuric acid.

NaI(s) + H2SO4(l) NaHSO4(s) + HI(g)

8HI(g) + H2SO4(l) 4I2(s) + H2S(g) + 4H2O(l)

9 a) Add a few drops of dilute aqueous solution of sodium hydroxide to an aqueous solution of the salt.

A pale blue precipitate forms.

b) Warm the salt with dilute aqueous sodium hydroxide solution.

Ammonia gas that turns moist red litmus paper blue is liberated.

c) Add an aqueous solution of barium chloride followed by dilute hydrochloric acid to an aqueous solution of the salt.

A white precipitate (barium sulphate) forms.

d) Heat the salt in a boiling tube. The water vapour evolved turns blue cobalt(II) chloride paper pink.

10 a) Warm each sample with dilute aqueous solution of sodium hydroxide.

Ammonium sulphate gives a gas that turns moist red litmus paper blue (ammonia).

(NH4)2SO4(s) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l) + 2NH3(g)

Sodium sulphate just dissolves.

b) Conduct a flame test.

Potassium nitrate gives a lilac flame.

Sodium nitrate gives a golden yellow flame.

10

To

pic

16

U

nit 5

3

c) Add dilute aqueous solution of sodium hydroxide to the aqueous solution of each sample.

Copper(II) chloride solution gives a pale blue precipitate.

Cu2+(aq) + 2OH–(aq) Cu(OH)2(s)

Iron(II) sulphate solution gives a green precipitate.

Fe2+(aq) + 2OH–(aq) Fe(OH)2(s)

d) Add dilute hydrochloric acid to each sample.

Magnesium carbonate gives a gas that turns limewater milky (carbon dioxide).

MgCO3(s) + 2HCl(aq) MgCl2(aq) + H2O(l) + CO2(g)

Magnesium chloride just dissolves.

e) Add concentrated sulphuric acid to each sample.

Sodium bromide gives a reddish brown vapour that forms dense white fumes with aqueous ammonia (hydrogen bromide + bromine vapour). The vapour condenses to a reddish brown liquid.

NaBr(s) + H2SO4(l) NaHSO4(s) + HBr(g)

2HBr(g) + H2SO4(l) Br2(g) + SO2(g) + 2H2O(l)

Sodium iodide gives a purple vapour that condenses to a black solid (iodine).

NaI(s) + H2SO4(l) NaHSO4(s) + HI(g)

8HI(g) + H2SO4(l) 4I2(s) + H2S(g) + 4H2O(l)

f) Add dilute hydrochloric acid to each sample.

Potassium sulphite gives a colourless gas that turns acidified aqueous solution of potassium dichromate green (sulphur dioxide).

K2SO3(s) + 2HCl(aq) 2KCl(aq) + H2O(l) + SO2(g)

Potassium sulphate just dissolves.

11 a) i) A white precipitate forms.


iii) AgCl(s) is readily soluble in aqueous ammonia.

AgBr(s) is soluble in very concentrated aqueous ammonia.

AgI(s) is insoluble in aqueous ammonia.

b) i) Cl2 0

HOCl +1

HCl –1

ii) Tap water contains chloride ions.

11

To

pic

16

U

nit 5

3

12 —

13 a) A: Potassium sulphite

B: Potassium chloride

C: Sulphur dioxide

D: Barium sulphite

b) i) K2SO3(aq) + 2HCl(aq) 2KCl(aq) + H2O(l) + SO2(g)

ii) Ba2+(aq) + SO32–(aq) BaSO3(s)

iii) 3SO2(aq) + Cr2O72–(aq) + 2H+(aq) 3SO4

2–(aq) + 2Cr3+(aq) + H2O(l)

c) A colourless solution would form.

d) Any soluble carbonate / sulphate

e) i) Neutralization

ii) SO2(g) + 2NaOH(aq) Na2SO3(aq) + H2O(l)

f)

14 a) A — ammonium sulphate

B — ammonia

C — barium sulphate

b) D — potassium bromide

E — bromine

F — sulphur dioxide

15 a) i) Iron(II) ion

ii) Iron(II) hydroxide is oxidized to iron(III) hydroxide on standing in air.

iii) A green precipitate (FeCO3(s)) forms.

b) i) Calcium ion

ii) Ca2+(aq) + 2OH–(aq) Ca(OH)2(s)

c) Add dilute aqueous ammonia to each solution.

Zinc ion gives a white precipitate that is soluble in excess dilute aqueous ammonia.

Lead(II) ion gives a white precipitate that is insoluble in excess dilute aqueous ammoinia.

d)

12

To

pic

16

U

nit 5

3

16 a) Calcium sulphate

The brick-red flame in flame test indicates that compound D contains calcium ion.

Compound D is insoluble in water, so it should be either calcium sulphate or calcium carbonate.

It does not react with dilute hydrochloric acid, so it should be calcium sulphate.

b) i) Copper(II) hydroxide / copper(II) oxide

B dissolves to give a blue solution, so it should contain copper(II) ion.

As B is insoluble in water, it might be a carbonate, hydroxide or an oxide. As no gas is evolved in its reaction with dilute hydrochloric acid, it should not be a carbonate.

ii) Cu(OH)2(s) + 2H+(aq) Cu2+(aq) + 2H2O(l) / CuO(s) + 2H+(aq) Cu2+(aq) + H2O(l)

c) i) A contains ammonium ion.

NH4+(aq) + OH–(aq) NH3(g) + H2O(l)

ii) Chloride ion

It gives a white precipitate with acidified aqueous solution of silver nitrate.

NH4Cl(aq) + AgNO3(aq) AgCl(s) + NH4NO3(aq)

d) i) Bromide ion

ii) The solution turns yellow-brown.

Chlorine is a stronger oxidizing agent than bromine. It oxidizes bromide ion to bromine.

Cl2(g) + 2Br–(aq) 2Cl–(aq) + Br2(aq)

e) Conduct a flame test.

Sodium gives a golden yellow flame.

17 a) Blue

b) i) Lead(II) chloride

ii) Pb2+(aq) + 2Cl–(aq) PbCl2(s)

c) CuCO3(s) / MgCO3(s) + 2H+(aq) Cu2+(aq) / Mg2+(aq) + H2O(l) + CO2(g)

d) A pale blue precipitate forms.

Mg2+(aq) + OH–(aq) Mg(OH)2(s) white precipitate

Cu2+(aq) + OH–(aq) Cu(OH)2(s) pale blue precipitate

Copper(II) hydroxide dissolves in excess NH3(aq) to give a deep blue solution. The white precipitate remains.

Cu(OH)2(s) + 4NH3(aq) [Cu(NH3)4]2+(aq) + 2OH–(aq)

e) Sodium ion

13

To

pic

16

U

nit 5

3

18 A — H2SO4(aq) B — Na2SO3(aq) C — NaOH(aq) D — BaCl2(aq) E — Fe(NO3)2(aq) F — NaOCl(aq)

A and B gives a gas that turns aqueous solution of potassium dichromate green (sulphur dioxide). Thus, A and B should be H2SO4(aq) and Na2SO3(aq).

SO32–(aq) + 2H+(aq) SO2(g) + H2O(l)

A and F give a gas that turns moist blue litmus paper red and then bleaches it (chlorine). Thus, A and F should be H2SO4(aq) and NaOCl(aq).

OCl–(aq) + 2H+(aq) + Cl–(aq) Cl2(g) + H2O(l)

Hence it can be deduced that A is H2SO4(aq), B is Na2SO3(aq) and F is NaOCl(aq).

A and D give a white precipitate. Thus, D is BaCl2(aq). It gives BaSO4(s) with A (H2SO4(aq)).

Ba2+(aq) + SO42–(aq) BaSO4(s)

Hence D is BaCl2(aq).

C and E give a white precipitate. Thus, C and E should be NaOH(aq) and Fe(NO3)2(aq).

Fe2+(aq) + 2OH–(aq) Fe(OH)2(s)

When C is mixed with A, only heat is liberated. Thus, it can be deduced that a neutralization occurs.

H+(aq) + OH–(aq) H2O(l)

Hence C is NaOH(aq) and E is Fe(NO3)2(aq).

14

To

pic

16

U

nit 5

3

19 The anions can be classified into two groups:

• Br–(aq) ion forms insoluble silver salts;

• CO32–(aq) ion and SO4

2–(aq) ion form insoluble barium salts. They can be precipitated from a solution by adding aqueous solution of barium nitrate.

Barium carbonate is soluble in dilute hydrochloric acid while barium sulphate is not.

BaCO3(s) + 2H+(aq) Ba2+(aq) + CO2(g) + H2O(l)

solution containing Br–(aq), CO3

2–(aq), SO42–(aq)

precipitate of AgBr(s)

precipitate of BaSO4(s)

Step 3 — add HCl(aq)

Step 2 — add Ba(NO3)2(aq)

Step 1 — add AgNO3(aq)

solution containing CO3

2–(aq), SO42–(aq)

solution containing BaCO3(s), BaSO4(s)

solution containing CO3

2–(aq)

20 Test Observation Deduction

Smelling / warming some solution in a test tube

characteristic smell of ammonia The tube contains NH3(aq).

Add NH3(aq) to give test tubes containing the other five solutions

one of the solutions forms a brown precipitate that dissolves in excess NH3(aq) to give a colourless solution

The tube contains AgNO3(aq).

two of the solutions form white precipitates that dissolve in excess NH3(aq) to give colourless solutions

The tubes contain ZnSO4(aq) and ZnCl2(aq).

Add AgNO3(aq) to four test tubes containing solutions other than NH3(aq)

two of the solutions form white precipitates

The tubes contain HCl(aq) and ZnCl2(aq).Hence the solution that gives observable changes with both NH3(aq) and AgNO3(aq) is ZnCl2(aq).The solution that gives observable changes with NH3(aq) only is ZnSO4(aq).The solution that gives observable changes with AgNO3(aq) only is HCl(aq).The remaining solution is Na2SO4(aq).

15

To

pic

16

U

nit 5

3

21 Divide the samples into two groups: sodium compounds and ammonium compounds

Conduct a flame test using the samples.

Only the three sodium compounds give a golden yellow flame.

OR

Warm the samples with NaOH(aq).

Only the two ammonium compounds give a gas that turns moist red litmus paper blue (ammonia).

Distinguish between the sodium compounds

Add HCl(aq) to the three sodium compounds.

Sodium carbonate gives a colourless gas that turns limewater milky (carbon dioxide).

