International Chemistry Olympiad

IChO 1978

Toruń, Poland · 5 theory problems with complete solutions

  1. Problem 1: Analysis of a chromium ore

    a) A chromium ore which does not contain water, consists of: \(\ce{Fe(CrO2)2}\), \(\ce{Mg(CrO2)2}\), \(\ce{MgCO3}\), and \(\ce{CaSiO3}\).

    b) It was found by analysis the ore contains 45.6 % of \(\ce{Cr2O3}\), 7.98 % of \(\ce{Fe2O3}\), and 16.12 % of \(\ce{MgO}\).

    c) When the ore was treated with a concentrated hydrochloric acid, chromium compounds being present in the ore did not react with the acid.

    d) When the reaction was finished, the ore was thoroughly washed with water (till the reaction with \(\ce{Cl-}\) was negative) and the solid residue was dried to a constant mass.

    Problems:

    1.1 Write stoichiometric and ionic equations for the reactions taking place when the ore is treated with the hydrochloric acid as given in c).

    1.2 Calculate:

    – the content of the compounds (in mass %) present in the ore,
    – amounts of substances of the compounds present in the ore.

    1.3 Calculate the content of \(\ce{Cr2O3}\) (in mass %) in the dried residue obtained according to d).

    1.4 A glass tube was filled with a sufficient amount of granulated \(\ce{CaO}\), the total mass of the filled tube having been 412.02 g. A gas formed by the reaction as given in c), was dried and then transmitted through the glass tube. Calculate the mass of the glass tube with its filling after the reaction was finished.

    Relative atomic masses: \(A_{\mathrm r}(\ce{Cr})=52.01;\quad A_{\mathrm r}(\ce{Fe})=55.85;\quad A_{\mathrm r}(\ce{Mg})=24.32;\quad A_{\mathrm r}(\ce{Ca})=40.08;\)

    \(A_{\mathrm r}(\ce{Si})=28.09;\quad A_{\mathrm r}(\ce{C})=12.01;\quad A_{\mathrm r}(\ce{O})=16.00.\)

    Official English text of the problem.

  2. Problem 2: Water hardness caused by Fe2+ and Ca2+

    A sample of water under investigation had 10° of temporary hardness and 10° of permanent hardness. Hardness of the water was caused by cations \(\ce{Fe^{2+}}\) and \(\ce{Ca^{2+}}\) only.

    A volume of \(10.00\ \mathrm{dm^3}\) of the water was at disposal. From this volume \(100.00\ \mathrm{cm^3}\) were taken for further procedure. The water was oxidised with a \(\ce{H2O2}\) solution and then precipitated with an aqueous ammonia solution. A brown precipitate was dried and after an appropriate heating \(0.01432\ \mathrm{g}\) of an anhydrous product was obtained.

    Problems:

    2.1 Calculate the molar ratio of \(\ce{Fe^{2+}}\) : \(\ce{Ca^{2+}}\) in the water under investigation.

    2.2 In another experiment, \(10.00\ \mathrm{dm^3}\) of the water was used again. The temporary hardness caused by cations \(\ce{Ca^{2+}}\) was removed first and the permanent hardness caused by cations \(\ce{Fe^{2+}}\) was removed by addition of \(\ce{Na3PO4}\). Calculate the mass of the precipitate (in its anhydrous form) on the assumption that only one half of cations \(\ce{Fe^{2+}}\) was oxidised to \(\ce{Fe^{3+}}\) in \(10.00\ \mathrm{dm^3}\) of the water analysed. Calculation should be made with an accuracy of one hundredth. Give the molar ratio in integers.

    1° of hardness = \(10\ \mathrm{mg}\) \(\ce{CaO}\) in \(1\ \mathrm{dm^3}\) of water.

    Relative atomic masses:

    \[ A_r(\ce{Ca})=40.08;\qquad A_r(\ce{Fe})=55.85;\qquad A_r(\ce{C})=12.01; \] \[ A_r(\ce{H})=1.01;\qquad A_r(\ce{P})=31.00;\qquad A_r(\ce{O})=16.00. \]

    Official English text of the problem.

  3. Problem 3: Chromium plating

    Chromium plating is usually made by electrolysis in a solution of chromic acid. The chromium plated objects form the cathode. The anode is an alloy that is inert under given conditions, i. e. it does not react either chemically or electrochemically.

    An electrolytic cell was filled with 100.0 dm\(^{3}\) of an aqueous solution which contained 0.230 kg of chromium acid anhydride in 1 dm\(^{3}\) of the solution.

    In electrolysis a current of 1500 A passed through the electrolyte for 10.0 hours. After electrolysis an increase of the mass of the cathode was 0.679 kg.

    The ratio of gas volumes

    \[ \frac{V_{\mathrm C}}{V_{\mathrm A}}=1.603 \]

    where \(V_{\mathrm C}\) is a volume of gases evolved at the cathode, whereas that marked as \(V_{\mathrm A}\) is the volume of gases which are evolved at the anode. Both volumes were measured at the same conditions.

    Problems

    3.1 What part of the total charge (in %) was used for a deposition of 0.679 kg of chromium?

    3.2 Calculate:

    a) the volume ratio of both gases (at STP) which are evolved as by-products at the cathode and anode,

    b) current efficiency for the corresponding reactions taking place separately at the cathode and anode when the gases are evolved.

