Chapter 1-8 Worksheet Solutions

8/9/2019 Chapter 1-8 Worksheet Solutions

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Worksheet 1.1: Solutions

Precision, accuracy andsignificant figures

No. Answer

1 B and D . Measured quantities have an inherent uncertainty.

2 a N b P c B

3 Averaging a set of results reduces the effects of random errors associated with takingmeasurements.

4 a 3

b 4

c 3

d 4

5 a 1.407 × 10 2 b 5.005 × 10 3 c 9.800 × 10 2 d 7.5 × 10 –3

6 a 7.80 b 6.01 × 10 2 c 9.83 × 10 1 d 6.00 × 10 -4

7 a 6 × 10 3 b 6.000 × 10 3

8 a 6.7 × 102

b 0.30 or 3.0 × 10 –1

c 2.0 d 4.4 × 102

9 a 44.6 b 358.2

10

12722

= 60.2 g (The answer should be given to 3 significant figures. The 12 is an exact

number, and is therefore not relevant to the significant figure count.)

Page 1© Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2008.

This page from the Chemistry Dimensions 2, Teacher’s Resource may be reproduced for classroom use.


2/38

Worksheet 1.2: SolutionsCalculations involving gasesand solutions

No. Answer

1 a n(C6H12O6) = =

5.00

180.0

m

M mol

n(O) = 6 × n(C6H12O6) =180.0

6 ×5.00

m

= 0.167 mol

b n(H 2O2) = =8.36

34.0 M mol

n(atoms) = 4 × n(H 2O2)

N (atoms) = n × N A = 4 ×8.36

34.0 × 6.02 × 10 23 = 5.92 × 10 23

2 c1 × V 1 = c2 × V 2, ∴ V 2 =

c1 × V 1

c2

×=

0.30 35.0

0.090 = 117 mL

∴ V (H 2O) added = 117 – 35 = 82 mL

3 pV = nRT ∴ p =

0.49 × 8.31 ×

3.0

327 = 4.4 × 10 2 kPa

4 M = 4 × 10 –22 × 6.02 × 10 23 = 2 × 10 2 g mol –1

5



T V = constant ∴

1T 1V =

2

2T V ∴ V 2 =

1

21T

T V × =260

325600 × = 750 mL

6 pV = nRT and n =

M m

∴ pV = M m

RT ∴ V = pM mRT

=0.4421039731.80.62

×

× × = 22.1 L

7 c(NaCl) = M21

01558270

...

.V M

mV n

=×

=×

=

0.671.01.2

2

112

×=

×=

cV c

V = 1.79 L

∴ 0.79 L of water must be added.8

pV = nRT and n = M

m ∴ pV =

M

m RT ∴ M =

pV mRT

=2.06 × 8.31 × 300

20 × 16 = 16 g mol –1

The gas is methane.

9 a

005250501

1

221 .

.c

V cV

×=

×= = 75 mL

b 1.50% m/v means 1.5 g of ammonia in 100 mL of solutionWe require x g of ammonia in 150 mL.

∴ x = g2.25100150 1.50 =×


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Worksheet 1.2: SolutionsCalculations involving gasesand solutions

No. Answer

10 p × V = constant ∴ p1V 1 = p2V 2 ∴ V 2 =



2

11

pV p ×× 143.8

=687

199= 496 mL

11 pV =

M mRT

∴ M = pV mRT

= 3101228992993187780

−××

×

.

.. × = 159 g mol –1

The halogen must be bromine (Br 2).

12 600 ppm = 600 mg in 1.0 L = 600000 μ g in 1.0 L = 600 000 ppb

13 n(Pb(NO 3)2) =

331.23.5

= M m

= 0.0106 mol

c(Pb(NO 3)2) =0.060

0.0106=

V n

= 0.177 M

c2 =2

11

V V c ×

=30

10177.0 × = 0.059 M

14 c1 =

1

22

V V c × 0.108 × 300

30 = = 1.08 M

n(Na 2CO 3) = c × V = 1.08 × 0.030 = 0.0324 molm(Na 2CO 3) = n × M = 0.0324 × 106.0 = 3.4 g

15 2.0% m/v means 2.0 g of glucose in 100 mL of solution.∴ in 50 mL of solution, 0.50 × 2.0 gTotal m(glucose) = (0.50 × 2.0) + (0.50 × 6.0) = 4.0 g

4.0 g of glucose in a volume of 300 mL gives a concentration of 1003004.0

× = 1.3% m/v

16

2

22

1

11

T V p

T V p

= ∴ 760298

273190750

21

2112

×==

pT T V p × ×

V = 172 mL

17 c2 =

2

1V ×1V

c =

16 × 50

100 = 8% m/v

8% m/v = 8 g/100 mL = 80 g L –1 = 80 000 mg L –1 = 8.0 × 10 4 ppm

18 a 2.33 g in 398 mL ∴ 2.33 ×

398100

in 100 mL = 0.585% m/v

b n(AgNO 3) = M m

=9.169

33.2 mol

c(AgNO 3) =V

n =

398.09.169

33.2×

= 3.45 × 10 –2 M


4/38

Worksheet 1.3: SolutionsStoichiometry

No. Answer

1 n(SO 2) =

5.240.50

=

M V

V = 2.04 mol

n(H 2SO 4) = n(SO 2)m(H 2SO 4) = n × M = 2.04 × 98.1 = 2.00 × 10 2 g

2 n(H 3PO 4) =

98.01000

=

M

m = 10.2 mol

n(NO 2) = 5 × n(H 3PO 4) = 51.0 molm(NO 2) = n × M = 51.0 × 46.0 = 2.35 × 10 3 g

3 n(NaHCO 3) = c × V = 2.86 × 0.0299 = 0.0855 moln(H 2SO 4) =

1

2 × n(NaHCO 3) = 0.0428 mol

c(H 2SO 4) =0.02380.0428

=n

V = 1.80 M

4 a 2NaN 3(s) → 2Na(s) + 3N 2(g)

b n(NaN 3) =65.075

=

M

m

n(N 2) =23

× n(NaN 3)

V (N 2) =3.1010.65

30331.87523

×

×××=

pnRT

= 43 L

5 m(Br 2) = density × volume = 3.12 × 10.0 = 31.2 g

n(Br 2) =159.831.2

=

M m

= 0.195 mol

n(NaOH) =10

4 × n(Br 2) = 0.488 mol

c(NaOH) =0.2390.488

=

V

n = 2.04 M

6 n(H 2SO 4) =

98.119.56

=

M m

= 0.1994 mol

n(NH 3) = 2 × n(H 2SO 4) = 2 × 0.1994 = 0.3988 mol

V (NH 3) = p

nRT =

0.3988 × 8.31 × 360

2.99 × 101.3 = 3.94 L




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Stoichiometry

No. Answer

7 MgCO 3(s) → MgO(s) + CO 2(g)

n(MgO) =40.3100

=

M m

= 2.48 mol

n(CO 2) = n(MgO) = 2.48 molV (CO 2) = n × V M = 2.48 × 22.4 = 55.6 L

8 a CaCO 3(s) + 2HNO 3(aq) → Ca(NO 3)2(aq) + CO 2(g) + H 2O(l)

b n(CaCO 3) =100.110.0

=

M

m = 0.100 mol

n(HNO 3) = c × V = 0.500 × 0.100 = 0.0500 mol

0.100 mol of CaCO 3 will react with 0.200 mol of HNO 3, ∴ HNO 3 is the limiting reagent.n(CO 2) =