Sodium hypochlorite gives a gas that turns moist blue litmus paper red and then bleaches it (chlorine).

Hence the remaining sample is sodium chloride.

Distinguish between the ammonium compounds

Add acidified BaCl2(aq) to the aqueous solution of the two ammonium compounds.

Only the aqueous solution of ammonium sulphate gives a white precipitate (BaSO4(s)).

16

To

pic

16

U

nit 5

4


In-text activities



• Add aqueous bromine to each compound.

Propene turns the yellow-brown aqueous bromine colourless quickly but propan-1-ol does not.

• Add acidified dilute aqueous solution of potassium permanganate to each compound.

Propene turns the purple permanganate solution colourless quickly but propan-1-ol does not.

• Add phosphous pentachloride to each compound.

Propan-1-ol gives steamy fumes but propene does not.

b) Treat each compound with the Lucas reagent.

Propan-2-ol gives cloudiness in 5 minutes.

Propan-1-ol shows no observable change.

c) Warm each compound with iodine and an aqueous solution of sodium hydroxide.

Pentan-2-ol gives a bright yellow precipitate but pentan-3-ol does not.

2 Warm each compound with acidified aqueous solution of potassium permanganate.

Pentan-1-ol is oxidized to pentanal and then pentanoic acid.

The purple permanganate ion is reduced to colourless manganese(II) ion.

Pentan-2-one resists oxidation.



• Treat each compound with 2,4-dinitrophenylhydrazine.

CH3CH2CHO gives a yellow to red precipitate but CH3CH2COOH does not.

• Treat each compound with the Tollens’ reagent.

CH3CH2CHO gives a silver mirror but CH3CH2COOH does not.

• Warm each compound with acidified aqueous solution of potassium dichromate.

CH3CH2CHO turns the orange dichromate solution green but CH3CH2COOH does not.

17

To

pic

16

U

nit 5

4

b) Any one of the following:

• Treat each compound with the Tollens’ reagent.

CH3(CH2)3CHO gives a silver mirror but C2H5COC2H5 does not.


CH3(CH2)3CHO turns the orange dichromate solution green but C2H5COC2H5 does not.

c) Any one of the following:

• Treat each compound with 2,4-dinitrophenylhydrazine.

C4H9COCH3 gives a yellow to red precipitate but C4H9CH(OH)CH3 does not.


C4H9CH(OH)CH3 turns the orange dichromate solution green but C4H9COCH3 does not.

2 a) A, B and C are either aldehydes or ketones because they give an orange precipitate with 2,4-dinitrophenylhydrazine.

B and C are aldehydes as they give a silver mirror with the Tollens’ reagent.

b) A C2H5 CH3C

O

B and C C3H7 H3C CH

CH3

HC

O

HC

O


a)Test reagent Test result Deduction

K2Cr2O7 / H3O+ positive functional group(s) that may be present:

— OH, — CHO

2,4-dinitrophenylhydrazine negative functional group(s) absent:— CHO, C=O

Aqueous bromine positive functional group(s) that may be present:— C=C, — C C —

b) Any three of the following:

H2C=CHCH2CH2OH

H3CCH=CHCH2OH

H2C=CHCH(OH)CH3

H3CCH=C(OH)CH3

18

To

pic

16

U

nit 5

4


Treat each liquid with the Tollens’ reagent.

Only ethanal gives a silver mirror.

Treat the remaining four liquids with 2,4-dinitrophenylhydrazine.

Only propanone gives a yellow to red precipitate.

Mix the remaining three liquids with an aqueous solution of sodium hydrogencarbonate.

Only ethanoic acid gives effervescence.

Mix the remaining two liquids with phosphorus pentachloride.

Only ethanol gives steamy fumes.

The remaining liquid is pentyl ethanoate.


Solvent A

The solubility of X in A is high at high temperatures but low at low temperatures.


1 Substances 1, 2 and 4 are present.

2 a) Each amino acid distributes itself between the water in the paper fibres and the developing solvent.

Amino acids that are more soluble in the developing solvent travel up more quickly. Thus, the amino acids separate.

b) Spray a developing agent such as ninhydrin onto the paper. This converts the amino acids into coloured spots on the chromatogram.

c) Rf value for B = 5.5 cm7.0 cm

= 0.79


1 a) Carry out simple distillation. Collect the liquid boiling off between 70 °C and 90 °C.

b) Cyclohexanol and phosphoric acid.

c) Transfer the distillate to a separating funnel.

Add saturated sodium chloride solution and shake. Allow the two layers to separate. Run off the lower aqueous layer.

Run the top cyclohexene layer into a small conical flask. Add some anhydrous calcium chloride. Allow to stand until the cyclohexene becomes clear.

Decant the cyclohexene into a clean flask and re-distil it, collecting the liquid which distils off between 81 °C and 85 °C.

19

To

pic

16

U

nit 5

4

2 a) Pour the reaction mixture over crushed ice and stir until the product solidifies.

Collect the product by filtration under reduced pressure after all the ice melts.

Wash the product with a little ice-cold ethanol (keep this wash liquid for (c)).

b) To re-crystallize the product, dissolve it in the minimum amount of hot ethanol.

Filter the mixture while hot.

Allow the filtrate to cool. Collect the crystals by filtration under reduced pressure and dry them.

c) Evaporate the wash liquid from (a) to 1 cm3.

Use a fine capillary table to put a spot of the solution on one end of a fine layer of silica coated onto a glass plate.

After drying, place the glass plate in a jar lined with a piece of filter paper dipping in a solvent (an ethoxyethane-hexane mixture).

Methyl 3-nitrobenzoate can be seen under ultraviolet light or by exposing the glass plate to iodine vapour.

Methyl 2-nitrobenzoate, a minor product, may appear as a yellow spot on the silica layer.

STSE Connections (page 92)

1 Pros of using distillation for water purification

• Distillation is often used as the preferred water purification method in developing nations or areas where the risk of waterborne disease is high, due to its unique capabilities to remove bacteria and viruses from drinking water.

• Distilled water will not cause scales in water-using appliances like coffee maker, iron or clothes steamer.

Cons of using distillation for water purification

• Distillation is expensive.

• Distillation removes every possible impurity from water, including minerals and salts. The human body is incapable of surviving on distilled water for a prolonged period of time, as it does not contain the much needed composition of minerals. These minerals are very much needed by the body and a deficiency is bound to cause some ailments.

• The content of salts and minerals also give water a peculiar and welcoming 'taste'.

2 Two other water treatment methods:

• boiling;

• portable filters providing various degrees of protection.

References

http://www.allaboutwater.org/distillation.html

http://www.buzzle.com/articles/distillation-of-water.html

http://campushealth.unc.edu/index.php?option=com_content&task=view&id=599&Itemid=65

20

To

pic

16

U

nit 5

4



1 Reagent Observation with A Observation with B

2,4-dinitrophenylhydrazine a yellow to red precipitate forms a yellow to red precipitate forms

Warm with Tollens’ reagent silver mirror forms on the wall of reaction vessel

no observable change

Warm with iodine and an aqueous solution of sodium hydroxide

no observable change a bright yellow precipitate forms

2 Heat each compound with acidified aqueous solution of potassium dichromate under reflux.

Pentan-1-ol is oxidized to pentanal and then pentanoic acid.

Orange dichromate ion is reduced to green chromium(III) ion.

Pentan-2-one resists oxidation.

3 a) Warm each chemical with acidified aqueous solution of potassium permanganate.

CH3CHO is oxidized to CH3COOH. Purple permanganate ion is reduced to colourless manganese(II) ion.

CH3CHO + [O] CH3COOH

CH3COCH3 shows no observable change.

b) Add 2,4-dinitrophenylhydrazine to each chemical.

CH3CH2COCH3 gives a yellow to red precipitate but CH3CH2COOCH3 does not.

H2N

H

NO2

H

NO2 + H2O

precipitate

NO2NO2CH3

N+ C N N

CH3CH2

C

CH3

O

CH3CH2

c) Mix each chemical with sodium hydrogencarbonate solution.

Effervescence occurs for HOOCCH2CH2CH3 but not for HCOOCH2CH2CH3.

HOOCCH2CH2CH3 + NaHCO3 +Na–OOCCH2CH2CH3 + H2O + CO2

d) Any one of the following:

• Add 2,4-dinitrophenylhydrazine to each chemical.

(CH3)2CHCHO gives a yellow to red precipitate but (CH3)2CHOH does not.

H

(CH3)2CH

H

NO2

H

NO2 + H2O

precipitate

NO2NO2

N C N NH2N+C

H

O

(CH3)2CH

21

To

pic

16

U

nit 5

4

• Add Tollens’ reagent to each chemical and warm.

(CH3)2CHCHO gives a silver mirror on the wall of the reaction vessel but (CH3)2CHOH does not.

C H + 2[Ag(NH3)2]+ + 3OH–(CH3)2CH (CH3)2CH

O

C O–

silver mirror

O

+ 2Ag + 4NH3 + 2H2O

• Add ethanoic acid to each chemical and warm.

(CH3)2CHOH gives a sweetish smell but (CH3)2CHCHO does not.

CH3COOH + (CH3)2CHOH CH3COOCH(CH3)2 + H2O ester

• Add PCl5 to each chemical.

(CH3)2CHOH gives steamy fumes but (CH3)2CHCHO does not.

(CH3)2CHOH + PCl5 (CH3)2CHCl + POCl3 + HCl steamy fumes

e) • Any one of the following:

Add aqueous bromine to each chemical.

(CH3)2C=C(CH3)2 turns the yellow-brown aqueous bromine colourless quickly but does not.

CH3CH3 CH3

CH3CH3 CH3CH3

CH3

CH3CH3

CH3CH3

CC + Br Br C

Br

C

OHin water

C

Br

C

Br

+ + HBr

• Add acidified dilute aqueous solution of potassium permanganate to each chemical.

(CH3)2C=CH(CH3)2 turns the purple permanganate solution colourless quickly but does not.

CH3

CH3CH3

CH3

CH3CH3

CH3CH3

CC + [O] + H2O C C

OHOH

4 a) Paper chromotography

b) Sample A – one black spot

Sample B – three colour spots (yellow, green and blue)

22

To

pic

16

U

nit 5

4

5 First place some of the dry solids in a glass capillary tube.

Then tie the tube to a thermometer with a rubber band. The solids in the tube should be level with the bulb of the thermometer.

Place this set-up together with a stirrer in an oil bath and heat gently with stirring.

