National German competition 2004

Volume 10
Preface

To become a member of the German IChO-team you have to be successful in 4 rounds.

The problems to be solved in the 1st round are sent to all highschools. To solve the problems
the students may use all resources available, e.g. textbooks etc.
All the students who solve about 70% will receive the problems of the 2nd round, which are to
be solved in the same way as mentioned above. These problems are the most difficult ones in
the whole competition.
The top 60 of the participants of the 2nd round are invited to the 3rd round, a one-week
chemistry camp. Besides lectures, excursions to chemical plants or universities and cultural
events there are two written theoretical tests of 5 hours each.
The top 15 of the 3rd round are the members of the 4th round, a one-week practical training.
There are two written five-hour tests - one theoretical and one practical - under the same
conditions as at the IChO. Here the team is elected.

Contact addresses:

IPN z.H. Dr.Bünder tel: +431-880-5013 (3168)

Olshausenstraße 62 fax: +431-880-5468
24098 Kiel email: buender@ipn.uni-kiel.de

StD.Wolfgang Hampe tel: +431-79433

Habichtweg 11
24147 Klausdorf email: Hampe@t-online.de

Association to promote the IChO

(Association of former participants and friends of the IChO)
Christoph Jacob tel. +6101-33100
Erlenweg 4 email: chjacob@web.de
61138 Niederdorfelden

Internet address : www.fcho.de

2
Contents

Part 1: The problems in the four rounds

First round (problems solved at home) ..................................... 6

Second round (problems solved at home) ..................................... 9
Third round, test 1 (time 5 hours)................................................ 14
Third round, test 2 (time 5 hours)................................................ 25
Fourth round, theoretical test (time 5 hours) ................................. 34
Fourth round, practical test (time 5 hours) ................................. 45

Part 2: The solutions for the problems of the four rounds

First round ........................................................................................... 50

Second round........................................................................................... 54
Third round, test 1 ........................................................................... 62
Third round, test 2 .......................................................................... 72
Fourth round, theoretical test .................................................................. 80

Part 3: Appendix

Tables about the history of the IchO and ranking .................................. 89

3
4
 Part 1

The problem set of the four rounds

 Problems Round 1

First Round (homework)

Problem 1-1: Gases in cakes or elsewhere
Salt of hartshorn can be found in recipes for ginger bread as baking-powder - a remedy to
rise dough. It is a mixture of ammonium hydrogencarbonate and ammonium carbamate.
In books you often find other compositions.
a) Give other specifications of the composition of salt of hartshorn and the place you found it.

Upon heating at 180°C both compounds decompose. Consider salt of hartshorn as a mixture
of equal amounts of ammonium hydrogencarbonate and ammonium carbamate.

b) Write a balanced reaction equation of the decomposition. Calculate the maximum

increase of volume of dough when 1 g of salt of hartshorn decomposes at 180° C
(p=1.013 bar).

The temperature is a bit higher in the next question. At very high temperatures molecular
⎯→
hydrogen decomposes into atoms. The equilibrium constant for the reaction H2 ←⎯ 2 H at
3000 K has the value Kp = 2.51·10-2 bar.
Imagine a system containing hydrogen only with a total pressure of p = 980 hPa.
c) Calculate the partial pressure of the atomic hydrogen in equilibrium.
d) Calculate the density of the gas (kg/m3) in equilbrium under the conditions given.

Problem 1-2: Sea water

Students have to investigate sea water. They have to determine the concentrations of
calcium-, magnesium-, sodium-, chloride- and sulfate ions.
To simplify the question assume that the amount of carbonate ions is neglectible and that
there are no other ions in sea water.
The following investigations are made:
1. 10 mL of see water pass a cation exchanger. Afterwards the solution is titrated with a
solution of sodium hydroxide (c = 0.500 mol/L).
Mean value of consumption: 11.76 mL.
2. 10 mL of see water are diluted to give 100 mL. 10 mL of this diluted solution are titrated
with a solution of silver nitrate (c = 0.086 mol/L).
Mean value of consumption: 6.21 mL.
3. An indicator buffer tablet is added to 10 mL of sea water. The solution is titrated with a
solution of EDTA (c = 0.05 mol/L).
Mean value of consumption: 12.60 mL.
4. An excess of ammonium oxalate is added to 100 mL of sea water. The precipitate is
filtered off and dissolved in hot diluted sulfuric acid. The resulting solution is titrated with
potassium permanganate (c = 0.02 mol/L).

6
 Problems Round 1

Mean value of consumption: 24.00 mL.

a) Calculate the concentration of the ions mentioned above (g/L). Show your way of
reasoning.
Give balanced equations for the reactions performed. Use „RH“ as a symbol for the cation
exchanger loaded with acid, „H4Y“ for EDTA or Y4- for its anion.
b) In performing the titration with silver nitrate (investigation 2.) some droplets of potassium
chromate are used as an indicator. Explain how it works.
c) How can the students recover the silver from the silver chloride precipitate (investigation
2.)? What is the value of the silver gained from treatment of 5 L solution of a silver salt
(c = 0.2 mol/L)? Price of silver: see newspapers.
d) Draw a line-bond formula of the anion of EDTA (Y4-) and a structural formula of the
calcium-EDTA complex.
Explain the colour change at the equivalence point in investigation 3.
e) In the inland it’s not always possible to get sea water. The following salts are available for
a teacher: NaCl, Na2SO4·10 H2O, CaCl2·6 H2O, MgCl2· 6 H2O, MgSO4·7 H2O.
Calculate how to compose a portion of „sea water“ of the composition you found above
using all or some of the salts available.

Problem 1-3: A present

Eight substances were given to a school as a gift: Ag, Fe, Cu, Mg, AgNO3, Fe(SO4)2,
Cu(NO3)2 and Mg(NO3)2. Unfortunately the labels on the bottles could not be read.

A B C D
A student had to investigate the content of the bottles 1
by preparing solutions of the salts (marked 1, 2, 3 ,4) 2
and adding the metals (marked A, B, C, D) to each 3
4
solution. In case of reaction he wrote „+“ otherwise
„-“ in his table.

After finishing his task he was happy until he turned

a bottle with concentrated sulfuric acid over.
Only two pieces of his table shown beside remained.
Other students laughed at him but the next day he came,
showed the right result and said there were even two
pieces of information he did not need.

a) Assign the name of the compounds to A, B, C, D, 1, 2, 3 ,4 and show your way of arguing.

7
 Problems Round 1

b) Which are the two pieces of information the student did not need?

Problem 1-4 Dangerous food

Stones of fruit and bitter almonds contain a compound A that decomposes upon acidic
hydrolysis into hydrocyanic acid among other compounds. When this fact became evident in
the 19th century it caused great surprise - and was frightening too.
Compound A has the molecular formula C20H27O11N.
An precise analysis of the hydrolysis products showed that besides hydrocyanic acid glucose
and benzaldehyde are formed.
a) Write a balanced reaction equation of the acidic hydrolysis of compound A.

Further investigation showed A to be a glycoside. Glycosides are compounds widespread in

plants consisting of a sugar component and a non sugar component. These two component
are connected by a glycosidic bond - similar to an ether bond.
b) Draw the structural formula of A. Don’t specify the sugar component, write it as molecular
formula [CxHyOz]. Thus stereochemistry can be ignored.

8
 Problems round 2

Second Round (homework)

Problem 2-1
Willi Wusel is used to keep his nails and screws in glasses. One day, he discovers a for-
gotten glass with a rusty nail where rain has got in in his garden. When he takes out the nail,
a reddish brown layer of Fe(OH)3 can be found at the glass. In order to clean the glass, he
continuously adds hydrochloric acid (that he has found in his shed) to the water in the glass
until the layer has dissolved completely. Three quarters of the glass (0.2 L) are filled with a
yellow liquid then.
He determines the pH of the solution with pH paper. The pH is 2.0. The concentration of the
hydrochloric acid that he has found in the shed is 0.5 mol/L. He puts a paper clip of copper
into the yellow solution. After some time, the colour of the solution has changed.
a) What is the pH of hydrochloric acid found in the shed?
b) Show that the following equation is approximately valid for the concentration of chloride
ions in the solution after the dissolution of the layer:
KL
c(Cl-) = 3 · · c(H+)3 + c(H+)
K 3W
c) Calculate the concentrations of chloride and iron(III) in the yellow solution. What was the
amount of rust (in mmol and mg) that the reddish brown layer consisted of?
d) How much hydrochloric acid (in mL) did Willy need to dissolve the layer?
e) How does the colour of the solution change? Give reasons for your answert. Determine
the concentrations of the iron- and copper ions.
Useful formalas and values:
solubility product of Fe(OH)3: KL = 2.0⋅10-39 mol4⋅L-4
ionic product of water: KW = 1.0⋅10-14 mol2⋅L-2
E0 (Fe3+/Fe2+) = 0.771 V E0 (Cu2+/Cu) = 0.345 V
The temperature is summer-like (27 °C).

Problem 2-2
In a nitrogen atmosphere, 85.0 mg of an unknown reddish brown metal iodide are put into a
cylindrical metal tube having a diameter of 12.0 mm and a length of 18.3 cm. The following
values for temperature and pressure are measured: ϑ = 25.0 °C, p = 1.013 bar. Afterwards,
the tube is closed and heated to a constant temperature of 450 °C. There is a continuous
pressure change in the tube until there is a constant pressure value of 3.346 bar. After
opening the metal tube that is still hot, a violet vapour escapes; the initial substance has
disappeared.
a) Determine the original metal iodide assuming that all gaseous substances show an ideal
behaviour.

9
 Problems round 2

One of the gas-phase reactions that have been investigated the best is the reaction of
hydrogen with iodine: k1
H2 + I2 k 2 HI
-1

The following rate constants result from kinetic measurements at different temperatures:
temperature [K] k1 [L⋅mol-1⋅s-1] k -1 [L⋅mol-1⋅s-1]
400 8.37⋅10-12 3.25⋅10-14
500 2.48⋅10-7 1.95⋅10-9
600 2.38⋅10-4 2.97⋅10-6
700 3.22⋅10-2 5.61⋅10-4
800 1.27 2.85⋅10-2
b) Is the reaction exothermal or endothermal? Give reasons. What principle has been
applied?
c) Calculate the reaction enthalpy and the reaction entropy of the formation of HI assuming
that these values are independent of temperature in the investigated range .
d) Caulate the degree of dissociation of HI at 6 K. How does it change with pressure
assuming the validity of the ideal gas law?

Elemental iodine is only poorly soluble in water. In the presence of iodide ions, however, the
solubility in water increases considerably. This can be attributed to the formation of triiodide
anions I3- :
⎯→
I2 + I- ←⎯ I3-
A certain amount of I2 is shaken together with CS2 and an aqueous KI-solution of the con-
centration c0(KI) = 31.25⋅10-3 mol/L until there is an establishment of equilibrium. Then, the
concentration of I2 is determined by titration with Na2S2O3. In the CS2-phase, it is 32.33 g/L
and in the aqueous solution, it is 1.145 g/L. The coefficient od distribution for I2 between CS2
and water is 585.
e) Calculate the equilibrium constant for the formation of triiodide anions.

Problem 2-3
a) Draw the three-dimensional structure of the complex ions that can be found in the
compounds A – H and find out which of them are chiral. To increase clarity and facilitate
drawing, you can use abbreviations for the chelate ligands, N–––N for 2,2’-bipyridine , O–
––O for oxalate, N–––O for glycinate and O O
N N for EDTA.
O O

[please write down, however, the complete structural formula of complex H ]

The coordination geometry of the metal must be clearly readable .
A: trans-[CuII(gly)2] B: [Zn(gly)2] C: [Ca(EDTA)]2-
D: [Cu(bpy)2]ClO4 E: K3[Fe(C2O4)3] F: K2[Cu(C2O4)2]
G: [CoIII{cis-CoIII(NH3)4(µ-OH)2}3]6+ H: [Pt(meso-1,2-(NH2)2C6H10)BrCl]

10
 Problems round 2

Explanations:
gly = glycinate bpy: 2,2’-bipyridine 1,2-(NH2)2C6H10 1,2-diaminocyclohexane;
concerning G: the µ-OH-groups bind the outer Co-ions to the central ones.
b) How many different types of achiral monodentate ligands (L1, L2,…) are needed at least to
obtain a chiral complex with an octahedrally coordinated metal ion (M)?
c) What is the composition of the complex in problem b)? Draw all possible structures of this
complex. Please write down which complexes are chiral and which are not chiral.

The following equation describes the synthesis of a ligand and a derived complex:
1. KNH2 / NH3 (l) NaNO2 , HBr, Cu0-Pulver
2. [O] CuBr ∆
P Q R S
C5H5N

d) Complete the synthesis scheme. P is a toxic, weak base that has an unpleasant smell and
can be mixed with water.
e) A surplus of S reacts with iron(II)-sulfate; complex T forms. What is the composition of this
complex? Draw the structure(s) of all isomers. Draw as well the orbital scheme for the d-
orbitals.

The investigation of the gradual formation of complex T shows that the equilibrium constant
for the last step of complex formation is higher than the equilibrium constant for the
penultimate step.
f) Write down a possible explanation that considerates the electronic situation of the
complexes.

Problem 2-4
Silicon dioxide occurs in nature in its crystalline (e.g. mountain crystal) as well as in its
amorphous (e.g. opal) form. SiO2-resources can be used for the production of water glasses.
SiO2 (quartz sand) is molten with much Na2CO3. The glass that has formed is dissolved in
water at 160°C under pressure. If such a water glass is analyzed by 29Si-NMR-spectroscopy,
mostly 5 groups of signals can be found in the spectrum.
a) Why does gaseous SiO2 that is analogous to CO2 not exist under normal conditions?

b) Give examples for the appplication of water glasses.

c) Which 5 principal Si-environments can be assigned to the 29Si-NMR-signals of water
glasses?

Another possibility of using SiO2 is its reduction to Si that is used e.g. for the production of
solar cells in its purified form. It can as well be processed to important basic chemicals.
Elemental Si reacts with chloromethane at about 300°C in the presence of catalysts. The
main products that form are methylchlorosilanes (CH3)nSiCl4-n (n = 1, 2, 3).

11
 Problems round 2

d) Write down coordinated reaction equations for the reactions of these 3 chlorosilanes with
an excess of water.
e) What are the hydrolysis products of (CH3)2SiCl2 and what is their technical sifnificance?

Silicon that is an element of the fourth main group is mostly tetravalent. Its oxidation number
is +4 in compounds. There are only a few examples of stable compounds with bivalent Si
having the oxidation number +2. Compound E is an example of such a compound. E is
produced by the condensation of 1,2-ethanedial with 2 equivalents of A (A = primary amine,
C4H11N, 2 signals in the 13C-NMR-spectrum) to B. B reacts with 2 equivalents of Li to form C.
Then, C reacts with SiCl4 to form D and 2 LiCl. The reduction of D with potassium results in
E.

f) Write down the equations for all the mentioned reactions. Write down the structural
formulas of the compounds A, B, C, D,and E. Conclude from the electronic structure of E
the reason for the extraordinary stability of the compound.
.
It is true that the oxidation number of Si is +4 in a different group of interesting Si-com-
pounds, but the Si-atom has additional coordinated donor atoms and reaches the coor-
dination number 6 (compounds in which the coordination number that is to be expected
according to the octet rule is exceeded are called hypercoordinated compounds). The
compounds Y and Z are such compounds and can be obtained by the following reaction:

N N
SiCl4 ZnF2
Y Z
-2X - ZnCl2
OH HO

Z crystallizes from acetonitrile as monoclinic crystals. 1 mol of acetonitrile crystallizes as well

in 1 mol of Z. The unit cell of Z has the following dimensions: a = 12.3826 Å, b = 10.8405 Å,
c = 13.8507 Å, β = 98.800°, α = γ = 90°.

g) Write down the structural formulas of Y and Z . What is X?

h) Why can Si exceed the octet rule in contrast to carbon?
i) Write down the value of n (n = number of molecules of Z and acetonitrile that are in the
unit cell, n = even whole number, assuming that every non hydrogen atom requires a
space of 15 to 19 Å3 . Calculate the density of Z.

12
 Problems round 2

Problem 2-5
The natural substance X that is extremely important for human beings can synthesized the
following way:
COOEt

HC NH COCH3
CH3COCOOH ZnCl2 , H+, ∆ CH2O, (CH3)2NH 3 H2O
A B C COOEt
D E X
- NH3, - CO2 - H2O - (CH3)2NH - CO2, -2 EtOH
- CH3COOH

Substance C that is a byproduct of this reaction (total formula: C8H7N) and E are natural
degradation products of X and can be found in human faeces. If C is oxidized in the body it
will form the product Cox that can be found in urine and is an important reactive intermediate
in an industrial process: The important dye F forms in an alkaline medium from two
molecules of Cox.
The results of the elemental analysis of the initial compound A are the following values:
66.62% of carbon, 7.47% of hydrogen and 25.91% of nitrogen; The molecular peak is at m/z
=108 in the mass spectrum..

a) Write down the structural formulas of the compounds A – E, X, Cox and F.

b) What is the reaction mechanism of the reaction from B to C?

The red dye Z has a structure that is analogous to that of F. Its total formula C16H8O2S2 can
be obtained synthetically from anthranilic acid:

COOH
NaNO2 , H+ Na2S2 Zn, CH3COOH
2 2G H 2I
- N2
NH2

1. KOH, NaNH2
Cl CH2COOH 2. H2O ∆ O2
I K L M Z
- CO2 - H2O

c) Write down the structural formulas of the substances G – M and Z.

13
 Problems round 3 test 1

Third Round, Test 1

A formulary and a periodic table were made available for the two tests of the third round.

Problem 3-1
A) Concentrated sulfuric acid was added to the following salts.
In which case can a redox reaction not be expected?
a) NaNO3 b) Na2S2O3 c) NaI d) Na3PO4 e) Na2C2O4 f) NaClO3

B) Which of the following elements has the lowest second ionization energy?
a) Be b) K c) Cs d) S e) Ba

C) Which of the following ions has the smallest ionic radius?

a): Ag+ b) F- c) H- d) Al3+ e) Na+

D) Which of the following particles has the most negative redox potential?
a): F2 b): Ag c): Na d): Li+ e): C

E) Which of the following empirical formulas represents exactly 5 isomers?:

a): C4H9Cl b): C7H16 c): C6H6 d): C3H7Br

F) Which of the following empirical formulas represents the highest number of isomers?
a): C4H9Cl b): C7H16 c): C6H6 d): C3H7Br

G) Which of the following images does not represent (R,R)-2-chlorine-3-aminobutanoic acid?

a) NH2 b) COOH
Cl COOH H3C H
COOH
e)
H CH3 H Cl
NH2
H Cl
H

NH2 H NH2
c) H3C Cl d) H
H H
CH3
H Cl
H2N COOH HOOC CH3

H) Which of the following molecules has the smallest dipole moment?

a) CO2 b) SiO2 c) SO3 d) SF6 e) UF6 f) XeF6

14
 Problems round 3 test 1

I) In which of the following alcohols is the hydroxyl group oxidized to the aldehyde group the
fastest (e.g. with dichromate)?

OH OH

OH OH OH
Cl Cl
a) b) c) d) e)

J) Which of the following compounds is the strongest base?

H
N

NH2
N
N H
a) b) c) d)

Problem 3-2 Unknown solutions

In 7 test tubes with numbers from 1 to 7, there are dilute aqueous solutions of the following
substances:
CuSO4 AgNO3 NaCl Na2CO3 NaOH HI H2SO4
In the test tubes X and Y, there are dilute aqueous solutions of an unknown substance each.
The following reactions could be observed each time two of the nine solutions were mixed:

1 2 3 4 5 6 7 X Y
1 * / / / p(b) p(bl) p(bl) p(b) /
2 * p(w) / / / / / /
3 * / p(y) p(b) p(wy) p(y) p(w)
4 * / / gas / /
5 * / gas / p(b)
6 * / pun. p(bl)
smell
7 * / p(bl)
X * p(b)
Y *
Meanings of the abbreviations:
p(w) = white precipitate p(wy) = white-yellow precipitate
p(y) = yellow precipitate p(b) = brown precipitate
p(bl) = light blueprecipitate gas = gas development
pun. sm = pungent smell / = no reaction observed
a) Write down which substance can be found in which test tube.
b) Which kinds of substances are X and Y?
c) Write down the reaction equations and ionic notations that are the beses of all the
observations that are expected to happen.

15
 Problems round 3 test 1

Problem 3-3 Iron-EDTA complexes

In an aqueous solution, iron(III)-ions react with EDTA (ethylenediaminetetraacetic acid = Y)
almost quantitatively. A [FeY]- forms primarily. This complex is sterically tense and continues
reacting with OH- to form the complexes B [FeY(OH)]2- (lg K1 = 6,45) and C [FeY(OH)2]3- (lg
K2 = 4.53). It comes to a steric relaxation. The coordination number of iron is 6 in all these
complexes.
[FeY]- + OH- [FeY(OH)]2- lg K1 = 6.45
2- - 3-
[FeY(OH)] + OH [FeY(OH)2] lg K2 = 4.53

OOC COO
N EDTA (Y)
N
OOC COO

a) Calculate the content of [FeY(OH)] 2- and [FeY(OH)2] 3- ions in a solution still containing
0.001 mol/dm3 [FeY] - . The pH-value of the solution is 8.
Why is the complex-formation constant K2 considerably lower than K1?

Instead of [FeY(OH)2]3-, Ph-OH complexes of the type [FeY(OH)(PhO)]3- form in the presence
of phenols. Almost no [FeY(PhO)2]3- forms, however, in an excess of phenol. Ligands like 8-
hydroxychinoline or 1,2-dihydroxybenzene-derivates, however, can as well displace the
second hydroxo ligand from the initial compound C.
b) Sketch a complex [FeY(OH)(PhO)] 3-, having the reaction behaviour described.
c) Why is [FeY(OH)2] 3- much more paramagnetic than [Fe(CN)6] 3-?

The exchange of a hydroxo ligand of C by a ligand like phenolate, thioalcoholate or

thiocyanate results in the formation of intensively coloured (red or blue) complexes. In
contrast to that, C is coloured only very weakly (orange-yellow).
d) What is the reason for the intensive colours of the phenolato-, thioalcoholato- und
thiocyanato-complexes?

This intensive colour of the complexes of the type [FeY(OH)(PhO)]3- can be used for
UV/VIS-spectroscopic determination of phenol contents.
In an instruction for the preparation of such a photometry solution you can read the following:
„ 25 mL of 0.4 M EDTA-solution and 2 mL of 1.0 M Fe(ClO4)3-solution are mixed and analyte-
solution is added to this mixture. The pH-value has to become 8 with NH3 and HClO4. Then,
the mixture is filled up to 50 ml and the extinction is measured in a 1 cm cuvette.
The results of the measurements are analyzed the following way: The amount of phenol in
the added analyte is calculated from the extinction (E) according to the equation
E = b·c + a (c in mg)
The values of the factors a and b are the following: for thymol: a = 0.013, b = 0.0048 mg-1
for vanillin: a = 0.008, b = 0.0245 mg-1

16
 Problems round 3 test 1

OH
OHC OH

thymol vanillin

According to the method mentioned above, thymol is determined in an extract of a natural

substance. 1 g of a raw substance is dissolved in methanol and filled up to 5 mL. 1 mL of this
solution is added as an analyte to the photometry solution described above. The
measurement of the extinction (at λmax = 540 nm), however, is carried out in a 5 cm cuvette.
The extinction is: E = 1.505.
e) Determine the content of thymol (mass fraction and mole fraction) in the extract of a
natural substance that has been analyzed.
f) Write down the apparently molar coefficient of extinction ε´ for the complex
[FeY(OH)(Thymolat)] 3-. Why is this an "apparent" coefficient of extinction and not the real
molar coefficient of extinction of this complex?
Why is the value b for vanillin considerably higher than that for thymol, when it can be
assumed that the real molar coefficients of extinction of the phenolato-complexes scarcely
differ?

Problem 3-4 Niobium-Sulfur-Clusters

When niobium(V)ethoxide (Nb(OCH2CH3)5 = Nb(OEt)5 ) reacts with hexamethyldisilathiane (
(CH3)3Si-S-Si(CH3)3 = (Me3Si)2S ) and tetraethylammoniumchloride in acetonitrile (CH3CN)
as a solvent, an interesting niobium-sulfur compound (1) forms in the form of black crystals
according to the following reaction equation:
CH3CN
Nb(OEt)5 (Me3Si)2S NEt4Cl (NEt4)4[Nb6S17] 3 CH3CN Me3SiOEt Me3SiCl
(1)
hexamethyldisilathiane: ρ = 0.846 g/ml niobium(V)ethoxide: ρ = 1.268 g/ml

a) Balance the reaction equation.

b) What is the driving force for this reaction ?
c) Calculate an experimental assay, if 5.000 g of the compound (1) should be obtained at a
yield of 75%. Take into consideration that there is 1.4 times the amount of the
components (Me3Si)2S and NEt4Cl with regard to Nb(OEt)5. What are the necessary
amounts of substances in gram (for NEt4Cl) or rather in millilitre (for Nb(OEt)5 and
(Me3Si)2S).
d) What are the oxidation numbers of niobium and sulfur in the initial compounds and in
compound (1) ?

17
 Problems round 3 test 1

Compounds like e.g. (1) can be of use as a brilliant initial compound for the production of
clusters. A cluster is a compound with metal-metal bonds in addition to metal-nonmetal
bonds. In a simple representation of clusters, mostly only the metal atoms are considered.
M3M’3 is such a cluster compound in which M and M’ are two different metals. In this M3M’3--
cluster the metal atoms should be octahedrally arranged.
e) Draw all possible isomers of the M3M’3 -cluster

Problem 3-5 Products of Electrolysis

A potassium-chloride solution is being electrolyzed for two hours at 80°C in a primitive
electrolytic cell. The voltage is 6 V and the intensity of the electric current is 2 A. The cathode
consists of a rectangular iron plate having the size of 20 cm · 30 cm. A spiral titanium wire
(diameter: 2 mm) having a length of 20 m is used as an anode.
a) What is the power of the electrolytic cell? What is the curent density (electric current per
electrode surface)?

After electrolysis, CO2 is led through the solution until it is saturated. Afterwards, water is
carefully being evaporated. A white residue remains. The test for chlorate is positive.
According to powder diffractometry, the residue consists altogether of three salts..
b) What substances have to be reckoned on as well in this residue?

Analysis 1: 1 g of this residue is being dissolved in water, acidified with nitric acid that
produces a slight gas formation and titrated with a 0.1 molar AgNO3-solution. consumption:
18.80 ml.

Analysis 2: 1 g of this residue is heated to 600 °C (the substance mixture melts), cooled
down again and the mass is determined once again: 0.95 g.

Analysis 3: A powder diffractogram of this melting residue shows that one component of the
initial substance is still present and that the other two components, however, have been
transformed into two new salts.

Analysis 4: The 0.95 g of the melting residue are as well being dissolved in water and
acidified with nitric acid. A slight gas formation can as well be observed. Then, it is being
titrated with a 0,1-molar AgNO3-solution. consumption: 33.05 mL.

c) Write down all reaction equations that are important for these analyses. Which two salts
have disappeared and which two salts have newly formed?
d) Determine the mass fractions of the three salts of the initial solid and those of the three
salts of the melting residue.