    If you find any disproportion between the data calculated and those given in the task, try to explain what process would take place in the electrolytic cell which has not been considered till now.

    Write the corresponding summary equation for the reactions at electrodes and correct your previous calculations if possible.

    Official English text of the problem.

  4. Problem 4: Thermal decomposition of ethane in a closed vessel

    A vessel of a volume of \(5.0\ \mathrm{dm^3}\) was filled with ethane at a temperature of 300 K and normal pressure and sealed. The vessel with the gas was then heated and the pressure in it was measured at distinct temperatures. The following data were found:

    \(T\) (K) Pressure \(p\), measured (kPa) Pressure \(p'\), calculated (kPa)
    300 101.25
    500 169.20
    800 276.11
    1000 500.48

    Problems:

    4.1 Calculate the pressure \(p'\) of ethane in the vessel according to the ideal gas law equation and fill in the values in a free column in the above table.

    \[ (R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}) \]

    4.2 Explain the differences between theoretical value \(p'\) and those (\(p\)) obtained by measurements.

    4.3 Write the chemical equation for the reaction which takes place probably in the vessel at higher temperatures.

    4.4 Calculate the value for the conversion degree \(\alpha\) of ethane and that for equilibrium constant \(K_p\) of the reaction that takes place at temperatures of 800 and 1000 K.

    4.5 The ratio of equilibrium constant \(K_p\) at two different temperatures is according to van’t Hoff’s equation equal to:

    \[ \ln\frac{K_1}{K_2} = \frac{\overline{\Delta H}}{R} = \left(\frac{1}{T_2}-\frac{1}{T_1}\right) \]

    \[ (\ln = 2.303\log) \]

    Calculate the mean value \(\overline{\Delta H}\) for reaction heat in the temperature range of 800 – 1000 K.

    4.6 What influence will have an elevation of temperature and pressure on the conversion degree of ethane?

    4.7 Calculate the relative error of the calculation.

    Official English text of the problem.

  5. Problem 5: Liquid compound X from coal tar

    A certain liquid organic compound X (being present in coal tar) with a mass of 1.06 g was burned to produce 0.90 g of water and 3.52 g of carbon dioxide. Its vapours were 3.79 times as dense as nitrogen.

    The compound X was oxidised by a hot mixture of \(\ce{CrO3 + H2SO4}\). A colourless crystalline substance A was isolated from the reaction mixture. It was soluble in an aqueous solution of \(\ce{NaOH}\) or \(\ce{NaHCO3}\).

    Compound A when heated loses water and converts to compound B. Condensation of compound B with phenol in the presence of \(\ce{H2SO4}\) or \(\ce{ZnCl2}\) yields a substance Y which is very often used as an acid-base indicator.

    Both compound A and compound B when heated with an access of 1-butanol (some drops of a concentrated \(\ce{H2SO4}\) solution are added) gives the same liquid compound C.

    If accepted that a carbon atom shows a tendency to form four bonds in organic compounds, it is possible to write formally two different formulas for the compound X. The formulas written in this way do not correspond, however, to the chemical structure of the molecule X because up to date nobody has succeeded in the preparation of the two hypothetical isomers.

    Haayman and Witbaut carried out in 1941 an ozonisation of the compound X in a \(\ce{CH3Cl}\) solution. After hydrolysis of ozonides, it was found that the water layer contains three different organic compounds in a molar ratio of D : E : F = 3 : 2 : 1. Only two of them formed new compounds G and H by a mild oxidation, the third one remains unchanged under these conditions but the effect of stronger oxidising agents as \(\ce{H2O2}\) for example, results in forming a well known liquid compound I with a characteristic sharp smell.

    A sample of a pure anhydrous compound G was dissolved in an aqueous 1-molar solution of \(\ce{H2SO4}\) and the resulting solution was titrated with a volumetric 0.05-molar \(\ce{KMnO4}\) solution. An amount of 0.288 g of substance G required 25.6 \(\mathrm{cm^3}\) of the \(\ce{KMnO4}\) solution.

    Problems:

    5.1 Write the summary formula for the compound X.

    5.2 Based on the information and data in the task, write chemical equations for the reactions by which products A, B, and C are formed.

    5.3 Give the name for compound Y and write the equation of its synthesis. Write its structural formula and colour in both acidic and basic solutions.

    5.4 Write two formal structural formulas for the compound X as well as a more correct structural formula according to the latest findings.

    5.5 Write the chemical equation for the ozonolysis of compound X by which the fact can be explained why compounds D, E, and F are after hydrolysis of ozonides present in water layer in a molar ration of 3 : 2 : 1.

    5.6 Write chemical equations for the reactions of formation of compounds G, H, and I.

    5.7 Write both formal structural formulas used before, and the more correct modern structural formula of another liquid organic compound if you know that the compound is also present in coal tar and it is a derivative of compound X. What name of a known chemist is connected in the history with the formula of this basic compound? What are the products obtained by its ozonolysis?

    5.8 Write the names of the substances X, Y, A – I under the corresponding compounds in the equations.

    Note:

    Use in your calculation:

    \(M(\ce{C}) = 12\ \mathrm{g\,mol^{-1}};\quad M(\ce{O}) = 16\ \mathrm{g\,mol^{-1}};\quad M(\ce{H}) = 1\ \mathrm{g\,mol^{-1}};\quad M(\ce{KMnO4}) = 158\ \mathrm{g\,mol^{-1}}.\)

    Official English text of the problem.