6 a) CH3CH2CH2CH2OH CH3CH2CH(OH)CH3

(CH3)2CHCH2OH

b) When treated with a solution of anhydrous zinc chloride in concentrated hydrochloric acid, cloudiness appears in 1 minute for alcohol X.

Cloudiness appears in 5 minutes for the secondary alcohol CH3CH2CH(OH)CH3.

The reaction mixture of CH3CH2CH2CH2OH and (CH3)2CHCH2OH remains clear.

7 a) Reagent Observation with X Observation with Y

Aqueous solution of sodium hydrogencarbonate

Effervescence occurs no observable change

2,4-dinitrophenylhydrazine no observable change a yellow to red precipitate forms

Acidified aqueous solution of potassium dichromate

no observable changeaqueous solution of potassium

dichromate turns from orange to green

b) i) CH2COOCH2CH3

ii)

H2C CH3O C

O

CHO

23

To

pic

16

U

nit 5

4

8 a) Test reagent Test result Deduction

K2Cr2O7 / H3O+ positive functional group(s) that may be present:

— OH, — CHO

Tollens’ reagent negative functional group(s) absent:— CHO

Aqueous bromine positive functional group(s) that may be present:C=C, C C

b)

HH

CH2OHH3C

CC

HH3C

CH2OHH

CC

9 Suppose we have 100 g of compound X, so there are 58.8 g of carbon, 9.80 g of hydrogen and 31.4 g of oxygen.

Carbon Hydrogen Oxygen

Mass of element in the compound

58.8 g 9.80 g 31.4 g

Number of moles of atoms that combine

58.8 g12.0 g mol–1 = 4.90 mol

9.80 g1.0 g mol–1 = 9.80 mol

31.4 g16.0 g mol–1 = 1.96 mol

Simplest ratio of atoms 4.90 mol1.96 mol

= 2.509.80 mol1.96 mol

= 5.001.96 mol1.96 mol

= 1.00

Simplest whole number ratio of atoms

2.50 x 2 = 5.00 5.00 x 2 = 10.0 1.00 x 2 = 2.00

∴ the empirical formula of D is C5H10O2.

Let (C5H10O2)n be the molecular formula of D.

Relative molecular mass of D = n(5 x 12.0 + 10 x 1.0 + 2 x 16.0) = 102n

i.e. 102n = 102.0 n = 1

∴ the molecular formula of D is C5H10O2.

When D undergo hydrolysis, the compound E produced is an alcohol.

E gives a pale yellow precipitate G (iodoform) when warmed with iodine in sodium hydroxide solution. It can

be deduced that E is a secondary alcohol with the CH3C

OH

H

group.

24

To

pic

16

U

nit 5

4

As the molecular formula of E is C3H8O. Hence it can be deduced the structure of E is CH3CH3 C

OH

H

.

D contains 5 carbon atoms while E contains 3. Thus, F should have 2 carbon atoms. Hence it can be deduced that F is CH3COO–Na+.

To summerize, D, E, F and G are as follows:

D — CH3COOCH(CH3)2 1-methylethylethanoate

E — (CH3)2CHOH propan-2-ol

F — CH3COO–Na+ sodium ethanoate

G — CHI3 iodoform

The equation for the reaction of D with sodium hydroxide solution is as follows:

CH3COOCH(CH3)2 + NaOH CH3COO–Na+ + (CH3)2CHOH

10 a) Butanal gives a yellow to red precipitate when treated with 2,4-dinitrophenylhydrazine.

b) Test Deduction(s)

I Treat with 2,4-dinitrophenylhydrazine X and Y may either be aldehydes or ketones.Z is neither an aldehyde nor a ketone.

II Warm with Tollens’ reagent Y is an aldehyde.X and Z are not aldehydes.

III Shake with acidified aqueous solution of potassium dichromate

Y and Z may be primary / secondary alcohols or aldehydes.X is a ketone or tertiary alcohol.

From Tests I and III, it can be deduced that X is a ketone.

The structural formula of X is CH3CH2COCH3.

From Test II, it can be deduced that Y is an aldehyde.

The structural formula of Y is (CH3)2CHCHO.

From Tests I, II and III, it can be deduced that Z is a primary / secondary alcohol.

The structural formula of Z is H2C=CHCH2CH2OH or

H

CH2OH

H

H3C

CC or

H

CH2OH

H3C

H

CC or

OH

11 a) A gives positive results when treated with the Tollens’ reagent. It can be deduced that A is an aldehyde with the –CHO functional group.

A decolorizes aqueous bromine readily. It can be deduced that A contains either a –C=C– bond or –C C– bond.

25

To

pic

16

U

nit 5

4

b) As the molecular formula of A is C9H8O, besides the aromatic ring (–C6H5 / –C6H4) and the –CHO functional group, the formula of the other group should be C2H2 / C2H3. Thus, A probably contains a –HC=CH– bond.

The structural formula of A is either

CHCHOHC

or

CHO

CH2CH

.

CHCHOHC

exhibits geometrical isomerism:

H

CHOH

CC

CHO

HH

CC

CHO

CH2CH

exhibits position isomerism:

CHO

CH2CH

CHO

CH2CH

CHO

CH2CH

12 —

13 —

14 a) Isolate the crude 1-bromobutane from the reaction mixture by distillation.

b) The distillate collected has an organic and an aqueous layer. Use a teat pipette to remove the aqueous upper layer.

Transfer the organic layer to a separating funnel and add concentrated hydrochloric acid. Stopper and shake well. This removes the unreacted butan-1-ol as it is soluble in the acid.

Allow the two layers to separate. Run the organic layer into a conical flask.

Shake the crude 1-bromobutane with sodium hydrogencarbonate solution in another separating funnel. This removes the excess hydrochloric acid and other acid impurities.

Allow the layers to separate and run the organic layer into a conical flask.

Add anhydrous sodium sulphate to dry the 1-bromobutane. Swirl the mixture until the liquid is clear.

26

To

pic

16

U

nit 5

4

Decant the dried product into a pear-shaped flask. Carry out fractional distillation. Collect the fraction with the boiling point range of 1-bromobutane. This removes any remaining organic impurities.

15 a) Reflux condenser

b) COOCH3

+ NaOH

COO–Na+

+ CH3OH

c) Add dilute sulphuric acid and filter.

d) Dissolve the crude sample in the minimum amount of hot water.

Filter the mixture while hot to remove the insoluble impurities.

Allow the filtrate to cool. Collect the crystals by filtration and dry them.

e) Determine the melting point of the product to see whether the product has a sharp melting point.

Compare the melting point obtained with data book value.

16 —

17 —

18 a) Each component in an orange drink sample distributes itself between the water in the paper fibres and the developing solvent.

Components that are more soluble in the developing solvent travel up more quickly. Thus, the components separate.

b) Only the spot from D travels the same distance and have the same colour as the spot from A.

c) i) Similarity

Any one of the following:

• Spots are placed on a medium.

• Solvent is allowed to soak up the medium.

• Different components travel different distances.

• Separation is by differences in distribution between the mobile and stationary phases.

Difference

Any one of the following:

• The medium is different.

• The medium is spread on a piece of glass.

ii) Permanent felt tip pen ink does not dissolve in water.

27

To

pic

16

U

nit 5

4

iii) Rf value for the felt tip pen ink = 4496

= 0.46

Rf value for the prohibited dye = 4598

= 0.46

iv) To be certain whether the spots are the same / different.

OR

To allow more accurate comparison between the spots.

19 a) Band A

Each component in the mixture has its own equilibrium between adsorption onto the surface of the adsorbent and solubility in the solvent.

The better adsorbed ones travel more slowly.

b) Band C

It is the component that is the most soluble in the water.

Thus, it moves the greatest distance from the origin relative to the solvent front.

20 —

21

Experment title: Experiment A — Testing methanal with Tollens’ reagent

Hazardous chemical, procedure or equipment involved

Nature of hazard Safety precautions Source of information

Methanal flammable and irritant

• wear safety glasses and protective gloves• carry out the experiment inside a fume

cupboard• keep the reagent bottle tightly closed• keep the reagent bottle well away from

naked flames• dispose of into appropriate waste bottle


Material Safety Data Sheet (MSDS)

Tollens’ reagent irritant and explosive

• wear safety glasses and protective gloves• Tollens’ reagent becomes explosive on

evaporation, therefore wash away the solution immediately after use

• dispose of into appropriate waste bottle inside the fume cupboard

Material Safety Data Sheet (MSDS) and previous instruction

Hot water may cause burns if spilt

• use water at 60 °C previous instruction

Continued on next page

28

To

pic

16

U

nit 5

4

Experment title: Experiment A — Preparing 2-chloro-2-methylpropane

Hazardous chemical, procedure or equipment involved

Nature of hazard Safety precautions Source of information

Methylpropan-2-ol flammable and harmful

• wear safety glasses and protective gloves• avoid skin contact with the chemical• keep the reagent bottle tightly closed• keep away from naked flames• dispose of into appropriate waste bottle


International Chemical Safety Cards (ICSCs)

2-chloro-2-methylpropane

flammable • wear safety glasses and protective gloves• keep the reagent bottle tightly closed• keep away from naked flames• dispose of into appropriate waste bottle



Concentrated hydrochloric acid

corrosive • wear safety glassses and protective gloves

• carry out this activity inside a fume cupboard as concentrated hydrochloric acid gives off harmful fumes

• wash affected areas with plenty of water if spilt on the skin or clothes; report the accident to the teacher

• wash the container of concentrated hydrochloric acid with great care; do not add water to the acid directly

• dispose of into appropriate waste bottle inside the fume cupboard


Separating funnel contents of the separating funnel may spill if not handled properly

• hold the stopper and tap when inverting the funnel to mix the contents

• open the tap regularly to release any pressure

• when standing in the fume cupboard, keep the funnel in an upright position safely with stands and clamps

previous instruction

29

To

pic

16

U

nit 5

4

22 Add aqueous bromine / acidified dilute aqueous solution of potassium permanganate to each liquid.

Only cyclohexene turns the yellow-brown aqueous bromine / purple permanganate solution colourless quickly.

Warm each of the remaining liquids with a mixture of ethanol and dilute aqueous solution of silver nitrate.

Only bromoethane gives a creamy precipitate.

Add aqueous solution of sodium hydrogencarbonate to each of the remaining liquids.

Only ethanoic acid gives effervescence.

Warm each of the remaining liquids with acidified aqueous solution of potassium dichromate.

Only butan-1-ol turns the orange dichromate solution green.