18
 Problems round 3 test 1

Problem 3-6 Saponification

Skrabal investigated the saponification of an ethylidene diacetate as an aqueous 0.1 M
solution and catalyzed by protons (HCl-concentration = 0.05 mol/L) at 25 °C.
(CH3COO)2CHCH3 + H2O ⎯→ 2 CH3COOH + CH3CHO
He obtained the following series of measurements:
t [min] 0 240 660 1400 2093 3403 6369
c(CH3COOH) [mol/L] 0.02160 0.04570 0.06495 0.09395 0.11520 0.14475 0.17915

All values of the calculation have to be provided with SI-units!

a) Find out by a diagram if it is a zero- or first-order reaction (note: both reaction orders have
to be checked for!).
b) Give a classic example of a zero- as well as a first-order reaction.
c) Determine the values of the rate constant and the half life.

Let`s have a look at the ester saponificationof ethyl acetate in the presence of alkali.
A 0.02 N solution of ethyl acetate has been saponificated with a 0.02 N caustic soda solution.
After 25 minutes, 73 % of it has been saponificated. The second-order rate law for equal
initial concentrations of the reactants can be applied to this reaction.
d) Write down the reaction equation of the ester saponification. How would you determine
the turnover of the reaction?
e) Determine the rate constant of this ethyl-acetate saponification.
f) Determine the half life of this saponification. When is a turnover of 99% obtained?

Problem 3-7 Redox equilibria

Solid iron(III)-bromide smells of bromine. Thus, it decomposes according to the following
equation whereat an equilibrium establishes:

2 FeBr3 2 FeBr2 + ½ Br2

In an aqueous solution as well, Fe3+-ions react with halide ions. 0.01 mol sodium halide per
litre of solution is dissolved in an acid, aqueous solution of 1 mol/L iron(III)-perchlorate at
25°C and without air. In addition to that, the following values are given:
E0(Fe3+ + e- ⎯→ Fe2+) = 0.77 V
E0(X2 + 2e- ⎯→ 2 X-) = 1.36 V for chlorine/chloride, 1.07 V for bromine/bromide and
0.54 V for iodine/iodide.
The solubilities of the halides in water should be regarded as negligibly low and thus,
their concentrations should be regarded as being constant.
a) Calculate for all three reactions (addition of chloride, bromide and iodide to the Fe3+ -
solution) the concentration of free Fe2+ and write down the pFe2+-value for every reaction.

19
 Problems round 3 test 1

b) In which form can the Fe3+- and. Fe2+-ions be obtained in a dilute acid, aqueous solution?
c) Why is the pH-value for the reaction above so important, although neither H+ nor OH- or
H2O take part in the reaction? Give a reason for this importance and write down 2 reaction
equations at least proving a pH-Wert-dependence of these redox systems.

The influence of the pH-value has to be noticed especially when iron is to be transformed
into the oxidation number +VI. Thus, iron(III)hydroxide can be oxidized with chlorine to
ferrate(VI) having a structure that is similar to that of chromate in highly concentrated cold
caustic soda lye. If potassium ferrate(VI) is dissolved in pure water, it will decompose again,
however, into iron(III)-hydroxide.
d) Write down balanced reaction equations for these reactions.

Problem 3-8 Peptide chain

a) What is the stereochemical configuration (according to Fischer) of almost all natural amino
acids ?
b) There is a peptide consisting of 50 molecules of alanine. Both stereoisomers of alanine
are possible. How many stereoisomers of the peptide are possible?
c) Determine the absolute configuration (according to the Cahn-Ingold-Prelog-nomenclature)
of every stereocentre of the following amino acids:

HS

COOH
H2N COOH N
H H H2N COOH

isoleucine proline cysteine

The "Edman degradation", e.g., is used for the analysis of the amino-acid sequence of a
peptide. It has the following mechanism:

20
 Problems round 3 test 1

Ph
Ph
NH
N C S
phenylisothiocyanate C S
Ph
+
NH
NH2 N
O S
H3C CH
H3C CH
C O NH
C O
H3C
NH phenylthiohydantoine
NH
HCl/H2O
H3C CH NH2
H3C CH

C O H3C CH
C O
NH C O
NH

H3C CH NH
H3C CH
C O H3C CH
C O
C O
phenylthiocarbamoyl-
peptide peptide
shorter peptide

reaction step 1 reaction step 2

The result of the two reaction steps is a one amino acid shorter peptide. The chromatogra-
phic analysis of phenylthiohydantoine shows which amino acid is found in phenylthiohydan-
toine. The amino-acid sequence of an unknown peptide can be determined by several
repititions of this process.
In a modified process 1, a little (!) isocyanate (R-N=C=O) is used as well, apart from
isothiocyanate. Isocyanate, however, stops reacting at reaction step 1. Both reaction steps
are repeated a lot of times in a reaction vessel, so that the peptide is (sometimes completely)
degradated. Afterwards, a mass-spectrometric analysis of the reaction solution is carried out.
Chromatographic methods have shown that an unknown peptide X only contains the amino
acids glycine and alanine:
H O
OH H2N
H2N OH
O CH3
glycine (Gly) alanine (Ala)
The analysis of the unknown peptide X according to process 1 resulted in mass peaks at the
following m/z-values in the mass spectrum:
521.5 450.5 402.4 393.4 336.3 331.3
279.3 274.3 217.2 208.2 160.2 89.1
It could be found out that the compounds belonging to the m/z-conditions did not contain s-
atoms.
d) Identify the compounds belonging to the m/z-conditions listed up above. (The explicit
structural formula needn`t be drawn, amino acids can be abbreviated by their three-letter-
code, e.g. NH2-ala-ala-COOH for the dipeptide alanine-alanine)
e) Determine the sequence of the unknown peptide X.

21
 Problems round 3 test 1

Problem 3-9 Mannich-reaction

In the following, the mechanism of the Mannich-reaction is explained by an example.
In the first step, an amine reacts with an aldehyde to form an iminium ion that reacts
with a ketone.
Then the nitrogen atom is protonated again. The reaction with an additional formaldehyde
molecule results in the formation of a further iminium ion leading to the product in an
intramolecular cyclization:

O
O H3COOC COOCH3
O H+
H3COOC COOCH3 + 2 + CH3NH2
H H N
CH3

CH3NH2 + H+ CH3NH3+

O H CH3
+
+ CH3NH3+ N H2O
H H
+
H H
iminium-ion

O OH
H3COOC COOCH3 H3COOC COOCH3

H H

+ N
+

H CH3

+
O OH

H3COOC COOCH3 H3COOC COOCH3

-H+

CH2 CH2

N N
H H CH3
CH3

A Mannich-reaction can as well be carried out with the following initial products:

O O
H H+ H3C
H + + CH3NH2 N
O

O
A
a) Draw the reaction mechanism with all intermediates (see above) leading to product A.

22
 Problems round 3 test 1

In the following, product A takes part in further reactions:

H3C O
N
strong base, +
e. g. NaOEt MeO OMe
Bx- Cy
- H+ - OMe-
O
NaBH4

H3C
N CO2CH3 + HOOC
Dz

O - H2O

O C

Ez
b) Draw the structural formulas of the compounds Bx- , Cy, Dz and Ez
It’s enough if you draw one possible isomer of every Bx- , Cy, Dz and Ez.
c) How many isomers can be obtained in this synthesis from the compound Ez ? Which
types of isomers appear in this synthesis?

Problem 3-10 Nitro compounds

According to the pH-value, the nitration of A results in compound B or C.

H3C CH3 H3C CH3 H3C CH3

N N N
+ NO2+ + NO2+ NO2

NO2 - H+ - H+
CH3 CH3 CH3

C A B

a) Which of the two products forms in a weakly acid solution and which forms in a strongly
acid solution ?
Explain this reactive behaviour with the resonance formulas of the σ-complexes appearing
as transient states in the formation process of the two processes.

23
 Problems round 3 test 1

Compound B reacts with zinc powder in dilute hydrochloric acid to form D. Compound D
reacts with acrolein (propenal) in a strongly sulfuric medium and is oxidized (e.g. with Fe3+) to
form E (E = C12H14N2). E has two condensed aromatic ring systems. This reaction has the
reactive intermediate D´ (see illustration).

N(CH3)2

D´

CH3

b) Write down a balanced reaction equation for the reaction from B to D.

c) Draw the structural formulas of D and E.

24
 Problems round 3 test 2

Third round, test 2

Problem 3-11
1) Which of the following mixtures of substances is not a buffer mixture?
a) KH2PO4/H3PO4 2:1 b) CH3COOH/NaOH 2:1 c) NaH2PO4/Na2HPO4 1:1
d) CH3COONa/CH3COOH 3:1 e) CH3COOH/KOH 1:2

2) Which of the following buffer solutions (CH3COOH / CH3COONa) is the most acid one?
a) 1-molar / 1-molar b) 1-molar / 0,1-molar c) 0,1-molar / 0,1-molar
d) 10-9 -molar / 10-12 -molar e) 10-10 -molar / 10-20-molar

3) 100 mL of destilled H2O are mixed with 1 l of ethanol (96 vol-%). The volume contraction
is 1%. What is the concentration of water in the mixture?
a) 14.8 mol/L b) 7.78 mol/L c) 7.14 mol/L
d) 0.09 mol/L e) 0.1 mol/L f) 16.5 mol/L

4) Which of the following formulas of the Fischer-projection shows the enantiomer that can
be found only once?
F F I I Br

I Cl Cl Br Cl Br Br Cl F Cl

Br I F F I

a) b) c) d) e)

5) How many cis/trans-isomers can the compound 1,3,5,7-nonatetraene form?:

a) 2 b) 4 d) 8 c) 5 e) 10 f) über 10

6) Which of the following compounds of C4H6O has the highest steam pressure at 25°C?

O
OH

OH O
a) b) c) d)

7) Which of the following elements has several modifications at 25°C?

a) bromine b) argon c) phosphorus d) nitrogen e) sodium

25
 Problems round 3 test 2

8) Which of the following compounds is the strongest acid?

Cl NO2

O2N NO2

NH2 O2N NO2

CHO OH OH
a) b) c)
CHO

HO OH

OHC CHO
CH3

OH OH
d) e)

Problem 3-12 Hybridization

a) Draw the Lewis formula of the allene molecule (propadiene) and give the hybridization of
the single C-atoms.
b) How does the hybridization of the borine atom change in the following reaction?
BF3 + F− ⎯→ [BF4]−

c) Give the hybridization of the Fe-atom in

the following diamagnetic compounds:
[Fe(CO)5], [Fe(cyclopentadienide)2] = ferrocene
(illustration) Fe
Explain your decision with the
electron configuration of the Fe-atoms
of the compounds.

d) The degree of hybridization of C, N, O in the compounds methane, ammonia and water is

sp3. Give reasons for the following different bond angles:
in CH4: H-C-H = 109.4°, in NH3: H-N-H = 106.8°, in H2O: H-O-H = 104.5°.

e) Explain the following bond angles with the hybridization concept:

R O
O CH3
A B O
H H 117...119 SiH3 SiH3
104,5 C
144,1

Why do the hybridization states in the compounds A, B and C differ?

26
 Problems round 3 test 2

Problem 3-13 The "adamantane diamond“

Adamantane (C10H16), see illustration 1, is a hydrocarbon with a very high symmetry.
Assuming that only the bridgehead C-Atome take part in further reactions, adamantane can
be called as well "inflated sp3-hybridized C-atom". This molecule containing 2 such "inflated
C-atoms", 1,1´-biadamantane, more or less a "di-adamantane-ethane" would have the
structure shown in illustration 2.

illustration 1: adamantane illustration 2: "di-adamantane-ethane"

a) Calculate the "atomic radius" of the "inflated C-atom" assuming that all carbon atoms of
adamantane are ideally sp3-hybridized and thus have bond angles of 109.4°.The C-C-
bond distances in the adamantane backbone are 154.0 pm. These distances are
157.8 pm between the adamantane units.

Carbon, in the modification of diamond, crystallizes in a cubic lattice. Every C-atom is

tetrahedrally surrounded by 4 further C-atoms (see illustration 3).

C illustration 3: binding conditions of illustration 4: A

a the C-atoms in diamond; the
tetrahedral edge length a is 252.22
part of the
"adamantane-
pm diamond".
C A
C C A A A
C
b) Calculate the atomic radius of the C-atom in the diamond lattice.

According to the idea of an "inflated C-atom", all the real C-atoms of the diamond have to be
substituted by adamantane units (illustration 4, atom A corresponds to the adamantane-
backbone). The density of a real diamond is 3.514 g/cm3
c) Calculate the density of the "adamantane diamond".
d) Make a qualitative statement on the solidity of the "adamantane diamond" in contrast to
the real diamond.
e) Why does cubic boron nitride (BN)x have physical properties that are similar to those of
the real diamond?

Problem 3-14 Catalysis

Many subgroup elements react with carbon monoxide to form complexes. There is, e.g., the
binuclear cobalt complex [Co2(CO)8] having a Co-Co single bond that can be cleaved by
hydrogenation so that [CoH(CO)4] forms.

27
 Problems round 3 test 2

[CoH(CO)4] is a very strong acid (pKs = 1) forming the ion [Co(CO)4]- in an aqueous solution.
a) How many ligands (CO, H−) are bound to the Co-atom in these 3 complex compounds.
Give reasons for your answer.
b) What is the difference between the ionic dissociation of [CoH(CO)4] and that of nitric
acid?

[CoH(CO)4] can be used as a catalyzer for so-called "hydroformylation". An alkene reacts

with H2 and CO to form an aldehyde in the presence of a catalyzer. At an increased
temperature (90°C - 250°C), the complex [CoH(CO)4] is mostly decomposed into carbon
monoxide and A. (M(A) = 144 g/mol). A can form the π-complex B by coordination to ethene.
B transposes to the alkyl complex C. The next step is the coordination of further CO to the
Co-atom. D forms and transposes to the alkyl complex E that is cleaved by hydrogen into A
and an aldehyde F.
c) Write down the structural formulas of the compounds A to F. (It can be assumed that the
complexes B and D obey to the same "coordination rules" as [Co2(CO)8] and [CoH(CO)4]
do.)
d) How do you think can the cleavage by hydrogenation of C be stpped in order to obtain a
maximum yield of the aldehyde F?

[CoH(CO)4] can be used as well as a catalyst for the isomerisation of alkenes. 3-methyl-1-
butene, e.g., transforms into 2-methyl-2-butene in the presence of [CoH(CO)4].
e) Propose a mechanism for such an isomerisation reaction.
f) Give the isomerisation product(s) of 2-ethyl-1-butene as well of propene-3-ol.

Problem 3-15 Nuclear reactions

Radioactive decay reactions always take place according to a first-order rate law: N(t) =
N0e−kt. N0 is the number of atoms at the point of time t = 0 and k is the radioactive decay
constant. The half life t½ is the time needed for half the initial number of nuclei to
desintegrate.
a) Write down a decay law not containing any longer k but t½.

When the elements came into being, many radionuclides came into being as well. Some of
them like certain uranium- and thorium isotopes can still be found on earth because of their
longevity. Natural uranium consists of several isotopes. The most long-life ones are 238U
(99.275% , t½ = 4.468·109 a) and 235U (0.720%, t½ = 7.038·108 a). The other uranium isotopes
have considerably shorter half lifes. Thorium consists exclusively of the long-life isotope 232Th
(t½ = 1.405·1010 a). 237Np (t½ = 2.14·106 a) was an additional, relatively long-life isotope. It
has, however, already decayed. All these isotopes undergo α-decay.
b) At what point of time in the past were the fractions of both uranium isotopes the same?

A further isotope 234U with a fraction of about 0.005% can be found in natural uranium. It has
not remained since the date of origin of the earth, but is formed continuously by the decay of

28
 Problems round 3 test 2

one of the four isotopes mentioned above. A radioactive equilibrium has established at which
the concentration of 234U is constant, that means that the formation- and decay rates are the
same.
c) Which of the isotopes mentioned above is 234U produced of by a series of α- and β−-
decays? Write down the path of formation.
d) Calculate the half life of 234U.

The recovery of the noble gas radon is difficult because of its short half life (t½=3.825 d). It is
recovered from he radon isotope 226Ra (t½ = 1598 a).
226 K1 222 k2
Ra ⎯⎯→ Rn ⎯⎯→ 218Po
A certain amount of RaCl2 is dissolved in water. After the concentration of radon has reached
99% of the concentration in the radioactive equilibrium, radon is drained by pumping.
Because the half life of 226Ra is much longer than that of 222Rn, the concentration of radon
can be assumed as being constant. The following relation can be applied:
k
[222Rn] = [226Ra] 1 (1−e−k2t)
k2
e) After what period of time can radon be drained by pumping?
f) Why is insoluble RaSO4 not taken as a basis?

Problem 3-16 Silicon monoxide

Silicon dioxide (SiO2) is the most frequent mineral of the earth`s crust. If it is heated in a high
vacuum to high temperatures (>1000 °C), it will decompose into gaseous silicon monoxide
(SiO) and oxygen according to the following reaction equation:
2 SiO2 (s) 2 SiO (g) + O2 (g)
-15
At a temperature of 1300 °C, the equilibrium constant Kp for this reaction is Kp = 3.9 ·10
3
mbar .
The thermodynamic data for SiO2 (s), SiO (g) and O2 (g) at different temperatures can be
found in the following table:
T [°C] SiO2 (s) SiO (g) O2 (g)
∆Hf S ∆Hf S ∆Hf S
[kJ mol-1 ] [J mol-1K-1] [kJ mol-1 ] [J mol-1K-1] [kJ mol-1 ] [J mol-1K-1]
800 - 860.3 122.1 - 74.3 253.8 25.2 246.1
1000 - 846.4 134.2 - 67.1 260.0 32.3 252.1
1200 - 832.0 144.8 - 59.7 265.4 39.5 257.3

(The values refer to a standard pressure p0 of 1.013 bar)

a) Why is the formation enthalpy ∆Hf indicated for oxygen not zero, although oxygen is a
chemical element?

29
 Problems round 3 test 2

b) Calculate ∆HR, ∆SR and ∆GR for this reaction at temperatures of 800 °C, 1000 °C and
1200 °C. Is the reaction exothermal or endothermal?
c) Write down the equilibrium constant Kp for this reaction at 800 °C, 1000° C and 1200 °C .
d) Calculate the partial pressure of SiO that will result from the equilibrium, if solid SiO2 is
heated to 1300 °C in a high vacuum.
e) How can gaseous SiO be produced without oxygen being formed as well? Write down the
reaction equation supporting your proposal!

Problem 3-17 Solubility product

The solubility product KL of iron(II)-hydroxide is to be determined by potentiometry. A
galvanic cell is set up. One half cell consists of a slightly acid 0.01 molar Fe2+-solution in
which an iron rod is placed. A copper half cell with a 0,5-molar Cu2+-solution is used as a
reference half cell. A voltage of 0.830V can be read.
Then, the pH-value in the iron half cell is made exactlly 12 with caustic soda lye. But watch
out! No air must get into the half cell. A white precipitate can be observed in the solution. The
voltage changes to 1.090 V.
The experiment is carried out under standard conditions.
Standard redox potential: Cu/Cu2+: +0.340V

a) Write down the reaction equations for the processes in the reduction- as well as the
oxidation half cells and the total equation.
b) Calculate the standard redox potential of Fe/Fe2+.
c) What will happen if caustic sodium lye is added to the iron half cell? Write down the
reaction equation.
d) What would happen if caustic sodium lye was added to an open iron half cell? Write down
the reaction equation.
e) Calculate the solubility product KL of iron(II)-hydroxide. Write down as well the pKL-value
of Fe(OH)2.

Problem 3-18
Ester A (see illustration) forms by a mixture of 3,3-dimethyl butanone, butanedioic acid
diethylester and sodium ethanolate (CH3CH2O−Na+). The mole ratio is 1 : 1 : 1.

COO C2H5
A

HOOC

30
 Problems round 3 test 2

EtO-Na+ 3,3-dimethyl butanone

butanedioic-acid diethylester Z1 Z2

-EtO-Na+

A Z3

a) Write down the reaction mechanism of the formation of A from the initial substances
mentioned above (see scheme). Give the intermediate stages Z1, Z2 and Z3. Why is a
stoichiometric amount of the catalyst sodium ethanolate added?

A undergoes the following reactions:

OH OH
O
COOC2H5

A B C (R,S)- und (S,R)-isomer

O
Base +
OH
Br
O
O
H2O / OH-
A´ B´ C´

MnO4-
H2O

D E

B is very reactive and transforms very quickly into C. If A is transformed into A´, the more
stable product B´ will form. B´ can only transform into C´ after alkali reprocessing.
The oxidation of A´with dilute alkali permanganate solution, however, does not lead to C but
to D and finally to E.
b) Draw the structural formulas of the compounds B, A´, B´, C´, D and E. Please write down
if and which stereoisomers can occur.

31
 Problems round 3 test 2

Problem 3-19

The following illustration shows a skeletal rearrangement::

CH3 + H+ rearrangement rearrangement

A+ B+ C+
HO - H2O
- H+

CH3

R = phenyl

a) Draw the structural formulas of the compounds A+, B+ as well as C+ and indicate with an
arrow in the structure the migration direction of the compounds (each rearrangement
results in only one compound wandering! )
Note: compound B+ already contains a three-membered ring

b) What is the driving force for the rearrangement ?

If C+ does not immediately separate a proton, but first reacts with the water molecule
separated in the first step, D will form.
c) Draw D, mark all asymmetric C-atoms in D. How many stereoisomers does D have?

32
 Problems round 3 test 2

Problem 3-20
The composition of an organic compound consisting of C, H, O and N was determined by
quantitative elemental analysis resulted in the following values: 37.02% C; 2.22% H; 18.50%
N. The peak in the mass spectrum of the compound is m/Z = 227.
a) Determine the total formula of the compound that has to be found out.

The following illustration shows the IR-spectrum of the compound:

There are two strong absorption bands at 1544cm-1 and 1351cm-1.

b) Which group(s) in the molecule do these signals refer to? (Use the added IR-table.)

In addition to that, the 1H-NMR spectrum of the compound is shown in the following diagram:

c) Use the complete information to determine the structure of the compound. Draw the Lewis
formula.
d) What is the usual name of the compound and what particular physical property is it
famous for?

33
 Problems round 4 (theoretical)

Fourth round (theoretical problems)

(The same formulary as in the third round and a periodic table were made available to the students)

Problem 4-1 Copper(I)-bromide

Copper(I)-bromide is a poorly soluble salt (pKL = 7.4).
a) How much water is needed to dissolve 1 gram of CuBr completely?

Copper(I)-ions react with ammonia to form diamminecopper(I)-ions :

+NH3 +NH3
Cu+ -NH3
[Cu(NH3)]+ -NH3
[Cu(NH3)2]+ (1)
The equilibrium constants for the two steps of complex formation are lgK1 = 6.18 and lgK2 =
4.69.
An ammonia solution (c = 0.1 mol/L NH3) can be used for the dissolution of 1 g of CuBr.
b) What is the volume of this NH3-solution needed for dissolving CuBr?
c) Give the conditional solubility product of CuBr in the resulting solution.
KL(kond) = {c(Cu+) + c([Cu(NH3 )]+) + c([Cu(NH3)2]+)} · c(Br−)

Problem 4-2: Thermodynamics of Xylenes

Xylenes (dimethylbenzenes) are obtained from naphtha, the medium petroleum fraction.
They are very precious initial products for the production of plastics. Because the natural
content of xylenes is only about a few percent, naphta fractions are submitted to the so-
called reforming process in which aromatic hydrocarbons form by cyclization and
dehydrogenation.
When 1 mol of liquid p-xelene is combusted, 4551.4 kJ/mol are liberated at constant
pressure and standard conditions.
a) Write down a balanced reaction equation for this combustion reaction.
b) Calculate the free standard heat of formation ∆Hf0 of liquid p-xylene by means of the heat
of combustion.
(The result does not correspond to the value indicated in the table)

The industrial needs for different xylene isomers are very different. The needed amounts are
largest for p-xylene, followed by o-xylene amd m-xylene. The isomerization reactions of
xylenes that take place in the gas phase with Lewis acids as catalysts are of great technical
importance.
c) Calculate the free standard heat of formation ∆Hf0 and free standard reaction entropy ∆Sf0
for the transformation of o-xylene into p-xylene in the gas phase at T = 500 K, assuming
that the heat capacities are independent of temperature. Use the thermodynamic data
indicated below.

34
 Problems round 4 (theoretical)

d) Calculate the percentages of the three xylene isomers in the equilibrium mixture at T =
500 K.

In industrial processes, acid zeolite catalysts are used for isomerization reactions. They
contain long channels that have diameters chosen in a way that para-isomers can diffuse
quickly into and out of these catalysts while the other isomers have to remain in the catalyst.
In this way, a fraction of p-xylene of about 80% can be reached.
e) Which principle was made use of in the upper process?

necessary values:
∆Hf0 S0 Cp(l) Cp (g) Ts ∆Hevap
[kJ/mol] [J/(molK)] [J/(molK)] [J/(molK)] [K] [kJ/mol]
o-Xylol -24.4 246.0 187.7 171.6 417.0 36.2
m-Xylol -25.4 253.8 184.6 167.1 412.3 35.7
p-Xylol -24.4 247.2 182.2 167.4 411.4 35.7
CO2 -393.5
H2O -285.8

Problem 4-3: Kinetics of ethyl iodide

a) Calculate the solubility product of AgI at 25oC from the values in table.
b) Calculate the solubility product at 75oC. Could the result be expected?

The thermal decay of ethyl iodide into hydrogen iodide and ethene is to be investigated – that
means that the rate constant is to be determined. A sample of ethyl iodide is heated to 600 K
for a certain period of time. HI that has formed is absorbed by a substance binding HI
selectively. Ethyl iodide that is left is acidified with 200 mL of 1M nitric acid at 25 oC and 50
ml of 1M NaOH are added to it. Then, 0.1 g of AgNO3 is added to this mixture. A silver
electrode is put into the solution and the potential is measured against a calomel electrode
(εKalomel = 0.283 V). The values of the emf depending on time at bei 600 K can be found in
table 2.

c) Calculate the equilibrium constant of the decay reaction at 600 K and give reasons for the
selective absorption of HI being necessary.
d) Give the potential of the silver electrode in dependence on the iodide concentration.
e) Which reaction order does the decay have? Give the iodide concentration in dependence
on time.
f) Insert the result of e) into the result of d) and calculate the emf (εKalomel - εAg) in
dependence on time. Simplefy this expression. Determine the rate constant and the initial
amount of ethyl iodide in g on the basis of an appropiate plotting of the values of table 2.
Note: Determine gradient and axis intercept graphically!