21

× n(HNO 3) = 0.0250 mol

V (CO 2) = n × V M = 0.0250 × 24.5 = 0.613 L

9 n(As) =

9.7457.6

=

M m

1

2

= 0.0877 mol

n(NaOH) = c × V = 0.779 × 0.250 = 0.195 mol1 As: 3 NaOH, ∴ 0.0877 mol of arsenic would react with 0.263 mol of sodium hydroxide∴ NaOH is the limiting reagent.

n(H 2) produced = × n(NaOH) = 0.0975 molm(H 2) = n × M = 0.0975 × 2.0 = 0.195 g

10 a Ba(OH) 2(aq) + 2HCl(aq) → BaCl 2(aq) + 2H 2O(l)

b n(Ba(OH) 2) =



M

m =

3.17100.8

= 0.0467 mol

n(HCl) = c × V = 1.886 × 0.120 = 0.226 mol0.0467 mol of Ba(OH) 2 would react with 0.0934 mol of HCl, ∴ HCl is in excessn(BaCl 2) = n(Ba(OH) 2) = 0.0467 mol

c(BaCl 2) = V n

= 120.00467.0

= 0.389 M


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Determining the molecularformula of a gas

No Answer

1 Na : B : O

Mole ratio23

8.22 :

8.105.21

:0.167.55

0.99 : 1.99 : 3.481 : 2 : 3.52 : 4 : 7

Empirical formula is Na 2B4O7.

2 C : H : O

Mole ratio 0.0556 :0.1

112.0 :



23

21

1002.61057.9 ×

×

0.0556 : 0.111 : 0.01593.5 : 7 : 17 : 14 : 2

Empirical formula is C 7H14O2.

3 As : O

Mole ratio9.74

7.75 :

0.16

3.24

1.01 : 1.522 : 3

Empirical formula is As 2O3.

EFM = 197.8 MF is (As 2O3)a , where a =EFMRMM

=395.6

197.8 = 2

Molecular formula is As 4O6.

4 C : H : O

Mole ratio0.163.53

:0.167.6

:0.120.40

3.33 : 6.67 : 3.331 : 2 : 1

Empirical formula is CH 2O.

EFM = 30 MF is (CH 2O) a , where a =RMM

EFM =

60

30 = 2

Molecular formula is C 2H4O2.

5 It absorbs water and so allows the mass of hydrogen in the sample to be determined.

6 It absorbs carbon dioxide and so allows the mass of carbon in the sample to be determined.


7/38


Determining the molecularformula of a gas

No Answer

7 m(C) in sample =

12.0

44.0 × 43.54 = 11.9 g

m(H) in sample =2.0

18.0 × 22.26 = 2.47 g

C : H

Mole ratio0.147.2

:0.129.11

0.99 : 2.47

1 : 2.52 : 5

Empirical formula is C 2H5.

8 The sample must vaporise at a temperature below that of the steam (below 100°C).

9 M =

pV

mRT =



310463.101

37331.80.16−

××

××

= 106 g mol –1

10 m(C) in sample =

12.0

44.0

× 27.83 = 7.590 g

m(H) in sample =2.0

18.0 × 11.38 = 1.264 g

m(O) in sample = 18.99 – 7.590 – 1.264 = 10.14 gC : H : O

Mole ratio0.16

14.10:

0.1264.1

:0.12

590.7

0.633 : 1.264 : 0.6341 : 2 : 1

Empirical formula is CH 2O.

M = pV mRT

= 2.17101.25

4738.316.212

××

××

= 89.99 g mol –1

EFM = 30 MF is (CH 2O) a , where a =RMM

EFM =

90

30 = 3

Molecular formula is C 3H6O3.

11 Mass spectrometry


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Worksheet 2.2: SolutionsGravimetric analysis

No. Answer

1 It was necessary to dry it so that water did not add any mass to the compound. Grinding ithelped it to dissolve.

2 n ((NH 4)2SO 4) =

132.11.00

= M

m = 7.57 × 10 –3 mol

3 Between steps 5 and 6, additional filtration with washing of the residue would have beenrequired to remove the insoluble components.

4 BaCl 2(aq) + (NH 4)2SO 4(aq) → BaSO 4(s) + 2NH 4Cl(aq)If the fertiliser was pure:n (BaCl 2) needed = n ((NH 4)2SO 4) = 7.57 × 10

–3mol

V (BaCl 2) =0.200

107.57 -3×=

c

n = 0.0379 L = 37.9 mL

5 The precipitating agent should be in excess to ensure complete precipitation.

6 The supernatant liquid was tested with BaCl 2 solution. No precipitate was observed.

7 Heat helps larger crystals of BaSO 4 form so it is better to wash it in hot water than cold.Washing removes any ions adsorbed onto the surface of the precipitate.

8 This means that the precipitate was dried, weighed, then dried and weighed again. If themasses are the same then constant mass has been achieved and there is no more water

present.

9 n ((NH 4)2SO 4) = n (BaSO 4) = =

1.63

233.4

m

M = 6.98 × 10 –3 mol

m ((NH 4)2SO 4) = n × M = 6.98 × 10 –3 × 132.1 = 0.923 g

% m/m ((NH 4)2SO 4) =00.1923.0

× 100 = 92.3%

10 • Errors in weighing

• Incomplete dissolution of fertiliser• Incomplete precipitation• Losses due to filtration• Not completely drying the precipitate




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Worksheet 2.3: SolutionsGravimetric analysis problems

No. Answer

1 Neither sodium nor nitrate ions form any insoluble compounds.

2 a i AgNO 3(aq) + NaCl(aq) → NaNO 3(aq) + AgCl(s)ii AgNO 3(aq) + KCl(aq) → KNO 3(aq) + AgCl(s)

b Because chloride ions are in both compounds, the chloride concentration is measuredcorrectly. The amount of chlorine is not equal to the amount of sodium or to the amountof potassium, therefore neither of these two metals can be analysed using this procedure.

3 Water for rinsing should be used sparingly. The more water used, the more precipitate thatwill dissolve and be lost.

4 There are many possible answers. For example:Mg(NO 3)2 → MgCO 3(s)AgNO 3 → AgI(s)CuSO 4 → Cu(OH) 2(s)

5 a Ba 2+(aq) + SO 42– (aq) → BaSO 4(s)b By simply removing the water, the positive ion is caught up in the remaining solid as well

as the sulfate ion, and any other ionic substances that are in the fertiliser.

6 Barium sulfate can form very fine particles. These particles will pass through normal filter paper.

7 a AgNO 3(aq) + LiCl(aq) → LiNO 3(aq) + AgCl(s)b n (AgNO 3) = c × V = 2.0 × 0.025 = 0.050 mol

0.02 mol of AgNO 3 reacts with 0.02 mol of LiCl, hence AgNO 3 is in excess.There is no problem with adding excess silver nitrate. In fact, it is important to add anexcess amount to ensure all the chloride is precipitated.

8 a C8H15Cl3 + 3AgNO 3(aq) → 3AgCl(s) + …

b n (AgCl) =4.143

478.0=

M

m = 0.00333 mol

n (C8H15Cl3) =3

1 × 0.00333 = 0.00111 mol

m (C8H15Cl3) = n × M = 0.00111 × 217.5 = 0.24 g

% purity =0.2

24.0 × 100 = 12%




10/38

Worksheet 2.3: Solutions Gravimetric analysis problems

No. Answer

9 a 2FeCl 3(aq) + … → Fe 2O3 + …

b n (Fe 2O3) =159.60.644

= M

m = 0.0040 mol

n (FeCl 3) = 2 × n (Fe 2O3) = 0.0080 molm (FeCl 3) = n × M = 0.0080 × 162.3 = 1.3 g

% purity FeCl 3 =0.23.1

× 100 = 65%

10 a Mass will be too high, ∴ the % P will be overestimated.b Precipitate will be lost, ∴ the % P will be underestimated.