Treat each of the remaining liquids with the Lucas reagent.

Only methylpropan-2-ol gives cloudiness in 1 minute.

Add the remaining liquid to blue cobalt(II) chloride paper.

The cobalt(II) chloride paper turns pink. The remaining liquid is water.

30

To

pic

16

U

nit 5

5


In-text activities

Discussion (page 117)

Some possible ways for minimizing possible sources of error in the experiment:

• Add a slight excess of precipitating agent to precipitate out all the phosphorus. (Avoid adding a large excess of precipitating agent as this increases chances of adsorption on the surface of the precipitate.)

• Use the minimum amount of solvent to wash the precipitate.

• Avoid splashes.

• Keep the precipitate in a desiccator when left for drying overnight.

• Weigh the precipitate and the filter paper rapidly.


Action Calculated result would be lower

than the true value

No effect on calculated result

Calculated result would be higher

than the true value

Step 1 The beaker for holding the 25.0 cm3 of sample solution had been previously washed with distilled water but not dried.

✔

Step 3 The mass of the empty filtering crucible was recorded incorrectly. The recorded mass was 0.100 g higher than the actual mass.

✔

Step 4 The beaker containing the reaction mixture was not rinsed after pouring.

✔

Step 4 The precipitate was not washed. ✔

Step 5 The dried precipitate was not kept in a desiccator while cooling.

✔

Step 1 The distilled water remaining in the beaker would not affect the amount of Ca2+(aq) ions in the sample solution.

Step 3 The mass of the precipitate obtained would be lower than the true value. Hence the calculated result would be lower than the true value.

Step 4 Some of the precipitate might remain in the beaker. So, the mass of the precipitate obtained would be lower than the true value.

31

To

pic

16

U

nit 5

5

Step 4 Some impurities might adhere to the precipitate. So, the mass of the precipitate obtained would be higher than the true value.

Step 5 The precipitate might take up moisture from the atmosphere. So, the mass of the precipitate obtained would be higher than the true value.


1 Zn in foot powder ZnNH4PO4 ZnP2O7

1.080 g of sample 0.368 g

Number of moles of ZnP2O7 = 0.368 g

239.4 g mol–1

= 1.54 x 10–3 mol = number of moles of Zn in foot powder

Mass of Zn in foot powder = 1.54 x 10–3 mol x 65.4 g mol–1

= 0.101 g

Percentage by mass of Zn in foot powder = 0.101 g1.080 g

x 100%

= 9.35%

2 a) Ba2+(aq) + SO42–(aq) BaSO4(s)

b) Add a little BaCl2(aq) to the mother liquor after filtering off the precipitate. If some sulphate is still present, a precipitate will form. Then add more precipitating agent to precipitate the remaining sulphate and re-filter.

c) To remove the impurities.

d) Mass of BaSO4 = (18.791 – 18.456) g = 0.335 g

Number of moles of BaSO4 = 0.335 g

233.4 g mol–1

= 1.44 x 10–3 mol = number of moles of sulphate

Mass of sulphate in fertilizer = 1.44 x 10–3 mol x 96.1 g mol–1

= 0.138 g

Percentage by mass of sulphate in fertilizer = 0.138 g0.220 g

x 100%

= 62.7%

e) Mass of (NH4)2SO4 in fertilizer = 1.44 x 10–3 mol x 132.1 g mol–1

= 0.190 g

Percentage by mass of (NH4)2SO4 in fertilizer = 0.190 g0.220 g

x 100%

= 86.4%

32

To

pic

16

U

nit 5

5

f) The following sources of error make the experimental result for the percentage by mass of sulphate in the fertilizer slightly lower than the true value:

• Not all sulphate is precipitated out as we may not have added enough barium chloride solution.

• As no ionic substance is completely insoluble in water, a little barium sulphate will remain dissolved in the solution.

• A little of the precipitate will be lost as we wash the precipitate.

• Little splashes, inefficient rinsing out of the beaker, and inefficient filtering can cause some loss of the precipitate.

The following sources of error make the experimental result for the percentage by mass of sulphate in the fertilizer slightly higher than the true value:

• Impurities may be present.

• The precipitate may not be dried completely and so residual moisture can add to its mass.


Some sources of error students may suggest:

• Bromide ions might be present in sea water.

• The burette, pipette and volumetric flask might hold or deliver volumes slightly different from the volumes indicated by the manufacturer.

• The level of eyes might not be on the same level as the graduation marks of the burette, pipette and volumetric flask when reading the graduation marks and thus leads to reading errors.

• It was difficult to tell exactly when the indicator changed colour.

• The aqueous solution of silver nitrate might be added too fast, not allowing enough time for the reaction between silver ions and chloride ions to take place. This would cause the addition of more aqueous solution of silver nitrate than required.

• Deciding when the titration end point is reached requires personal judgement and thus may lead to errors.

• Little splashes can cause errors.


1 a) MnO4–(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

b) An aqueous solution of potassium permanganate serves as its own indicator. Add the permanganate solution until the first appearance of a persistent pale pink colour.

c) MnO4–(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

0.0215 mol dm–3 0.190 g 28.8 cm3

Number of moles of MnO4– ions in 28.8 cm3 solution = 0.0215 mol dm–3 x

28.81 000

dm3

= 6.19 x 10–4 mol

33

To

pic

16

U

nit 5

5

According to the equation, 1 mole of MnO4– ions reacts with 5 moles of Fe2+ ions.

i.e. number of moles of Fe2+ ions in solution = 5 x 6.19 x 10–4 mol = 3.10 x 10–3 mol

Mass of iron in sample = 3.10 x 10–3 mol x 55.8 g mol–1

= 0.173 g

Percentage by mass of iron in sample = 0.173 g0.190 g

x 100%

= 91.1%

∴ the percentage by mass of iron in the sample of the iron wire is 91.1%.

d) Any one of the following:

• Impurities in the iron sample have no reaction with permanganate ions.

• There is no air oxidation of iron(II) ions to iron(III) ions.

2 a) 5H2O2(aq) + 2MnO4–(aq) + 6H+(aq) 5O2(g) + 2Mn2+(aq) + 8H2O(l)

10.0 cm3 0.0100 mol dm–3

35.0 cm3

500.0 cm3

(used) 25.0 cm3


35.01 000

dm3

= 3.50 x 10–4 mol

According to the equation, 2 moles of MnO4– ions react with 5 moles of H2O2.

i.e. number of moles of H2O2 in 25.0 cm3 diluted solution = 52

x 3.50 x 10–4 mol

= 8.75 x 10–4 mol

Number of moles of H2O2 in 500.0 cm3 diluted solution = 8.75 x 10–4 x 500.025.0

mol

= 1.75 x 10–2 mol

Mass of H2O2 in 500.0 cm3 diluted solution = 1.75 x 10–2 mol x 34.0 g mol–1

= 0.595 g

Concentration of original H2O2 solution = 0.595 g

10.01 000 dm3

= 59.5 g dm–3

∴ the concentration of the aqueous solution of hydrogen peroxide is 59.5 g dm–3.

b) The concentration of hydrogen peroxide in a rainwater sample is too low.

34

To

pic

16

U

nit 5

5


1 a) To minimize air oxidation of the iodide ions.

b) Use starch solution as an indicator.

The iodine added will react with the sulphur dioxide. Once all the sulphur dioxide has reacted, the excess iodine added will react with the starch to form a dark blue complex. This colour change indicates the end point.

c) SO2(g) + I2(aq) + 2H2O(l) SO42–(aq) + 2I–(aq) + 4H+(aq)

25.0 cm3 0.00450 mol dm–3

15.8 cm3

Number of moles of I2 in 15.8 cm3 solution = 0.00450 mol dm–3 x 15.8

1 000 dm3

= 7.11 x 10–5 mol

According to the equation, 1 mole of SO2 reacts with 1 mole of I2.

i.e. number of moles of SO2 in 25.0 cm3 solution = 7.11 x 10–5 mol

Mass of SO2 in 25.0 cm3 solution = 7.11 x 10–5 mol x 64.1 g mol–1

= 4.56 x 10–3 g = 4.56 mg

Concentration of SO2 in the sample = 4.56 mg25.0

1 000 dm3

= 182 mg dm–3

∴ the concentration of sulphur dioxide in the sample of white wine is 182 mg dm–3.

2 a) i) I2(aq) + 2S2O32–(aq) 2I–(aq) + S4O6

2–(aq)

ii) Number of moles of S2O32– ions reacted with the iodine liberated = 0.100 mol dm–3 x

24.61 000

dm3

= 2.46 x 10–3 mol

According to the equation, 1 mole of I2 reacts with 2 moles of S2O32– ions.

i.e. number of moles of I2 formed in Stage 1 = 2.46 x 10–3

2 mol

= 1.23 x 10–3 mol

b) i) Cl2(aq) + 2I–(aq) I2(aq) + 2Cl–(aq) 1.23 x 10–3 mol

According to the equation, 1 mole of Cl2 reacts with I– ions to give 1 mole of I2.

i.e. number of moles of Cl2 in bleaching powder = 1.23 x 10–3 mol

Mass of Cl2 in bleaching powder = 1.23 x 10–3 mol x 71.0 g mol–1

= 8.73 x 10–2 g

∴ the mass of ‘available chlorine’ in the bleaching powder is 8.73 x 10–2 g.

35

To

pic

16

U

nit 5

5

ii) Percentage by mass of Cl2 in bleaching powder = 8.73 x 10–2 g

0.285 g x 100%

= 30.6%

∴ the percentage by mass of ‘available chlorine’ in the bleaching powder is 30.6%.



1 a) Gravimetric analysis involves determining the mass of the species being analyzed or the mass of a compound chemically related to that species.

b) Any two of the following:

• Analyze the concentration of chloride ions in a sample solution by precipitating the chloride ions as silver chloride.

• Determine the phosphorus content in a sample of fertilizer by precipitating the phosphorus as magnesium ammonium phosphate (MgNH4PO4•6H2O).

• Determine the calcium content in a sample solution by precipitating the calcium ions as calcium oxalate.

2 Any five of he following:

• the sample type

• the sample size available

• concentration range of the species being analyzed

• the accuracy required

• separation that may be required to eliminate interferences

• the instruments available

• the cost

• the speed

3 a) A pipette / burette may hold / deliver a volume slightly different from the volume indicated by the manufacturer.

b) During a titration, a student who is insensitive to colour changes may use more reagent.

c) During a titration, a small excess of the reagent is added to cause an indicator to change colour.