35
 Problems round 4 (theoretical)

table 1 table 2
∆Hof [kJ/mol] ∆Gof [kJ/mol] E0 [V] t [h] emf [mV]
+ -
Ag + e → Ag 0.799 2 365
½ I2 + e - → I- 0.535 5 361
AgI(s) -61.84 -66.19 8 356
Ag+(aq) 105.79 14 347
-
I (aq) -56.78 19 340
Ag(s) 0.0
I2(s) 0.0
HI(g) 26.5 1.7
C2H4(g) 52.5 68.4
C2H5I(g) -8.1 19.2

Problem 4-4 Free molecule spectroscopy

Biological cells as well as subcellular units (organelles) are separated from each other by
lipid bilayers. In the recent years, fluorescence-free molecule spectroscopy has made
important contributions to the information about their structures and functions. This method
contains the integration of lipid-like fluorescent dye molecules (fluorophores like e.g. "DiO"
(3,3'-dioctadecyloxacarbocyanine perchlorate, structure: see next page) as markers in
membranes and the observation of their movements by the emitted fluorescence of the
single molecules by means of high-sensitive CCD-cameras. The membranes are mostly
lighted with intensive (kW/cm²) laser pulses of a duration of some ms.
50 µL of a 10 mM solution of DOPC (1,2-dioleoyl-sn-glycero-3-phosphocholine) in chloroform
are used for the modelling of natural membranes. After the addition of methanolic
DiOsolution, the mixture is filled up to100 µL with chloroform.
a) How many µL of a methanolic DiO-solution (10 µg/L) have to be added, so that 50 DiO-
molecules can be found per 100 µm² of lipid bilayer (free of chloroform)?
(note: every DOPC-molecule demands 0,64 nm² in the lipid membrane)

Within DOPC-bilayers, lipid molecules diffuse with a diffusion coefficient of D = 6·10-8 cm²/s
at room temperature. 10 ms are neded for the exposure and 25 ms are needed for the
selection process per photo. The averaged square of the covered distance ∆x can be
( ∆x )
2
described by the formula = 2 fD ∆ t . f is the degree of freedom (dimension) of the
T

movement and ∆t is the time distance between two photos. ( A T

is the time average of A.)
b) How long is the average distance (in µm) of the movement of a molecule between two
sequential photos?
c) How long is the distance after 4 photos?
d) How long does it take on average until a lipid molecule has "hiked through" the extentions
of a cell (10 µm) within the model membrane that has been investigated?

36
 Problems round 4 (theoretical)

After the absorption of a photon, a molecule can fluoresce while returning to the basic state.
It can, however, as well change into a longer-life excited state (triplet state). Within this state,
it can irreversibly chemically react and decoulourize. It is assumed that DiO decolourizes
exponentially under the selected conditions of excitation with a half life of 75 ms.
e) After how many photos are 10% of all DiO molecules that have initially been present in an
investigated part of a membrane still fluorescent?
Structure and molecular mass of DiO

M = 882 g/mol

Structure of DOPC

Problem 4-5 Solid-state chemistry: spinels

Spinels are a large group of compounds having the general formula AB2O4. The parent
compound, the mineral spinel, has the composition MgAl2O4. Its noble varieties are used as
precious stones.
The simplest way of producing spinels is the direct reaction of the single oxides:
MgO + Al2O3 ⎯→ MgAl2O4
The reaction proceeds with noticeable speed only at very high temperatures (> 1800 °C),
because it is only then that the mobility of the particles in the solid becomes sufficiently high.
Two blocks of the metal oxides AO andd B2O3 are compressed and heated for the
determination of the reaction mechanism of spinel formation.

wire

pressure AO B2O3 pressure

The interphase is marked e.g. by a thin wire. The reaction starts by ions diffusing through the
interphase. After some time, some spinels have formed. Conclusions about the reaction
mechanism can be drawn from the position of the wire.

37
 Problems round 4 (theoretical)

pressure AO AB2O4 B2O3 pressure

In principle, three basic mechanisms are possible: (i) only the cations A2+ and B3+ diffus
(ii) only the ions of AO diffuse (A2+ und O2–) (iii) only the ions of B2O3 diffuse (B3+ and O2–)
a) Decide for each of the three cases in which ratio the ions have to diffuse to keep
electroneutrality.
b) Decide for each of the three cases where in the AB2O4-layer the wire that has marked the
initial interphase can be found after the reaction has finished.
c) Why can the wire nevertheless be found sometimes at positions different from those
predicted in (b)?

Problem 4-6: Chemistry of phosphorus

Determine the compounds from A to K.
notes: the stoichiometric ratios are not indicated.
F: one equivalent of chloroalkane forms as a byproduct
J: J is a salt

A B

O2
H2O, KOH
(excess) P
P D
P
P

Cl2 EtOH

(EtO)3P EtOH C O2 E

PhMgBr PhMgBr
HCl (excess) (excess)

H 2O
H I
F G
Na, NH3(l)

K NH4Cl J

38
 Problems round 4 (theoretical)

Problem 4-7 Anionic polymerization

High degrees of polymerization can be obtained already in the beginning of free radical
polymerization. The degree of polymerization, however, ideally increases linearly with the
monomer turnover in the anionic-polymerization process (no termination reaction during
polymerization). In addition to that, the fast initiation reaction with carbanionic combounds
like sec-butyllithium guarantees a closed molecular-weight distribution. The (numerically
averaged) degree of polymerization Pn indicates the number of minomer units per polymer
molecule.
Different polysterene samples are to be synthsized by anionic polymerization by means of a
single solution, so that a monomer turnover of 25 %, 50 % and und 75 % can be obtained.

styrene

a) A reaction with 100 g of sterene and 1,54 g of sec-butyllithium is prepared.

Calculate the average degrees of polymerization (numerical average) of the samples for
monomer turnovers of 25, 50 und 75 %.
b) What is aimed at with polymerization is getting samples of polysterene that have the same
masses but different degrees of polymerization. Calculate the volume fractions of the
solutions at 25, 50 and 75 % of turnover, so that the masses of pure polymer (without
monomer or any solvent) are the same for all three degrees of polymerization. Ignore
changes in volume as a result of polymerization. Assign the volume fractions to the initial
solution (solution before the start of the reaction). Consider that the total residual solution
has to be taken and processed at the last sampling at 75 %.

Polymer analytics

The distribution of the molar mass of a synthesized polymer is analyzed by mass

spectrometry (matrix assisted laser desorption ionization, time-of-flight, MALDI-TOF-MS).
TheMALDI-TOF-method guarantees a minimal fragmentation of the polymer molecules. After
the substraction of the charge carrier (Ag+), the following frequency distribution can be
obtained:

molar mass M [g/mol] 890 994 1098 1202 1306

frequency of the molecules in 2.5 25 45 25 2.5
the mixture h [mol-%]

c) What will bethe name of the monomer of the polymer and which terminal group does the
polymer have, if sec-butyllithium is used as an initiator? Calculate the degree of
polymerization of each species.
d) Calculate from the table above two two average values of the molecular weight,

39
 Problems round 4 (theoretical)

∑h ⋅ M
i =1
i i
the numerical average Mn , defined as Mn = ∞

∑h i =1
i

∞ ∞

∑ wi ⋅ M i ∑h ⋅ M i i
2

and the weight average Mw , defined as Mw = i =1

∞
= i =1
∞

∑w i =1
i ∑h ⋅M
i =1
i i

(w is the weight fraction of the single molecules in the mixture).

Calculate the quotient from Mw and Mn. Calculate as well the quotient from Mn and Mw for
monodisperse beef insulin (single species at 5733,5 g/mol). So what does the quotient
Mw/Mn mean now ?

e) If the MALDI-TOF-spectrum is expanded, there will be a finer structuring of the single

peaks, like e.g. the simulation for the peak at 1098 g/mol (see illustration below). Give
reasons for the occurrence of the additional signals.

1.098 1.099 1.100 1.101 1.102 1.103 molar mass [g/mol]

Synthesis scheme
Apart from the usual linear polymers, other polymer structures are of great interest as well.
New structures can be synthesized by the combination of different methods.
f) Give the formulas of A, B and C. Make a rough sketch of C. Which current names does
product A have? Which complications that could be negative for the planned architecture
could appear in the last step?

CH3 O
1.
C O
K O O
CH3
O
2. CH3COOH Cl
n O A B C
pyridine

40
 Problems round 4 (theoretical)

Problem 4-8 Organic structural analysis

a) Which organic compound A that consists of three elements (chemical formula and name)
produces the following spectra?

mass spectrum

1
H- NMR- spectrum

b) Assign a cationic molecular fragment to each quite high peak (mass numbers 15, 27, 42,
43, 58) in the mass spectrum . Why is there only one peak in the 1H- NMR- spectrum?

If compound A reacts with itself in an acid medium, pinacol will mainly form in a reductive
coupling.
c) Write down the reaction equation and give the name of pinacol according to the IUPAC-
rules!

41
 Problems round 4 (theoretical)

Pinacol - pinacolone - rearrangement

If pinacol is distilled with highly concentrated sulfuric acid, pinacolon (2,2- Dimethyl-butan-3-
on) will form, water will be cleaved and there will be an intramolecular rearrangement.
d) Formulate the reaction equation as well as the reaction mechanism of the rearrangement.
e) A compound that is homologous to A produces the two spectra shown below. What kind
of compound is that?

1
H- NMR- spectrum

IR- spectrum

Problem 4-9 Heteroaromatics

The following molecules are examples of heteroaromatics. Heteroaromatics are aromatic
systems in which one or more carbon atoms are substituted by heteroatoms.
H
N O S

pyrrole
Pyrrol furan
Furan thiophene
Thiophen

42
 Problems round 4 (theoretical)

a) When is a pi-electron system aromatic? Give four conditions.

b) Draw resonance structures of pyrrole emphasising its aromatic property.
c) Order the given molecules according to increasing aromaticity. Give reasons for this
order.
d) In which position do you expect an electrophilic aromatic substitution of pyrrole? Give
reasons.
e) Give the products of the following reactions:

O
O Cl2, CH2Cl2 S CH3
SnCl4 , H3C Cl
A B

O O
H
N + H3C S CH3 H3C
N
H 3C O O- Br, AlCl3
C D

H
N CH2O/HNEt2
E
AcOH

Problem 4-10 Organic synthesis

a) Complete the following reaction scheme:

Cl2
catalyt. FeCl3
NaNH2

(3)
(1) (2) (catalyt. intermediate)

+
name of reaction

(4)

(5)

b) Give the name of the compound (2)

43
 Problems round 4 (theoretical)

c) What is the name of the reactive intermediate (3) ?

d) What is the name reaction of (3) with (4) to (5) called? What is the name of compound
(5)?

Iin the following, two molecules of the compound (6) react with each other in the name
reaction of d) to form one or several possible product(s).

Cl

produkt(s)

(6)
e) Draw the structural formula(s) of the possible product(s).

f) Complete the reaction scheme of the following reaction with endo-selectivity:

AlCl3

CO2Me

2 isomers

g) What is the stereochemical relation of the two isomers ?

44
 Problems round 4 (practical)

Fourth round (theoretical problems)

Practical problem 4-11 Synthesis

Procedure:

0.30 g of NaOH globules are added to 1.0 g of 2,3-dihydroxynaphthalene in 10 ml of water at

about 60°C, so that this mixture is a clear liquid.
This mixture is cooled down to room temperature (20...25°C) (e.g. with a cool water bath ).
0.65 mL of acetic anhydride are added (by injection) drop by drop while shaking the solution
softly. After some drops already, the product precipitates as a solid. After each further
addition of acetic anhydide, the cream has to be mixed well by powerful shaking. If
necessary, a glass rod can be used for shaking.
Then, 10 ml of distilled water are added to the mixture. 5 minutes later, it is kept in a warm
water bath at 60°C for about 10 minutes, until it shows a neutral or weakly acid reaction.
Afterwards, it is cooled down to 15°C (water bath) and kept at this temperature for about 10
minutes.
The product is sucked off, washed several times with a small amount of cold distilled water
and pressed dry on filter paper. About 10 mg (= sample 1) are withdrawn from the raw
product and dried in air.
The rest of the raw product is stirred in maximally 10 mL of about 50%- ethanol (that has to
be produced by yourself) in the heat and, if necessary, dissolved completely in the heat by
the addition of some 96%-ethanol. When it is cooled down, the product crystallizes in the
form of glittering placelets. For completion of the crystallization, an ice bath can be used for
cooling.
The purified product is sucked off, washed with a small amount of cold 50%-ethanol, pressed
dry on filter paper and dried in air (= sample 2).
Mass and melting point of sample two have to be determined.
A TLC is made for 2,3-dihydroxynaphthalene, sample 1 and sample 2 (the solvent acetone/n-
hexane 1:2 has to be produced by yourself). After drying, the thin-layer chromatogram is
contemplated under ultraviolet light. The substance stains are marked with a pencil and
wetted with dilute FeCl3-solution with a capillary.
The Rf-values have to be determined.

a) Give the reaction equation as well as the mechanism of the reaction.

b) Which compounds may form as by-products?
c) Illustration 1 (annexe) shows the 1H-NMR-spectrum and the 13C-NMR-spectrum of the
product that is to be synthesized. Draw the structure of the compound and assign the
marked signals of the 1H-NMR-spectrums to the hydrogen atoms.
d) In the past, ink was produced from iron salts and gallic acid (see illustration 2). How could
gallic acid be obtained at that time?

45
 Problems round 4 (practical)

e) Which analogy is there between the production of this gallic ink and the qualitative
reaction with FeCl3 that has been carried out by you?

HO O

illustration 2: gallic acid

HO OH

OH

f) Phenols like e.g. gallic acid were and are used in pyrotechnics for the production of
whistling device. What is the mixing ratio of gallic acid and potassium chlorate (KClO3) to
obtain only water, carbon dioxide and KCl?

Practical problem 4-12 Analysis

Procedure:

The content of ammonium chloride of a sample has to be determined by acid-base titration.

1) The NaOH-solution for the titration has to be produced by yourself. A certain amount of
solid NaOH has to be dissolved in about 100 mL of water (in a 250 mL gratuated flask)
and filled up to 250 mL, so that a 0.1-molar NaOH-solution forms. By at least two titrations
of a sample of about 100 to 150 mg of oxalic acid dihydrate HOOC-COOH · 2 H2O (oxalic
acid dihydrate has to be weighed accurately to 0.1 mg, transformed into an Erlenmeyer
flask and dissolved in about 50 mL of distilled water) against phenolphthalein, the factor
(titer) of caustic soda solution has to be determined.
Afterwards, a sample of ammonium chloride with an unknown concentration has to be
titrated with this caustic soda solution.
2) AT YOUR PLACE: Fill up the ammonium-chloride solution (in the second graduated flask,
marked as "analysis") to the mark.
IN THE HOOD: 10 mL of the formaldehyde solution have to be put into the Erlenmeyer
flask, filled up with about 50 to 100 mL of distilled water and 3 drops of phenolphthalein
solution have to be added to this mixture.
AT YOUR PLACE: The dilute formaldehyde solution has to be neutralized exactly with
caustic soda solution. Then, an aliquot (20 or 25 mL) of the ammonium-chloride solution is
added to this mixture. The mixture is titrated until there is a new change in colour (the
colour should remain at the equivalence point for one minute at least).
The titration will be repeated to obtain a reproduceable analytical result.
a) Give the ammonium-chloride content of your sample (in mg).
b) Give all reaction equations that are important for this analysis.

46
 Problems round 4 (practical)

c) Sketch the titration curve for i) the titration of ammonium chloride with caustic soda
solution (pKS(NH4+) = 9.25) and ii) the titration of ammonium chloride with caustic soda
solution in the presence of formaldehyde.
d) Which preconditions does an acid-base indicator has to fulfill to be appropiate for this
titration?
e) Oxalic-acid dihydrate is an original titer in this problem. Which properties must a
substance have to be used as an original titer?
f) Which of the following compounds can not be used as original titers?:
NH4SCN, KIO3, Zn, Mg(NO3)2, K2Cr2O7, NaCl
g) Illustration 3 shows the formula of phenolphthalein as a colourless compound in an acid
solsution. Draw the structural formula of phenolphthalein as a red compound in a basic
solution.
OH

illustration 3: phenolphthalein, colourless form O

HO

47
 Problems round 4 (practical)

48
 Answers

Part 2

Answers

49
 Answers Round 1

Answers Round 1

Solution to problem 1-1

a) You find (i) ammonium hydrogencarbonate NH4HCO3
(ii) ammonium carbonate (NH4)2CO2
(iii) a mixture of (i) and (ii).
b) NH4HCO3 ⎯→ NH3 + CO2 + H2O
NH4CO2NH2 ⎯→ 2 NH3 + CO2

A mixture of 1 mol of ammonium hydrogencarbonate (M = 79.06 g/mol) and 1 mol of

ammoniumcarbamate (M = 78.08 g/mol) weighs 157.14 g and releases upon heating
6 mol of gases.
1 g of salt of hartshorn provides n = 6·(1/157.14) mol of gases upon heating.

6 ⋅ 1 ⋅ 8.314 ⋅ ( 180 + 273 ) 3

Using p·V = n·R·T : V = 5
m = 1.42·10-3 m3
157.14 ⋅ 1.013 ⋅ 10
V = 1.42 dm3
p2( H )
c) p(H) + p(H2) = 0.98 bar and = 2.51·10-2 bar
p( H 2 )
⇐ p2(H) - 2.51·10-2 bar · (0.980 bar – p(H)) = 0 p(H) = 0.145 bar

d) From c) you get p(H2) = 0.835 bar.

1 mol of the mixture contains 0,145/0.98 mol of H and 0.835/0.98 mol of H2.
Mean value of the molar mass:
(0.145 ·1.008 + 0.835 · 2.016) ·10 -3 kg·mol -1
M (Wasserstoff, 3000 K, 0.98 bar) =
0.98
M (Wasserstoff, 3000 K, 0.98 bar) = 1.87 · 10-3 kg·mol-1.
With p·V = n·R·T, n = m/M and ρ = m/V you get

p ⋅M 105 Pa ⋅ 1.87 ⋅ 10 −3 kg ⋅ mol −1

m/V = ρ = , ρ = , ρ = 7.35·10-3 kg/m3 .
R ⋅T 8.314JK −1mol −1 ⋅ 3000K

Solution to problem 1-2

a)
investigation 2: investigation 1: investigation 3: investigation 4:
concentration of Cl- total concentration of total concentration of concentration of Ca2+
ions cations Ca2+ and Mg2+ ions ions
1442443 1442443 1442443
Ø Ø Ø
concentration of SO42- concentration of Na+ concentration of Mg2+
investigation 1:
50
 Answers Round 1

RH + Na+ + H2O ⎯→ RNa + H3O+

2 RH + Ca2+ (Mg2+) + 2 H2O ⎯→ R2Ca (R2Mg) + 2 H3O+
n(H+) = n(Na+) + 2· n(Ca2+) + 2· n(Mg2+)
n(H+) = V(NaOH)·c(NaOH) in 10 mL of sea water n(H+) = 0.500·11.76·10-3 mol
for 1 L of sea water: c(Na+) + 2· c(Ca2+) + 2· c(Mg2+) = 0.588 mol/L

investigation 2:
Ag+ + Cl- ⎯→ AgCl
for 1 L of sea water: n(Cl-) = 0,086 ⋅ 6,21 ⋅ 10 −3 mol = 0.534·10-3 mol
c(Cl-) = 0.534 mol/L

investigation 3:
Y 4- + Ca2+ ⎯→ CaY2- and Y 4- + Mg2+ ⎯→ MgY2-
in 10 mL of sea water: n(Ca ) + n(Mg2+) = 0.05·12.60·10-3 mol = 0.63·10-3mol
2+

c(Ca2+) + c(Mg2+) = 0.063 mol/L

investigation 4:
Ca2+ + (COO-)2 ⎯→ Ca(COO)2
Ca(COO)2 + 2 H3O+ ⎯→ Ca2+ + (COOH)2 + 2 H2O
5 (COOH)2 + 2 MnO4- + 6 H3O+ ⎯→ 2 Mn2+ + 10 CO2 + 14 H2O
n(Ca2+) = n((COO-)2) = 52 ·n(MnO4-)
in 100 mL of sea water n(Ca2+) = 5
2 ·0.02·24.00·10-3 mol
c(Ca2+) = 0.012 mol/L

Calculation
investigation 2: c(Cl-) = 0.534 mol/L, with M(Cl-) = 35.45 g/mol 18.93 g/L Cl-

investigation 1: 0.588 mol/L = c(Na+) + 2· c(Ca2+) + 2· c(Mg2+) = c(Cl-) + 2·c(SO42-)

2·c(SO42-) = (0.588 – 0.534) mol/L = 0.054 mol/L
c(SO42-) = 0.027 mol/L, with M(SO42-) = 96.07 g/mol 2.59 g/L SO42-

investigation 4: c(Ca2+) = 0.012 mol/L, with M(Ca2+) = 40.08 g/mol 0.48 g/L Ca2+

investigation 3: c(Ca2+) + c(Mg2+) = 0.063 mol/L und c(Ca2+) = 0.012 mol/L

c(Mg2+) = 0.063mol/L – 0.012 mol/L = 0.051 mol/L
with M(Mg2+) = 24.31 g/mol 1.24 g/L Mg2+
c(Na+) + 2· c(Ca2+) + 2· c(Mg2+) = 0.588 mol/L and
2·c(Ca2+) + 2·c(Mg2+) = 2·0.063 mol/L c(Na+) = 0.462 mol/L
with M(Na+) = 22.99 g/mol 10.62 g/L Na+

b) Silver chloride (white) and silver chromate (reddish brown) are sparingly soluble salts.
However, the solubility of silver chloride is smaller than that of silver chromate. So,

51
 Answers Round 1

during investigation 2 silver chloride precipitates first. The reddish brown silver chromate
does not precipitate till silver chloride has precipitated nearly completely. This is the
reason silver chromate can serve as an indicator for the end of the precipitation of silver
chloride.

c) Additon of other less noble metals:

2 AgCl(s) + Zn(s) ⎯H⎯ ⎯
2O
→ 2 Ag(s) + Zn2+(aq) + 2 Cl-(aq)
m(Ag) = 107,9 g/mol · 5 L · 0,2 mol/L = 107.9 g
Price of silver on March 3rd 2003: 124 €/kg, value of 107.9 g of silber = 13.4 €

d) The anion of EDTA O

(-OOC-CH2)2 N-CH2-CH2-N (CH2-COO-)2
is hexacoordinated. - O
N
The structure of the complex shown in the O O-
second figure has the metal ion in the middle.
- O O
N
O-
Upon reacting with the indicator ions ofMg2+ and
Ca2+ form a red complex while the free indicator O
is blue.
If indicator is added to a solution of ions of Ca2+
and Mg2+ a red solution forms. The red complex
disappears upon adding a solution of EDTA. That’s
why at the equivalence point a clear change of
colour from red to blue is visible.

e) To prove other proposals for the composition the molar masses of the salts given are
shown:
salt NaCl Na2SO4·10 H2O CaCl2·6 H2O MgCl2·6 H2O MgSO4·7 H2O
mol. mass 58.44 g/mol 322.05 g/mol 218.98 g/mol 203.21 g/mol 246.38 g/mol

c(Na+) = 0.462 mol/L 27.00 g/L NaCl

2+
c(Ca ) = 0.012 mol/L 2.63 g/L CaCl2·6 H2O
c(SO42-) = 0.027 mol/L 6.65 g/L MgSO4·7 H2O
2+
c(MgCl2·6 H2O) = c(Mg ) – c(MgSO4·7 H2O) = 0.051 mol/L – 0.027 mol/L
c(MgCl2·6 H2O) = 0.024 Mol/L 4.88 g/L MgCl2·6 H2O

Na2SO4·10 H2O is not used for this solution.

52
 Answers Round 1

Solution to problem 1-3

a) The key to the solution of this problem is the different number of reactions the metals
and the salts undergo. You need to know the standard electrode potentials.
Silver does not react with any of the given salts, copper with one of them (AgNO3), iron
with two (AgNO3, Cu(NO3)2) and magnesium with three (AgNO3, Cu(NO3)2, Fe(NO3)3) .
Magnesium nitrate reacts with none of the given metals, iron nitrate with one (Mg),
copper with two (Fe und Mg) and silver nitrate with three (Fe, Mg, Cu).
On the remaining pieces of paper you realize:
B, C, D react at least once ⇐ A = silver,
1, 3, 4 react at least once ⇐ 2 = magnesium nitrate,
1 does not react with silver (A) but twice otherwise ⇐ 1 = copper nitrate,
Cu(NO3)2 does not react with silver (A) nor with copper ⇐ B = copper,
copper (B) reacts with silver nitrate only ⇐ 4 = silver nitrate,
that leaves 3 = iron nitrate
Fe(NO3)3 (3) reacts witth magnesium only ⇐ D = magnesium
that leaves C = iron .

b) Results not used: A/3 and A/4 .

Solution to problem 1-4

a) Reaction equation for the hydrolysis of A:

C20H27O11N + 2 H2O ⎯→ 2 C6H12O6 + C6H5CHO + HCN

glucose benzaldehyde hydrocyanic acid

b) Structural formela of compound A

H

C6H5 C CN

O C12H21O10

glycosidic sugar component consisting of

bond two glucose moieties

53
 Answers round 2

Answers round 2

Solution to problem 2-1

c( H + )
a) pH = - lg with co = 1 mol·L-1 , pH = - lg 0.5 pH = 0.30
c0
b) c(Cl-) + c(OH-) = 3 · c(Fe3+) + c(H+)
KW
At pH = 2 c(OH-) = +
= 10-12 mol·L-1.
c( H )
As the concentration of Cl- is at least as large as the concentration of H+ ( = 10-2 mol·L-1)
the concentration of OH- can be neglected.
KL KW
c(Cl-) = 3 · − 3
+ c(H+) and c(OH-) =
c( OH ) c( H + )
K
⇐ c(Cl-) = 3 · 3L · c(H+)3 + c(H+)
KW
c) Chloride:
2 ⋅ 10 −39 mol 4 ⋅ L−4
c(Cl-) = 3 · · (10-2 mol·L-1)3 + 10-2 mol·L-1
( 10 −14 mol 2 ⋅ L−2 ) 3
c(Cl-) = 1,6 · 10-2 mol·L-1 .