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Worksheet 3.1: SolutionsRevision of acids and bases

No. Answer

1 a 2KHCO 3(s) + H 2SO 4(aq) → K 2SO 4(aq) + 2H 2O(l) + 2CO 2(g)b Fe 2O3(s) + 6HNO 3(aq) → 2Fe(NO 3)3(aq) + 3H 2O(l)c Ca(OH) 2(aq) + 2HCl(aq) → CaCl 2(aq) + 2H 2O(l)

2 a Ca(OH) 2(s) Ca 2+(aq) + 2OH – (aq) → OH 2b H2SO 4(l) + H 2O(l) → H3O+(aq) + HSO 4 – (aq)

HSO 4 – (aq) + H 2O(l) → H3O+(aq) + SO 42– (aq)c Ba(OH) 2(aq) + 2HCl(aq) → BaCl 2(aq) + 2H 2O(l)

3 a S2– and O 2– b H

2SO

4 and H

3PO

4

4 a HCl/Cl – and NH 4+/NH 3b H2O/OH – and OH – /O2–c H2CO 3/HCO 3 – /CO 32– and H 2SO 4/HSO 4 – /SO 42–

( H 2CO 3 decomposes to form H 2O and CO 2.)

5 a HSO 4 – (aq) + OH – (aq) → SO 42– (aq) + H 2O(l)b HSO 4 – (aq) + HCl(aq) → H2SO 4(aq) + Cl – (aq)

6 Different acid strengths (e.g. one strong, one weak) or different acid types (e.g. onemonoprotic, one diprotic)

7 a Methanoic acid is a weak acid that only partially ionises in water, producing few ions andso having low conductivity. Hydrochloric acid is a strong acid that ionises completely inwater, producing many ions and so having high conductivity.

b Only the O–H bond has sufficient polarity to make the hydrogen atom acidic. The threeC–H bonds are only slightly polar, so these hydrogen atoms are not acidic.

8 The solutions listed in order of decreasing pH are: NaOH is a strong base, ∴ high pH NH 3 is a weak base, ∴ pH > 7 but not too highH2O has pH = 7

CH 3COOH is a weak acid, ∴ pH < 7HNO 3 is a strong acid, ∴ low pHH2SO 4 is a strong, diprotic acid, ∴ pH < pH of HNO 3




12/38

Worksheet 3.1: Solutions Revision of acids and bases

No. Answer

9 a HNO 3 is a strong, monoprotic acid. It completely ionises in water.

( ) 1.3105.0log]O[Hlog pHM105.0=0.050=][HNO=]OH[

2.010

+310

2.03

+3

=×−=−=

∴ ×

∴ −

−

1410log]OH[log pH

10=1.0=0.502=][OH

1410

+310

0

=−=−=

b Ba(OH) 2 completely dissociates in water to produce 2 OH – ions per unit of Ba(OH) 2.

M10]OH[

10]OH[10

10][OH]OH[

14+3

14+3

0

14+3

∴

=∴

=×∴

=×

−

−

−−

×

−

−

10 n(H+) = c × V = 10 –3.0 × 0.0200 = 2.00 × 10 –5.0

4.010log]OH[log pH

100.200

102.00]OH[

4.010

+310

4.05.0

+3

=−=−=

=×

==

−

−

−

V n

A 10-fold dilution produces a change of one pH unit.

11 c1V 1 = c2V 2 c1 = [H 3O+] = 10 –pH = 10 –2.0 and c2 = [H 3O+] = 10 –pH = 10 –2.5 ∴ 10

–2.0

× 50.0 = 10 –2.5

× V 2 V 2 = 158 mL, therefore 108 mL must be added.

12 n(H+) = n(HCl) = c × V = 0.00100 × 20.00 × 10 –3 = 2.00 × 10 –5 moln(OH – ) = 2 × n(Ba(OH) 2) = 2 × c × V = 2 × 0.00100 × 20.00 × 10 –3 = 4.00 × 10 –5 mol∴ OH – is in excess by 2.00 × 10 –5 mol

10.7)10(2.00log]OH[log pH

M102.00]OH[

10]OH[105.00

10]OH[]OH[

mol105.001040.00

102.00]OH[

1110

+310

11+3

14+3

4

14+3

43

5

=×−=−=



∴

×=∴

=××

=×

×=×

×==

−

−

−−

−−

−

−

−

−

V n

∴


13/38

Worksheet 3.2: SolutionsAcid-base titrations

No. Answer

1 n(Na 2CO 3) = c × V = 0.0500 × 0.250 = 0.0125 molm(Na 2CO 3) = n × M = 0.0125 × 106.0 = 1.325 g1 Accurately weigh approximately 1.3 g sample of the primary standard.2 Transfer the sample to a 250.0 mL volumetric flask.3 Ensure complete transfer by washing with water.4 Dissolve the primary standard in water by shaking.5 Add water to make up the solution to the calibration mark, then mix again.

2 The sodium hydroxide is very concentrated. An exact concentration cannot be calculatedusing the data available.

3 The ratio of volumes is 20:5 or 1:4. This means the sodium hydroxide is four times moreconcentrated than the nitric acid. Therefore the concentration is 0.4 M.This ratio only works if the molar ratio in the balanced equation is 1:1.

4 a 2HNO 3(aq) + Na 2CO 3(aq) → 2NaNO 3(aq) + H 2O(l) + CO 2(g)b n(HNO 3) = c × V = 0.10 × 0.040 = 0.0040 mol

n(Na 2CO 3) =21

× n(HNO 3) = 0.0020 mol

V (Na 2CO 3) =20.0

0020.0=

cn

= 10 mL

5 a HCl(aq) + NH 3(aq) → NH 4+(aq) + Cl – (aq)b i At the equivalence point, some NH 4

+ exists. This is a weak acid, hence the pH will beabout 5.

ii After excess HCl is added, the pH is slightly greater than 1. (pH of the acid is 1. Theacid will be diluted by the increased volume due to the neutralised ammonia solution,giving a concentration less than 0.1 M and so a pH greater than 1.)

iii

6 CH 3COOH(aq) + NaOH(aq) → CH 3COONa(aq) + H 2O(l)n(NaOH) = c × V = 0.114 × 0.01545 = 0.001761 moln(CH 3COOH) = n(NaOH)

n(CH 3COOH) in original 25.0 mL = n(CH 3COOH) in aliquot ×0.20

100 = 0.00881 mol

c(CH 3COOH) =0.02500.00881

=

V n

= 0.352 M




14/38

Worksheet 3.2: SolutionsAcid-base titrations

No. Answer

7 Indirect titrations are performed when the substance to be analysed is:• insoluble• volatile ( so the titration method is not fast enough to carry out the analysis before the

concentration of the solution changes as evaporation occurs )• a weak acid or a weak base ( it is sometimes difficult to obtain sharp endpoints in these

titrations ).

8 a

b CH 3COO – is present at the equivalence point. This weak base will mean that the pH will be above 7 at the equivalence point.

c An indicator such as methyl red would be a poor choice because it does not change colouruntil pH 3, long after the equivalence point has been reached.