4 a) An aqueous solution of potassium permanganate is unstable as it tends to decompose slowly at room temperature.

b) Standardize an aqueous solution of potassium permanganate by titrating it against an aqueous solution of sodium oxalate. In an acidic solution, the oxalate ions are converted to oxalic acid. Potassium permanganate reacts with the oxalic acid according to the following equation:

5H2C2O4(aq) + 2MnO4–(aq) + 6H+(aq) 10CO2(g) + 2Mn2+(aq) + 8H2O(l)

36

To

pic

16

U

nit 5

5

Follow the steps below in the standardization process:

• Dissolve a known mass of sodium oxalate in dilute sulphuric acid.

• Heat the oxalate solution to 80 – 90 °C and titrate with the aqueous solution of potassium permanganate until a pale pink colour persists, showing that all the oxalic acid has reacted.

5 a) Iodine is a strong oxidizing agent. It reacts with other reducing agents in the environment.

b) i) IO3–(aq) + 5I–(aq) + 6H+(aq) 3I2(aq) + 3H2O(l)

ii) Number of moles of KIO3 in 250.0 cm3 of solution = 1.07 g

214.0 g mol–1

= 5.00 x 10–3 mol

Number of moles of I2 produced = 3 x 5.00 x 10–3 mol

Molarity of iodine solution = 3 x 5.00 x 10–3 mol

250.01 000 dm3

= 6.00 x 10–2 mol dm–3

6 Chloride ions react with silver ions according to the following reaction:

Ag+(aq) + Cl–(aq) AgCl(s)

Number of moles of AgCl obtained = 0.291 g

143.4 g mol–1

= 2.03 x 10–3 mol

Number of moles of Cl– ions = 2.03 x 10–3 mol

Mass of KCl in the sample = 2.03 x 10–3 mol x 74.6 g mol–1

= 1.51 x 10–1 g

Percentage by mass of KCl in the sample = 1.51 x 10–1 g

0.250 g x 100%

= 60.4%

∴ the percentage by mass of KCl in the sample is 60.4%.

7 a) Mg2+ ions will precipitate as Mg(OH)2.

b)

c) To remove the impurities.

37

To

pic

16

U

nit 5

5

d) sample of fertilizer MgNH4PO4•6H2O 3.150 g 2.085 g

Number of moles of MgNH4PO4•6H2O = 2.085 g245.3 g mol–1

= 8.50 x 10–3 mol

According to the formula of the precipitate, MgNH4PO4•6H2O, 1 mole of P gives 1 mole of the precipitate.

i.e. number of moles of P in sample = 8.50 x 10–3 mol

Mass of P in sample = 8.50 x 10–3 mol x 31.0 g mol–1

= 0.264 g

% by mass of P in sample = 0.264 g3.150 g

x 100%

= 8.38%

e) Mass of PO43– in sample = 8.50 x 10–3 mol x 95.0 g mol–1

= 0.808 g

Percentage by mass of PO43– in sample =

0.808 g3.150 g

x 100%

= 25.7%

f) Any two of the following:

• As no ionic substance is completely insoluble in water, a little magnesium ammonium phosphate will remain dissolved in the solution.

• A little of the precipitate will be lost as we wash the precipitate.

• Little splashes, inefficient rinsing out of the beaker, and inefficient filtering can cause some loss of the precipitate.

8 a) i) Chromate indicator

ii) When all the chloride ions are precipitated, the first excess aqueous solution of silver nitrate gives a reddish brown silver chromate precipitate with the chromate indicator. This signals the end point of the titration.

b) Ag+(aq) + Cl–(aq) AgCl(s) 0.0750 mol dm–3 25.0 cm3

19.8 cm3

250.0 cm3

(used) 25.0 cm3

Number of moles of Ag+ ions in 19.8 cm3 solution = 0.0750 mol dm–3 x 19.8

1 000 dm3

= 1.49 x 10–3 mol

38

To

pic

16

U

nit 5

5

According to the equation, 1 mole of Ag+ ions reacts with 1 mole of Cl– ions.

i.e. number of moles of Cl– ions in 25.0 cm3 diluted sea water = 1.49 x 10–3 mol

Number of moles of Cl– ions in 250.0 cm3 diluted sea water = 10 x 1.49 x 10–3 mol = 1.49 x 10–2 mol = number of moles of Cl– ions in 25.0 cm3 sea water sample

Concentration of Cl– ions in sea water sample = 1.49 x 10–2 mol

25.01 000 dm3

= 0.596 mol dm–3

= 0.596 mol dm–3 x 35.5 g mol–1

= 21.2 g dm–3

∴ the concentration of chloride ions in the sea water sample is 21.2 g dm–3.

9 a) H2O2(aq) + 2H+(aq) + 2e– 2H2O(l)

b) i) Use a pipetle to deliver the hydrogen peroxide solution to a conical flask.

Add dilute sulphuric acid to the conical flask.

Fill a burette with the aqueous solution of potassium permanganate.

Run the aqueous solution of potassium permanganate into the hydrogen peroxide solution until the first appearance of a persistent pale pink colour.

Repeat to obtain at least two consistent titres.

ii) 2MnO4–(aq) + 6H+(aq) + 5H2O2(aq) 2Mn2+(aq) + 8H2O(l) + 5O2(g)

0.0200 mol dm–3 (used) 10.0 cm3

18.2 cm3


18.21 000

dm3

= 3.64 x 10–4 mol

According to the equation, 2 moles of MnO4– ions react with 5 moles of H2O2.

i.e. number of moles of H2O2 ions in 10.0 cm3 solution = 52

x 3.64 x 10–4 mol

= 9.10 x 10–4 mol

Concentration of the diluted H2O2 solution = 9.10 x 10–4 mol

10.01 000 dm3

= 9.10 x 10–2 mol dm–3

Concentration of the undiluted H2O2 solution = 10 x 9.10 x 10–2 mol dm–3

= 0.910 mol dm–3

∴ the concentration of the undiluted H2O2 solution is 0.910 mol dm–3.

39

To

pic

16

U

nit 5

5

iii) Concentration of the undiluted H2O2 solution = 0.910 mol dm–3 x 34.0 g mol–1

= 30.9 g dm–3

Mass of H2O2 in 100 cm3 of solution = 30.9 g dm–3 x 100

1 000 dm3

= 3.09 g

∴ the undiluted H2O2 solution is unsuitable for treating paintings.

10 —

11 —

12 a) i) Any one of the following:

• Add aqueous solution of silver nitrate.

A yellow precipitate forms.

• Add aqueous chlorine and then an organic solvent.

A purple organic layer forms.

ii) Element Initial oxidation number Final oxidation number

Iodine +5 –1

Sulphur +4 +6

iii) The total change in oxidation number of S = 3 x (+2) = +6

The total change in oxidation number of I = –6

b) i) Pipette

ii) Starch solution

Dark blue to colourless

iii) 0.0100 mol dm–3 x 24.0

1 000 dm3 = 2.40 x 10–4 mol

iv) According to the equation, 1 mole of iodine reacts with 2 moles of Na2S2O3.

Number of moles of iodine = 2.40 x 10–4

2 mol

= 1.20 x 10–4 mol

v) Concentration of iodine solution = 1.20 x 10–4 mol

10.01 000 dm3

= 0.0120 mol dm–3

13 —

14 —

40

To

pic

16

U

nit 5

5

15 Place 25.0 cm3 of the water sample in a conical flask using a pipette.

Add 5 cm3 of dilute sulphuric acid and mix.

Place the conical flask in a boiling water bath for 10 minutes.

Add 5.00 cm3 of the aqueous solution of potassium permanganate to the water sample using a pipette.

After 10 minutes, add 5.00 cm3 of the aqueous solution of sodium oxalate using a pipette. Wait until the solution becomes colourless.

Fill a burette with the aqueous solution of potassium permanganate. Run the aqueous solution of potassium permanganate into the hot mixture until the first appearance of a persistent pale pink colour.

Initially, 5.00 cm3 of KMnO4(aq) and 5.00 cm3 of Na2C2O4(aq) are added to the water sample.

Number of moles of KMnO4 in 5.00 cm3 solution = 0.00200 mol dm–3 x 5.00

1 000 dm3

= 1.00 x 10–5 mol

Number of moles of Na2C2O4 in 5.00 cm3 solution = 0.00500 mol dm–3 x 5.00

1 000 dm3

= 2.50 x 10–5 mol

Under acidic conditions, the oxalate ions are converted to oxalic acid. Potassium permanganate reacts with the oxalic acid according to the following equation:

5H2C2O4(aq) + 2MnO4–(aq) + 6H+(aq) 10CO2(g) + 2Mn2+(aq) + 8H2O(l)

According to the equation, 5 moles of C2O42– ions react with 2 moles of MnO4

– ions.

If none of the permanganate ion is consumed by the oxidizable matter in the water sample, all the oxalate ions will react with the permanganate ions and no oxalate ion will remain in the mixture. Suppose V cm3 of aqueous solution of permanganate are required to react with the oxalate ions remaining in the mixture. It can be deduced that this amount of permanganate ions is consumed in oxidizing the oxidizable matter in the water sample.

Number of moles of MnO4– ions consumed in oxidizing matter in water sample

= 0.00200 mol dm–3 x V

1 000 dm3

= 2V x 10–6 mol

= 2V x 10–3 mmol

Number of millimoles of O2 required to oxidize the same amount of matter = 1.2 x 2V x 10–3 mmol

Mass of O2 required to oxidize the same amount of matter = 2.4V x 10–3 mmol x 32.0 mg mmol–1

= 7.68V x 10–2 mg

Permanganate index of water sample = 7.68V x 10–2 mg

25.01 000 L

= 3.07V mg of O2 per litre

41

To

pic

16

U

nit 5

6


In-text activities


a)

b) As the concentration of fluoride ions increases, more In(aq) (red-coloured indicator) reacts.

c) 1.50 mg dm–3

d) At zero absorbance, the concentraton of fluoride ions is 7.50 mg dm–3.

Number of moles of fluoride ions in 25.0 cm3 = 7.50 mg dm–3

19.0 g mol–1 x 25.0

1 000 dm3

= 9.87 x 10–6 mol

∴ the value of X is 9.87 x 10–6.


1 a)

CH2OH

C

H

H

C

H

CHO

C

H

H

C

H

Prop-2-en-1-ol Acrolein

b) The broad absorption at 2 500 – 3 300 cm–1 indicates the presence of an O–H bond (in acid).