Iron:
KL 2 ⋅ 10 −39 mol 4 ⋅ L−4
c(Fe3+) = = = 2 · 10-3 mol·L-1.
c( OH − ) 3 ( 10 −12 mol 2 ⋅ L− 2 ) 3

Rust (Fe(OH)3):
n(Fe(OH)3) = n(Fe3+) = c(Fe3+) · V
= 2·10-3 mol·L-1 · ¾ · 0.2 L = 0.3 mmol .
M(Fe(OH)3) = 106.88 g/mol
m(Fe(OH)3) = n(Fe3+) · M(Fe(OH)3 = 0.3 mmol · 106.88 g/mol ≈ 32 mg
d) Amount of Cl- in the yellow solution:
n(Cl-) = n(Cl-) · V = 1.6 · 10-2 mol·L-1 · 0.15 L = 2.4 mmol
The same amount must have been in the hydrochloric acid (c0 = 0,5 mol·L-1) added
before.
n(HCl)
∆V(HCl) = = 4,8 mL
c(HCl)
e) 2 Fe3+ + Cu ⎯→ 2 Fe2+ + Cu2+
yellow bluish green

Assuming the reaction is complete:

cbluish green (Fe2+) = cyellow(Fe3+) = 2 · 10-3 mol·L-1
cbluish green(Cu2+) = ½ · cyellow(Fe3+) = 1 · 10-3 mol·L-1
54
 Answers round 2

In the bluish green solution: E(Fe3+/Fe2+) = E(Cu2+/Cu)

R ⋅T c( Cu 2+ )
E = E0(Cu2+/Cu) + · ln
2⋅F c0
8.314 ⋅ 300
E = 0.345 V + V · ln 1 · 10-3 E = 0.256 V
2 ⋅ 96487
R ⋅T c( Fe 3+ )
E = E0(Fe3+/Fe2+ ) + · ln
F c( Fe 2+ )
c( Fe 3+ )
0.256 V = 0.771 V + 0.0259 V · ln
c( Fe 2+ )
c(Fe3+) = 2.31 · 10-9 · c(Fe2+)
c(Fe3+) = 4.63 · 10-12

Solution to problem 2-2

x
a) AIx ⎯→ A+ I2
2
p2 = p2 (N2) + p2 (I2).
T 723.15 K
p2 ( N 2 ) = p1( N 2 ) 2 = 1,013 ⋅ 105 Pa = 2,457 ⋅ 105 Pa ,
T1 298.15 K
p2(I2) = p2 – p2(N2) = 3.346⋅105 Pa – 2.457⋅105 Pa = 8.89⋅104 Pa.
pV 8.89 ⋅ 10 4 Pa ⋅ 2.070 ⋅ 10 -5 m 3
n( I 2 ) = = −1 −1
= 3.061⋅ 10 − 4 mol .
RT 8.314J ⋅ K mol ⋅ 723.15 K
m(I2) = n(I2)⋅M(I2) = 77,7 mg,
m(A) = m(AIx) – m(I2) = 7,3 mg
2
n(A) = n( I 2 )
x
m( A ) x m( A ) 7.3 ⋅ 10 −3 g
⇐ M ( A )= = ⋅ = x⋅ = 11.9 ⋅ x g / mol .
n( A ) 2 n( I 2 ) 2 ⋅ 3.061⋅ 10 − 4 mol
x = 1 M(A) =11.9 g/mol (C) ⇒ C isn’t a metal
x = 2 M(A) =23.9 g/mol (Mg) ⇒ MgI2 is not volatile
x = 3 M(A) = 35.8 g/mol (Cl) ⇒ Cl isn’t a metal
x = 4 M(A) = 47.7 g/mol (Ti) ⇒ TiI4 is volatile and decomposes into
the elements
x = 5 M(A) = 59.5 g/mol ⇒ there is no suitable element.
Metal iodides with more than 5 iod atoms are currently not to be known.

Solution: TiI4.

b) Für die Änderung der Konzentration von HI in der (kurzen) Zeit ∆t gilt:
∆c(HI) = (2k1c(H2)c(I2) – 2k –1c2(HI))∆t.

55
 Answers round 2

In equilibrium ∆c(HI) = 0 ⇐ k1c(H2)c(I2) – k –1c2(HI) = 0

c 2 ( HI ) k
K= = 1 .
c( H 2 ) c( I 2 ) k −1

T, K 400 500 600 700 800

K 258 127 80.1 57.4 44.6

The higher the temperature the lower die equilibrium constant ⇒ the reaction is
exothermic (prinziple of Le Chatelier).
c) Die Temperaturabhängigkeit der Gleichgewichtskonstante ist durch die van’t Hoffsche
Reaktionsisobare gegeben:
K ∆H ⎛ 1 1⎞
ln 2 = − ⎜⎜ − ⎟⎟ ,
K1 R ⎝ T2 T1 ⎠
so dass aus der Steigung der Geraden ln K = f (1/T) die Reaktionsenthalpie abgelesen
werden kann.
T, K 400 500 600 700 800
-1
1/T, K 2.50⋅10 2.00⋅10 1.67⋅10 1.43⋅10 1.25⋅10-3
-3 -3 -3 -3

ln K 5.55 4.85 4.38 4.05 3.80

5,6

5,4

5,2

5,0

4,8
ln K

4,6

4,4

4,2

4,0

3,8

3,6
0,0012 0,0014 0,0016 0,0018 0,0020 0,0022 0,0024 0,0026
-1
1/T, K
From values at 400 K and 800 K ∆Hr = – 11.7 kJ/mol.
∆Gr = – RT ln K
∆Sr = (∆Hr – ∆Gr )/T.

T, K 400 500 600 700 800

∆Gr, kJ/mol –18.5 –20.1 –21.9 –23.6 –25.3
∆Sr, J⋅K-1⋅mol-1 16.9 16.9 16.9 17.0 16.9

∆Sr = 16.9 J⋅K-1⋅mol-1.

56
 Answers round 2

d) Degree of dissociation of HI at 600 K = α. c(H2) = c(I2) = ½ αc0,

c(HI) = (1–α) c0,
c 2 ( HI ) 4α 2
K= = K = 80,1 ⇐ α = 0.955.
c( H 2 ) c ( I 2 ) 1 − α
The number of particles does not change ⇐ the equivalent constant does not depend on
the pressure.
c(I in CS2 )
e) K= 2 = 585.
c(I 2 in H 2O )
32.33 g / L
c(I2 in CS2 ) = = 0.127 mol / L ,
253.8 g / mol
0.127 mol / L
c( I 2 in H 2O ) = = 2.18 ⋅ 10 − 4 mol / L.
585
1.145 g / L
c(I 3- in H 2O ) = − 2.18 ⋅ 10 -4 mol / L = 4.29 ⋅ 10 -3 mol / L ,
253.8 g / mol
c( I − in H 2O ) = c 0 ( KI ) − c(I3- in H 2O ) = 3.125 ⋅ 10 -2 mol / L ⋅ 4.29 ⋅ 10 -3 mol / L
= 2.70 ⋅ 10 -2 mol / L.
c( I3− in H 2O )
K= −
= 7.3 ⋅ 10 2 L / mol .
c( I 2 in H 2O ) c( I in H 2O )

2-

N O

Solution to problem 2-3 O

Cu
N
Zn O O N
N Ca
a) N O O O N

(C: the five-times coordinated O

structure is also taken as correct,

A achiral B chiral C chiral
it is chiral too)

3-
+
N O 2-
O O
Cu N Cu
N O O
N Fe O O
O O
O

D achiral E chiral F achiral

6-
N

N N
Co
N O N
H H
O O N
Co Co
O O N H2N NH 2
Pt
N O N Br Cl
Co
N N
N

G chiral H chiral
57
 Answers round 2

b) 3
c) M(L1)2(L2)2(L3)2. with 6 stereoisomeres

L1 L1 L1 L2

L2 L1 L1 L2 L2 L3 L1 L3
Fe Fe Fe Fe
L2 L3 L3 L2 L2 L3 L1 L3
L3 L3 L1 L2

chiral achiral achiral

L3 L1

L1 L2 L2 L3
Fe Fe
L1 L2 L3 L2
L3 L1

achiral achiral

d)

N N NH2 N Br N N

P Q R S
2,2'-Bipyridin

e) T = [Fe(bpy)3] SO4. N N

N N N N
Fe Fe
N N N N
N N

f) The formation of [Fe(bpy)3]2+ in an aquous solution is a three step process:

bpy bpy bpy
[Fe(H2O)6]2+ [Fe(H2O)4(bpy)]2+ [Fe(H2O)2(bpy)2]2+ [Fe(bpy)3]2+
- 2 H2O - 2 H2O - 2 H2O
high-spin high-spin high-spin low-spin

In the third step a transition from high-spin- to low-spin takes place. The stabilisation
energy of the ligand field connected with this step favors the complexation and draws
attention to the equivalent constant.
Solution to problem 2-4
58
 Answers round 2

a) The Si atom is larger than the C atom, so a higher coordination number is preferred.
Si belongs to the 3rd period of the PSE so overlapping of π orbitals does hardly
contribute to a gain of bonding energy. Forming of 4 σ bondings with O is favored.
b) Hardening agent and impregnant in construction industry, paper manufacturing, ...
c) Q-Gruppen von Q0 bis Q4, also SiO44-,(Si-O-)SiO33-,(Si-O-)2SiO22-, Si-O-)3SiO-,(Si-O-)4Si
d) 2 (CH3)3SiCl + H2O ⎯→ (CH3)3SiOSi(CH3)3 + 2 HCl
m (CH3)2SiCl2 + m H2O ⎯→ [(CH3)2SiO]m + 2m HCl
n CH3SiCl3 + 1,5n H2O ⎯→ [CH3SiO1,5]n + 3n HCl
e) Silicones
f) A ist tert-Butylamin:

H3C

H2N CH3 = H2N

H3C

2 H2N
N NLi N N
O Cl
2 Li SiCl4 2K
Si Si
- 2 H2O -2 LiCl Cl -2 KCl
N NLi N N
O

B C D E

Compund E is aromatic

g) Y, Z = (R = Cl bei Y, R = F bei Z) X = HCl.

N R N

Si

O R O

h) The octet rule is strictly valid only in the 2nd period. As Si belongs to the 3rd period there
are d orbitals with a capacity of 18 electrons (principally). The Si atom provides empty
orbitals for further ligands/donor atoms.
i) Volume of a monoclinic unit cell: V(EZ) =a⋅b⋅c⋅sinβ = 1837 Å3,
C20H21F2N3O2Si, has 28 non-hydrogen atoms.
With n = 4 : 16.4 Å3 for a non-hydrogen atom (n = 2 is to small, n = 6 is to large).
The unit cell consists of 4 complex molecules Z and 4 molecules acetonitrile.
59
 Answers round 2

m( EZ ) n ⋅ ( M ( Z ) + M ( CH 3CN )) 4 ⋅ 401.5 ⋅ 1.66 ⋅ 10 −27 g

ρ= = = − 24 3
= 1.42 g / cm 3
V ( EZ ) V ( EZ ) 1.873 ⋅ 10 cm

Solution to problem 2-5

a)
CH3

N
H3C COOH CH3

NH2 N
NH N N
NH
H H
A B C (Indol) D

COOEt CH3 NH2

NH
COOEt O COOH

N N
H H

E X (Tryptophan)

OH O O
H
N

N N N
H H H
O

Cox (Indoxyl) F (Indigo)

60
 Answers round 2

b) Mechanismus der Fischer-Indolsynthese:

COOH
H3C COOH H2C COOH
H+
+
N NH NH2
+ NH2
NH NH2

COOH
COOH

- H+ NH NH2
NH2 N
H

∆
COOH
- NH3 - CO2
N N
H H

C (Indol)

c)
COOH
a)
COOH COOH COOH
S S
2 2 2
+
NH2 N N SH
HOOC
G H I
O
O
COOH COOH
COOH
SH S
S COOH S

I K L M
O

O
Z (Thioindigo)

61
 Answers round 3 test 1

Answers round 3 test 1

(They are explained in a more detailed way than expected from the pupils)

Solution to problem 3-1

A) d B) e C) d D) c E) a F) c G) b H) f I) d J) b

Solution to problem 3-2

a)
1 CuSO4 2 NaCl 3 AgNO3 4 H2SO4 5 HI 6 NaOH 7 Na2CO3

b) X = NH4I Y = CuCl2

c)
Cu2+ + 2I- ⎯→ CuI + 1/2 I2 N(b) [CuI2- is correct as well]
Cu2+ + CO32- ⎯→ CuCO3 N(bl) [Cu(OH)2·CuCO3 is correct as well]
Cu2+ + 2 OH- ⎯→ Cu(OH)2 N(bl)
Ag+ + Cl- ⎯→ AgCl N(w)
Ag+ + I- ⎯→ AgI N(y)
2Ag+ + CO32- ⎯→ Ag2CO3 N(wy)
2Ag+ + 2OH- ⎯→ Ag2O + H2O N(b) [AgOH is correct as well]
NH4+ + OH- ⎯→ NH3 + H2O pungent smell
2H3O+ + CO32- ⎯→ CO2 + 3H2O gas

Solution to problem 3-3

a) K1 = c(B)/[c(A)·c(OH)]
c(B) = K1 · c(A) · c(OH) = 106.45 · 0.001 · 10-6 = 0.002818 mol/L
4.53 -6
c(C) = K2 · c(B) · c(OH) = 10 · 0.002818 · 10 = 0.000095 mol/L
The major part of steric expansion already occurs at the formation of B from A.

b) O

OOC O
N OH
Fe
OOC N O
O
It must be a complex with
cis-monodentate ligands.

O
62
 Answers round 3 test 1

c) Cyanide is a very strong ligand, this is the reason why there is a high ligand-field splitting
of the d-orbitals in the octahedral field of hexacyanoferrate(III). Hexacyanoferrate(III)
occurs as a low-spin-complex (1 unpaired electron). The iron-EDTA-complexes are high-
spin-complexes (5 unpaired electrons). Thus, the O-donor-ligands produce a much
weaker ligand field than cyanide.

d) Iron(III) is a weak oxidizing agent; thiolates and phenolates are weak reducing agents.
They make possible a charge-transfer (partial redox process between ligand and central
ion). This process can easily be excited by light. This is the reason for the intensive colour
of CT-complexes.

e) E = 1.505 (5cm-cuvette) corresponds to E = 0.301, according to the original prescription

with a 1 cm-cuvette. E = b · c + a, c = (E-a)/b = (0.301-0.013)/0.0048 mg-1 = 60 mg of
thymol in the analyte. However, only 1 mL of 5 mL of the analyte-solution was added.
Thus, the sample (1 g) contained altogether 300 mg of thymol. The mass fraction is 30 %.
M(thymol) = 150,23 g/mol. 0,3 g = 2 · 10-3 mol (per gram of raw substance) or rather
0,002 mol per gram of raw substance.

f) E = c · d · ε´ ε´ = E/(c · d) The concentration of thymol in the photometry-solution was:

(2 · 10-3 mol/g and one fifth of it ( 4 · 10-4 mol) in 50 cm3 of photometry solution)
c = 8 · 10-6 mol/cm3. d = 5 cm, E = 1,505. The result is ε´ = 37625 cm2/mol.
ε´ is the apparent coefficient of extinction, because the real concentration of
[FeY(OH)(thymolate)]3- is not known.
The OH-group of vanillin is sterically less hindered than that of thymol. Thus, there is a
higher fraction of CT-complex in the photometry-solution and a higher relative extinction
b.

Solution to problem 3-4

a)
CH3CN
6 Nb(OEt)5 17 (Me3Si)2S 4 NEt4Cl (NEt4)4[Nb6S17] 3 CH3CN 30 Me3SiOEt 4 Me3SiCl
(1)

b) In the course of the reaction 35 particles form from 27 particles (30 particles, if the solvent
molecules CH3CN are taken into consideration as well). That means that the numer of
particles and the disorder increase. A higher disorder corresponds to a higher entropy
that is energetically more favourable, that means that it favours the reaction.
The nascent Si-O and Si-Cl-bonds are more stable than the S-Si-bond in the initial
compound.

63
 Answers round 3 test 1

c) At first, the relative molar masses of the compounds Nb(OEt)5, (Me3Si)2S, NEt4Cl and
(NEt4)4[Nb6S17]*3CH3CN have to be calculated:
M(Nb(OEt)5) = 318.21 g/mol M((Me3Si)2S) = 178.45 g/mol
M(NEt4Cl) = 165.45 g/mol M((NEt4)4[Nb6S17] · 3CH3CN) = 1745.7 g/mol
5g of (1) should be obtained from a yield of 75%.
100%
The following calculation has to be made at first: ⋅ 5.000g = 6.667g .
75%
6.667g
6,667g of (1) correspond to = 3.819 ⋅ 10 −3 mol .
1745.7g / mol
Thus, the following amounts are needed from the initial substances:

n in mol n in mol m in g V in mL
Nb(OEt)5 6·3.819·10-3 mol 0.02291 mol 7.291 5.750
(Me3Si)2S 1.4·17·3.819·10-3 mol 0.09089 mol 16.220 19.173
NEt4Cl 1.4·4·3.819·10-3 mol 0.02139 mol 3.538

(The volumes can be determined from the indicated densities.)

d) Nb has the oxidation number +V in the initial substance as well as in (1).

S has the oxidation number – II in the initial substance as well as in (1).

e) There are two isomers:

M M'
M M' M'
M' M
M' M'
M
M M
mer-Isomer
mer-isomer fac-Isomer
fac-isomer

Solution to problem 3-5

a) P = 6 V · 2 A = 12 W,
Cathode: surface = 2 · 0.2 m · 0.3 m = 0.12 m2; that results in a current density (cathode)
of 16.67 A/m2
Anode: surface = 20 m · 0.002 m · π = 0.1257 m2, current density (anode) = 15.91 A/m2

b) Unconverted KCI will probably form. In addition to that, KOH forms during the electrolysis.
It reacts with an excess of CO2 to form KHCO3.

64
 Answers round 3 test 1

c) KClO3 and KHCO3 are decomposed; KClO4 and K2CO3 form

acidification: H+ + HCO3- ⎯→ H2O + CO2
2 H+ + CO32- ⎯→ H2O + CO2
titration: Ag+ + Cl- ⎯→ AgCl
in the melt: 2 KHCO3 ⎯→ H2O + CO2 + K2CO3
4 KClO3 ⎯→ 3 KClO4 + KCl

d) M(KCl) = 74.55 g/mol; M(KClO3) = 122.50 g/mol; M(KClO4) = 138.50 g/mol;

M(CO2) = 44.01 g/mol; M(KHCO3) = 100.10 g/mol; M(K2CO3) = 138.20 g/mol;
M(H2O) = 18.01 g/mol.
Melting results in a decrease of mass of 0.05 g. (H2O and CO2 escape in the molar ratio
of 1:1) M(H2O)+ M(CO2) = 62.02 g/mol.
0.05 g correspond to 8.061·10-4 mol (H2O + CO2). They have formed from 0.0016122 mol
of (0.161 g of) KHCO3, thus 8.061·10-4 mol (0.111 g of) of K2CO3 have formed.
0.0188 L · 0.1 mol/L = 0.00188 mol of silver nitrate were consumed in the first titration.
This corresponds to 0.00188 mol of (0.140 g of) KCl.
Finally, the melting mixture contains 0.003305 mol of (0.246 g of) KCl.
The mass difference of the initial substance (1 g - 0.161 g - 0.140 g) = 0.699 g
corresponds to 0.0057061 mol of KClO3. That is 0.0042795 mol of (0.593 g of) KClO4 in
the melt. This mass corresponds to the mass difference as well (0.95 g - 0.111 g- 0.246
g) = 0.593 g.
initial substance: 0.161 g of KHCO3 0.140 g of KCl 0.699 gof KClO3 Σ = 1.00 g
melt: 0.111 g of K2CO3 0.246 g of KCl 0.593 g of KClO4 Σ = 0.95 g

Solution to problem 3-6

a) The concentration of ethylen diacetate is:
c((CH3COO)2CHCH3) = c0((CH3COO)2CHCH3) - 0,5·c(CH3COOH)
In order to check for a zero-order reaction, c - c0 is plotted against the time t in a diagram,
according to the rate law c = c0 - k0·t.
In order to check for a first-order reaction, c = c0 - k0·t is plotted against the time t in a
diagram, according to the rate law c = c0 ⋅ e − k1t .
t [min.] 0 240 660 1400 2093 3403 6369
t [s] 0 14400 39600 84000 125580 204180 382140
c (acid) [mol/L ] 0.0216 0.0457 0.06495 0.09395 0.1152 0.14475 0.17915
c(ethylen diacetate)[mol/L ] 0.0892 0.0772 0.0675 0.0530 0.0424 0.0276 0.0104
c0-c [mol/L ] 0 0.01205 0.0216750.036175 0.0468 0.061575 0.078775
ln (c0/c) 0 0.145 0.278 0.520 0.7437 1.172 2.147

65
 Answers round 3 test 1

Geschw indigkeitsgesetz 0. Ordnung Geschw indigkeitsgesetz 1. Ordnung

zero-order rate law first-order rate law

0,1 2,5
c(0)-c [mol/l]

0,08 2

ln{c(0)/c}
0,06 1,5
0,04 1
0,02 0,5
0 0
0 100000 200000 300000 400000 0 100000 200000 300000 400000
t [s] t [s]

Only the diagram for the first-order rate law corresponds to a straight line. Thus, the
reaction is a first-order or rather a pseudo first-order reaction (because of the high excess
of water).
b) zero-order reaction: heterogeously catalyzed reactions at surfaces; the absorbed
amount of substance is constant
first-order reactions: radioactive decay
c) The rate constants are calculated and the mean values are determined from the data of
the table above for ln(c0/c) and t by means of k1 = ln(t-1·c0/c):
t [s] 14400 39600 84000 125580 204180 382140
ln (c0/c) 0.145 0.278 0.520 0.7437 1.172 2.147
k1 [s-1] 1.0078·10 7.0299·10 6.1919·10 5.9224·10 5.7408·10 5.6175·10-6
-5 -6 -6 -6 -6

The first value at t = 14400 s is not considered for the calculation of the mean value,
because this value has a deviation that is too high. Thus, k 1 = 6.1·10-6 s-1, so that
t1/2 = k1-1·ln 2. This results in t1/2 = 113.6·103 s ≈ 31.56 h
d) CH3COOCH2CH3 + OH- ⎯→ CH3COO- + C2H5OH.
For the determination of the reaction turnover the reaction is stopped by the addition of
destilled water or rather strongly slowed down. Afterwards, the amount of still remaining
caustic sodium lye is quickly determined by titration.
e) After t = 25 min, 73 % of the ester were saponificated with c0 = 0.02 mol/l.
⇐ c(ester, 25 min) = c0 · (100-73)/100 = 5.4· 10-3 mol/L
The second-order rate law is valid for equal initial concentrations of the educts:
c-1 = k2 · t + c0-1.
k2 = 0.090 L/(mol·s) can be obtained by insertion.
f) The half lifet t1/2 results from the insertion of c = ½·c0 into the rate law:
(0.5·c0)-1 = k2 · t1/2 + c0-1 ⇐ t1/2 = (c0 · k2)-1 t1/2 = 555 s = 9.25 min
-1 -1 -1
(0.01·c0) = k2 · t99% + c0 ⇐ t99% = 99· (c0·k2) t99% = 54.9·103 s = 15.3 h.

Solution to problem 3-7

66
 Answers round 3 test 1

R ⋅T c( Ox )
a) E = E0 + ln
n ⋅ F c(Re d )
R ⋅ T c( Fe 3+ )
iron half cell: E = 0.77 V + ln
F c( Fe 2+ )
R ⋅T 1
halogen-halide half cell: E = E0 (halogen) + ln − 2
2 ⋅ F c( X ) /( 1 mol / L ) 2

c(Fe2+) = 0.01 mol/L - c(X-), c(Fe2+) = 1 mol/L - c(Fe3+).

Because of the standard redox potentials, the equilibrium is strongly on the left side ⇐
c(Fe3+) = 1 mol/L or rather c(Cl-) = 0.01 mol/L
E(Fe) = E(Cl) results in the following:
0,77 V + RT/F · ln [(1 mol/L)/c(Fe2+)] = 1,36 V + RT/F · ln (1/0.01)
⇐ c (Fe2+) = 10-12 mol/L pFe2+ = 12

The calculation for the system with bromide results in the following:
⇐ c (Fe2+) = 8.23 ·10 -8 mol/L pFe2+ = 7.08

Because of the standard potentials of the system with iodide, the equilibrium is strongly
on the right side:
c(Fe2+) = 0.01 mol/L - c(I-), c(Fe2+) = 1 mol/L - c(Fe3+).
0,77 V + RT/F · ln ((1 mol/L -c(Fe2+)/c(Fe2+)) = 0.54 V+ RT/F · ln (1/(0.01 mol/L - c(Fe2+)
⇐ c (Fe2+) = 9.99·10-3 mol/L pFe2+ = 2

b) These cations are hydrated; at simplest, they can be found as complexes: [Fe(H2O)6]2+ or
rather [Fe(H2O)6]3+.

c) The hydrated cations have an acidic effect. The different hydroxo-complexes have
different redox potentials. The redox potentials of the halogens as well depend very
strongly on the pH-value, because with an increasing pH, chlorine, bromine and iodine
undergo always more and more disproportionation reactions like e.g.
Cl2 + 2 OH- ⎯→ Cl- + ClO- + H2O.

d) 2 Fe(OH)3 + 3 Cl2 + 10 OH- ⎯→ 2 FeO42- + 6 Cl- + 8 H2O

4 K2FeO4 + 10 H2O ⎯→ 4 Fe(OH)3 + 8 K+ + 8 OH- + 3 O2.

Solution to problem 3-8

a) L-configuration (Fischer projection)
67
 Answers round 3 test 1

b) Every amino acid can have either a D- or an L-configuration. Thus, there are 250
stereoisomers of a peptide consisting of 50 molecules of alanine.

c) HS

s s R

COOH H2N COOH

s N
H
Cystein
cysteine
H2N COOH
H
proline The
Das S-atom
S-Atom has a higher
hat eine priority
höhere than
Priorität
isoleucine the O-atom of the COOH group
als die O-Atome der COOH-Gruppe

d) The peak at m/z = 89.1 can be attributed to

H O alanine. Concerning polypeptide fragments,
OH H 2 N notice that every formation of a peptide bond
H2N OH
results in a cleavage of water (M = 18.02 g/mol ).
O CH3
The addition of the molar mass of a second
75.07 gmol-1 89.09 gmol-1 amino acid (glycine or alanine) -18.02 g/mol -
glycine (Gly) alanine (Ala) (water) shows that there is only the mass peak
for the dipeptide ala-ala. By further trying out the
following peptides form. Their masses appear in the mass spectrum:

CH3 O O CH3 O
H H
N H2N N
H2N OH N OH
H
O CH3 H O CH3
alanine alanine glycine alanine alanine
molar mass: 160.17 gmol-1 molar mass: 217.22 gmol-1

H O CH3 O O H O CH3 O
H H H H
N N H2N N N
H2N N OH N N OH
H H H
O H O CH3 H O H O CH3
glycine glycine alanine alanine glycine glycine glycine alanine alanine
molar mass: 274.27 gmol-1 molar mass: 331.33 gmol-1

CH3 O H O CH3 O
H H H
N N N
H2N N N OH
H H
O H O H O CH3
alanine glycine glycine glycine alanine alanine
molar mass: 402.40 gmol-1

The remaining mass peaks can be obtained from the phenyl carbamoyls of the peptides:
Ph-NH-CO-NH-ala-COOH 208.2 g/mol
68
 Answers round 3 test 1

Ph-NH-CO-NH-ala-ala-COOH 279.3 g/mol

Ph-NH-CO-NH-gly-ala-ala-COOH 336.3 g/mol
Ph-NH-CO-NH-gly-gly-ala-ala-COOH 393.4 g/mol
Ph-NH-CO-NH-gly-gly-gyl-ala-ala-COOH 450.5 g/mol
Ph-NH-CO-NH-ala-gly-gly-gly-ala-ala-COOH 521.5 g/mol
e) X = NH2-ala-gly-gly-gly-ala-ala-COOH

Solution to problem 3-9

a)
CH3NH2 + H+ CH3NH3+

O
H +
CH3
H N H2O
H + CH3NH3+
H
+
O
O
H
O OH

+
O OH O
H O O
H
+ N
+
H H -H+ H H
H CH3
N N
H H CH3
CH3

OH +H+
O

O
H H
+ +
H3C N H 3C N H
H
+
H3C N H
H H O
H
+
OH O
H3C
N
-H+
H H =
H 3C N H3C N

H H O

69
 Answers round 3 test 1

b)
H3C H3C O H3C
N N N
strong base, + CO2CH3
e. g. NaOEt C MeO OMe

- H+ - OMe-
O B1 O O
C1/2
(2 diastereomers)
H3C H3C
N N

C
H3CO2C
O O
B2
C3/4 (2 diastereomers)

NaBH4

H3C
H3C N
HOOC CO2CH3
N CO2CH3 +
Dy

- H2O
O
OH
O C D1/2/3/4
(4 diastereomers)
H3C
N
(8 diastereomers)

E1/2/3/4/5/6/7/8
H3CO2C
OH
D5/6/7/8
(4 diastereomers)

c) Altogether 8 isomers can be obtained. There are two enantiomers of compound Bx- ;
compound Cy has four isomers (2 diastereomeric pairs of enantiomers). There are eight
isomers each of compound Dz and Ez (4 diastereomeric pairs of enantiomers). Thus, the
different types of isomerism are stereoisomers, strictly speaking enantiomers and
diastereomers.