9 a i Water

ii Solution going into it—hydrochloric acidiii Solution going into it—sodium carbonateiv Water

One method to try with this style of question is to picture a very exaggerated amount ofwater in the piece of glassware and to ask whether this will affect the experiment in any way.b 2HCl(aq) + Na 2CO 3(aq) → 2NaCl(aq) + H 2O(l) + CO 2(g)

n(Na 2CO 3) = c × V = 0.100 × 0.0183 = 0.00183 moln(HCl) = 2 × n(Na 2CO 3) = 2 × 0.00183 = 0.00366 mol

n(HCl) in original 10.0 mL = n(HCl) in aliquot ×250.0

20.00 = 0.00366 ×

250.0

20.00 = 0.0458 mol

c(HCl) = 0.01000.0458

=V n

= 4.58 M

10 2HNO 3(aq) + Mg(OH) 2(aq) → Mg(NO 3)2(aq) + 2H 2O(l)n(Mg(OH) 2) = c × V = 0.10 × 0.0250 = 0.00250 moln(HNO 3) = 2 × n(Mg(OH) 2) = 2 × 0.00250 = 0.00500 mol

c(HNO 3) =



0135.000500.0

=

V n

= 0.370 M


15/38

Worksheet 3.3: SolutionsA back titration

No. Answer

1 a CaCO 3(s) + 2HCl(aq) → CaCl 2(aq) + H 2O(l) + CO 2(g)b HCl(aq) + NaOH(aq) → NaCl(aq) + H 2O(l)

2 a n(HCl) initially = c × V = 1.020 × 40.00 × 10 –3 = 4.080 × 10 –2 molb n(NaOH) = c × V = 0.275 × 25.56 × 10 –3 = 7.029 × 10 –3 molc n(HCl) unreacted = n(NaOH) = 0.275 × 25.56 × 10 –3 = 7.029 × 10 –3 mold n(HCl)reacting = n(HCl) initially – n(HCl)unreacted

= 4.080 × 10 –2 – 7.029 × 10 –3 = 3.377 × 10 –2 mol

e n(CaCO 3) =1

2 × n(HCl) reacting =

1

2 × 3.377 × 10 –2 = 1.689 mol

f m(CaCO 3) = n × M =1

2 × 3.377 × 10 –2 × 100.1 = 1.690 g

g % CaCO 3 in marble =



marble)()CaCO( 3

m

m ×

1100 1.690

1.740 = ×

1100

= 97.13%

3 Solid sodium hydroxide absorbs water from the atmosphere. Solid samples may therefore bedamp and therefore impure. Sodium hydroxide solutions react with carbon dioxide in theatmosphere, decreasing the concentration of the solution.

2NaOH(aq) + CO 2(g) → Na 2CO 3(aq) + H 2O(l)

4 CO 2 is an acidic oxide. It reacts with NaOH. If the CO 2 was not removed, more NaOH

would be required for the titration, leading to an increased value for the unreacted HCl, andhence a decreased value for the CaCO 3 percentage.

5 Calcium carbonate is not soluble.The carbonate ion is a weak base that gives an indistinct endpoint in a direct titration.

6 a This would dilute the acid. More would be required to react with the marble, giving anincreased percentage for the carbonate.

b This would dilute the NaOH. More would be required to react with the HCl, giving anincreased value for the unreacted HCl. This, in turn, would give a decreased value for thereacting HCl, and so a decreased percentage for the carbonate.

c This has no effect. Flasks should be rinsed with water.

7 n(CO 2) =

2968.31760765 3

××

××=

−

RT

pV 1095.0101.3 × = 0.00394 mol

n(CaCO 3) = n(CO 2) = 0.00394 molm(CaCO 3) = n × M = 0.00394 × 100.1 = 0.394 g

% CaCO 3 in marble =marble)(

)CaCO( 3mm

×1

100 0.3940.411

= ×1

100 = 95.9%

8 Small amounts of gas may have been lost in the collection process.Some carbon dioxide may remain dissolved in the reaction solution.


16/38

Worksheet 4.1: Solutions Oxidation numbers and redoxequations



No. Answer

1 Examples: a NF 3 b N2O5 c N2 d NH 3

2 a +6 b +6 c +4 d –1

3 a Not redoxb Oxidant = I 2O5, reductant = COc Not redoxd Oxidant = Hg 2+, reductant = N 2H4 e Oxidant = NO 3

– , reductant = H 2Sf Oxidant = NO 2, reductant = NO 2

4 a +6 b +2 c +7

5 C = 0 H = +1 O = –2

6 a Oxidation: CH 3CH 2OH(aq) + H 2O(l) → CH 3COOH(aq) + 4H +(aq) + 4e – Reduction: Cr 2O7

2– (aq) + 14H +(aq) + 6e – → 2Cr 3+(aq) + 7H 2O(l)Redox: 3CH 3CH 2OH(aq) + 2Cr 2O72– (aq) + 16H +(aq)

→ 3CH 3COOH(aq) + 4Cr 3+(aq) + 11H 2O(l)

b Oxidation: 2I – (aq) → I2(aq) + 2e – Reduction: BrO 3 – (aq) + 6H +(aq) + 6e – → Br – (aq) + 3H 2O(l) Redox: 6I – (aq) + BrO 3 – (aq) + 6H +(aq) → 3I 2(aq) + Br – (aq) + 3H 2O(l)

c Oxidation: Fe 2+(aq) → Fe 3+(aq) + e – Reduction: MnO 4

– (aq) + 8H +(aq) + 5e – → Mn 2+(aq) + 4H 2O(l) Redox: 5Fe 2+(aq) + MnO 4 – (aq) + 8H +(aq) → 5Fe 3+(aq) + Mn 2+(aq) + 4H 2O(l)

d Oxidation: H 2O2(aq) → O2(g) + 2H +(aq) + 2e – Reduction: MnO 4

– (aq) + 8H +(aq) + 5e – → Mn 2+(aq) + 4H 2O(l) Redox: 5H 2O2(aq) + 2MnO 4 – (aq) + 6H +(aq) → 5O 2(g) + 2Mn 2+(aq) + 8H 2O(l)

e Oxidation: H 2S(g) → S(s) + 2H +(aq) + 2e – Reduction: Cr 2O72– (aq) + 14H +(aq) + 6e – → 2Cr 3+(aq) + 7H 2O(l)

Redox: 3H 2S(g) + Cr 2O72– (aq) + 8H +(aq) → 3S(s) + 2Cr 3+(aq) + 7H 2O(l)


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Recovering silver fromsolution

No. Answer

1 a AgNO 3 solutionb NaCl solutionc deionised water

2 a Ag +(aq) + Cl – (aq) → AgCl(s)b Ag +(aq) + CrO 4

– (aq) → AgCrO 4(s)

3 n (NaCl) = c × V = 0.100 × 0.02000 = 0.002000 moln (AgNO 3) = n (NaCl) = 0.002000 mol

c (AgNO 3) =0.019320.002000=

V n = 0.1035 M

4 There is uncertainty in the pipette and burette volumes, and in the judging of the endpoint.

5 Dry and weigh the sample until the mass was constant.

6 a 2Ag +(aq) + Cu(s) → 2Ag(s) + Cu 2+(aq)b Silver ion

7 n (AgNO 3) = c × V = 0.1035 × 0.02000 = 0.002070 mol

n (Ag) = n (AgNO 3) = 0.002070 molm (Ag) present = n × M = 0.002070 × 107.9 = 0.2234 g

% recovery =



presentmassrecovered mass

× 100 =2234.0206.0

× 100 = 92.2%

8 • Reaction between the Ag + ions in solution and the Cu spiral was incomplete.• Some silver remained adhering to the copper spiral.• Small amounts of silver were lost in the transfer and filtering process.

9 S2O42– (aq) + 2H 2O(l) → 2SO 32– (aq) + 4H +(aq) + 2e –

10 2H 2O(l) → O2(g) + 4H +(aq) + 4e –


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Worksheet 4.3: SolutionsRedox titrations

No. Answer

1 a I2(aq) + 2e – → 2I – (aq)b If iodine is present a blue colour will be present. When the iodine has all reacted (i.e. the

endpoint has been reached), the blue colour will disappear.