The absorption at 1 680 – 1 750 cm–1 indicates the presence of a C=O bond.

Thus, acrylic acid was formed.

42

To

pic

16

U

nit 5

6

2 Suppose we have 100 g of either compound X or Y, so there are 40.0 g of carbon, 6.70 g of hydrogen and 53.3 g of oxygen.



40.0 g 1.70 g 53.3 g


40.0 g12.0 g mol–1 = 3.33 mol

6.70 g1.0 g mol–1 = 6.70 mol

53.3 g16.0 g mol–1 = 3.33 mol


= 1.006.70 mol3.33 mol

= 2.013.33 mol3.33 mol

= 1.00

∴ the empirical formula of either compound is CH2O.

Let (CH2O)n be the molecular formula of either compound.

Relative molecular mass = n(12.0 + 2 x 1.0 + 16.0) = 30n

i.e. 30n = 60.0 n = 2

∴ the molecular formula of either compound is C2H4O2.

The absorption at 1 700 cm–1 indicates the presence of a C=O bond.

The absorption at 3 350 cm–1 indicates the presence of an O–H bond.

Thus, it can be deduced that compound Y is a carboxylic acid while compound X is either an aldehyde, a ketone or an ester.

The structure of compound Y is OHH3C C

O

.

As compound X contains two oxygen atoms, thus it should be an ester.

The structure of compound X is CH3CH O

O

.

3 a) The absorption at 3 230 – 3 670 cm–1 indicates the presence of an O–H bond.

Thus, compound Y is an alcohol. As compound Y is formed from propanal, it can be deduced that compound Y is propan-1-ol.

Compound A is an ester formed by the reaction between compounds X and Y. Thus, compound X is propanoic acid (formed from propanal).

The structural formulae of the compounds are as follows:

Compound A CH3CH2 CH2CH2CH3C O

O

43

To

pic

16

U

nit 5

6

Compound X OHCH3CH2 C

O

Compound Y CH3CH2CH2OH

b) K2Cr2O7 / H3O+

Propanal compound X (propanoic acid) heat

1 LiAlH4 / ethoxyethane Propanal compound Y (propan-1-ol) 2 H3O

+

c) As a catalyst


1 a) The peak at m/e = 72

b) The peak at m/e = 43

CH3CO+ ion

c)

m/e = 43molecular ionm/e = 72

CH3CO CH2CH3

+CH3CO+ + •CH2CH3

•

2 a) m/e Ion

100 [H2C=CHCH2OOCCH3]•+

57 H2C=CHCH2O+

43 CH3CO+

b)

CH2CHC

m/e = 43m/e = 100

fragment lost

+

CH3CO+ + •O•

CH3CHCH2H2C CH2O

O


1 The difference between 44 and 28 is 16. Thus, the peak at m/e = 28 is formed from the loss of an oxygen atom from the molecular ion. The peak at m/e = 28 is due to the CO+ ion.

Hence the gas is carbon dioxide.

2 a) The difference between 46 and 29 is 17. Thus, the peak at m/e = 29 is probably due to the loss of a hydroxyl radical.

44

To

pic

16

U

nit 5

6

b) m/e Ion

28 CO+

29 CHO+

45 CHO2+

c) C OHH

O



• Mix phosphorus pentachloride with each compound.

Only CH3CH2CH2OH gives steamy fumes.


Only CH3CH2CH2OH turns the orange dichromate solution green.

b) i) The IR spectrum of compound A should have the following absorptions:

1 000 – 1 300 cm–1, indicating the presence of a C–O bond;

1 680 – 1 750 cm–1, indicating the presence of a C=O bond.

The IR spectrum of compound B should have the following absorption:

3 230 – 3 670 cm–1, indicating the presence of an O–H bond.

ii) Confirm the identify of a sample by comparing its spectrum with those of known compounds.

c) i) The molecular ion peak of CH3COOCH3 is at m/e = 74. The base peak is at m/e = 43, due to the CH3CO+ ion.

The difference between 74 and 43 is 31. The peak at m/e = 43 is probably due to the loss of a •OCH3 fragment.

ii) •

m/e = 43m/e = 74CH3CO OCH3

+CH3CO+ + •OCH3

2 a) The difference between 74 and 57 is 17. Thus, the peak at m/e = 57 is probably due to the loss of a hydroxyl radical.

b) The difference between 74 and 45 is 29. Thus, the peak at m/e = 45 is probably due to the loss of an ethyl radical.

c) An ethyl ion

d) O–H bond

45

To

pic

16

U

nit 5

6

e) In the mass spectrum, the peak at m/e = 57 suggest the loss of a hydroxyl radical from the molecular ion. Thus, a hydroxyl group is likely to be present in compound X.

The peak at m/e = 29 is probably due to the CH3CH2+ ion. This suggests the presence of an ethyl group

in compound X.

The infrared absorption at 1 680 – 1 750 cm–1 suggests the presence of a C=O bond.

Hence, a possible structure of compound X is C OHCH3CH2

O

.



1 Physical property measured Analytical method based on measurement of the property

Mass gravimetric analysis

Volume volumetric analysis

Absorption of electromagnetic radiation colorimetry / infrared spectroscopy

Mass-to-charge ratio mass spectrometry

2 a) Propan-2-ol H C

H

H

C

H

OH

C

H

H

H

Propanone H C

H

H

C

O

C

H

H

H

b) X: C=O bond

Y: O–H bond

c) Spectrum 1 is the spectrum of propanone.

Spectrum 2 is the spectrum of propan-2-ol.

3 a) A — sample inet

B — ionization chamber

C — ion detector

D — magnetic field

b) i) The peak at the m/e which equals the relative molecular mass of the sample

ii) The strongest detector signal which is set to 100%

46

To

pic

16

U

nit 5

6

4 a) Let CxHy be the molecular formula of A.

The peak at m/e = 42 is the molecular ion peak.

Relative molecular mass of A = 42.0 = 12.0x + 1.0y x = 3 and y = 6

∴ the molecular formula of A is C3H6.

b) The difference between 42 and 27 is 15, so it is probable that a methyl radical is lost from the molecular ion, producing the peak at m/e = 27.

The structural formula of hydrocarbon A is CH3CH=CH2.

5 a) The peak at m/e = 91 corresponds to the C6H5CH2+ ion.

The peak at m/e = 65 corresponds to the C5H5+ ion.

b)

m/e = 65

CH2+

+ +

rearrangement

m/e = 91 m/e = 91

–HC CH

6 a) A colorimeter is an instrument for measuring the amount of light absorbed by a sample when a beam of light passes through the sample.

Prepare a series of standard solutions of the species concerned. Using light of the selected colour (that the species absorbs the most), record the absorbance of each solution. Plot the absorbance against the concentration of the solutions.

Shine the light of the selected colour through the sample solution under test. Record the absorbance.

From the calibration curve, read off the concentration of the species concerned in the sample solution according to the absorbance recorded.

b) Concentration of H2O2 in rainwater sample = 1.55 x 10–6 mol dm–3 x 0.2780.410

= 1.05 x 10–6 mol dm–3

47

To

pic

16

U

nit 5

6

7 a) Concentration of the stock solution = 0.100 g

250.01 000 dm3

= 0.400 g dm–3

Mass of curcumin in 10.0 cm3 of stock solution = 0.400 g dm–3 x 10.0

1 000 dm3

= 4.00 x 10–3 g

Volume of standard 3 containing 4.00 x 10–3 g of curcumin = 4.00 x 10–3 g

1.00 x 10–2 g dm–3

= 4.00 x 10–1 dm3

= 400 cm3

Volume of water added to 10.0 cm3 of stock solution = (400 – 10.0) cm3

= 390 cm3

b) Curcumin concentration in the solution = 8.90 x 10–3 g dm–3

Mass of curcumin in 250.0 cm3 solution = 8.90 x 10–3 g dm–3 x 250.01 000

dm3

= 2.23 x 10–3 g = 2.23 mg

Curcumin content in the food sample = 2.23 mg24.5 g

= 0.0910 mg per gram of food sample

8 a) Ethyl ethanoate CH3COOCH2CH3

Butan-1-amine CH3CH2CH2CH2NH2

b) In the spectrum of compound A, the absorption at 3 350 – 3 500 cm–1 is due to the N–H bond. Hence compound A is butan-1-amine.

In the spectrum of compound B, the absorptions at 1 000 – 1 300 cm–1 and 1 680 – 1 750 cm–1 are due to the C–O bond and the C=O bond. Hence compound B is ethyl ethanoate.

9 a) The spectrum of X shows a strong absorption at about 1 700 cm–1, indicating the presence of a C=O bond.

The broad absorption bond at about 3 000 cm–1 is characteristic of an O–H bond.

b) X is a carboxylic acid.

Let CnH2n+1COOH be the molcular formula of X.

Molar mass of X = 60.0 g mol–1

= 12n + 2n + 1 + 12.0 + 2 x 16.0 + 1.0 n = 1

Thus, the structural formula of X is C OHCH3

O

.

48

To

pic

16

U

nit 5

6

10 a) i) Ester functional group

Carbon-carbon double bond

ii)

O

O

iii) C12H14O2

b) Stereoisomers have their atoms linked in the same way, but they differ in the spatial arrangement of their atoms.

Structural formula of compound B:

H3C

CH2

CC

OC

O

CH3

H

In compound A, the methyl groups are on opposite sides of the double bond.

In compound B, the methyl groups are on the same side of the double bond.

c)

H3C HOH2C

CC

OHC

O

CH3H

d) i) The absorption at 1 000 – 1 300 cm–1 is due to the C–O bond.

The absorption at 1 680 – 1 750 cm–1 is due to the C=O bond.

ii) 2 500 – 3 300 cm–1 / 3 230 – 3 670 cm–1

O–H bond is not present in compound A.

11 a)

H3C CH3

CH3

CCH3CHCH2OH

CH3CH2 O CH2CH3 O CH3CH3CH2CH2

CH3

CH3CH2CH2CH2OH CH3CH2CHCH3

OH

OH

49

To

pic

16

U

nit 5

6

b) The broad absorption at 3 230 – 3 670 cm–1 indicates the presence of an O–H bond (alcohol).

As compound X has a pair of enantiomers. Thus, it should have a chiral carbon.

Hence the structure of compound X is as follows:

CH3CH2CHCH3

OH

chiral carbon

12 —

13 a) Pentan-2-one CH3CH2CH2COCH3

Pentan-3-one CH3CH2COCH2CH3

b) The peak at m/e = 86

c) The difference betwen 86 and 71 is 15, so it is probable that a methyl radial is lost from the molecular ion, producing the peak at m/e = 71.