70
 Answers round 3 test 1

Solution to problem 3-10

a) weakly acid: B, strongly acid (NMe2-group protonated): A
H3C CH3 H
N +
H3C N CH3
+ H+

CH3
CH3

H H H
+ + +
H3C N CH3 H3C N CH3 H3C N CH3
+ NO2+ H +
H C H
H NO2
+ + NO2 NO2
+
H3C N CH3 C C
ortho-
attack
CH3 CH3 CH3

H H H
CH3 + +
+
+ NO2+ H3C N CH3 H3C N CH3 H3C N CH3

+ +
C C
NO2 NO2 NO2
+
meta- H C H
H
attack
CH3 CH3 CH3

The ortho attack (concerning the dimethylammonia group) results in an extremely

unfavourable resonance structure in which the C-atom has a further positive formal
charge apart from the nitrogen atom that is already positively charged anyway. This
resonance structure is so unfavourable, that exclusively the meta product forms in a
strongly acid solution. Such a resonance structure does not appear in the formation of
such a product.

b), c)

NMe2 NMe2 NMe2

NO2 NH2 N

+ 3 Zn, + 6 HCl
- 3 ZnCl2, - 2 H2O
D E
Me Me Me

71
 Answers round 3 test 2

Answers round 3 test 2

Solution to problem 3-11

1) e 2) b 3) c 4) c 5) d 6) c 7) c 8)c

Solution problem 3-12

a)
H2C C CH2
2
sp sp sp2

b) sp2 ⎯→ sp3

c) [Fe(CO)5]: dsp3 (Fe(0) has 8 valence electrons, diamagnetic: only paired valence
electrons. It follows: 1 free 3d-orbital, free 4s-orbital, 3 free 4p-orbitals. These orbitals are
occupied with 5 electron pairs of the ligands.)
[Fe(Cp)2]: d2sp3 (Fe(+II), diamagnetic, has 2 free 3d-orbitals, free 4s-orbital as well as 3
free 4p-orbitals accepting 12 electrons from the ligands.)

d) A free electron pair needs more space than a single bond to a (H-) atom. Thus, with an
increasing number of free electron pairs at the atom X, the H-X-H bond angles become
smaller.
e) A: angles almost tetrahedral angles, sp3-hybridized at the O-atom because of 4 electron
pairs (2 free electron pairs + 2 single bonds)
B: angles approximately 120°, sp2-hybridization at the O-atom, because one of the free
electron pairs occupies a p-orbital to be in conjugation with the π-electron system of the
aromatic.
C: angle between 120° and 180° indicates a partial double-bond property of the two Si-
O-bonds; hybridization between sp2 and sp.

Solution to problem 3-13

a) The radius of the "inflated C-atom" is
154 pm composed of half the distance
between the units and the distance
between the bridgehead C-atoms and
the centre of the adamantane cage
154 pm (this distance is the same as the C-C
Z
bond length in the adamantane
backbone because of the symmetry
of the cage). RAD = ½ ·157.8 pm +
Z = zentroide 154.0 pm = 232.9 pm.
RAD = 232.9 pm
RAD
72
 Answers round 3 test 2

b) The edge length a of the tetrahedron is the same as the plane diagonal of the auxiliary
construction "cube". In this cube, the atomic radius RC is a quarter of the space diagonal.

2 RC

C 4 RC = (3b2)1/2, a = (2b2)1/2, b2 = a2/2

4 RC = (3a2/2)1/2, RC = (3a2/2)1/2 / 4
C a RC = (3 · (252.22 pm)2/2)1/2 / 4
C RC = 77.2263 pm

C edge length b

c) The ratio of the volumes of the adamantane unit to the C-atom is

VAD / VC = RAD3/RC3 = 27.429. The ratio of the masses (adamantane unit to C-atom) is MAD
/ MC = (10·MC + 12· MH) / MC = 11.007.
ρAD = MAD / VAD, ρC = MC / VC results from ρ = m/V
The following equation results from the transposition of the formula: ρAD = ρC · (MAD / MC) /
(VAD / VC)
If the ratios of the masses of the units as well as those of the unit volumes are inserted,
ρAD = 1.41 g/cm3 will be obtained.

d) The adamantane diamond should have a considerably lower hardness than the real
diamond because of the smaller "bond density" (C-C bonds per volume unit) between the
units.

e) isomorphic, isoelectronic

Solution to problem 3-14

a) Co(0) with 4 ligands and a Co-Co bond as well as Co(+I) with 5 ligands and Co(-I) with 4
ligands fulfill the 18-electrons rule; thus they have a noble-gas shell.
b) When [CoH(CO)4] dissociates, the oxidation number of the hydrido-ligand (-I) increases to
+ I (proton), that of the Co-central particle decreases from +I to -I. This is a redox
reaction.
c)
O O O O O
O O
CH2
C C C C C O C
H2 C
Co C O
H
H Co C O Co C O Co C O Co C O
H
C
C C C C
A B C D O E F
O O O O

73
 Answers round 3 test 2

d) The formation of D and the transformation to E are more likely to take place by the use of
a high excess of CO.
e)
OC
CO

OC Co
OC
CO
+
CO
OC Co H

CO oder

H H2
C CH3 C CH3
H3C H3C
- H
CH3 CH3

[Co(CO)4] [Co(CO)4]
f)

CHO

from 2-ethyl-1-butene from propene-3-ol

Solution to problem 3-15

− t / t½
a) N (t ) = N 0 2
b)
N0( 238 ) ⋅ 2 −t / T238 = N0( 235 ) ⋅ 2 −t / T235
N0( 235 )
2 −t / T238 +t / T235 =
N0( 238 )

t ⎛⎜ 1 − 1 ⎞ = log N0( 235 )

T T ⎟ 2 N0( 238 )
⎝ 235 238 ⎠

N N
log 2 0( 235 ) ln 0( 235 )
N0( 238 ) N0( 238 )
t= =
1 1
− ln 2 ⎛⎜ 1 − 1 ⎞
T235 T238
⎝ T235 T238 ⎟⎠
t = −5.9 ⋅ 109 a

c) Mass numbers only change by alpha-decay, namely by 4 units.

The decay path is the following:.
238 α 234 β 234 β 234
U ⎯⎯→ Th ⎯⎯→ Pa ⎯⎯→ U

74
 Answers round 3 test 2

d) The formation rate of 234U from 238U is the same as the decay rate of 234U: (all
intermediate steps cancel each other out!)
N 234 k 234 = N 238 k 238
ln 2 ln 2
N 234 = N 238
T234 T238
N 234
T234 = T238
N 238
T234 = 2.3 ⋅ 105 a
e) In the radioactive equilibrium, (t >> 0) the term in brackets equals 1. The time t results
from the following:
1 − 2 − t / T Rn = 0.99
t ⋅ ln 2
ln 0.01 = −
TRn
t = 25.4 d
f) In insoluble RaSO4, nascent radon would be trapped in the solid. In an aqueous system,
Rn is dissolved in water and evaporates slowly. It can be just simply drained by pumping.

Solution to problem 3-16

a) The standard formation enthalpies ∆H0f (at a temperature of 298.15 K and a pressure of
1013 mbar) of the elements are zero by definition. The formation enthalpy of oxygen,
however, is indicated at a different temperature.
b) The following formulas are valid for the different temperatures:
∆HR = 2 ∆Hf (SiO) + ∆Hf (O2) – 2 ∆Hf (SiO2)
∆SR = 2 S (SiO) + S (O2) – 2 S (SiO2)
∆GR = ∆HR – T ∆SR
The following values are obtained (notice the units, convert the temperatures into
Kelvinin!):
∆HR[kJ mol-1] ∆SR [J mol-1K-1] ∆GR [kJ mol-1]
T = 800 °C 1597,2 509,5 1050,4
T = 1000 °C 1590,9 503,7 949,7
T = 1200 °C 1584,1 498,5 849,8
The reaction is (strongly) endothermal at all temperatures (because ∆HR> 0)

c) The dimensionless, thermodynamic equilibrium constant Kth can be calculated for the
single temperatures according to ∆G = -RTlnKth
The equilibrium constant Kp for the given reaction results from the following:
Kp = Kp · p0∆n = Kth · p03 = Kth · (1013 mbar) 3

75
 Answers round 3 test 2

with ∆n = 3, because only gaseous species are taken into consideration.

The following numerical values are obtained:
Kth (800 °C) = 7.3 · 10-52 Kp (800 °C) = 7.6 · 10-43 mbar -3
-39
Kth (1000 °C) = 1.1 · 10 Kp (1000 °C) = 1.1 · 10-30 mbar -3
Kth (1200 °C) = 7.4 · 10-31 Kp (1200 °C) = 7.7 · 10-22 mbar -3

d) According to the reaction equation, p(SiO) = 2·p(O2). The following can be determined:
p(O2) = x and p(SiO) = 2x
2 2 3
Thus, Kp = p(SiO) · p(O2) = (2x) · x = 4 x
Kp
so: x = 3 = 9.9 ·10-6 mbar p(SiO) = 2x = 2.0 ·10-5 mbar
4
(note: the pressure is quite low, thus the reaction has to be carried out under high-
vacuum conditions, so that as few other gases as possible are present as impurities)

e) Gaseous SiO can be produced in a comproportionation reaction by heating a mixture of

solid Si and solid SiO2:

Si(solid) + SiO2 (solid) ⎯→ 2 SiO(g) O2 (g) (temperature > 1100 °C)

Gaseous SiO can as well produced by leading gaseous oxygen above solid Si at high
temperatures:
2 Si(solid)) + O2 (g) ⎯→ 2 SiO(g) (temperature > 1100 °C)

(this method, however, needs a very careful dosing of oxygen)

Solution to problem 3-17

a) iron is oxidized: Fe ⎯→ Fe2+ + 2 e-
copper is reduced : Cu2+ + 2 e- ⎯→ Cu .
Fe + Cu2+ ⎯→ Fe2+ + Cu

b) Nernst equation:

0.830V = ∆E 0 +
R ⋅T
⋅ ln
(
c Cu 2+ ) ∆E 0 = 0,830V −
R ⋅T
⋅ ln
0.5
= 0.780V
2⋅F (
c Fe 2+ ) 2⋅F 0.01
∆E0 = E0red − E0ox E0ox = E0red − ∆E 0 = 0.340V − 0.780V = −0.440V

c) By the addition of caustic soda solution the concentration of OH- -ions increases in the
solution. Fe(OH)2 is a poorly soluble salt, so that it starts to appear as a white precipitate

76
 Answers round 3 test 2

when the concentration of OH- -ions is increased. So, the concentration of Fe2+ is highly
decreased.
Fe2+(aq) + 2 OH-(aq) ⎯→ Fe(OH)2(s)

d) If air was added, iron(II)-hydroxide would immediately be oxidized to iron(III)-hydroxide.

4 Fe(OH)2 + O2 + 2 H2O ⎯→ 4 Fe(OH)3

R ⋅T
e) potential of the copper half cell: E red = +0.340 V + ⋅ ln 0.5 = 0.331 V
2⋅F
potential of the iron half cell: E ox = E red − ∆E = 0.331 V − 1.09 V = −0.759 V
concentration of Fe2+:
R ⋅T
− 0.759 V = −0.440 V + ⋅ ln[ c( Fe 2+ ) /( 1mol / L )]
2⋅F
c( Fe 2+ ) = 1.58 ⋅ 10 −11 molL−1
solubility product:
K L = c( Fe 2+ ) ⋅ c 2 ( OH − ) = 1.58 ⋅ 10 −11 molL−1 ⋅ ( 10 −2 molL−1 ) 2 = 1.58 ⋅ 10 −15 mol 3 L−3
pKL = 14.8

Solution to problem 3-18

EtO-
a) COOEt
COOEt

O
O Z1
-EtOH

EtOOC
EtOOC

COOEt

COOEt
Z3 O
Carboxylic acid that has formed O -EtO- Z2
reacts with ethanolate to form
ethanol and the carboxylic-acid O EtOOC

anion. The elimination reaction

(lactonic cleavage), that is an
equilibrium reaction, is shifted COOEt
to the side of the products.

HOOC

77
 Answers round 3 test 2

b)

O COO C2H5
COO C2H5

B A’
B
A´
HOOC EtOOC

OH OH
O COO C2H5
COO COO

bzw.
B’ O OH
OOC
EtOOC
C’
bzw.
B´
O C´

OH OH COO OH OH
C2H5
COO COO

D O bzw. OH
OOC
EtOOC

O
E

B: enantiomers: (R,R); (S,S) A´: trans-isomer remains

B´: enantiomers: (R,R); (S,S) C´: enantiomers: (R,S); (S,R)
D: enantiomers: (R,R); (S,S) E: enantiomers: (R,R); (S,S)

Solution to problem 3-19

a) In the first step, water is eliminated. The carbenium ion that has formed undergoes two
[1,2]-Wagner-Meerwein-rearrangements. In the last step, H+ is eliminated:
R R
R
CH3 CH3
+ H+ + rearrangement CH3
HO C +
- H2O C

A+ B+
rearrangement

R
R

H3C
H3C - H+ +
C
H
78
C+
 Answers round 3 test 2

b) The steric relaxation of the initial compound that contains two highly tightened four-
membered rings is the driving force for the rearrangement.

c)

*
*
*

OH

D has 8 stereoisomers.
4 pairs of enantiomers: (R,R,R) (S,S,S) (R,R,S) (S,S,R)
(R,S,S) (S,R,R) (R,S,R) (S,R,S)

Solution to problem 3-20

a) oxygen: 100% - 37.02% - 2.22% - 18.50% = 42.26%
total formula: C37.02/12H2.22/1O42.26/16N18.50/14 C3.08H2.22O2.64N1.32
C7H5O6N3
b) The two strong bands in the range around 1540cm-1 and 1350cm-1 indicate one or
several nitro groups (-NO2). Because there is no indication of other functional groups
with oxygen or nitrogen in the IR-spectrum, three nitro groups have to be found in the
molecule according to the total formula.
c) The 3H-singlet in the H-NMR indicates an isolated methyl group that is only weakly
screened. The further 2 H-atoms are still in the aromatic range, but they are badly
screened as well by the neighbouring NO2-groups and thus shifted to a lower level. The
two aromatic H-atoms are chemically equivalent. That indicates a symmetric
arrangement of the nitro groups and the methyl group at the aromatic ring.

CH3
O2N NO2

NO2

d) The compound is: TNT (2,4,6-trinitrotoluene). It is highly explosive.

79
 Answers round 4

Answers round 4

Solution to problem 4-1

a) pKL = 7.4 KL = 3.98·10-8 mol2/L2 Ln = (KL)1/2 = 2.00·10-4 mol/L
M(CuBr) = 143.45 g/mol Lm = 0.0286 g/L V(H2O) = 34.9 L

b) lgK1 = 6.18 K1 = 1513561 L/mol lgK2 = 4.69 K1 = 48978 L/mol

assumption: because of the high numerical values for K1 and K2, Cu+ can be found in an
ammoniacal solutions nearly exclusively as [Cu(NH3)2]+.
The following equations can be applied:
1) K1K2 = c([Cu(NH3)2]+)/ c(Cu+)c(NH3)2
2) KL = c([Cu(NH3)2]+)·c(Cu+), because c(Br-) = c([Cu(NH3)2]+)
3) c(NH3) = 0.1 mol/L - 2c([Cu(NH3)2]+)
The insertion of 2) and 3) into 1) results in the following equations:
c(NH3)2 - {0.05/(0.25 - K1K2KL)}·c(NH3) + 0.0025/(0.25 - K1K2KL) = 0
c(NH3) = 9.12·10-4 mol/L
Without any simplification the following equations can be applied:
1) K1K2 = c([Cu(NH3)2]+)/ c(Cu+)c(NH3)2 K1 = c([Cu(NH3)]+)/ c(Cu+)c([NH3)
2) KL = {c([Cu(NH3)2]+) + c([Cu(NH3)]+) + c(Cu+)}·c(Cu+)
3) c(NH3) = 0.1 mol/L - c([Cu(NH3)]+) - 2c([Cu(NH3)2]+)
Insertion and rearrangement lead to the following equation:
(K1K2 - 4·K12K22KL)·c(NH3)4 + (K1 - 0.2·K1K2 + 4·K12K2KL)·c(NH3)3 + (1 - 0.2·K1 + 0.01·K1K2 -
KLK1)·c(NH3)2 + (0.01·K1 - 0.2)·c(NH3) + 0.01 = 0 (term 1 = 0)
The insertion of the approximate result c(NH3) = 9.12·10-4 mol/L as an initial value results
in the following:
term 1 = 27.09
approximation: c(NH3)1st experiment = 9.20·10-4 mol/L term 1 = 16.61
c(NH3)second experiment = 9.30·10-4 mol/L term 1 = 2.84
c(NH3)third experiment = 9.32·10-4 mol/L term 1 = 0.0004
The following values can be obtained from c(NH3) = 9.32·10-4 mol/L:
c(Cu+) = 7.61·10-7 mol/L
c(Br-) = 0.052 mol/L. that is 7.50 g/L of CuBr.
133.3 ml of 0.1-molar NH3-solution are needed for dissolving 1 g of CuBr.

c) KL(cond) = c(Br−)2 = c(dissolved CuBr)2 KL(cond) = (0.0523 mol/L)2 = 2.74·10-3 mol2/L2

80
 Answers round 4

Solution to problem 4-2

a) C8H10 + 10,5 O2 ⎯→ 8 CO2 + 5 H2O.

b) According to a), the combustion enthalpy is:

∆Hc = 8 ∆Hf0(CO2) + 5 ∆Hf0(H2O) - ∆Hf0(p-Xylol)
∆Hf0(p-Xylol) = 8 ∆Hf°(CO2)+5 ∆Hf°(H2O)- ∆Hc
= 8·(-393.5 kJ/mol) + 5·(-285.8 kJ/mol) + 4551 kJ/mol
= -25.6 kJ/mol

c) The formation enthalpies of the gaseous xylenes at T = 500 K can be calculated from the
standard values according to the following scheme:
∆Hf(g.500 K) = ∆H0f + Cp(l)·(Ts – 298.15 K) + ∆Hdil+ Cp(g) ·(500 K – Ts).
Analogous to the upper equation. the following equation can be applied to the entropies
of the gaseous xylenes:
S(g,500 K) = S0 + Cp(l)·ln(Ts/298.15 K) + ∆Hdil/Ts + Cp(g) ·ln(500 K /Ts).

∆Hf(g. 500 K). [kJ/mol] S(g. 500 K). [JK-1mol-1]

o-Xylol: 48.4 426.9
m-Xylol 46.0 432.6
p-Xylol 46.8 425.3
The reaction enthalpy of the isomerization reaction can be calculated the following way :
ortho-xylene ⎯→ para-xylene
∆Hr(500 K) = 46.8 kJ/mol - 48.4 kJ/mol = -1.6 kJ/mol
∆Sr(500 K) = 425.3 JK-1mol-1 - 426.9 JK-1mol-1 = -1.6 JK-1mol-1

d) The reaction enthalpies for meta-Xylol ⎯→ para-Xylol can be obtained according to the
same way:
∆Hr(500 K) = 46.8 kJ/mol - 46.0 kJ/mol = 0.8 kJ/mol
∆Sr(500 K) = 425.3 JK mol - 432.6 JK mol = -7.3 JK-1mol-1
-1 -1 -1 -1

The following values are the changes in free enthalpy for the two reactions:
ortho-xylene → para-xylene ∆Gr(500 K) = ∆Hr(500 K) – 500 K·∆Sr(500 K) = -0.8 kJ/mol
meta-xylene → para-xylene ∆Gr(500 K) = ∆Hr(500 K) – 500 K·∆Sr(500 K) = 4.5 kJ/mol
The equilibrium constants for the isomerizations can be calculated from the upper values:
c( para )
K ( ortho → para ) = = e − ∆GR /( RT ) = 1.21
c( ortho )
c( para )
K ( meta → para ) = = e − ∆GR /( RT ) = 0.34
c( meta )
The composition of the equilibrium mixture at 500 K is:
para-xylene: 21 % meta-xylene: 62 % ortho-xylene: 17 %.

81
 Answers round 4

e) Because the product p-xylene is faster removed, the equilibrium is shifted to the side of p-
xylene. This is the result of the mass action law.

Solution to problem 4-3

a) Division of the total reaction into two parts:
AgI ⎯→ Ag + ½ I2 (K1) Ag + ½ I2 ⎯→ Ag + + I- (K2)
+ -
AgI ⎯→ Ag + I (KL) KL = K1·K2
K1 is calculated from thermodynamic data, K2 is calculated from electrochemical data:
K1 = e- ∆G(AgI)/ (R . 298 K)
∆G2 = (εAg+/Ag – εI2/I- ) . F K2 = e- ∆G2/ (R · 298 K)
. -17 2
KL = 8.56 10 (mol/L)
b) Application of van’t Hoff`s equation:
ln K(75) = ln K(25) - ∆rH1/R . (1/(348 K) – 1/(298 K))
mit ∆rH1 = ∆H(I-) + ∆H(Ag+) - ∆H(AgI) = 110.88 kJ/mol
K(75) = 5.31 . 10-14 (mol/L)2
Explanation with Le Chatellier: the solubilization process is endothermal.
c) ∆rG2(298) = ∆G(C2H4) + ∆G(HI) - ∆G(C2H5I) = 50.9 kJ/mol
∆rH2(298) = ∆H(C2H4) + ∆H(HI) - ∆H(C2H5I) = 87.1 kJ/mol
Van’t Hoff`s equation (reaction at 600 K):
ln K(600) = -∆rG2(298)/ (R . 298K) – ∆rH2 / R (1/(600 K) – 1/(298 K)) = 0.06
HI has to be removed to prevent a back reaction.
d) ε = εo + RT/F ln [Ag+] = εo + RT/F ln KL – RT/F ln [I-]
e) It is a first-order reaction. c(I-) = c0I-)· e-kt c0(I-) is as well the initial concentration of C2H5I.
f) The insertion results in the following:
ε = εKalomel – εo – RT/F ln KL + RT/F ln [I]0 – RT/F kt
If emf is plotted against t. the gradient is k = 1.6 . 10-5 s-1
The axis intercept results in c0(I-) = 0.077 M
m(C2H5I) = 0.25 L . 0.077 mol/L . M(C2H5I) = 3 g?

Solution to problem 4-4

a) 100µm² of lipid bilayer contain N DOPC = 2 ⋅ 100µm 2 0,64nm 2 = 3.13 ⋅ 108 DOPC-
molecules. (it must be factor 2 because it is a bilayer). Thus, for every DiO-molecule,
there are A = nDOPC nDiO = 3.13 ⋅ 108 50 = 6.25 ⋅ 10 6 DOPC –molecules.
⇐ nDiO = nDOPC A = VDOPC ⋅ c DOPC A = 50 µl ⋅ 10mM 6,25 ⋅ 10 6 = 8.0 ⋅ 10 −14 mol .
⎛ µg ⎞
VDiO = n DiO c DiO = n DiO ( c DiO
w
M DiO ) = 8,0 ⋅ 10−14 mol ⎜ 10 882g ⎟ = 7,1µl
⎝ l ⎠

82
 Answers round 4

b) The following formula is needed for the calculation of the

'
2 ∆t
distance: ( ∆x ) T
= 2fD∆t ' = 2 ⋅ 2 ⋅ 6 ⋅ 10−8 cm 2 s ⋅ ( 25 + 10 ) ms = 0,84µm 2 (f = 2)

∆t '
Thus, ∆x ∆t ' = (∆x )2 T
= 0.91µm (mit ∆t ' = 35ms ).

c) The lenght of the distance is:

' '
2 4 ∆t 2 ∆t
∆x 4 ∆t ' = ( ∆x ) T
= 2fD ⋅ 4∆t ' = 2 ( ∆x ) T
= 1,82µm .

d) The time needed by the lipid molecule is:

'
2 ∆t
( ∆x ) (10µm )
2

∆t ' = T = = 4,2s
2fD ( 2 ⋅ 2 ⋅ 6 ⋅10 −8
cm 2 s )

e) The concentration of the molecules decreases exponentionally with time, that means
according to first order:

N (t ) = N 0 ⋅ 2 = N 0 ⋅ (e ln 2 )
−t τ 1 2 −t τ 1 2 −t ⋅ln 2 τ 1 2
= N0 ⋅ e .

t ln (N (t ) N0 ) ln 0.1
Thus, it follows: N (t ) N 0 = 0,1 : =− =− = 3.3
τ1 2 ln 2 ln 2
After t = 3.3 ⋅ τ 1 2 = 3.3 ⋅ 75ms = 250ms of exposure time, 10% of all molecules are still
fluorescent, that means after maximally 250ms 10ms = 25 photos.