2 a C6H4O2(OH) 4(aq) + I 2(aq) → C6H4O4(OH) 2(aq) + 2I – (aq) + 2H +(aq)b n (I2) = c × V = 0.10 × 0.0250 = 0.0025 mol

n (C6H4O2(OH) 4) = n (I2) = 0.0025 mol

c (C6H4O2(OH) 4) =0.01780.0025

=V

n = 0.140 M

3 a CH 3CH 2OH(aq) + H 2O(l) → CH 3COOH(aq) + 4H+(aq) + 4e

–

Cr 2O72– (aq) + 14H +(aq) + 6e → 2Cr 3+(aq) + 7H 2O(l)b 2Cr 2O72– (aq) + 16H +(aq) + 3CH 3CH 2OH(aq) → 4Cr 3+(aq) + 11H 2O(l) + 3CH 3COOH(aq)c n (Cr 2O72– ) = c × V = 0.200 × 0.01945 = 0.00389 mol

n (CH 3CH 2OH) =23

× n (Cr 2O72– ) = 1.5 × 0.00389 = 0.00584 mol

n (CH 3CH 2OH) in original = n (CH 3CH 2OH) in aliquot ×250.0

20.00 = 0.00584 ×

250.0

20.00 = 0.0730 mol

c (CH 3CH 2OH) =0.025

0.0730=

V

n = 2.92 M

4 a Fe 2+(aq) → Fe 3+(aq) + eMnO 4 – (aq) + 8H +(aq) + 5e – → Mn 2+(aq) + 4H 2O(l)

b MnO 4 – (aq) + 8H +(aq) + 5Fe 2+(aq) → Mn 2+ + 4H 2O(l) + 5Fe 3+(aq)c The pink colour of the MnO 4 – persists when the endpoint has been reached.d n (MnO 4

– ) = c × V = 0.110 × 0.0128 = 0.00141 moln (Fe 2+) = 5 × n (MnO 4 – ) = 5 × 0.00141 = 0.00704 mol

c (Fe 2+) =0.020

0.00704=

V

n = 0.352 M




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Worksheet 4.3: SolutionsRedox titrations

No. Answer

5 a Fe(s) + 2HCl(aq) → FeCl 2(aq) + H 2(g)b Fe(s) → Fe 2+(aq) + 2e –

2H +(aq) + 2e – → H2(g)c HCl(aq) + NaOH(aq) → NaCl(aq) + H 2O(l)

n (HCl) initially added = c × V = 1.08 × 0.100 = 0.108 moln (NaOH) = c × V = 1.00 × 0.0459 = 0.0459 moln (HCl) remaining after reaction with iron = n (NaOH) = 0.0459 moln (HCl) reacting with iron = 0.108 – 0.0459 = 0.0621 mol

n (Fe) =



2

1 × n (HCl) =

2

0.0621 = 0.0311 mol

m (Fe) = n × M = 0.0311 × 55.8 = 1.732 g

% Fe =)nail(

)Fe(m

m × 100 =

2.871.732

× 100 = 60.4%


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Worksheet 5.1: SolutionsAnalysis of iron in iron ore

No. Answer

1 An absorbance spectrum would be created by testing the absorbance of the complex ionsolution at various wavelengths. The wavelength of maximum absorbance would be chosen(provided this wavelength was not also absorbed by other components in the ore solution).

2 To relate the absorbance of the solution to its concentration.

3 Absorbance of 0.400 corresponds to a concentration of 0.0125 M.0.0125 mol in 1.0 L means 0.00125 mol in 100.0 mL.m (Fe) = n × M = 0.00125 × 55.8 = 0.0698 g

% Fe =)o()Fe(

rem

m 0.0698

0.100 × 100 = × 100 = 69.8%

4 MnO 4 – (aq) + 8H +(aq) + 5e – → Mn 2+(aq) + 4H 2O(l)Fe2+(aq) → Fe 3+(aq) + e – MnO 4 – (aq) + 8H +(aq) + 5Fe 2+(aq) → Mn 2+(aq) + 4H 2O(l) + 5Fe 3+(aq)

5 n (MnO 4 – ) = c × V = 0.0335 × 19.75 × 10 –3 moln (Fe) = n (Fe 2+) = 5 × n (MnO 4 – )m (Fe) = n × M = 5 × 0.0335 × 19.75 × 10 –3 × 55.8 = 0.1846 g

% Fe =



)o()Fe(

rem

m × 100 =

268.01846.0

× 100 = 68.9%

6 The determined value would be lower, because less permanganate solution would berequired for titration (since it will only react with the Fe 2+, not the Fe 3+).

7 a Fe 2+(aq) + 2OH – (aq) → Fe(OH) 2(s) b Fe 3+(aq) + 3OH – (aq) → Fe(OH) 3(s)

8 A pure sample of known composition could not be obtained because:• the precipitate may contain a mixture of iron(II) and iron(III) hydroxides• the iron hydroxide precipitates are not stable when heated.

9 2Fe → 2Fe(OH) 2 → Fe 2O3

n (Fe2O

3) =

M

1.08

159.6

m = = 0.00677 mol

n (Fe) = 2 × n (Fe 2O3)m (Fe) = n × M = 2 × 0.00677 × 55.8 = 0.755 g

% Fe =)o()Fe(

rem

m 0.755

1.01 × 100 = × 100 = 74.8%

10 Precipitates other than iron hydroxide may have formed, leading to a larger precipitate mass.

11 a Spectroscopic b Gravimetric c Gravimetric d Spectroscopic

12 Atomic absorption spectroscopy of a solution prepared by dissolving ore in acid can be used.


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Colorimetric determination ofmanganese in steel

No Answer

1 a Mn 2+(aq) + 4H 2O(l) → MnO 4 – (aq) + 8H +(aq) + 5e – b IO 4 – (aq) + 2H +(aq) + 2e – → IO 3 – (aq) + H 2O(l)c 2Mn 2+(aq) + 3H 2O(l)+ 5IO 4 – (aq) → 2MnO 4 – (aq) + 6H +(aq) + 5IO 3 – (aq)

2 Straight line graph of y = 0.076 x.

3 2.56 ppm = 2.56 mg L –1

4

n(MnO 4 –

) = 118.9 M 102.56 3−×

=m

moln(Mn) = n(MnO 4 – )

m(Mn) = n × M = 54.9118.9

102.56 3×

× −

= 1.18 × 10 –3 g

% Mn m/m = 100890.2

1018.1100

)sample()Mn( 3

××

=×

−

mm

= 0.0409%

5 Steel is mostly composed of iron. Step 1 oxidised the iron to iron(II) then to iron(III).

6 8.0 ppm MnO 4 – = 8.0 mg per 1.0 L (1000 mL)∴ 0.8 mg of MnO 4 – in 100 mL

∴

0.8 × 10 − 3

118.9 mol of MnO 4 – in 100 mL

∴

0.8 × 10 − 3

118.9 mol of KMnO 4 in 100 mL

∴

0.8 × 10 − 3

118.9× 158.0 g of KMnO 4 in 100 mL

∴ 1.1 × 10 –3 g of KMnO 4

7 Everything added to the sample (nitric acid, potassium persulfate, periodate and phosphoricacid) should be added in the same proportions, in case any of these substances absorb lightin the region being analysed.

8 400 nm is in the blue–violet region, which is close to the colour of MnO 4 – and therefore willnot be absorbed strongly. 520 nm is closer to green, and will be more strongly absorbed.