The CH3CO+ ion is responsible for the peak at m/e = 43.

d) The compound is pentan-2-one.

The compound is not pentan-3-one because its mass spectrum should have a peak at m/e = 57 due to the CH3CH2CO+ ion.

The peak at m/e = 71 is produced by this fragmentation:

[CH3CH2CH2CO CH3]•+ CH3CH2CH2CO+ + •CH3

molecular ion m/e 86

The peak at m/e = 43 is produced by this fragmentation:

[CH3CH2CH2 COCH3]•+ CH3CO+ + •CH2CH2CH3

14 a) CH3CH2CH2CH2CH3

CH3

CH3CHCH2CH3

H3C CH3

CH3

CH3

C

b) The base peak is at m/e = 57. This shows that alkane A can give a stable C4H9+ ion during

fragmentation.

This rule out the structure CH3CH2CH2CH2CH3 because the C4H9+ ion formed from it by losing a methyl

radical is a primary carbocation. The primary carbocation has a low stability and the intensity for this ion should be low.

50

To

pic

16

U

nit 5

6

[CH3CH2CH2CH2 CH3]•+ CH3CH2CH2CH2

+ + •CH3

Alkane A is not

CH3

CH3CHCH2CH3. The molecular ion of

CH3

CH3CHCH2CH3 can give a stable secondary C4H9+

ion by losing a methyl radical, and also a stable secondary C3H7+ ion by losing an ethyl radical.

•

•

+

m/e = 57

CH3CHCH2CH3

CH3

+

CH3CHCH2CH3 + •CH3

+

m/e = 43

CH3CHCH2CH3

CH3

+

CH3CH + •CH2CH3

CH3

H3C CH3

CH3

CH3

C is consistent with the mass spectrum. Its molecular ion can undergo fragmentation

readily to give a stable tertiary C4H9+ ion.

•H3C CH3

CH3

CH3

C H3C CH3

CH3

C

m/e = 57

+

+ •CH3

+

15 a) i) The difference between 134 and 105 is 29, so it is probably that an ethyl radical is lost from the molecular ion, producing the peak at m/e = 105.

Thus, the

CO+

ion is responsible for the peak at m/e = 105.

The C6H5+ is responsible for the peak at m/e = 77.

ii) COCH2CH3 CO+

+ •CH2CH3

+

m/e = 105

•

b) i) COCH2CH3 CH(OH)CH2CH3

1 LiAlH4 / ethoxyethane

2 H3O+

51

To

pic

16

U

nit 5

6

ii)

HO

CH2CH3 CH2CH3

OHH H

C C

c) i) CH(OH)CH2CH3 CH=CHCH3

conc. H2SO4

or conc. H3PO4

heat

) Geometr cal somer sm

) Restr cted rotat on about a carbon-carbon double bond.

Two d fferent atoms / groups on each carbon atom of the carbon-carbon double bond.

16 —

17 a) Suppose we have 100 g of compound A, so there are 69.8 g of carbon, 11.6 g of hydrogen and 18.6 g of oxygen.


Mass of element n the compound

69.8 g 11.6 g 18.6 g

Number of moles of atoms that comb ne

69.8 g12.0 g mol–1 = 5.82 mol

11.6 g1.00 g mol–1 = 11.6 mol

18.6 g16.0 g mol–1 = 1.16 mol

S mplest rat o of atoms

5.82 mol1.16 mol

= 5.0211.6 mol1.16 mol

= 10.01.16 mol1.16 mol

= 1.00

∴ the emp r cal formula of compound A s C5H10O.

b) The molecular on peak s at m/e = 86. Thus, the relat ve molecular mass of compound A s 86.0.

Let (C5H10O)n be the molecular formula of compound A.

Relat ve molecular mass of A = n(5 x 12.0 + 10 x 1.0 + 16.0) = 86n

.e. 86n = 86.0 n = 1

∴ the molecular formula of compound A s C5H10O.

c) Compound A g ves an orange prec p tate w th 2,4-d n trophenylhydraz ne. Thus, compound A s e ther an aldehyde or a ketone.

Compound A g ves no observable change w th ac d f ed aqueous solut on of potass um d chromate. Thus, compound A s a ketone, not an aldehyde.

Poss ble structural formulae of compound A:

CH3CH2CCH2CH3

O

CH3CCH2CH2CH3

O

CH3CCH(CH3)2

O

52

To

pic

16

U

nit 5

6

d) i) The C2H5+ ion is responsible for the peak at m/e = 29.

The CH3CH2CO+ ion is responsible for the peak at m/e = 57.

ii) The structure of compound A is as follows:

CH3CH2CCH2CH3

O

18 The molecular ion peak of each alcohol is at m/e = 74.

The major peak at m/e (M – 45) is due to the C2H5+ ion.

The major peak at m/e (M – 43) is due to the loss of a C3H7 radical from the molecular ion.

It can be deduced that alcohol B has a propyl group. Thus, its structural formula is CH3CH2CH2CH2OH.

It can be deduced that alcohol A has an ethyl group. Thus, its structural formula is CH3CH2CH(OH)CH3.

19 a) Suppose we have 100 g of compound X, so there are 77.8 g of carbon, 7.40 g of hydrogen and 14.8 g of oxygen.



77.8 g 7.40 g 14.8 g


77.8 g12.0 g mol–1 = 6.48 mol

7.40 g1.00 g mol–1 = 7.40 mol

14.8 g16.0 g mol–1 = 0.925 mol

Simplest ratio of atoms

6.48 mol0.925 mol

= 7.017.40 mol0.925 mol

= 8.000.925 mol0.925 mol

= 1.00

∴ the empirical formula of compound X is C7H8O.

i) The m/e of the molecular ion of compound X is 108.

ii) C6H5+ ion

b) The broad absorption at 2 500 – 3 300 cm–1 is due to an O–H bond (acid).

In the mass spectrum of compound X, the molecular ion peak is at m/e = 108. Thus, its relative molecular mass is 108.

Let (C7H8O)n be the molecular formula of compound X.

Relative molecular mass of X = n(7 x 12.0 + 8 x 1.0 + 16.0) = 108n

i.e. 108n = 108 n = 1

∴ the molecular formula of compound X is C7H8O.

As compound X can undergo oxidation to give an acid, it can be deduced that compound X is an alcohol or an aldehyde.

53

To

pic

16

U

nit 5

6

As the mass spectrum of compound X has a peak at m/e = 77 due to the C6H5+ ion, it can be deduced

that compound X contains a phenyl group.

Beside the phenyl group (–C6H5), the formula of the other group should be C2H3O. Thus, compound X probably has a CH2OH group attached to the phenyl group.

The structure of compound X is as follows:

CH2OH

The structure of compound Y is as follows:

COOH

20 Suppose we have 100 g of compound A, so there are 66.7 g of carbon, 11.1 g of hydrogen and the rest is oxygen (i.e. 22.2 g).



66.7 g 11.1 g 22.2 g


66.7 g12.0 g mol–1 = 5.56 mol

11.1 g1.0 g mol–1 = 11.1 mol

22.2 g16.0 g mol–1 = 1.39 mol


= 4.0011.1 mol1.39 mol

= 7.991.39 mol1.39 mol

= 1.00

∴ the empirical formula of compound A is C4H8O.

In the mass spectrum of compound A, the molecular ion peak is at m/e = 72. Thus, its relative molecular mass is 72.0.

Let (C4H8O)n be the molecular formula of compound A.

Relative molecular mass of A = n(4 x 12.0 + 8 x 1.0 + 16.0) = 72n

i.e. 72n = 72.0 n = 1

∴ the molecular formula of compound A is C4H8O.

The difference between 72 and 57 is 15. Thus, the peak at m/e = 57 is probably due to the loss of a methyl radical from the molecular ion.

The peak at m/e = 43 may be due to the CH3CO+ ion or the C3H7+ ion.

54

To

pic

16

U

nit 5

6

The strong IR absorption at 1 720 cm–1 is due to a C=O bond.

Hence the structure of compound A may be

CH3CH2COCH3 or CH3CH2CH2CHO.

21 The absorption at 1 680 – 1 750 cm–1 indicates the presence of a C=O bond.

The molecular ion peak of compound X is at m/e = 58. Thus, its relative molecular mass is 58.0.

Other than the oxygen atom (relative atomic mass = 16.0), there is probably 3 carbon atoms and 6 hydrogen atoms in compound X.

Hence compound X is probably an aldehyde or a ketone.

The peak at m/e = 29 in the mass spectrum is due to the C2H5+ ion. Hence compound X probably contains

an ethyl group.

Thus, the structure of compound X is CH3CH2CHO.

55

To

pic

16

U

nit 5

7


In-text activities


1 Titrating with aqueous solution of silver nitrate / precipitation using an aqueous solution of silver nitrate in gravimetic analysis

2 Precipitation using oxalate ions in gravimetric analysis

3 Treating the sample with molybdenum blue to obtain a blue-coloured species and measuring the absorbance of the species using a colorimeter

4 Titrating the acid with standard aqueous solution of sodium hydroxide

5 Titrating the ethanoic acid with standard aqueous solution of sodium hydroxide

6 Titrating the sulphur dioxide with standard aqueous iodine


Some peaks in the chromatogram of the athlete’s sample coincide with peaks in the standard. It is likely that the weight lifter was using steroids.


1 In 1 m3 of air, there is 365106 m3 of CO2.

Number of moles of CO2 in 1 m3 of air = 365 x 10–6 m3

0.0240 m3 mol–1

= 1.52 x 10–2 mol

Mass of CO2 in 1 m3 of air = 1.52 x 10–2 mol x 44.0 g mol–1

= 0.669 g = 669 000 x 10–6 g = 669 000 µg

∴ the concentration of carbon dioxide in the air is 669 000 µg m–3.

2 Number of moles of NO2 in 1 m3 of air = 50.0 mg46.0 g mol–1

= 1.09 x 10–3 mol

Volume of NO2 in 1 m3 of air = 1.09 x 10–3 mol x 0.0240 m3 mol–1

= 2.62 x 10–5 m3

Concentration of NO2 (in ppm) = 2.62 x 10–5 m3

1 m3 x 106 ppm

= 26.2 ppm

∴ the concentration of nitrogen dioxide in the sample of air is 26.2 ppm.