Solution to problem 4-5

a) (i) 3 ions A2+ to the right, 2 ions B3+ to the left; (ii) one ion A2+ to the right and one ion O2–
to the right; (iii) two ions B3+ and three ions O2– to the left.
b) (i) 3 A2+ move to the right, these are three formula units AB2O4, two B3+ move to the left,
this is one formular unit AB2O4. The wire would divide the AB2O4-layer in the ratio of 3:1
and would be on the right side at 75%; (ii) AB2O4 forms only at the right side of the wire,
the wire would be situated extremely on the the left side; (iii) AB2O4 only forms on the left
side of the wire, the wire would be situated extremely on the right side.
c) 1. Several of the basic mechanisms can take place simultaneously ; 2. the metals can
possibly change their oxidation numbers because of diffusion – electrons and metal will
diffuse separately then.
83
 Answers round 4

Solution to problem 4-6

A = P4O10 B = PH3 C = PCl3 D = (EtO)3PO
E = POCl3 F = (EtO)2POH O H = PPh3
I = Ph3PO J = NaPPh2 G= K = Ph2PH
(EtO)2P H

Solution to problem 4-7

a)
Li

C8H8 C4H9Li
M = 104.15 g/mol M = 64.06 g/mol

styrene sec-butyllithium

c( Styrol )
= 40
c(sec − Butyllitium

⇒ degree of polymerization Pn at a turnover of 25%: Pn = 10

at a turnover of 50%: Pn = 20
at a turnover of 75%: Pn = 30

b) vx is the volume fraction of the umpteenth sampling

0.25 . v1 = 0.5 . v2 = 0.75 . v3 mit v3 = 1 – v1 – v2
. . .
0.25 v1 = 0.75 – 0.75 v1 – 0.75 v2 v1 = 0.75 – 0.75 . v2
0.5 . v2 = 0.75 – 0.75 . v1 – 0.75 . v2 0.75 . v1 = 0.75 – 1.25 . v2

⇐ v1 = 0.55 v2 = 0.27 v3 = 0.18

Polymer analytics
c) difference of the molar masses of two neighbouring polymer molecules:
e.g. 994 g/mol – 890 g/mol = 104 g/mol
⇐ sterene is the most common monomer with the calculated molecular weight
calculation of the degree of polymerization e.g. for the species at 1098 g/mol:
1098 g/mol : 104 g/mol = 10,56
species [g/mol] 890 994 1098 1202 1306
Pn 8 9 10 11 12

84
 Answers round 4

Concerning a degree of polymerization of 10, there are the

H
following masses per mole for the initial- and final groups of
the polymer:
1098 g/mol – 1040 g/mol = 58 g/mol
only one proton remains as a final group of the polymer while
sec-butyllithum is an initiator-entity grouping (sec-Butyl 8-12
~ 57 g/mol).
∞ ∞ ∞
d) ∑ hi ⋅ M i = 1098 g/mol;
i =1
∑ hi = 1;
i =1
∑h ⋅M
i =1
i i
2
= 1213175.2 g2/mol2

Mn = 1098 g/mol; Mw = 1104.9 g/mol ⇐ Mw / Mn = 1.00

the quotient for beef insulin is: Mn = Mw = 5733.5 g/mol ⇐ Mw / Mn = 1
Mw / Mn is a measure for the range of a molecular-weight distribution.
(The sample is monodisperse for Mw / Mn = 1; usually, the Mw / Mn – values of radical
polymerizations are around 2, while values around 1.01 are not seldom for anionic
polymerization)
e) Large molecules are more likely to contain one or more rather heavy isotopes of carbon
(13C) or hydrogen (2H). There are as well heavy polymer molecules in the mixture.
(The shifting of the lightest peak from 1098 to almost 1099 is caused by the proton mass
that is a little bit higher than 1 g/mol, because M(1H) = 1,008 g/mol and M(12C) = 12,000
g/mol.)
f)
H2 H2
C O C O
C H C
H2 H2
n n O
A = polyethylene oxid (or B
polyethylene glycol, polyoxirane ...)

O
e.g.
brush-like structure
O

O O

O O

CH2 CH2

H2C H2C
C

m-1

Hydrogen transports from the polyethylene-oxide chain to the growing polymer radical are
possible by the radical course of the reaction. In this way, undesirable crosslinkings may
form.

85
 Answers round 4

Solution to problem 4-8

a) It is acetone

b)
mass number 15 27 42 43 58
+
+ + + +
species CH3 CH3 – C CH3 – C – CH3 , CH3 – C = O O
+
CH2 – C = O
H3C CH3
There is only one signal in the 1H- NMR-spectrum, because all hydrogen atoms are
chemically equivalent in A.

c)
OH
O O
+ +2H

OH
The systematic name of pinacolone is 2,3-dimethylbutane-2,3-diol.

d) An oxonium ion forms by the protonation of a OH-group. This ion transforms into a
carbenium-ion and water is cleaved. By 1,2-shifting of an alkyl-anion, a resonance-
stabilized carbenium-ion forms that transforms into pinacolone by release of a proton.
H
+
O H - H2O
H3C H3C +
C +
C
OH OH OH

Oxonium-Ion Carbenium-Ion - H+

Pinakolon
O
e) butanone

Solution problem 4-9

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 Answers round 4

a) cyclic, conjugated, planar, 4n+2 electrons

b)
H H+ H+ H+ H+
N N N N N
- -
CH HC
- -
CH CH

c) first variant: furan < pyrrole < thiophene

(decreasing electronegativity of the heteroatom favours the delocation of a free electron
pair)
second variant: thiophene < furan < pyrrole
(a different orbital size of the S-atom could be unfavourable for the conjugation)

d)
H H H H+
N N H N H H
E+ an C2/C5 N
+
HC
E E E
+
CH

H H H+
N N N
E+ an C3/C4 +
CH

E E
H H

Î electrophilic aromatic substitution at C2/C5 is favoured, because of a higher

delocation of the positive charge
e)
Cl O O O
S CH3 H
A N N+
B H3C
C -
O-

H3C S CH3 H
N
D
E N CH3
CH3

CH3

87
 Answers round 4

Solution problem 4-10

a) - d)
Cl
Cl2
cataly. FeCl3
NaNH2

- HCl
aryne / benzyne
(3)
chlorbenzene (2)
(1) (reactive intermediate)

Diels-Alder-
eeaction +

anthracene (4)

(5) trypticene
e) 2-chlorobutadiene reacts with itself in a Diels-Alder reaction. Altogether four different
products can form:
Cl Cl

Cl Cl

Cl
Cl

Cl
Cl

Cl
Cl

Cl Cl

Cl
Cl Cl
Cl

f) and g)

AlCl3

CO2Me CO2Me

CO2Me

The two compounds are enantiomers.

88
 IChO: Theoretical test

Part 3

36. International Chemistry Olymmpiad

Kiel, Friday 29. July

Theoretical Test
Problem 1: Thermodynamics
For his 18th birthday party in February Peter plans to turn a hut in the garden of his parents
into a swimming pool with an artificial beach. In order to estimate the costs for heating the
water and the house, Peter obtains the data for the natural gas composition and its price.
1.1 Write down the chemical equations for the complete combustion of the main
components of natural gas, methane and ethane, given in Table 1. Assume that
nitrogen is inert under the chosen conditions.
Calculate the reaction enthalpy, the reaction entropy, and the Gibbs energy under
standard conditions (1.013·105 Pa, 25.0°C) for the combustion of methane and ethane
according to the equations above assuming that all products are gaseous.

The thermodynamic properties and the composition of natural gas can be found in Table 1.

1.2 The density of natural gas is 0.740 g L-1 (1.013·105 Pa, 25.0°C) specified by PUC, the
public utility company.
a) Calculate the amount of methane and ethane (in moles) in 1.00 m3 of natural gas
(natural gas, methane, and ethane are not ideal gases!).
b) Calculate the combustion energy which is released as thermal energy during the
burning of 1.00 m3 of natural gas under standard conditions assuming that all products
are gaseous. (If you do not have the amount from 1.2a) assume that 1.00 m3 natural
gas corresponds to 40.00 mol natural gas.)
According to the PUC the combustion energy will be 9.981 kWh per m3 of natural gas if
all products are gaseous. How large is the deviation (in percent) from the value you
obtained in b)?

The swimming pool inside the house is 3.00 m wide, 5.00 m long and 1.50 m deep (below
the floor). The tap water temperature is 8.00°C and the air temperature in the house
(dimensions given in the figure below) is 10.0°C. Assume a water density of ρ = 1.00 kg L-1
and air behaving like an ideal gas.

89
 IChO: Theoretical test

1.3 Calculate the energy (in MJ) which is required to heat the water in the pool to 22.0°C
and the energy which is required to heat the initial amount of air (21.0% of O2, 79.0%
of N2) to 30.0°C at a pressure of 1.013·105 Pa.

In February, the outside temperature is about 5°C in Northern Germany. Since the concrete
walls and the roof of the house are relatively thin (20.0 cm) there will be a loss of energy.
This energy is released to the surroundings (heat loss released to water and/or ground
should be neglected). The heat conductivity of the wall and roof is 1.00 W K-1 m-1.
1.4 Calculate the energy (in MJ) which is needed to maintain the temperature inside the
house at 30.0°C during the party (12 hours).

1.00 m3 of natural gas as delivered by PUC costs 0.40 € and 1.00 kWh of electricity costs
0.137 €. The rent for the equipment for gas heating will cost him about 150.00 € while the
corresponding electrical heaters will only cost 100.00 €.
1.5 What is the total energy (in MJ) needed for Peter’s “winter swimming pool” calculated
in 1.3 and 1.4? How much natural gas will he need, if the gas heater has an efficiency
of 90.0%?
What are the different costs for the use of either natural gas or electricity? Use the
values given by PUC for your calculations and assume 100% efficiency for the electric
heater.
Table 1: Composition of natural gas
Chemical
mol fraction x ∆fH0·( kJ mol-1)-1 S0·(J mol-1 K-1)-1 Cp0·(J mol-1 K-1)-1
Substance

CO2 (g) 0.0024 -393.5 213.8 37.1

N2 (g) 0.0134 0.0 191.6 29.1

CH4 (g) 0.9732 -74.6 186.3 35.7

C2H6 (g) 0.0110 -84.0 229.2 52.5

H2O (l) - -285.8 70.0 75.3

H2O (g) - -241.8 188.8 33.6

O2 (g) - 0.0 205.2 29.4

90
 IChO: Theoretical test

Equation:
J = E · (A · ∆t)-1 = λwall · ∆T · d -1
J energy flow E along a temperature gradient (wall direction z) per area A and time ∆t
d wall thickness
λwall heat conductivity
∆T difference in temperature between the inside and the outside of the house

Problem 2: Kinetics at catalyst surfaces

Apart from other compounds the exhaust gases of an Otto engine are the main pollutants
carbon monoxide, nitrogen monoxide and uncombusted hydrocarbons, as, for example,
octane. To minimize them they are converted to carbon dioxide, nitrogen and water in a
regulated three-way catalytic converter.
2.1 Complete the chemical reaction equations for the reactions of the main pollutants in
the catalyst.

To remove the main pollutants from the exhaust gas of

an Otto engine optimally, the -value is determined by
an electro-chemical element, the so called lambda
probe. It is located in the exhaust gas stream between
engine and the three-way catalytic converter.
The lambda value is defined as
amount of air at the inlet
λ= .
amount of air necessary for complete combustion

w: λ-window
y: conversion efficiency (%)
z: Hydrocarbons
2.2 Decide the questions on the answer sheet concerning the λ probe.

The adsorption of gas molecules on a solid surface can be described in a simple model by
using the Langmuir isotherm:
K⋅p
θ=
1+ K ⋅ p
where θ is the fraction of surface sites that are occupied by the gas molecules, p is the gas
pressure and K is a constant.
The adsorption of a gas at 25 °C may be described by using the Langmuir isotherm with
K = 0.85 kPa-1.
2.3 a)Determine the surface coverage θ at a pressure of 0.65 kPa.
2.3 b)Determine the pressure p at which 15 % of the surface is covered.

91
 IChO: Theoretical test

2.3 c) The rate r of the decomposition of gas molecules at a solid surface depends on the
surface coverage (reverse reaction neglected): r = k· θ
Give the order of the decomposition reaction at low and at high gas pressures
assuming the validity of the Langmuir isotherm given above (products to be
neglected).
2.3 d) Data for the adsorption of another gas on a metal surface (at 25°C)

3000

2500

2000 x axis: p · (Pa)-1

y axis: p·Va-1 · (Pa cm-3)-1
1500
y axis Va is the gas volume that has
1000 been adsorbed.

500

0
0 200 400 600 800 1000 1200
x axis

If the Langmuir isotherm can be applied, determine the gas volume Va,max needed for
a complete coverage of the metal surface and the product K·Va,max.
Hint: Set θ = Va / Va,max .

Assume that the catalytic oxidation of CO on a Pd surface with equal surface sites proceeds
in the following way:
In a first step adsorbed CO and adsorbed O2 form adsorbed CO2 in a fast equilibrium,
k1
CO (ads.) + 0.5 O2 (ads.) CO2 (ads.)
k -1

In a slow second step, CO2 is then desorbed from the surface:

CO2 (ads.) ⎯⎯→

k2
CO2 (g)
2.4 Derive the formula for the reaction rate of the CO2(g) - formation as a function of the
partial pressures of the reaction components.
Hint: Use the Langmuir isotherm with the proper number of gas components
K i ⋅ pi
θ(i) = j: relevant gas components
1+ ∑K
j
j ⋅ pj

92
 IChO: Theoretical test

Problem 3: Monovalent alkaline earth compounds?

In the past there have been several reports on compounds of monovalent calcium. Until
recently the nature of these “compounds” was not known but they are still of great interest to
solid state chemists.
Attempts to reduce CaCl2 to CaCl have been made with
(a) Calcium (b) Hydrogen (c) Carbon
3.1 Give the corresponding reaction equations that could potentially lead to the formation
of CaCl.

After an attempt to reduce CaCl2 with the stoichiometric 1:1 molar amount of Ca one obtains
an inhomogeneous grey substance. A closer look under the microscope reveals silvery
metallic particles and colorless crystals.
3.2 What substance are the metallic particles and the colorless crystals?

When CaCl2 is attempted to be reduced with elemental hydrogen a white product forms.
Elemental analysis shows that the sample contains 52.36 m/m% of calcium and 46.32
m/m% of chlorine.
3.3 Determine the empirical formula of the compound formed!

When CaCl2 is attempted to be reduced with elemental carbon a red crystalline product
forms. The molar ratio of Ca and Cl determined by elemental analysis is n(Ca):n(Cl)=1.5:1.
During the hydrolysis of the red crystalline substance the same gas is evolved as during the
hydrolysis of Mg2C3.
3.4 a) Show the two acyclic constitutional isomers of the gas that is formed by
hydrolysis.
b) What compound is formed by the reaction of CaCl2 with carbon?
(Provided that monovalent calcium does not exist.)

As none of these attempts lead to the formation of CaCl more consideration has to be given
as to the hypothetical structure of CaCl. One can assume that CaCl is likely to crystallize in a
simple crystal structure.
It is the radius ratio of cation r(Mm+) and anion r(Xx-) of salts that often determines the crystal
structure of a particular compound as shown for MX compounds in the table below.

Coordination Surrounding of Radius ratio Structure type estimated

number of M X rM//rX ∆LH0 for CaCl
3 Triangular 0.155-0.225 BN - 663.8 kJ mol-1
4 Tetrahedral 0.225-0.414 ZnS - 704.8 kJ mol-1
6 Octahedral 0.414-0.732 NaCl - 751.9 kJ mol-1
8 Cubic 0.732-1.000 CsCl - 758.4 kJ mol-1

93
 IChO: Theoretical test

∆LH0(CaCl) is defined for the reaction Ca+(g) + Cl-(g) ⎯→ CaCl(s)

3.5a) What type of structure is CaCl likely to have?
[r(Ca+) ≈ 120 pm (estimated), r(Cl-) ≈167 pm)]

Not only the lattice energy ∆LH0 for CaCl is important for the decision whether CaCl is
thermodynamically stable or not. In order to decide whether it is stable to decompositon into
its elements, the standard enthalpy of formation ∆fH0 of CaCl has to be known.
3.5b) Calculate the value of ∆fH0 (CaCl) with the aid of a Born-Haber-cycle.

heat of fusion ∆fusionH0(Ca) 9.3 kJ mol-1

ionization enthalpy ∆1. IEH(Ca) Ca ⎯→ Ca+ 589.7 kJ mol-1
ionization enthalpy ∆2. IE H(Ca) Ca+ ⎯→ Ca2+ 1145.0 kJ mol-1
heat of vaporization ∆ vap H0(Ca) 150.0 kJ mol-1
dissociation energy ∆dissH(Cl2) Cl2 ⎯→ 2 Cl 240.0 kJ mol-1
enthalpy of format. ∆fH0(CaCl2) -796.0 kJ mol-1
electron affinity ∆EAH(Cl) Cl + e- ⎯→ Cl- - 349.0 kJ mol-1

To decide whether CaCl is thermodynamically stable to disproportionation into Ca and CaCl2

the standard enthalpy of this process has to be calculated. (The change of the entropy ∆S is
very small in this case, so its influence is negligible.)
3.6 Does the disproportionation of CaCl take place from a thermodynamic point of view?
Base your decision on a calculation!

Problem 4: Determining atomic masses

The reaction of the element X with hydrogen leads to a class of compounds that is
analogous to hydrocarbons. 5.000 g of X form 5.628 g of a molar 2:1 mixture of the
stoichiometric X-analogues of methane and ethane, respectively.
4.1 Determine the molar mass of X from this information. Give the chemical symbol of X,
and the 3D-structure of the two products.

The following more complex case is of great historical interest.

The mineral Argyrodite is a stoichiometric compound that contains silver (oxidation state +1),
sulphur (oxidation state -2) and an unknown element Y (oxidation state +4). The ratio
between the masses of silver and Y in Argyrodite is m(Ag) : m(Y) = 11.88 : 1. Y forms a
reddish brown lower sulfide (oxidation state of Y is +2) and a higher white sulfide (oxidation
state of Y is +4). The coloured lower sulfide is the sublimate obtained by heating Argyrodite
in a flow of hydrogen. The residues are Ag2S and H2S. To convert 10.0 g of Argyrodite
completely, 0.295 L of hydrogen are needed at 400 K and 100 kPa.
4.2 Determine the molar mass of Y from this information. Give the chemical symbol of Y,
and the empirical formula of Argyrodite.
The atomic masses are correlated with spectroscopic properties.
94
 IChO: Theoretical test

To determine the vibrational frequency ν~ expressed in wave numbers of chemical bonds in

IR spectra chemists use Hooke's law which focuses on the frequency of the vibration
(attention to units!):
1 k
ν~ = ⋅
2π c µ
ν~ vibrational frequency of the bond, in wavenumbers (cm-1)
c speed of light
k force constant, indicating the strength of the bond (N m-1= kg s-2)
3m( A)m(B)
µ reduced mass in AB4, which is given by µ =
3m( A) + 4m(B)
m(A), m(B) the masses of the two bond atoms

The vibrational frequency of the C-H bond of methane is known to be 3030.00 cm-1. The
vibrational frequency of the Z-analogue of methane is known to be 2938.45 cm-1. The bond
enthalpy of a C-H bond in methane is 438.4 kJ mol-1. The bond enthalpy of a Z-H bond in the
Z-analogue of methane is known to be 450.2 kJ mol-1.
4.3 Determine the force constant k of a C-H bond using Hooke's law.
Estimate the force constant k of a Z-H bond, assuming that there is a linear
proportionality between force constant and bond enthalpy.
Determine the atomic mass of Z from this information.
Give the chemical symbol of Z.

Problem 5: Biochemistry with Thermodynamics

Structure of ATP4 –
NH2

N
N

N
O O O N

-
O P O P O P O
O

O -
O -
O - H H

H H
OH OH

Shifting chemical equilibria with ATP:

Animals use free energy from the oxidation of their food to maintain concentrations of ATP,
ADP, and phosphate far from equilibrium. In red blood cells the following concentrations
have been measured:

95
 IChO: Theoretical test

c(ATP4-) = 2.25 mmol L-1 c(ADP3-) = 0.25 mmol L-1

c(HPO42-) = 1.65 mmol L-1
Free energy stored in ATP can be released according to the following reaction:
⎯→
ATP4- + H2O ←⎯ ADP3- + HPO42- + H+ ∆G°’= -30.5 kJ mol-1 (1)
As the pH is close to 7 in most living cells, biochemists use G°’ instead of G°. The
standard state of G°’ is defined as having a constant pH of 7. In equations with G°’ and K’
for reactions at pH=7 the concentration of H+ is therefore omitted. Standard concentration is
1 mol L-1.
5.1 Calculate the actual G’ of reaction (1) in the red blood cell at 25°C and pH = 7.

In living cells many so-called “anabolic” reactions take place, which are at first sight
thermodynamically unfavourable because of a positive G. The phosphorylation of glucose
is an example:
⎯→
glucose + HPO42- ←⎯ glucose 6-phosphate2- + H2O ∆G°’= +13.8 kJ mol-1 (2)

5.2 Calculate first the equilibrium constant K' of reaction (2) and then the ratio c(glucose 6-
phosphate) / c(glucose) in the red blood cell in chemical equilibrium at 25°C and pH =
7.

To shift the equilibrium to a higher concentration of glucose 6-phosphate, reaction (2) is

coupled with hydrolysis of ATP:
hexokinase
4- ⎯→
glucose + ATP ←⎯ glucose 6-phosphate2- + ADP3- + H+ (3)
5.3 Calculate ∆G°’ and K’ of reaction (3). What is now the ratio
c(glucose 6-phosphate) / c(glucose) in the red blood cell in chemical equilibrium at
25°C and pH = 7?

ATP synthesis:
An adult person ingests about 8000 kJ of energy (∆G’) per day with the food.
5.4 a) What will be the mass of ATP that is produced per day if half of this energy is used
for ATP synthesis? Assume a ∆G’ of -52 kJ mol -1 for reaction (1), and a molecular
weight of 503 g mol-1 for ATP.
b) What mass of ATP does the human body contain on average if the mean lifetime of
an ATP molecule until its hydrolysis is 1 min?
c) What happens to the rest of the free energy, which is not used for ATP synthesis?
Mark on the answer sheet.

In animals the energy obtained by the oxidation of food is used to pump protons out of
specialized membrane vesicles, the mitochondria. ATP-synthase, an enzyme, will allow
protons to re-enter the mitochondria if ATP is simultaneously synthesized from ADP and
phosphate.

96
 IChO: Theoretical test

5.5 a) How many protons (H+) are in a spherical mitochondrium with a diameter of 1 µm at
pH = 7?
b) How many protons have to enter into each of the 1000 mitochondria of a liver cell via
the ATP-synthase to allow the production of a mass of 0.2 fg of ATP per cell?
Assume that 3 protons have to enter for the synthesis of 1 molecule of ATP.

Problem 6: Diels-Alder Reactions

The Diels-Alder reaction, a concerted [4+2]-cycloaddition between a diene and an olefin to
yield a cyclohexene, was discovered in 1928 here in Kiel. Prof. Otto Diels and his coworker
Kurt Alder mixed p-benzoquinone with an excess of cyclopentadiene and obtained the
following result:
O O

20°C
20 °C
+ [A]

C11H10O2
O O
B
6.1 Draw the structure of A (without stereochemical information).

The Diels-Alder reaction is a concerted, one-step reaction that proceeds with high
stereospecificity. For example, only a single stereoisomer C is formed in the following
reaction

H CN
CN CN H
+ =
CN CN H
CN
CN H CN
C
Not formed

If you use the E-isomer of the alkene instead, you will obtain two other stereoisomers D1
and D2.
6.2 Give the structures of D1 and D2.

Accordingly, in the original reaction (formation of B from cyclopentadiene and

benzoquinone) Diels and Alder found only one of the following six conceivable
stereoisomers of B (see next page).

97
 IChO: Theoretical test

Hints:
- keep the stereospecific formation of C in mind
and CN
- the sterically less hindered isomer forms. CN
C

H O O O
H H H H H

H H H H H H
O O O
1 2 3

H O O O
H H H H H

H H H H H H
O O O
4 5 6

6.3 Which single isomer of the six stereoisomers 1-6 of B shown above did they isolate?

After prolonged heating (15h, 120°C) of the originally isolated stereoisomer B (melting point
mp: 157°C), Diels and Alder obtained two new stereoisomers E (mp: 153°C) and F (mp:
163°C). Equilibration of B with a catalytic amount of a strong base at 25°C gave a further
stereoisomer G (mp: 184°C).

B E + F
10% 20% 70%

B G
60% 40%
6.4 Decide the questions on the answer sheet concerning the Diels-Alder reaction.
Hint: You do not need to know, which of the six stereoisomers 1 - 6 (shown above)
corresponds to either E, F or G in order to anwer this question.

The Diels-Alder reaction plays also an important role in the following reaction sequence.

98
 IChO: Theoretical test

OMe

O CO2Me strong base

strong base strong base
strong base OMe
+ I K L
MeO CO2Me ∆
C12H16O5 C11H12O4

∆ OMe

- CO2 CO2Me
- MeOH

6.5 Draw the structures for I, K and L.

Hints: - K has only one methyl group.
- L is the Diels-Alder adduct of K and the alkene shown.

Problem 7: Stereochemistry in Drugs

The Cahn-Ingold-Prelog rules are used to specify the stereochemistry of molecules.
7.1 Order the groups on the answer sheet according to their priority in the Cahn-Ingold-
Prelog (CIP)-system.

Pseudoephedrine 1 is a constituent in many common drugs against colds, e.g. in nasal

sprays.

7.2 Mark the stereocenters in 1 with an * on the answer sheet.

Order the substituents on each stereocenter in 1 according to their priority and
determine their absolute configuration (R or S).
7.3 Draw a Newman or a sawhorse representation of 1.
Draw a Fischer representation of 1.

Treatment of 1 with acidic permanganate solutions under mild conditions yields the stimulant
Methcathinone 2:

99
 IChO: Theoretical test

7.4 Draw the stereochemically correct structure of 2 and a balanced redox equation of the
reaction. Indicate in your equation the particular oxidation number on all atoms which
undergo a change in their formal oxidation numbers.

The treatment of 2 with LiAlH4 results exclusively in compound 3, which differs from 1 in its
melting point.

7.5 a) Draw the stereochemically correct structure of 3.

7.5 b) Decide the statements on the answer sheet concerning isomers.
7.5 c) Draw a structural model to rationalize the exclusive formation of 3 from 2.

Problem 8: Colloids
The combination of an inorganic and an organic component on a nanometer scale yields
materials with excellent properties. Thus the synthesis of hybrid nanoparticles is of interest.
(T = 298.15 K throughout whole problem)

Solution A is an aqueous solution of CaCl2 with a concentration of 1.780 g L-1.

Solution B is an aqueous solution of Na2CO3 with a concentration of 1.700 g L-1.
pKa1(H2CO3) = 6.37 pKa2(HCO3-) = 10.33
8.1 Calculate the pH of solution B using reasonable assumptions.

100 mL of solution A and 100 mL of solution B are mixed to form solution C. Solution C is
adjusted to pH 10. A precipitate forms.
Ksp(Ca(OH)2) = 6.46·10-6 mol3 L-3 Ksp(CaCO3) = 3.31·10-9 mol2 L-2

8.2 Show by calculation for each of the compounds Ca(OH)2 and CaCO3 whether it can be
found in the precipitate or not.

In a similar experiment 100 mL of solution A additionally contain 2 g of a copolymer

consisting of two water soluble blocks: a poly(ethylene oxide) block and a poly(acrylic acid)
block:

H
H
COOH
C O
C
H C
H 68 C H
H 8 H
H H

100
 IChO: Theoretical test

The polymer does not undergo any chemical reaction (except protolysis of the acid) and yet
has a strong effect: after mixing of the two solutions (A+B) no precipitate can be observed.
Small calcium carbonate particles with the polymer chains attached to their surface form.
The attached polymers prevent further crystal growth and the hybrid particles remain in
solution.
8.3. Circle the block of the polymer (on the answer sheet) that attaches to the surface of
the growing calcium carbonate crystal.
To characterize the hybrid particles they are separated from the preparation solution and
transferred into 50 mL of an aqueous NaOH solution (c(NaOH) = 0.19 mol L-1). The solution
is diluted by the addition of 200 mL of water. Assume that the new solution contains only the
hybrid particles and no additional calcium or carbonate ions. All acidic groups participate in
the acid-base equilibrium.
• For the new solution, a pH of 12.30 is measured.
• In electron microscopy you only can see the inorganic particles (not the polymer):
Spherical particles of 100 nm diameter are observed.
• The molar mass of the hybrid particles (inorganic and organic part together) is
determined to be M = 8.01.108 g moL-1
• The charge of the particles is found to be Z = - 800 (number of unit charges).
(pKa(COOH, copolymer) = 4.88)
8.4 How much of the initial amount of polymer (2 g) can still be found in the hybrid
particles?