9 If all absorbances were shifted by the same quantity, it would not have made a difference.

10 Atomic absorption spectrometry could be used.




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Worksheet 5.3: SolutionsAnalysing mass spectra

No. Question Answer

1 Vaporisation, ionisation and deflection

2 a The vaporised sample is bombarded with high-energy electrons to produce positive ions by knocking electrons off the sample atoms or molecules.

b Positive ions moving in a magnetic field are deflected to varying degrees depending ontheir charge to mass ratio. The greater the charge, the greater the deflection; the greaterthe mass, the smaller the deflection. Thus ions are separated by mass.

3 Peak at mass 20 is , 21 is 101020 Ne + 21 Ne + and 22 is 10

22 Ne Peak at mass 36 is , 38 is20

36 Ar +20

38 Ar and 40 is 2040 Ar +

4 A r = Σ (RIM × abundance fraction)Let a be the percentage abundance of the lighter isotope.∴ 121.75 × 100 = (120.90 × a ) + (122.90 × (100 – a ))∴ a = 57.5%

5 a M = CH 3CH 2CH 2CH(OH)CH 3+ b M – 1 = CH 3CH 2CH 2C(OH)CH 3

+ c M – 15 = CH 3CH 2CH 2CH(OH) +d M – 43 = CH(OH)CH 3+

6 Compound I. The spectrum shows prominent peaks at mass-to-charge ratios of 57 and 29.These would correspond to the masses of fragments CH 3CH 2CO

+ and CH 3CH 2+. These

fragments are likely to form from compound I. Compound II would be expected to formfragments at mass-to-charge ratios of 71 (CH 3CH 2CH 2CO +), 43 (CH 3CH 2CH 2+) and 15(CH 3+).

7 Comparison of the mass spectrum of an unknown substance with computer files of massspectra of many compounds allows the unknown substance to be identified. Thefragmentation pattern can be used for ‘fingerprinting’ a substance (in much the same as thefingerprint region of the infrared spectrum is used).




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Worksheet 5.4: Solutions Spectroscopic analysis oforganic compounds I



No. Answer

1 a i 5ii 2

b i 3:2:2:2:1ii 9:1

c 2,2-dimethylpropanal (peaks in the ratio 9:1)d The fingerprint regions of the infrared spectra of the unknown could be compared with

that of 2,2-dimethylpropanal.

2 a i Infrared radiation is absorbed by molecules that move from ground state to excitedvibrational energy states. The energy is absorbed as bonds within the molecules bendand stretch.

ii The peak at 1710 cm –1 suggests a C=O bond, present in both structures. The –OHgroup in compound II would be expected to produce a peak in the 2500–3000 cm –1

range. The absence of this peak suggests that the unknown is compound I.b i Radio waves

ii Compound I would produce a spectrum with two peaks with the area ratio 1:1.Compound II would produce a spectrum with three peaks with the area ratio 3:2:1.The unknown is therefore likely to be compound I (methyl ethanoate).

3 Compound I would produce a spectrum with four peaks with the area ratio 3:2:2:3.Compound II would produce a spectrum with two peaks with the area ratio 6:4 or 3:2.The unknown is therefore compound II, 3-pentanone.

4 Notice that the peak at chemical shift 1.2 ppm, caused by the CH 3 nuclei, is split into three peaks, called a triplet. The peak at chemical shift 2.5, caused by the CH 2 nuclei, is split intofour peaks, called a quartet.The splitting follows an n + 1 rule. A peak will be split n + 1 times by n adjacent nuclei. Inthe case of the peak at 1.2 ppm, the CH 3 nuclei have two neighbouring protons (CH 2), so the

peak is split 3 (i.e. 2 + 1) times (a triplet). For the peak at 2.5 ppm, the CH 2 nuclei have threeneighbouring protons (CH 3), so the peak is split 4 (i.e. 3 + 1) times (a quartet).

5 a

i3ii 3

b i 3:2:3ii 3:2:3

c i One single, one triple and one quartetii One single, one triple and one quartet

d The chemical shift values could be used to distinguish between them. For example, theCH 3CO group (shift 2.1) and the RCOOCH 2R group (shift 4.1).


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Worksheet 6.1: Solutions Amino acid analysis using TLC



No. Answer

1 R f (W) = 0.5, R f (X) = 0.35, R f (Y) = 0.2 and R f (Z) = 0.1

2 W could be leucine or isoleucine, X could be proline or tyrosine or valine, Y is alanine or possibly threonine and Z is arginine.

3 R f (W) = 0.7, R f (X) = 0.8, R f (Y) = 0.5 and R f (W) = 0.55

4 W could be leucine or isoleucine, X is proline, Y is alanine and Z is arginine.

5 Some components may not separate in a given solvent. Using a second solvent makes itmore likely that separation can be achieved. The double development also allows for alonger development time, increasing the chance of resolution of component spots.

6 If the spot is below the solvent level, the amino acids will dissolve into the solvent, ratherthan travel up the filter paper.

7 The amino acids are colourless, and the spots developed using ninhydrin fade quickly inlight. It is therefore necessary to mark their positions quickly.

8 Perspiration on the skin contains amino acids. These could interfere with the analysis if theywere transferred to the plate.

9 Separation of the components occurs because the components adsorb (stick) to the stationary phase with different strengths. The stronger the adsorption, the more slowly the componentmoves as the mobile phase (the solvent) sweeps over the stationary phase (the plate).Components undergo continuous adsorption and desorption (unsticking) to differingdegrees. The components move at different rates depending on their strength of adsorptionand the tendency to desorb.


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GLC (Gas liquidchromatography)

No. Answer

1 A primary alkanol is one with the –OH functional group on carbon 1.

2 It must constantly be heated so that the mixture being analysed remains gaseous at all times.

3 It means that the sample is heated very quickly so that it enters the column as vapour.

4 This shows whether or not the machine has exactly the same parameters for each analysis. Ifthe 1-hexanol peak occurs at the same retention time and the same height each time, themachine is properly calibrated.

5 The 1-hexanol peak was constant in terms of height and retention time.

6 Oxygen is too reactive and with the heat could have oxidised the alkanol mixture.

7 They are not easily vaporised and are more susceptible to decomposing with heat.

8 Dispersion forces

9 Ethanol is a smaller molecule. Hence, the dispersion forces formed between ethanol and the

stationary phase are weaker than the bonds formed between 1-butanol and the stationary phase. Ethanol will spend less time adsorbed to the stationary phase and will pass throughthe column more quickly.

10 Retention time is qualitative; peak height is quantitative.

11 m (C4H9OH) = density × volume

n (C4H9OH) =



M

m =

74.11030.806 3−××

= 3 × 10 –5 mol

12 Ethanol, 1-butanol and 1-pentanol

13 A peak height of 45 was obtained for 1-butanol in 3 mL of mixture.A peak height of 21 was obtained for a 3 µL (3 × 10 –5 mol) sample of 1-butanol.

∴ n (C4H9OH) =2145

× 3 × 10 –5 = 6 × 10 –5 mol

14 CH 3CH 2CH 2CH 2CH 2OH

15 Due to its branched structure 2-propanol would have slightly weaker dispersion forces than1-propanol and so would have been marginally more weakly adsorbed. It would thereforehave had a retention time slightly less than 2.43 min.