56

To

pic

16

U

nit 5

7


1 Normally, oxygen is transported from the lungs to cells via red blood cells. These cells contain a special protein called haemoglobin which is able to combine with oxygen to form oxyhaemoglobin.

Oxyhaemoglobin is fairly unstable and decomposes in the intercellular spaces to release free oxygen and haemoglobin. The oxygen is then available to carry out metabolic reactions in cells, reactions from which the body obtains energy.

If carbon monoxide is present in the lungs, this sequence is disrupted. Carbon monoxide bonds with haemoglobin to form a more stable compound than oxyhaemoglobin. It has much less tendency to break down. As a result, cells are unable to obtain the oxygen they need for metabolism and energy production dramatically decreases.

2 In 2009, a group of researchers reported their study examining the efficacy of restricting access to charcoal in preventing suicides from carbon monoxide poisoning by charcoal burning in Hong Kong. All charcoal packs were removed from the open shelves of major retail outlets in the intervention region for 12 months; in the control region, charcoal packs were displayed as usual. The suicide rate from charcoal burning was reduced by a statistically significant margin in the intervention region but not in the control region. They observed no significant change in the suicide rate using other methods in either location.

Their results strongly suggested the efficacy of limiting access to retail charcoal in preventing suicide by charcoal burning. The fact that the overall suicide rate also fell suggested that the intervention did not prompt a switch to alternative methods of suicide.

References

http://science.jrank.org/pages/1212/Carbon-Monoxide-Physiological-effects.html

http://bjp.rcpsych.org/cgi/reprint/196/3/241.pdf


1 Conduct the test after the driver has thoroughly rinsed his mouth with water. A positive result probably indicates that the driver has drunk.

Ethanol is soluble in water. The concentration of ethanol in the breath will drop after the driver has rinsed his mouth.

2 Students supporting the idea may suggest problems associated with excessive drinking of alcohol.

Students opposing the idea may suggest the following points:

• in small amounts alcohol makes people relaxed;

• a positive effect of drinking alcohol is the thinning of blood and hence being less prone to heart attacks;

• moderate drinking is accepted by the society.

3 Students supporting the idea may suggest:

• problems associated with excessive drinking of alcohol;

• we should protect our young people and stop the glorification of alcohol on television;

• the advertising of tobacco has been banned.

57

To

pic

16

U

nit 5

7

Students opposing the idea may suggest:

• a ban would infringe our basic human rights;

• moderate drinking is beneficial to health;

• the action would have a great impact on the advertising business.



1 a) The mobile phase is the carrier gas.

The stationary phase particles are coated onto the inside of the column.

The sample is swept along the column by a carrier gas.

Components in the sample that dissolve more readily in the liquid stationary phase move more slowly in the carrier gas stream. Separation of components in the sample results.

b) Time

c) Any one of the following:

• Checking whether foodstuffs are tempered with cheaper alternatives.

• Air monitoring in polluted environments, such as gas quality in mines.

• Testing of urine samples from athletes for banned substances.

• Determining the concentration of a certain component in a sample.


• From incomplete burning of household and industrial wastes

• Exhaust from cars and burning wood

b) Use gas chromatography to separate the species of interest in a complex mixture and a mass spectrometer as a detector to quantify the species as it elutes from the chromatographic column.

Instead of obtaining a full mass spectrum, the mass spectrometer is set to monitor only m/e values of ions of the species of interest. Then the concentration of the species based on peak heights at the m/e values can be obtained.

3 The orange dichromate crystals turn green.

When the breath of a drunken person is passed through, the following chemical reaction takes place:

2Cr2O72–(aq) + 3CH3CH2OH(aq) + 16H+(aq) 4Cr3+(aq) + 3CH3COOH(aq) + 11H2O(l)

Green chromium(III) ions are formed.

4 Develop latent fingerprints with iodine fumes.

The lipids present in the fingerprint deposit give a yellow brown coloration upon the absorption of iodine vapour.

58

To

pic

16

U

nit 5

7

5 Any one of the following:

• Blood glucose test

• Blood cholesterol test

• Phosphorus content in urine

6 One peak in the chromatogram of the athlete’s sample coincides with that of amphetamine.

7 a) i) Rf value is the retention ratio, the distance a component travels relative to the solvent.

ii) Retention time is the time a component held in a column.

b) i) Role of the gas — mobile phase / to carry the sample through the chromatographic column

Role of the liquid — stationary phase

ii) Trace with two peaks drawn

iii) Measure the area under each peak.

Find the total area of the peaks.

Percentage composition of a component = area of peak of the componenttotal area of peaks

x 100%

8 —

9 —

10 In 1 m3 of air, there are 29 900 µg of NO2.

Number of moles of NO2 in 1 m3 of air = 29 900 x 10–6 g46.0 g mol–1

= 6.50 x 10–4 mol

Volume of NO2 in 1 m3 of air = 6.50 x 10–4 mol x 24 000 cm3 mol–1

= 15.6 cm3

∴ there are 15.6 cm3 of NO2 in 106 cm3 of air.

i.e. the concentration of NO2 is 15.6 ppm by volume.

11 a) Number of molecules of a pollutant per million (106) molecules in air.

b) In 1 m3 of air, there is 12106 m3 of CO.

Molar volume of gas at room temperature and pressure = 24.0 dm3 mol–1

= 24 000 cm3 mol–1

= 0.0240 m3 mol–1

∴ number of moles of CO in 1 m3 of air = 12 x 10–6 m3

0.0240 m3 mol–1

= 5.0 x 10–4 mol

Mass of CO in 1 m3 of air = 5.0 x 10–4 mol x 28.0 g mol–1

= 1.4 x 10–2 g = 14 mg

∴ the concentration of carbon monoxide in the sample of car exhaust is 14 mg m–3.

59

To

pic

16

U

nit 5

7

c) Incomplete combustion of car fuels produces carbon monoxide.

d) Carbon monoxide is toxic.

12 a) SO2(g) H2O2(aq)

SO42–(aq)

BaClO2(aq) BaSO4(s)

240 dm3 0.0700 g

Number of moles of BaSO4 = 0.0700 g

233.4 g mol–1

= 3.00 x 10–4 mol

According to the sequence shown, 1 mole of SO2 produces 1 mole of BaSO4.

i.e. number of moles of SO2 in 240 dm3 air sample = 3.00 x 10–4 mol

Mass of SO2 in 240 dm3 air sample = 3.00 x 10–4 mol x 64.1 g mol–1

= 1.92 x 10–2 g = 19 200 µg

Concentration of SO2 in air sample = 19 200 µg240 dm3 x 1 000 dm3

1 m3

= 8.00 x 104 µg m–3

∴ the concentration of sulphur dioxide in the air sample is 8.00 x 104 µg m–3.

b) Volume of SO2 in air sample = 3.00 x 10–4 mol x 24.0 dm3 mol–1

= 7.20 x 10–3 dm3

Concentration of SO2 (in ppm) = 7.20 x 10–3 dm 3

240 dm3 x 106 ppm

= 30.0 ppm


• Pressed-wood products made with urea-formaldehyde

• Combustion sources (e.g. gas appliances, kerosene heaters)

b) In 1 m3 of air, 0.10106 m3 of formaldehyde is allowed.

Volume of formaldehyde allowed in a 360 m3 room = 0.10106 x 360 m3

= 3.6 x 10–5 m3

Molar volume of gas at room termperature and pressure = 24.0 dm3 mol–1

= 24 000 cm3 mol–1

= 0.0240 m3 mol–1

Number of moles of formaldehyde allowed in a 360 m3 room = 3.6 x 10–5 m 3

0.0240 m3 mol–1

= 1.5 x 10–3 mol

Mass of formaldehyde allowed in a 360 m3 room = 1.5 x 10–3 mol x 30.0 g mol–1

= 0.045 g

∴ the maximum allowable formaldehyde in a room of volume 360 m3 is 0.045 g.

60

To

pic

16

U

nit 5

7

c) Highly irritating to the eyes / respiratory tract

d) High performance liquid chromatograph

14 a) I2(aq) + 2S2O32–(aq) 2I–(aq) + S4O6

2–(aq)

b) Titrate the iodine liberated in the mixture against a standard aqueous solution of sodium thiosulphate until the colour of the mixture changes to pale yellow.

At this point, add a few drops of starch solution, producing a dark blue colour. Add the sodium thiosulphate solution dropwise until the mixture becomes colourless.

c) I2(aq) + 2S2O32–(aq) 2I–(aq) + S4O6

2–(aq) 1.00 mol dm–3

40.0 cm3

Number of moles of S2O32– ions reacted with the iodine liberated = 1.00 mol dm–3 x

40.01 000

dm3

= 4.00 x 10–2 mol

According to the equation, 1 mole of I2 reacts with 2 moles of S2O32– ions.

i.e. number of moles of I2 reacted with S2O32– ions

= 4.00 x 10–2

2 mol

= 2.00 x 10–2 mol = number of moles of I2 liberated by ozone and iodide ions

Volume of ozone = 2.00 x 10–2 mol x 24.0 dm3 mol–1

= 0.480 dm3

Percentage by volume of ozone in the ozonized oxygen = 0.480 dm 3

10.0 dm3 x 100%

= 4.80%

15 a) i) Oxidation state of sulphur in: Oxidation state of iodine in:

SO2 = +4 I2 = 0

SO42– = +6 I– = –1

ii) Sulphur is oxidized.

The oxidation state of sulphur increases from +4 to +6.

b) i) From brown to colourless

ii) Number of moles of iodine used = 0.0100 mol dm–3 x 16.201 000

dm3

= 1.62 x 10–4 mol

iii) According to equation 1, 1 mole of iodine reacts with 1 mole of SO2.

i.e. number of moles of SO2 in 50.00 cm3 of wine = 1.62 x 10–4 mol

61

To

pic

16

U

nit 5

7

iv) Concentration of SO2 in wine = 1.62 x 10–4 mol

50.001 000 dm3

= 3.24 x 10–3 mol dm–3

v) Concentration of SO2 in wine = 3.24 x 10–3 mol dm–3 x 64.1 g mol–1

= 0.208 g dm–3

vi) The concentration of SO2 in the wine is between 0.01 g dm–3 and 0.25 g dm–3. Thus, the wine can be preserved.

16 a) Cellulose acetate

b) Polypropene

Documents

Topic 16Unit 53 - STMGSSintranet.stmgss.edu.hk/STUDENT/SUBJECTS/Chemistry/CHEM_TB_AN… · Topic 16Unit 53 Suggested answers to in-text activities and unit-end exercises In-text activities