8.5. Calculate which modification of Modification density

calcium carbonate has been formed. Calcite 2.71 g cm-3
Vaterite 2.54 g cm-3
Aragonite 2.95 g cm-3

A periodical system and the following list of constants and useful formulas were provided.

101
 IChO: Theoretical test

Constants and useful formulas

f p n µ m k M G T
femto pico nano micro milli kilo mega giga tera
10-15 10-12 10-9 10-6 10-3 103 106 109 1012

Gas constant R = 8.314 J K-1 mol-1 Faraday constant F = 96485 C mol-1

Use as standard pressure: p = 1.013·105 Pa
Use as standard temperature: T = 25°C = 298.15 K
Avogadro’s number NA = 6.022·10 mol-1 23
Planck constant h = 6.626·10-34 J s
Speed of light c = 3.00·108 m s-1

∆G = ∆H - T ∆S ∆G = - nEF
product of c( products)
∆G0 = - RT·lnK ∆G = ∆G0 + RT·lnQ with Q =
product of c(reactands)
∆H(T1) = ∆H0 + (T1 - 298.15 K)·Cp (Cp = constant)

Ea
−
Arrhenius equation k = A ·e R ⋅T

Ideal gas law pV = nRT

RT c
Nernst equation E = E0 + ⋅ ln ox
nF c red

Bragg’s law nλ = 2d·sinθ

P0
Beer- Lambert Law A = log = ε·c·d
P
F
p= F = ma
A
4 3
V(cylinder) = πr2h A(sphere) = 4πr 2 V(sphere) = πr
3

1J=1Nm 1 N = 1 kg m s-2 1 Pa = 1 N m-21 W = 1 J s-1

1C=1As

102
 IChO: Solution to the theoretical test

Solutions to the Theoretical Test

Solution to problem 1
1.1 a) methane: CH4 + 2 O2 ⎯⎯→ CO2 + 2 H2O
b) ethane: 2 C2H6 + 7 O2 ⎯⎯→ 4 CO2 + 6 H2O
Thermodynamic data for the equations:
∆H0 = [2 · (-241.8) - 393.5 - (-74.6)] kJ mol-1 = -802.5 kJ mol-1
∆S0 = [2 · (188.8) + 213.8 – 186.3 - 2 · 205.2] J mol-1 K-1 = -5.3 J mol-1 K-1
∆G0 = -802.5 kJ mol-1 – 298.15 K · (-5.3 J mol-1 K-1) = -800.9 kJ mol-1

Methane: ∆H0 = -802.5 kJ mol-1 ∆S0 = -5.3 J mol-1 K-1 ∆G0 = -800.9 kJ mol-1
∆H0 = [6 · (-241.8) - 4 · 393.5 - 2 · (-84.0)] kJ mol-1 = -2856.8 kJ mol-1
∆S0 = [6·188.8 + 4·213.8 – 2·229.2 - 7·205.2] J mol-1 K-1 = +93.2 J mol-1 K-1
∆G0 = -2856.8 kJ mol-1 – 298.15 K · (93.2 J mol-1 K-1) = -2884.6 kJ mol-1
Ethane: ∆H0 = -2856.8 kJ mol-1 ∆S0 = +93.2 J mol-1 K-1 ∆G0 = -2884.6 kJmol-1

1.2 a) m = ρ·V  = 0.740 g L-1 · 1000 L = 740 g

Mav = ∑ i x ( i )M ( i ) = 0.0024 · 44.01 g mol-1 + 0.0134 · 28.02 g mol-1
+ 0.9732 · 16.05 g mol-1 + 0.011 · 30.08 g mol-1
= 16.43 g mol-1
ntot = m (Mav)-1 = 740 g · (16.43 g/mol)-1 = 45.04 mol
n(i) = x(i) · ntot n(CH4) = x(CH4) · ntot = 0.9732 · 45.04 mol = 43.83 mol
n(C2H6) = x(C2H6) · ntot = 0.0110 · 45.04 mol = 0.495 mol

1.2 b) Energy of combustion, deviation:

Ecomb.(H2O(g)) = ∑ i n( i )∆ c H °( i ) = 43.83 mol · (-802.5 kJ mol-1) + 0.495 mol · 0.5 · (-2856.8 kJ mol-1)
= -35881 kJ
Ecomb. = -35881 kJ
Deviation from PUC
EPUC(H2O(g)) = 9.981 kWh m-3 · 1 m3 · 3600 kJ (kWh)-1 = 35932 kJ
deviation: ∆E = (Ecomb.(H2O(g)) - EPUC(H2O(g)) ·100% · [Ecomb.(H2O(g))]-1
=
(35881 kJ - 35932 kJ) · 100% · (35881 kJ)-1 = -0.14%
deviation = -0.14 %

1.3 Energy for heating the water:

Volume of water: Vwater = 22.5 m3
nwater = Vwater ρwater (Mwater)-1 = 22.5 m3 · 106 g m-3 · (18.02 g mol-1)-1 =1.249·106 mol
Ewater = nwater · Cp · ∆T = 1.249·106 mol · 75.30 JK-1 mol-1 · 14 K = 1316 MJ Ewater = 1316 MJ
Energy for heating the air
Volume of the house is: Vair = 15 m · 8 m · 3 m + 0.5 · 15 m · 8 m · 2 m = 480 m3
nair = pV · (RT)-1 = 1.013·105 Pa · 480 m3 · (8.314 J (K mol)-1 · 283.15 K)-1 = 2.065·104 mol Cp(air)
= 0.21 · 29.4 J (K mol)-1 + 0.79 · 29.1 J (K mol)-1 = 29.16 J (K mol)-1
Eair = nair · Cp(air) · ∆T = 2.065·104 mol · 29.17 J (K mol)-1 · 20 K = 12.05 MJ Eair = 12.05 MJ

103
 IChO: Solution to the theoretical test

1.4 Energy for maintaining the temperature:

surface area of the house:
Ahouse = 3 m · 46 m + 8 m · 2 m + ((2 m)2 + (4 m)2)1/2 · 2 · 15 m = 288.16 m2
Heat conductivity: λwall = 1 J (s K m)-1
Energy flux along a temperature gradient (wall thickness d = 0.2 m)
J = Eloss (A · ∆t)-1 = λwall · ∆T · d -1
Eloss = 288.16 m2 · (12·60·60 s) · 1 J (s K m)-1 · 25 K · (0.2 m)-1 = 1556 MJ Eloss = 1556 MJ

1.5 Total energy and costs:

total energy: Etot = Ewater + Eair + Eloss = 1316 MJ + 12 MJ + 1556 MJ = 2884 MJ
total energy Etot = 2884 MJ
2884 MJ corresponds to 2.884·106 kJ · (3600 s h-1 · 9.981 kJ s-1 m-3 · 0.9)-1 = 89.18 m3
volume of gas V = 89.18 m3
2884 MJ correspond to a cost of: 0.40 €m-3 · 89.18 m3 = 35.67 €
rent for equipment: 150.00 €
total cost of gas heating = 185.67 €

2884 MJ correspond to a cost of

2.884·106 kJ · 0.137 € · (3600 s h-1·1 kJ s-1 h)-1 = 109.75 €
rent for equipment: 100.00 €
total cost of electric heating = 209.75 €

Solution to problem 2
2.1 Reaction equations:
2 CO + O2 ⎯→ 2 CO2
2 NO + 2 CO ⎯→ N2 + 2 CO2
2 C8H18 + 25 O2 ⎯→ 16 CO2 + 18 H2O

2.2 Questions concerning the λ probe: true false no decision

possible
If the λ-value is in the range of the λ-window, carbon monoxide
and hydrocarbons can be oxidised at the three-way catalytic
converter. x † †
With λ > 1, carbon monoxide and hydrocarbons can be oxidised at
the three-way catalytic converter. x † †
With λ < 0.975, nitrogen oxides can be reduced poorly † x †

0.85kPa −1 ⋅ 0.65kPa
2.3 a) Surface coverage: θ= θ = 0.356 or 35.6 %
1 + 0.85 ⋅ 0.65

2.3 b) Pressure at which 15% of the surface is covered:

K ⋅p θ
θ = ⇔ K·p = θ + θ ·K·p ⇔ p · (K - θ ·K)= θ ⇔ p=
1+ K ⋅ p K −θ ⋅K
θ = 0.15 p = 0.21 kPa

104
 IChO: Solution to the theoretical test

2.3 c) Orders of decomposition:

order of the decomposition reaction at low gas pressures 1
order of the decomposition reaction at high gas pressures 0
notes:
K⋅p 1
r = k ⋅θ = k ⋅ , p low ⇒ p << ⇒ r = k ⋅ K ⋅ p reaction order 1.
1+ K ⋅ p K
1
p high ⇒ p >> ⇒ r = k reaction order 0
K
2.3 d) Gas volume Va,max and product K·Va,max:
1 1 Va,max 1 p p
= +1 = ⇒ + =
θ K⋅p Va K ⋅ Va,max Va,max Va
1
slope: = 1.9 cm −3 ⇒ Va,max = 0.53 cm3
Va,max
1
intercept: = 6·102 Pa cm-3 ⇒ K·Va,max = 1.7⋅10-3 Pa-1 cm3
K ⋅Va,max
2.4 Equation for reaction rate:
The information given in the text leads directly to r = k 2 ⋅ θ CO2
The law of mass action for the first step of the mechanism is given by
1 1
k1 k1
θ CO2 = ⋅ θ co ⋅ θ o2 , (2) ⇒ r = k2 ⋅ ⋅ θ co ⋅ θ o2 .
k -1 2
k -1 2

The Langmuir isotherm gives:

KCO ⋅ pCO K O2 ⋅ pO2

θ CO = and θ O2 =
1 + KCO2 ⋅ pCO2 + KCO ⋅ pCO + KO2 ⋅ pO2 1 + K CO2 ⋅ pCO2 + K CO ⋅ pCO + K O2 ⋅ pO2

( )2
1
k K CO ⋅ pCO ⋅ K O2 ⋅ pO2
r = k2 1 .
(1 + KCO )2
k −1 3
2
⋅ pCO2 + K CO ⋅ pCO + K O2 ⋅ pO2

Solution to problem 3
3.1 Chemical equations: (a) CaCl2 + Ca ⎯→ 2 CaCl
(b) 2 CaCl2 + H2 ⎯→ 2 CaCl + 2 HCl
(c) 4 CaCl2 + C ⎯→ 4 CaCl + CCl4

3.2 silvery metallic particles: Ca

colorless crystals: CaCl2
Note: CaCl cannot be obtained by a conventional solid state reaction of Ca and CaCl2

3.3 Empirical formula: 100 % –(m/m% Ca + m/m% Cl) = m/m% X

100 % –(52.36% + 46.32%) = 1.32% X
mol% of Ca = 52.36 m/m% / M(Ca)
= 52.36 m/m% / 40.08 g mol-1 = 1.31 mol%
mol% of Cl = 46.32 m/m% / M (Cl) mol% of X = 1.32 % X / M (H)
-1
= 46.32 m/m% / 35.45 g mol = 1.32 % X / 1.01 g mol-1
= 1.31 mol% = 1.31 mol%
n(Ca) : n(Cl) : n(H) = 1 : 1 : 1 empirical formula CaClH

105
 IChO: Solution to the theoretical test

Notes: The reaction of CaCl2 with hydrogen does not lead to CaCl. The hydride CaClH is formed
instead. The structure of this compound was determined by X-ray structure analysis which is not a
suitable method to determine the position of light elements like hydrogen. Thus, the presence of
hydrogen was missed and CaClH was thought to be CaCl for quite a long time

3.4 a) Structures only:

H H
H C C CH3
C C C
H H

3.4 b) Empirical formula of the compound formed: Ca3C3Cl2

Notes: If the ratio of n(Ca):n(Cl) = 1.5 : 1 [or better = 3 : 2 which can be rewritten as CaCl2·2Ca2+ =
Ca3Cl24+ ] is given and the reduction product must contain a C34- anion which needs two Ca2+ cations
for electroneutrality, the composition Ca3C3Cl2 will follow.

3.5 a) Structure type CaCl likely to have: r(Ca+)/r(Cl-) = 120 pm/167 pm = 0.719
NaCl CsCl ZnS BN no decision possible
x † † † †

3.5 b) ∆fH0(CaCl) with a Born-Haber-cycle:

∆LH0

Summing up of all the single steps of the Born-Haber-cycle:

∆fH0 (CaCl) = ∆sublH0(Ca) + ∆1. IEH(Ca) + ½ ∆dissH(Cl2) + ∆EAH(Cl) + ∆LH(CaCl)
= (159.3 + 589.7 + 120 - 349.0 - 751.9) kJ mol-1
∆ fH0(CaCl) = -231.9 kJmol-1

3.6 Stability to disproportionation:

2 CaCl ⎯→ CaCl2 + Ca
∆H = ∆ fH (CaCl2) - 2 ∆ fH (CaCl) = -796.0 kJ mol-1 + 463.8 kJ mol-1 = -332.2 kJ mol-1
0 0

disproportionation yes no no decision possible, more information needed

† x †

106
 IChO: Solution to the theoretical test

Solution to problem 4
4.1 Atomic mass of X, symbol of X, structures:
1) X + 2 H2 ⎯→ XH4
2) 2 X + 3 H2 ⎯→ X2H6
I) 5.0 g = [n1(X) + n2(X)] · M(X)
II) 5.628 g = n1(XH4) · [M(X) + 4·1.01 g mol-1] + n2(X2H6) · [2M(X) + 6·1.01 g mol-1]
III) n1(XH4) = 2n2(X2H6)
III,I) → I’) 2n1(X) · M(X) = 5.0 g
III,II) → II’) n1(X) · [2M(X) + 7.07 g mol-1] = 5.628 g
I’,II’) → VI) (5.0 g) · [2M(X)]-1 = (5.628 g) · [2M(X) + 7.07 g mol-1]-1
M(X) = 3.535 g mol-1 · (5.628 g)-1 ·[(5.0 g)-1-(5.628 g)-1]-1
M(X) = 28.14 g mol-1
atomic mass of X M(X) = 28.14 g mol-1 chemical symbol of X: Si

3D structures of the two products: H H

H
Si H
Si H Si
H H
H
H H

4.2 Atomic mass of Y and empirical formula of Argyrodite:

AgaYbS0.5·a + 2·b + b H2 ⎯→ 0.5a Ag2S + b YS + b H2S
I) 10 g = n(AgaYbS0.5·a + 2·b) · [a·107.87 g mol-1 + b·M(Y) + (0.5·a + 2·b)·32.07 g mol-1]
p ⋅V( H2 ) 100kPa ⋅ 0.295 ⋅ 10 −3 m 3
II) n(H2) = n(H2) =
RT 8.314JK − 1mol −1 ⋅ 400K
n(H2) = 8.871·10-3 mol n(AgaYbS0.5·a + 2·b) = b-1 · 8.871·10-3 mol
a ⋅ 107.87gmol −1
III) 11.88 = a·107.87 gmol-1 = 11.88·b·M(Y)
b ⋅ M (Y )
II,I) →II’) b·10 g·(8.871·10-3 mol)-1 = a ·107.87 g mol-1 + b·M(Y) + (0.5·a + 2b)·32.07 g mol-1
b·1127 g mol-1 = a ·107.87 g mol-1 + b·M(Y) + (0.5·a + 2b)·32.07 g mol-1

III,II’)→IV) b·1127 g mol-1 = 11.88·b·M(Y) + b·M(Y) + (0.5·a + 2b)·32.07 g mol-1

11.88 ⋅ b ⋅ M (Y )
b·1127 g mol-1 = 11.88·b·M(Y) + b·M(Y) + (0.5· + 2b)·32.07 g mol-1
107.87gmol −1
M(Y) = 72.57 g mol-1
molar mass M(Y) = 72,57 g mol-1
M(Y) = 72.57 g mol-1 → III a:b = 8:1
chemical symbol of Y: Ge empirical formula of Argyrodite: Ag8Ge

4.3 The force constants of a C-H bond:

1 3M ( C ) ⋅ M ( H )
k(C-H) = [2π·c·ν~ (C-H)]2· ⋅
N A 3M ( C ) + 4M ( H )
1 3 ⋅ 12.01⋅ 1.01
= [2π·3·1010 cm·s-1·3030 cm-1]2· −1
⋅ gmol −1
6.022 ⋅ 10 mol 23 3 ⋅ 12.01 + 4 ⋅ 1.01
-1
k(C-H) = 491.94 N m

107
 IChO: Solution to the theoretical test

The force constants of a Z-H bond:

∆ H( Z − H )
k(Z-H) = k(C-H) · b
∆ b H( C − H )
= 491.94 N m-1·450.2 kJ mol-1·[438.4 kJ mol-1]-1
k(Z-H) = 505.18 N m-1

The molar mass and symbol of Z:

3M ( Z ) ⋅ M ( H ) k( Z − H ) ⋅ N A
=
3M ( Z ) + 4M ( H ) [ 2π ⋅ c ⋅ν~( Z − H )] 2
−1
4 ⎛⎜ [ 2π ⋅ c ⋅ν~( Z − H )] 2 1 ⎞⎟
M(Z) = ⋅⎜ −
3 ⎝ k( Z − H ) ⋅ N A M ( H ) ⎟⎠
−1
4 ⎛ [ 2π ⋅ 3 ⋅ 10 10 ⋅ 2938.45 ] 2 1 ⎞⎟
M(Z) = ⋅⎜ − g mol-1
3 ⎜ 505180 ⋅ 6.022 ⋅ 10 23 1.01 ⎟⎠
⎝
molar mass of Z M(Z) = 72.68 g mol-1
chemical symbol of Z Ge
Note: Even if the students find different values (± 2) due to different ways of rounding, they will be able
to find Ge as Z has to be an analogue of carbon.

Solution to problem 5
5.1 Actual ∆G’ of reaction (1):

o
c(ADP3-)/(1 molL-1).c(HPO42-)/(1 molL-1)
∆G’ = ∆G ’ + RT ln
c(ATP4-)/(1 molL-1)

= -30500 J mol-1 + 8.314 J mol-1 K-1 · 298.15 K · ln (0.00025 · 0.00165 / 0.00225)

= -30.5 kJ mol-1 – 21.3 kJ mol-1 = -51.8 kJ mol-1
∆G’ = -51.8 kJ mol-1

5.2 Equilibrium constant K' of reaction (2), ratio c(glucose 6-phosphate) / c(glucose):
Go' = -RT·lnK’ K’ = e- G°’/RT
= e-13800 J/mol / (8.314 J/(mol K) · 298.15 K)
= 0.0038
c(glucose 6-phosphate)/(1 mol L-1)
K' =
c(glucose)/(1 mol L-1).c(HPO42-)/(1 mol L-1)

c(glucose 6-phosphate) = K’ · c(HPO42-)·(1 mol L-1)-1

c(glucose)
= 0.0038 · 0.00165 = 6.3 · 10-6
K’ = 0.0038 c(glucose 6-phosphate) = 6.3 · 10-6
c(glucose)

5.3 ∆G°’ and K’ of reaction (3), ratio c(glucose 6-phosphate) / c(glucose):

∆G°’(3) = ∆G°’(1) + ∆G°’(2)

= -30.5 kJ mol-1 + 13.8 kJ mol-1 = -16.7 kJ mol-1

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 IChO: Solution to the theoretical test

∆G°’ = -RT·lnK’
K’ = e-∆G°’/RT = e16700 J/mol / (8.314 J/(mol K) · 298.15 K) = 843

K’ = c(glucose 6-phosphate).c(ADP3-)

c(glucose).c(ATP4-)

c(glucose 6-phosphate) c( ATP 4 − )

= K’ ·
c(glucose) c( ADP 3 − )
= 843 · 2.25 mmol L-1 / 0.25 mmol L-1 = 7587

5.4 a) Mass of ATP produced per day:

Energy available for ATP synthesis: 8000 kJ day-1 · 0.5 = 4000 kJ day-1
Energy required for synthesis of ATP: 52 kJ mol-1
Amount of ATP produced: 4000 kJ day / 52 kJ mol-1 = 76.9 mol day-1
-1

Mass of ATP produced: 76.9 mol day-1· 503 g mol-1 = 38700 g day-1 = 38.7 kg day-1
mday-1 = 38.7 kg day-1

5.4 b) Mass of ATP in the human body:

Average lifetime: 1 day = 1440 min 1 min = 1440–1 day
Mass of ATP in the body: 38.7 kg day-1/(1440 min day–1) · 1 min = 26.9 g mbody
= 26.9 g

5.4 c) What happens to the rest of the free energy? Mark one correct answer:
It is used to reduce the entropy of the body. †
It is released from the body in the O-H bonds of the water molecule and
the C=O bonds of the carbon dioxide molecule. †
It is used to regenerate the state of the enzymes which act as catalysts
in the production of ATP. †
It heats the body of the person. x

5.5 a) How many protons are in a spherical mitochondrium with a diameter of 1 m at pH=7?
V = 4/3 π·r3 = 4/3 π (0.5 · 10-6 m)3 = 5.2 · 10-19 m3 = 5.2 · 10-16 L
c = 10-7 mol L-1
n = V · c · NA = 5.2·10-16 L · 10-7 mol L-1 · 6.022·1023 mol-1 n = 31

5.5 b) How many protons have to enter a mitochondrium?

Number of ATP molecules:
m( ATP ) ⋅ N A 0.2 ⋅ 10 −15 g ⋅ 6.022 ⋅ 1023 mol −1
n(ATP) = = = 239400 Number
M ( ATP ) 503 g mol −1
of H+ per cell n(H+per cell) = n(ATP) ·3 = 718300
Number of H per mitochondrium: n(H+mit) = n(H+per cell)/ 1000 = 718
+
n(H+mit) = 718

109
 IChO: Solution to the theoretical test

Solution to problem 6
6.1 Structure of A:
O

[A]:

O
6.2 Structures of D1, D2 only:

CN
D1: (1) D2:
CN (1)

CN CN
alternatively, the following structures are also correct:
H H
CN CN

CN CN
H H
Note: The two compounds are enantiomers

6.3 Correct structure of B (circle only one):

1 2 3 4 5 6

Notes: The Diels-Alder reaction gives products with an endo-stereochemistry. The preference of
this configuration was outlined in problem 6.2, structure C. As shown in structure C this endo-
configuration is characterized by the two H atoms and the CH2-bridge of the bicyclic system being
on the same side of the ring. Only structures 1 and 2 of the six stereoisomers have an endo,endo
stereochemistry. All other isomers have at least one exo configuration. In structure 1 the three
rings form a U-shaped molecule which is sterically more hindered than structure 2 which has a
zig-zag structure

6.4 Decide the questions concerning the Diels-Alder reaction.

true false no decision possible
The Diels-Alder reaction is reversible x † †
The formation of B in the original reaction is
thermodynamically controlled † x †
B is thermodynamically more stable than E † x †
E is thermodynamically less stable than F x † †
G is an enantiomer of B † x †
G is thermodynamically more stable than F † † x

110
 IChO: Solution to the theoretical test

6.5 Structures of I, K, L only:

I K L

O O O
CO2Me OMe
O O
OMe
CO2Me CO2Me
CO2Me

Notes next page

from the initial olefin
bonds formed by DA
after loss of MeOH
OMe OMe
OMe - MeOH - CO2 X
OMe X OMe K
CO2Me CO2Me CO2Me CO2Me
C13H16O3 L X = CO2 X = CO2
diene must result from los s of CO2
C15H20O6 C11H12O4
product from the initial DA cyclohexene adduct
must be a 1,3-diene

X O O - MeOH OH O
K CO2Me CO2Me
CO2Me CO2Me CO2Me CO2Me
K I C12H16O5 tautomer
X = CO2 lactone = transesterification from Michael Addition with
from the initial
cyclohexenone with loss of MeOH subsequent loss of MeOH

Solution to problem 7
7.1 Fill in < or > (A < B means A has a priority lower than B) :
O
SCH3 > P(CH3)2 O <
CH3

CCl3 CH2Br
<

7.2
OH
(s)
CH3
* *
(s)
NHCH3

1
Ph
highest priority lowest priority
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→
OH CH(NHCH3)CH3 Ph H

NHCH3 CH(OH)Ph CH3 H

111
 IChO: Solution to the theoretical test

7.3 Newman projection or sawhorse projection of Fischer projection of 1:

1:
CH3
or
CH3
HO H MeHN H
or
MeHN H
HO H
Ph

Ph
(Me = CH3)

7.4 Equation with oxidation numbers and stereochemically correct structure of 2:

7.5a) Structure of 3 (correct stereochemistry):

7.5b) Statements concerning isomers:

true false
1 and 3 are stereo-isomers x †
1 and 3 are enantiomers † x
1 and 3 are diastereomers x †
1 and 3 are conformational isomers † x

7.5c) Draw a structural model to rationalize the exclusive formation of 3 from 2

Notes: Attack of hydride occurs from the sterically least hindered side.
Full points will also be given for an explanation using the formation of a hydrogen bond.
1 point will be given for any representation indicating the attack of hydride on the correct face of the
carbonyl group, i.e.