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Worksheet 6.3: Solutions A summary of analyticaltechniques



No. Question Answer

1 Procedure Procedure is part of the technique of

GA VA SP CHPreparation of a standard solution orsolutions

Selection of wavelength to be used foranalysis

Production of a precipitate

Determination of R f or R t values

In some forms the sample to be tested mustfirst be vaporised

A calibration curve is plotted using peakareas of samples of known concentration

2 Analysis required Analysis could be conducted using

GA VA SP CH

To detect the presence of a C=O group in amolecule

To test a water sample for the presence ofchromium

To check the amount of ammonia in awindow cleaner

To determine the ratio of 16O to 18O in an icesample

To determine the level of iron in an iron oresample

To identify a performance-enhancing drug inurine samples

To separate the pigments in plant leafextracts


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Worksheet 6.3: Solutions A summary of analyticaltechniques



No. Question Answer


AAS Mass Infrared UV-vis NMRRadiation is absorbed byelectrons.

Spectral information is given inchemical shift values.

The sample for analysis is in avaporised state.

Radiation is absorbed by nuclei.

Transitions occur between atomicelectronic energy levels.

Spectral information is given inwave numbers.

Transitions occur between nuclearspin energy levels.

The sample for analysis is insolution.

It does not involve the absorptionof radiation by the sample.


TLC HPLC GLCThe stationary phase is a solid.

The mobile phase is a liquid.

Used for volatile organic compounds.

Results are presented as a chromatogram forwhich the retardation factor ( R f ) of each

component is found.

It relies on the principle of separation ofcomponents by selective adsorption to astationary phase.

R f values of known samples are compared withthose of unknown samples in order to identifythe unknown.

Results are presented as a chromatogramshowing a series of peaks at times taken for thecomponents to travel a set distance.


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Worksheet 7.1: SolutionsNaming organic compounds

No. Answer

1 a 2-fluoropropeneb propanec 3-methylbutanoic acidd 2-methyl-1-propanole trichlorofluoromethane

2

3 a C n H2n + 2 b

Cn

H2n

c C n H2n + 1OHd C n – 1H2n – 1COOH

4 a

b Butane has the higher boiling point, because the more linear molecule means thatmolecules pack well together, giving greater contact between adjacent molecules andhence stronger dispersion forces. Stronger forces mean a higher boiling point. The

branched molecules of 2-methylpropane do not pack together very well.

5 1-pentene, 2-pentene, 2-methyl-1-butene, 2-methyl-2-butene, 3-methyl-1-butene,dimethylpropene




29/38

Worksheet 7.1: SolutionsNaming organic compounds

No. Answer

6




30/38

Worksheet 7.2: Solutions Organic reaction pathways



No. Answer

1 Compound A: ethene CH 2CH 2 Compound B: ethanol CH 3CH 2OHCompound C: ethanoic acid CH 3COOHCompound D: propane CH 3CH 2CH 3 Compound E: 1-chloropropane CH 3CH 2CH 2ClCompound F: 1-propanol CH 3CH 2CH 2OHCompound G: 1-propyl ethanoate CH 3COOCH 2CH 2CH 3

2 Test with bromine solution. Compound A (unsaturated) will decolourise bromine, while D(saturated) will not.

3 C is acidic and will produce a solution with pH less than 7. G is neither acidic nor basic; itssolution will be neutral.

4 B and F belong to the primary alkanol homologous series. F is larger than B and so will havelarger dispersion forces between molecules, leading to a higher boiling point.

5 Isomers

6 Compound L is an acid (i.e. contains an acidic functional group).

7 Compound H: 1-butene CH 2CHCH 2CH 3 Compound I: 2-butene CH 3CHCHCH 3 Compound J: 1-chlorobutane CH 3CH 2CH 2CH 2ClCompound K: 2-chlorobutane CH 3CH(Cl)CH 2CH 3Compound L: 1-butanol CH 3CH 2CH 2CH 2OHCompound M: butanoic acid CH 3CH 2CH 2COOHCompound N: polybut-2-ene …-CHCH 3CHCH 3CHCH 3CHCH 3-…


31/38


Spectroscopic analysis oforganic compounds II

No. Answer

1 m (C) in sample = m (C) in the CO 2 produced =

44.012.0

× 1.32 = 0.36 g

m (H) in sample = m (H) in H 2O produced =18.02.00

0.36

12.0

× 0.72 = 0.080 g

m (O) in sample = m (sample) – m (C) – m (H) = 0.60 – 0.36 – 0.080 = 0.16 g

C : H : O = : 0.080

1.0: 0.16

16.0 = 0.03 : 0.08 : 0.01 = 3 : 8 : 1

The empirical formula (EF) of the compound is C 3H8O.

2 From the mass spectrum, RMM is 60. EFM = 60. The molecular formula of the compound isalso C 3H8O.

3 A peak at M – 15 is often due to the loss of a methyl group. The molecule probably containsa methyl group or groups.

4 I CH 3CH(OH)CH 3 II CH 3CH 2CH 2OHIII CH 3CH 2OCH 3

5 Compound I would have 2 peaks only (the CH 3 nuclei and the CH(OH) nucleus). The other

two compounds would each produce three peaks.6 a Compound I would have three peaks in the ratio 6:1:1.

Compound II would have four peaks in the ratio 3:2:2:1.Compound III would have three peaks in the ratio 3:2:3.

b The peak at chemical shift 1.2 ppm is due to the CH 3 nuclei. The neighbouring carbonhas one 1H nucleus, so the peak is split in two ( n + 1). The peak at chemical shift 3.9 ppmis due to the CH nucleus. The neighbouring carbons have six 1H nuclei, so the peak issplit in seven ( n + 1).

7 Compare the fingerprint regions of the infrared spectra of the unknown and 2-propanol, orcompare the fragmentation patterns of the mass spectra of the unknown and 2-propanol.

8 m (C) in sample = m (C) in the CO 2 produced =

44.012.0

× 2.78 = 0.758 g

m (H) in sample = m (H) in H 2O produced =18.02.00

× 1.14 = 0.127 g

m (O) in sample = m (sample) – m (C) – m (H) = 1.560 – 0.758 – 0.127 = 0.675 g

C:H:O =0.758

12.0: 0.127

1.0: 0.675

16.0 = 0.063:0.127:0.042 = 1.5:3.0:1.0 = 3:6:2

The empirical formula of the compound is C 3H6O2.




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Worksheet 7.3: Solutions Spectroscopic analysis oforganic compounds II



No. Answer

9 From the mass spectrum, RMM is 74. EFM = 74, so the molecular formula of the compoundis also C 3H6O2.

10 Isomer I Isomer II Isomer III

Structural formulaCH 3COOCH 3 HCOOCH 2CH 3 CH 3CH 2COOH

Number of peakson the 1H NMRspectrum

2 3 3

Pattern of peaksplitting on the 1H

NMR spectrum

Two single peaks One single peak, onequartet and onetriplet

One quartet, onetriplet and one single

peak

11 No. Both isomers II and III fit the pattern.

12 The chemical shift values would be needed to identify the types of 1H nuclei present.For example, if the single peak has a shift of 11–12 (due to the 1H of the OH), isomer III

would be correct. If the single peak has a shift of 8–9 (due to the1

H of the –COOH), thenisomer II would be correct.

13 The peak in the 1700–1750 cm –1 region is typical of the C=O group, present in both isomersII and III. The peak in 2500–3000 cm –1 region is probably due to the OH of the carboxylicacid group, suggesting that isomer III is the correct structure.

14 Isomer III is acidic, and so would be expected to produce a solution with pH less than 7, andreact with a solution of Na 2CO 3 to produce carbon dioxide gas. Isomers I and II are estersand are not acidic. They may have a distinctive fruity smell.


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Worksheet 8.1: SolutionsProtein structure and function

No. Answer

1 a Refer to figure 8.2.1 on page 171 of the coursebook. Note that the COOH and NH 2 groups are attached to the same carbon atom.

b Refer to the diagrams of the twenty amino acids shown on the examination data booklet.

2

3

4 a 1b 2

c 4d 999 (in general n – 1)

5 a i –NHCH 2CONHCH 2CONHCH 2CO–ii –NH(CH 2)3CONH(CH 2)3CONH(CH 2)3CO–iii –CH 2CH 2CH 2 –

b i Condensationii Condensationiii Addition

c A protein has a complex sequence of many different amino acids. It does not use a singlemonomer like this.

6

The oppositely charged dipole ends along the chain align, as the chain forms a helix.




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Worksheet 8.1: Solutions Protein structure and function



No. Answer

7 a An enzyme is a biological catalyst. It is a protein with a specific 3D structure. It is thisstructure/shape that allows it to catalyse a particular reaction.

b For example: lactase is required for the digestion of milk; sucrase for the breakdown ofsucrose and catalase to break down peroxide molecules.

c See the lock and key diagram of enzyme functioning (figure 8.2.8) on page 174 of thecoursebook.

8 ‘Inhibitors’ or blockers can act by either filling the active site on the enzyme, thus making itno longer a catalyst, or by changing the shape of the active site so that it no longer performsits usual function.

9 This is an example of denaturation of the protein (casein) in the milk. Denaturation is a breakdown of the complex tertiary structure of the protein.

10 Strong base, ultraviolet light and other organic molecules can all cause denaturation.

11 Several examples are outlined on page 193 of the coursebook.


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Worksheet 8.2: Solutions DNA structure and function



No. Answer

1 a Deoxyribonucleic acidb Ribonucleic acidc The nitrogenous bases, adenine, thymine, cytosine, guanine and uracild Polymerase chain reaction—the laboratory process in which a DNA fragment is

replicated many times (amplified)

2 a A sequence of three nitrogenous bases on the DNA strand. This sequence usually codesfor a particular amino acid.

b The specific bonding that occurs between certain base pairs on opposing strands of DNAin the double-stranded molecule. These pairs are A and T, and C and G.

c The copying of a double-stranded DNA molecule to produce another identical, double-stranded DNA molecule.

d The combination of a deoxyribose sugar (in DNA) or ribose sugar (in RNA) moleculewith a nitrogenous base and a phosphate molecule. A nucleotide is the basic ‘building’

block of nucleic acid polymers.

3 Refer to figure 8.3.4 on page 177 of the coursebook.Base + sugar → nucleoside

4 Refer to figure 8.3.4 on page 177 of the coursebook.Phosphate + nucleoside → nucleotide

5 Refer to figure 8.3.4 on page 177 of the coursebook.… + nucleotide + nucleotide → polynucleotide

6 The section contains 400 bases. 40 adenine molecules must base pair with 40 thyminemolecules, leaving 320 molecules. Since each cytosine must pair with a guanine molecule,there must be 160 of each molecule present.

7 a Section A: (from top to bottom) GGGCTATTAG; Section B: TTACGAAGGTb The C–G link involves 3 hydrogen bonds and so is stronger than the A–T link which has

2 hydrogen bonds. Section A has more C–G links than section B, so section B would beeasier to separate (i.e. require less energy).

8 a Translationb Transcriptionc Replication

9 a 18 (9 per strand)b 4 (note that 5 sugar-phosphate bonds already exist within the nucleotides)c 66 (3 per amino acid. In practice, more are needed for the ‘stop’ and ‘start’ of the

translation process.)


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Worksheet 8.3: Solutions Biochemical fuels



No. Some search results and comments

1 Biofuel Generation/source Chemical structure Energy productionEthanol It is produced by

fermentation, the actionof yeast enzymes onglucose under anaerobicconditions.

CH 3CH 2OH Burnt in air, or mixedwith petrol forcombustion:CH 3CH 2OH(l) + 3O 2(g)

→ 2CO 2(g) + 3H 2O(g)Reacted with oxygen ina fuel cell

Biogas It is formed whenorganic matter decays inan oxygen-free(anaerobic) environment.The biogas is generated

by the action of micro-organisms on organicmolecules.

CH 4 CH 4(g) + 2O 2(g)→ CO 2(g) + 2H 2O(g)

Biodiesel It is formed whentriglycerides are reactedwith concentratedsodium hydroxide andmethanol. The sodiumhydroxide breaks theester bond in thetriglyceride, allowing themethanol to form an esterwith the long fatty acidmolecule.

Refer to figure 8.4.5on page 187 of thecoursebook.

biodiesel + O 2(g)→ CO 2(g) + H 2O(g)

2 Brazil has used ethanol blends for many years. Sugarcane is often their source of ethanol.The energy from the ethanol needs to outweigh the cost of producing the crops. Corn is a

common starting point in the US. There are several steps to purifying the ethanol. Purdueyeast offers the possibility of using cellulose (cheap forest waste) to get glucose to getethanol.

3 The Sunshine tip and Windermere piggery are examples of biogas production. The main product is methane, which can be burnt to produce energy.

4 There are several Australian ‘recipes’ for biodiesel. Biodiesel can be used in normal dieselengines. Unrefined vegetable oil can only be used in blends with normal diesel.Rapeseed (also known as canola) is a common oil chosen. Waste oil from fish and chipshops can be used also.European countries actually sell blends of biodiesel and diesel.


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Preparation and analysis ofaspirin

No. Answer

1 Analgesic: pain relieverAntipyretic: reduces feverAnti-inflammatory: reduces production of prostaglandins that lead to inflammation

2 Used by the ancient Greeks and by the American Indians, who chewed the leaves.

3 a Carboxylic acidb Hydroxyc Ester

4 See figure 8.5.2 on page 189 of the coursebook.

5

6 a It is a catalyst.b It is used in many ester-forming reactions.

7 a The binder holds the tablet together so that it is a single, solid tablet.b The binder must be harmless, stable and edible—starch is suitable for this.

8 a m (acetylsalicylic acid) = 325 × 10 6 mg = 325 × 10 3 g

n (acetylsalicylic acid) =180

10325

3×

=

M

m

n (salicylic acid) = n (acetylsalicylic acid)

= n × M = 18010325 3×

× 138 = 249 × 103 g = 249 kg

b The process is far from 100% efficient. Reaction is incomplete and isolation of the product is difficult because the solubility of the aspirin makes it hard to crystallise outfrom the reaction solution.

9 Stomach bleeding may occur. In addition, the blood is thinned too much, making it unlikelyto congeal and so interfering with the clotting process.

10 Methyl salicylate (oil of wintergreen) is produced. See figure 8.5.2 on page 189 of thecoursebook.




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Worksheet 8.5: Solutions A biomolecules glossary

No. Answer

1 a Similar to figure 8.1.3 on page 168 of the coursebook.i Hydroxyii Disaccharide

b Similar to figure 8.2.3 on page 171 of the coursebook.i Amino and carboxyii Dipeptide

c See figure 8.3.4 on page 177 of the coursebook.i Hydroxy and aminoii Nucleotide

d Similar to figure 8.4.4 on page 186 of the coursebook.i Hydroxy and carboxy

ii Triglyceride

2 a Glucose moleculesb Amino acidsc Nucleotides

3 It will have a different number of beads, different types of beads and a different sequence of beads.

4 A different number of beads and a different sequence of A, T, G and C beads.

5 See figure 8.2.4a on page 172 of the coursebook.

6 See figure 8.3.5 on page 178 of the coursebook.

7 a Covalentb Hydrogen bonding

8 a Hydrogen bondb Ionicc Covalent (disulfide)d Dispersion forces

9 Heat, change in pH, addition of detergent, UV light, organic compounds

10 Refer to the Chem Snippet on page 173 of the coursebook for examples.

Chapter 1-8 Worksheet Solutions

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