112
 IChO: Solution to the theoretical test

Solution to problem 8
8.1 pH of solution B:
c(HCO3− ) /(1 mol L−1 ) ⋅ c (OH− ) /(1 mol L−1 ) 10 −14
Kb2 = (1) Kb2 =
c(CO23− ) /(1 mol L−1 ) 10 −10.33
Kb2 = 2.14·10-4 Kb1 = 2.34·10-8

Since Kb2 >> Kb1, only one protonation step of the CO32- has to be considered.
c(HCO3-) = c(OH-) = x and c(CO32-) = c0(CO32-) – x
1.700 g L−1
c0(Na2CO3) = -1
c0(Na2CO3) = c0(CO32-) = 0.016 mol L-1
105.99 g mol
x 2 /(1 mol L−1 )
Kb2 = 2-
(1) x = c(OH-) = 1.75·10-3 mol L-1
(c 0 (CO3 ) - x)
pH = 11.2

8.2 Ca(OH)2 , CaCO3 in the precipitate?

M(CaCl2) = 110.98 g mol-1 pH = 10 , c(OH-) = 10-4 mol L-1
1.700 g L−1 1.780 g L−1
c0(Na2CO3) = c(CaCl2) =
105.99 g mol-1 ⋅ 2 110.98 g mol-1 ⋅ 2
c0(Na2CO3) = 8.0·10-3 mol L-1 (0.5) c(CaCl2) = c0(Ca2+) = 8.0·10-3 mol L-1
Calculations for Ca(OH)2:
c(OH-)2 ·c0(Ca2+) = 8·10-11 mol3 L-3 < 6.46·10-6 mol3 L-3 = Ksp(Ca(OH)2) no precipitate
Calculations for CaCO3:
(regarding proteolysis: 1 point)
c(HCO3− ) ⋅ c (OH− ) K b2
Kb2 = 2−
, c(HCO3-) = · c(CO32-)
c(CO3 ) c(OH − )
c(HCO3-) = 2.14 · c(CO32-) and c(HCO3-) + c(CO32-) = c0(Na2CO3)
2.14· c(CO32-) + c(CO32-) = 8.0·10-3 mol L-1
Initial concentration of CO32- in solution C: c(CO32-) = 2.55·10-3 mol L-1
Initial concentration of Ca2+ in solution C: c(Ca2+) = 8.0·10-3 mol L-1
hence c(CO3 ) · c(Ca ) = 2.04·10 mol L > 3.31.10-9 mol2 L-2
2- 2+ -5 2 -2
= Ksp(CaCO3) precipitate
Ca(OH)2 will be found in the precipitate yes † no x
CaCO3 will be found in the precipitate yes x no †

113
 IChO: Solution to the theoretical test

8.3 Circle the block that attaches to the CaCO3 crystal:

H
H
COOH
C O
C
H C
H 68 C H
H 8 H
H H

Notes: Both polymer blocks are hydrophilic. The acrylic acid block will preferably bind to the crystal
since it is more polarized and additionally charged. The polymer binds to the surface at positions
where there is an excess of calcium ions on the surface of the ionic crystal.

8.4 How much of the initial amount of polymer (2 g) can still be found in the hybrid particles?
⎯→
RCOOH + OH- ←⎯ RCOO- + H2O pKb = 9.12
pH and pKa lead to the total concentration of COOH groups in the solution:

c(COO-) = x c(COOH) = c0(COOH) – x x = c0(OH-) – c(OH-)

50 mL
c0(OH− ) = 0.19 mol L−1 c0(OH-) = 0.038 mol L-1
250 mL
c(OH-) = 10-1.7 mol L-1 = 0.02 mol L-1 (0.5) x = 0.018 mol L-1
(c 0 (COOH) − x) /(1 mol L−1 ) ⋅ c(OH − ) /(1 mol L−1 )
Kb =
x /(1 mol L−1 )
K b x ⋅ (1 mol L−1 ) ⎛ 0.018 ⋅10−9.12 ⎞
c 0 (COOH) = +x (1) c0(COOH) = ⎜ + 0.018 ⎟mol⋅L−1
c(OH − ) ⎜ 0.02 ⎟
⎝ ⎠
c0(COOH) = 0.018 mol·L-1
(Or as pH >> pKa : c0(COOH) = c(COOH)+ x ≈ x )
c 0 (COOH)
Total concentration of polymer chains c(polymer) =
8
M(polymer) = M(C160O84H306) = 3574.66 g mol-1
(0.5) (0.5)
m(polymer) = c(polymer)·V·M(polymer)

c 0 (COOH) ⋅ V ⋅ M(polymer ) 0.018 ⋅ 0.250 ⋅ 3574 .66

m(polymer) = = g = 2.0 g
8 8

8.5 Modification of CaCO3:

The charge of the particles is caused by the number of protolized COOH groups per particle.
c(COO-) ≈ c0(COOH), ≈ 1
Z
Number of COOH groups per particle: NCOOH = NCOOH = 800
α
NCOOH
Number of polymer chains per particle: : Npolymer = = 100
8
The number of polymers per particle indicates the mass of polymer per particle. Thus, the mass of the
calcium carbonate particle can be calculated:
M (CaCO3 particle) = M (total particle) – Npolymer ·M(polymer)
M (CaCO3 particle) = 8.01·108 g mol-1 – 100 · 3574.66 g mol-1
M (CaCO3 particle) = 8.01·108 g mol-1
Mass of one CaCO3 particle: m(CaCO3particle) = M (CaCO3 particle) · NA-1

114
 IChO: Solution to the theoretical test

4
and with the volume of the spherical particle (V = ⋅ π ⋅ r 3 ) the density can be calculated:
3
m( CaCO3 particle ) 3 ⋅ m( CaCO3 particle )
ρ(CaCO3) = =
V ( CaCO3 particle ) 4π ⋅ r 3
3(M (total particle) - N polymer ⋅ M(polymer) )
=
N a ⋅ 4π ⋅ r 3
3 ⋅ 8.01 ⋅ 10 8 g mol -1
= −6 3
= 2.54 g cm-3
N A ⋅ 4π ( 5 ⋅ 10 cm )
The modification of calcium carbonate is Calcite † Vaterite x Aragonite †

115
 IChO: Practical test

36th IChO Practical Problems

Problem 1. Two-Step Organic Synthesis of 2,2-Bis(p-phenyleneoxyacetic

acid)propane (Bisphenol A bis(carboxymethyl)ether)

Introduction
In the first step the sodium salt of bisphenol A results as an intermediate from the alkaline hydrolysis
of a polycarbonate. By adding an acid this salt is converted into the free 2,2-bis(4-
hydroxyphenyl)propane (bisphenol A).

O
CH3
O
O C O

CH3
n
CH3
+
NaOH, H2O H3O , H2O n HO C OH

CH3

In the second step bisphenol A reacts with sodium chloroacetate to form the phenolic ether,
bisphenol A bis(carboxymethyl)ether.

CH3
+
ClCH2COONa, NaOH, H2O H3O , H2O
HO C OH

CH3

C CH2 CH3

HO O C O O

CH3 H2C C

OH

• In each step the product has to be isolated. (Drying and weighing will be done by the organizer.)
• For the product of step 2 three melting point tubes have to be filled. (Filling of the melting point
tubes in step 1 will be done by the organizer.The melting points will be determined by the
organizer.)
• When the organizer receives your labelled beaker A of step 1 you will get 2.00 g of bisphenol A as
starting material for the second step.
• Answer the additional questions on the answer sheet P1.
• Do not remove the Ceran plate from the magnetic stirrer.

116
 IChO: Practical test

Procedures
Step 1 Preparation of bisphenol A by alkaline hydrolysis of a polycarbonate
Preparation:
• Put the pre-weighed 2.54 g of polycarbonate (No. 1), 4.0 g of sodium hydroxide (No. 5) and 3 mL
of demineralized water into a 100 mL Erlenmeyer flask with ground-glass joint.
• Close the flask with a plastic plug and swirl it gently so that the solution does not contact the
ground-glass joint. For aeration open the plastic plug occasionally. Strong heating can be
observed, as the sodium hydroxide partially dissolves.
• Remove the plastic plug after swirling for about 4 minutes, add a magnetic stirring bar and put the
flask onto a heating plate. Put a reflux condenser above the neck of the flask. Use a Teflon
coupling as a connection between flask and condenser. Fix the apparatus tightly to a stand.
• Finally, add 20 mL of ethanol (No. 2) through the condenser while stirring the reaction mixture.
• Heat the reaction mixture under reflux for 60 minutes. In the beginning adjust the thermostat of the
heating plate to maximum. When the mixture starts boiling reduce the heat carefully, so that the
mixture is kept under gentle reflux.
• A white precipitate is formed on heating.

During this waiting period you are highly advised to start working on the analytical chemistry
experiment.

Isolation:
• Stop heating after one hour, allow the reaction mixture to cool down to ambient temperature,
remove the condenser, add 25 mL of demineralized water and transfer the reaction mixture into a
400 mL beaker. Rinse the Erlenmeyer flask with 25 mL of demineralized water and add this to the
contents of the beaker.
• Finally, fill up to 150 mL with demineralized water.
• If the reaction mixture is not clear, the mixture has to be filtered over fibre glass into an
Erlenmeyer flask.
• Add slowly 15 mL of hydrochloric acid (No. 3) stirring the mixture simultaneously with a glass rod.
A rather oily or sometimes crystalline precipitate is formed.
• Ask your instructor for some seed crystals of bisphenol A (No. 27) in order to accelerate the
crystallization.
• Stir the reaction mixture thoroughly with the glass rod. For a quantitative crystallisation continue
stirring from time to time till the supernatant aqueous solution is nearly clear.
• Collect the crude product by vacuum filtration, wash it twice with 10 mL portions of demineralized
water and transfer it quantitatively into the tared and labelled beaker A.
• For drying and weighing deliver your labelled beaker A into the instructor room.
• Afterwards you will get a small jar filled with 2.00 g of bisphenol A (No. 28), your starting material
of the second step.

• On delivery of your product and on receipt of the starting material you have to sign. Even
if you do not have any bisphenol A, please bring the empty beaker A to the instructors’
room in order to get the starting material for step 2.

117
 IChO: Practical test

Step 2 Reaction of Bisphenol A with Chloroacetic Acid forming 2,2-Bis(p-phe-

nyleneoxyacetic acid)propane (Bisphenol A bis(carboxymethyl)ether)
Preparation:
• Pour all the bisphenol A (No. 28) you have received from the organizer when you had finished
step 1 into a 100 mL Erlenmeyer flask with ground-glass joint.
• Add 10 mL of an aqueous sodium-hydroxide solution (No. 6), 1 mL of demineralized water and a
magnetic stirring bar.
• Put the flask onto a heating plate. Put a reflux condenser above the neck of the flask. Use a
Teflon coupling as a connection between flask and condenser. Fix the apparatus tightly to a stand.
• Heat the reaction mixture with gentle stirring until a clear solution is formed.
• Remove the heating plate and the condenser and add 5.0 g of the sodium salt of chloroacetic acid
(No. 4) to the reaction mixture.
• After reconnecting the flask with the reflux condenser, heat the mixture to reflux with vigorous
stirring for 30 min.
• Initially a clear solution is formed on heating. In some cases a colorless solid precipitates. If the
complete mixture becomes solid in the course of the reaction, heating must be stopped.
• After that, 50 mL of ethanol (No. 2) are added carefully through the condenser (beware of sudden
boiling!). The mixture is stirred and heated under reflux for 5 minutes. A colourless solid
precipitates, or the crystallisation which has already started is completed.
Isolation:
• After leaving it to cool down for 5 minutes, transfer the reaction mixture with another 50 mL of
ethanol (No. 2) quantitatively to a beaker. The reaction mixture should be stirred vigorously.
• The magnetic stirring bar is removed and the reaction mixture is filtered through a suction filter.
Solids which separate in the filtrate are rejected. Rinse the beaker with 10 mL of ethanol (No. 2).
The precipitate is washed twice with 10 mL portions of ethanol (No. 2). (The filtrate must be
disposed of in the organic solvent waste!)
• Transfer the precipitate quantitatively into a beaker, add a stirring bar and dissolve it in 150 mL of
demineralized water. The mixture must be stirred vigorously. Larger lumps of the solid must be
crushed with the spatula.
• If the solution is not clear, it has to be filtered over a folded filter paper into an Erlenmeyer flask.
• The slow addition of 5 mL of hydrochloric acid (No. 3) to the stirred reaction mixture results in the
formation of a white precipitate.
• Collect the crude product by vacuum filtration, wash it twice with 10 mL portions of demineralized
water and transfer it quantitatively into the tared and labelled beaker B.
• Take a small sample of the product with a micro spatula, crush it and dry it on a shard. Fill three
melting point tubes with the homogenized, dried sample. For a close-packed and 5 mm high filling
use the 75 cm glass tube and the measure.
• Put all three melting point tubes into the test tube B, which is labelled with your student
code, and give it together with your labelled beaker B with the product to the instructor. On
delivery you have to sign.

118
 IChO: Practical test

Problem 2. Qualitative and Quantitative Analysis of a Superconductor

Introduction
Superconductors based on lanthanum cuprate (La2CuO4) have the general composition of LaxM(2-
x)CuO4 (M = Ca, Sr, Ba).

This problem consists of two parts:

- the qualitative determination of the alkaline earth metal(s)
- the quantitative determination of lanthanum and copper.

Read the burette as accurately as possible. Report your results on the answer sheets.
Answer the additional questions and write the results with adequate accuracy.
The qualitative and quantitative parts of this experiment may be done in any order.

Procedures
2.1 Qualitative determination of the alkaline earth metal(s) (If the hood is occupied start with the
titration 2.2)

In this experiment you have to use the superconductor as a solid (LaxM(2-x)CuO4; No. 14).
At the beginning, lanthanum has to be separated as an insoluble residue.
All steps must be carried out in the hood!

Dissolve the complete sample in a beaker in about 5 mL of perchloric acid (No. 22) by heating the
mixture. Add 5 mL of demineralized water afterwards.
Cool down the solution until it is lukewarm.

Add about 5 mL of demineralized water and then ammonia solution (No. 17), until the reaction mixture
shows a basic reaction. Lanthanum precipitates as hydroxide and copper forms an intense blue-
coloured tetraammine complex. The precipitate is filtered off and washed with a small amount of
demineralized water.

An excess of ammonium-carbonate solution (No. 18) is added to the filtrate and the mixture is being
boiled for some minutes. The alkaline earth metal(s) will precipitate as carbonate(s). The precipitate is
filtered off and washed a few times with a small amount of demineralized water.

Then, the precipitate is dissolved in acetic acid (No. 16). Add sodium acetate (No. 9) and an excess of
potassium-dichromate solution (No. 23). In the presence of barium, yellow BaCrO4 precipitates. After
boiling the mixture for one minute barium chromate is filtered off and washed with demineralized
water.
(If there is no barium chromate precipitation, proceed in a way as if there were precipitation.)

Ammonia solution (No. 17) is added to the clear filtrate until it is basic. Add an excess of ammonium-
carbonate solution (No. 18) and boil the mixture for some minutes. In the presence of strontium and/or
calcium, white carbonate(s) precipitate(s).

The precipitate is filtered off and washed a few times with demineralized water.
119
 IChO: Practical test

Then it is dissolved in a mixture of about 2 mL of demineralized water and a few drops of hydrochloric
acid (No. 3). This solution is devided between two test tubes:
• Saturated calcium-sulfate solution (No. 21) is added to one of the test tubes. In the presence
of strontium a small amount of white strontium sulfate precipitates. To accelerate the precipitation,
you can grind the inner surface of the test tube with a glass rod for a few minutes.
• Ammonium-sulfate solution (No. 20) is added to the second test tube. In the presence of
strontium and/or calcium, white sulfate(s) precipitate(s). The precipitate is filtered off and washed
with a very small amount of demineralized water.
1 mL of ammonium-oxalate solution (No. 19) is added to the filtrate. In the presence of calcium,
white calcium oxalate precipitates after a few minutes.

Preparation of the superconductor parent solution

There is a superconductor solution (LaxM(2-x)CuO4 in perchloric acid; No. 13) in a volumetric
flask.
Fill it up with demineralized water to a volume of 250.0 mL. From now on this solution is called
“parent solution”.

2.2 Quantitative determination of the total content of lanthanum and copper

Transfer 25.00 mL of the parent solution into an Erlenmeyer flask.
Add about 5-6 piled spatula of sodium acetate (CH3COONa; No. 8) and 2 micro spatula of xylenol
orange indicator (No. 15) to this solution and make up with demineralized water to a volume of about
75 mL.
The pH-value has to be about pH 6 before the determination, otherwise add more sodium acetate.
Titrate the solution with Na2-EDTA solution (No. 7). The color of the solution changes from light violet
to intensely light-green. (In between, the color changes a few times.)

Repeat this procedure as many times as necessary.

2.3 Quantitative determination of the copper content

Transfer 25.00 mL of your parent solution (No. 13) into the 100 mL volumetric flask and fill up with
demineralized water to a volume of 100.0 mL.
For each titration, transfer 25.00 mL of this solution into an Erlenmeyer flask and add sodium
hydroxide solution (No. 6), until the solution shows an alkaline reaction. During this procedure, a blue
precipitate for

120
 About the history of the IChO

About the history of the International Chemistry-Olympiads

The idea of chemistry olympiads was born 1968 during an Czechoslovakian national
olympiad that was attended by observers from Poland and Hungary. These three countries
participated in the first IChO 1968 in Prague. The participating countries of the following
years are shown in the table.

Participating Delegations
(in the alphabetical order of the German names)
(+ = host, + = participant, o = observer)

Country, Year → 6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 0 0 0 0 0 0
↓ 8 9 0 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5
Argentina + + + + + + + + + +
Armenia o
Australien o + + + + + + + + + + + + + + + + +
Austria + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Azerbaijan o o + + + + +
Belarus + + + + + + + + +
Belgium + + + + + + + + + + + + + + + + + + + + + + + + +
Brasil o o + + + + + +
Bulgaria + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Canada o o + + + + + + + + + + + + + + + + + + +
China + + + + + + + + + + + + + + + + + +
Chinese Taipei + + + + + + + + + + + + +
Croatia o o + + + + +
Cuba + o + + + + + + + + + + + + + + +
Cyprus o + + + + + + + + + + + + + + +
Czech Rep. + + + + + + + + + + + +
Czechoslovakia + + + + + + + + + + + + + + + + + + + + + + + +
Denmark + + + + + + + + + + + + + + + + + + + + + + +
DDR o + + + + + + + + + + + + + + + + + + +
Egypt o o + + +
Estonia + + + + + + + + + + +
Finland o + + + + + + + + + + + + + + + + + + + + + + + + + + +
France o + + + + + + + + + + + + + + + + + + + + + + + +
Germany o + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Greece + + + + + + + + + + + + + + + + + + + +
Hungary + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Iceland o o + + +
India o o + + + + + +
↑ Year → 6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 0 0 0 0 0 0
Country 8 9 0 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5

121
 About the history of the IChO

Country, Year → 6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 0 0 0 0 0 0
↓ 8 9 0 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5
Indonesia o + + + + + + + +
Iran + + + + + + + + + + + +
Ireland o o + + + + + + +
Israel o
Italy + + + + + o o + + + + + + + + + + + + + + + + + +
Japan o + +
Jugoslavia + + + + + + + + + + o
Kazakhstan o o + + + + + + +
Kenia o o
Korea + + + + + + + + + + + + +
Kuwait o o + + + + + + + + + + + + + + + +
Kyrgyzstan o o + + + + + +
Latvia + + + + + + + + + + + + + +
Lithuania + + + + + + + + + + + + + +
Malaysia o
Mexico + + + + + + + + + + + + +
Moldova o
Mongolia o o
Netherlands + + + + + + + + + + + + + + + + + + + + + + + + +
New Zealand + + + + + + + + + + + + +
Norway o + + + + + + + + + + + + + + + + + + + + + + +
Pakistan o
Peru o o +
Philippines o
Poland + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Portugal o o + +
Romania + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
GUS/Russ.Fed. + + + + + + + + + + + + +
Saudi Arabia o
Singapore o + + + + + + + + + + + + + + + +
Slovakia + + + + + + + + + + + +
Slovenia + + + + + + + + + + + + + +
Spain o + + + + + + + + +
Sweden + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Switzerland o + + + + + + + + + + + + + + + + + +
Tajikistan o o +
Thailand o + + + + + + + + + + + + + + +
↑ Year → 6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 0 0 0 0 0 0
Country 8 9 0 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5

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 About the history of the IChO

Country, Year → 6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 0 0 0 0 0 0
↓ 8 9 0 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5
Turkey o + o + + + + + + + + + + +
Turkmenistan o o o + + +
UdSSR + + + + + + + + + + + + + + + + + + +
Ukraine + + + + + + + + + + +
United
o o + + + + + + + + + + + + + + + + + + + + + +
Kingdom
United States o o + + + + + + + + + + + + + + + + + + + + +
Uruguay o o + + + + + +
Venezuela o o + + + + + + + + + + + +
Vietnam + + + + + + + + +

↑ Year → 6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 0 0 0 0 0 0
Country 8 9 0 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5
Number of 3 4 7 7 7 9 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 4 4 4 4 4 5 5 5 5 5 6
teams 2 2 2 2 1 3 4 7 8 0 1 2 6 6 6 8 0 3 8 1 2 5 7 7 1 3 4 7 9 1

Number of teams attending the IChO

60

50
Number of teams

40

30

20

10

0
1970 1975 1980 1985 1990 1995 2000 2005
Year of olympiad

123
 About the history of the IChO

Inofficial ranking since 1974

(set up by adding the points of the teams, up to position 50)

1974 1975 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985 1986 1987 1988

IChO held in RO H DDR CS PL SU A BG S RO D CS NL H FIN

1 SU SU DDR CS SU PL PL H CS RO D SU NL SU RC
. RO H SU SU PL SU D CS D SU CS CS PL RC D
. CS PL H H D RO DDR PL PL D SU D D RO USA
. H BG PL PL DDR CS H BG NL CS H A SU CS PL
5 PL RO A S CS A A A A H A NL A D GB
. DDR DDR RO A H S RO D SU A GB H USA F DDR
. BG S BG D A H BG DDR H F PL DDR H GB N
. YU CS CS DDR RO D CS RO BG DDR USA PL BG PL RO
. S A S RO S BG S SU DDR PL RO USA F H H
10 D* D D BG BG FIN FIN NL S NL DK F RO DDR SU
. YU YU YU TR DDR NL FIN F BG S GB CS NL I
. B B B FIN I S FIN GB NL RO GB USA NL
. B F N N FIN BG S BG BG
. I RO DK F N DDR A CS
15 * hors concours DK FIN BG S CDN S AUS
. YU S N FIN N FIN SGP
. I I I YU DK N F
. YU GR B B DK A
. YU GR FIN I FIN
20 B DK GR GR CDN
. C KWT C DK
. YU B C
. YU S
. CDN B
25 CH CH
. KWT KWT
(List of abbreviations see page 127)

124
 About the history of the IChO

1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000
IChO held in DDR F PL USA I N RC RUS CDN AUS T DK
1 DDR RC RC RC RC RC RC IR H SGP USA RC
. D PL RO H TPE GB IR RC D USA ROK RUS
. RC D H PL USA USA RO RUS TR ROK RC USA
. BG USA PL USA I A A A TPE RC IR H
5 SU CS NL A GUS SGP D D IR H RO TPE
. H RO USA GUS H ROK GB USA RUS RA H A
. PL F I D D TPE SK UA ROK RUS TPE SK
. RO A D RO CDN CZ TPE CZ RC AUS UA BY
. CS DDR N F SGP GUS I H SGP D PL VN
10 I H GB I CZ IR CZ RO PL GB AUS TR
. NL GB CS SGP A D RUS GB USA PL VN SGP
. GB I SU CS RO H H TPE UA A D D
. A AUS A AUS P RO AUS BY AUS RO RA ROK
. USA SGP AUS NL NZ DK SGP SGP CDN TPE BY IR
15 S NL DK DK ROK I F RA RO SK T CZ
. F N SGP ROK LV T TR TR A NL F FIN
. N DK CDN GB IR NZ PL F T IR TR T
. AUS T BG CH DK UA USA I EST UA SGP MEX
. CDN FIN F T AUS AUS DK AUS CZ VN IND GB
20 DK CDN S LV NL F RA ROK VN LT GB AUS
. FIN BG T NZ LT PL ROK EST F TR RUS IND
. B C CH S SK NL UA CDN S BY MEX CDN
. C S LV LT F SK LT T BY F A RA
. GR CH LT N C CDN T VN NZ I IRL UA
25 CH B FIN CDN GB LT NL SK LV T NZ PL
. KWT GR C SLO T S CH CH RA FIN I NZ
. KWT GR BG BG N BG NL SLO CZ CDN BG
. CY B TPE B BG S NZ GB CDN LT F
. CY B S FIN NZ DK SK S NL DK
30 SLO FIN FIN EST EST PL LT BG SK NL
. GR SLO LV CDN SLO I N BG B
. CY GR CH MEX MEX DK MEX KZ RO
. MEX MEX MEX N LV NL CH DK KZ
. N SLO SLO N IRL SLO CH LT
35 CH B LV CY N EST CZ CH
. YV CY CY BG MEX CY FIN SLO
. CY GR B S CH LV B EST
. KWT TR GR LT CY DK S S
. YV FIN E E NZ CY YV
40 C YV B FIN GR EST CY
. KWT KWT GR BG KZ LV HR
. C FIN YV E SLO I
. YV GR IRL YV RI
. C B B BR N
45 KWT RI KS E AZ
. KWT YV N IRL
. C RI RI E
. GR LV
. ROU GR
50 C BR
(List of abbreviations see page 127)

125
 About the history of the IChO

2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012
IChO held in IND NL GR D TPE ROK
1 RC RC RC RC
. ROK T IR ROK
. USA TPE ROK RUS
. RUS ROK T UA
5 IR A BY D
. TR UA RUS PL
. IND USA IND TPE
. AUS PL SGP H
. TPE IND D TR
10 T D TPE VN
. SGP IR UA IND
. PL H PL IR
. RO RUS CDN RO
. F CDN CZ LT
15 SK TR RO CZ
. H AUS KZ USA
. VN GB VN SGP
. CZ SGP EST CDN
. RA E GB AZ
20 BY SK AUS AUS
. C BY H KZ
. D VN SK GB
. GB FIN USA J
. UA F YV A
25 A LT IND BLR
. MEX CZ F SK
. DK KZ A T
. CDN LV I RA
. EST NL TR EST
30 RI RO AZ F
. HR RA MEX NZ
. I EST LT SLO
. N HR NL HR
. BG BG FIN LV
35 CY NZ HR NL
. KZ I J I
. B DK DK CH
. LT SLO RA FIN
. NZ N GR RI
40 CH YV LT S
. E MEX E BG
. FIN BR TM KS
. SLO S BR E
. NL RI BG GR
45 LV TM CH BR
. BR B NZ TM
. S IRL IS CY
. YV CH IRL YVA
. IRL C CY IRL
50 GR CY KS IS
(List of abbreviations see page 127)

126
 About the history of the IChO

List of abbreviations

A Austria KZ Kasakhstan
AUS Australia LV Latvia
AZ Azerbaijan LT Lithuania
B Belgium MEX Mexico
BG Bulgaria MGL Mongolei
BR Brazil N Norway
BY Belarus NL Netherlands
C Cuba NZ New Zealand
CDN Canada P Portugal
CH Switzerland PE Peru
CS Czechoslovacia PL Polen
CY Cyprus Republic RA Argentina
CZ Czech Republic RI Indonesia
D Germany RC China
DDR German Democratic Republic RO Romania
DK Denmark ROK South Korea
E Spain ROU Uruguay
EAK Kenya RUS Russian Federation
EST Estonia S Sweden
ET Egypt SGP Singapore
F France SK Slovakia
FIN Finland SLO Slowenia
GB United Kingdom SU Sowjet Union
GR Greece T Thailand
GUS Commonwealth of Independent States TJ Tadschikistan
H Hungary TM Turkmenistan
HR Croatia TPE Chinese Taipei
I Italy TR Turkey
IND India UA Ukraine
IR Iran USA United States of America
IRL Ireland VN Vietnam
IS Iceland WAN Nigeria
J Japan YU Yugoslavia
KS Kyrgistan YV Venezuela
KWT Kuwait

127