5 theoretical problems 3 practical problems

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1 6 th 5 theoretical problems practical problems

2 THE SIXTH INTERNATINAL CHEMISTRY LYMPIAD 1 10 JULY 197, BUCURESTI, RMANIA THERETICAL PRBLEMS PRBLEM 1 By electrochemical decomposition of water, there are in an electric circuit a voltmeter, platinum electrodes and a battery containing ten galvanic cells connected in series, each of it having the voltage of 1.5 V and internal resistance of 0. Ω. The resistance of the voltmeter is 0.5 Ω and the polarisation voltage of the battery is 1.5 V. Electric current flows for 8 hours, 56 minutes and 7 seconds through the electrolyte. Hydrogen obtained in this way was used for a synthesis with another substance, thus forming a gaseous substance A which can be converted by oxidation with oxygen via oxide to substance B. By means of substance B it is possible to prepare substance C from which after reduction by hydrogen substance D can be obtained. Substance D reacts at 180 C with a concentration solution of sulphuric acid to produce sulphanilic acid. By diazotization and successive copulation with p-n,n-dimethylaniline, an azo dye, methyl orange is formed. Problems: 1. Write chemical equations for all the above mentioned reactions. 2. Calculate the mass of product D.. Give the exact chemical name for the indicator methyl orange. Show by means of structural formulas what changes take place in dependence on concentration of H + ions in the solution. Relative atomic masses: A r (N) = 1; A r () = 16; A r (C) = 12; A r (H) = 1. 6

3 SLUTIN 1. N 2 + H 2 2 NH (A) NH N + 6 H 2 2 N N 2 2 N 2 + H 2 + 1/2 2 2 HN (B) H 2 S + HN N2 + H 2 (C) N H e - NH H 2 (D) 180 C NH H S H S NH 2 + H 2 + H S NH 2 + HN + HCl H S N N Cl H 2 H S N N + Cl - + N - HCl 180 C H S N=N N '-dimethyl amino -azo benzene sulphonic acid 6

4 2. m = M F z I t F = C mol -1 I b Eb Ep = = R + b R v i ( V) V = A 0.5Ω+(10 0.Ω) b - number of batteries, E b E p - voltage of one battery, - polarisation voltage, R v R i - resistance of voltmeter, - internal resistance of one battery -1 1 g mol m (H 2) = A 2167 s = 1 g C mol From equations: 1 g H 2 i. e. 0.5 mol H 2 corresponds 1 mol NH... 1 mol HN... 1 mol C6 H 5 N mol C 6H 5 NH 2 (D) The mass of product D: m = n M = 1 g C 6 H 5 NH 2. (-) S N N N H + (-) S N N - H+ H N (+) 65

5 PRBLEM 2 Substance G can be prepared by several methods according to the following scheme: G H E HCN D K A Cl 2 B NH + HCN NH C H F Compound A is 8.60 mass % carbon, 8.10 % hydrogen, and.0 % oxygen. It reacts with a freshly prepared silver(i) oxide to form an undissolved salt. An amount of 1.81 g of silver(i) salt is formed from 0.7 g of compound A. Compound D contains 5.5 mass % of carbon, 9.09 % of hydrogen, and 6.7 % of oxygen. It combines with NaHS to produce a compound containing 21.6 % of sulphur. Problems: 1. Write summary as well as structural formulas of substances A and D. 2. Write structural formulas of substances B, C, E, F, and G.. Classify the reactions in the scheme marked by arrows and discuss more in detail reactions B G and D E.. Write structural formulas of possible isomers of substance G and give the type of isomerism. Relative atomic masses: A r (C) = 12; A r (H) = 1; A r () = 16; A r (Ag) = 108; A r (Na) = 2; A r (S) = 2. SLUTIN 1. Compound A : R-C + Ag R-CAg + H 2 A : (C x H y z ) n x : y : z = : : = 1 : 2 : If n =, then the summary formula of substance A is: C H

6 M(A) = 7 g mol -1 A = -CH 2 -C Compound D: (C p H q r ) n p : q : r = : : = HN CH C 1 : 2 : 0.5 CH C IV (C) NH 2 (G) If n = 2, then the summary formula of substance D is: C 2 H. M(D) = g mol -1 C + NaHS CH H S Na D = -CH Reaction: The reduction product contains 21.6 % of sulphur _ C (A) _ C Cl (B) I _ C K _ C II Cl (B) (G) _ C NH _ C III (B) Cl (C) NH 2 67

7 _ C HN _ C IV (C) NH 2 (G) HCN CH _ CH CN V (D) (E) _ CN + H, H _ C VI (E) (G) _ CH NH + HCN _ CN VII (D) (F) NH 2 _ CN H, H + _ C VIII (F) NH 2 (C) NH 2. I - substitution reaction II - substitution nucleophilic reaction III - substitution nucleophilic reaction IV - substitution reaction V - additive nucleophilic reaction VI - additive reaction, hydrolysis VII - additive reaction VIII - additive reaction, hydrolysis 68

8 . CH C CH 2 CH 2 C position isomerism CH C CH 2 C H CH CH 2 C CH 2 structural isomerism H H C C d(+) stereoisomerism (optical isomerism) C C l(-) CH C racemic mixture 69

9 PRBLEM The following 0.2 molar solutions are available: A: HCl B: HS C: C D: Na E: Problems: C F: CNa G: 1. Determine the concentration of H + ions in solution C. 2. Determine ph value in solution A. HP H: H 2 S. Write an equation for the chemical reaction that takes place when substances B and E are allowed to react and mark conjugate acid-base pairs.. Compare acid-base properties of substances A, B and C and determine which one will show the most basic properties. Explain your decision. 5. Write a chemical equation for the reaction between substances B and G, and explain the shift of equilibrium. 6. Write a chemical equation for the reaction between substances C and E, and explain the shift of equilibrium. 7. Calculate the volume of D solution which is required to neutralise 20.0 cm of H solution. 8. What would be the volume of hydrogen chloride being present in one litre of A solution if it were in gaseous state at a pressure of kpa and a temperature of 7 C? Ionisation constants: C + H 2 C - + H + K a = H 2 C + H 2 HC - + H + K a = HC - + H 2 2- HS + H 2 2- C + H + K a = S + H + K a = HP + H 2 Relative atomic masses: A r (Na) = 2; A r (S) = 2; A r () = P + H + K a =

10 SLUTIN 1. C + H 2 C - + H + K a [CH C ][H ] [H ] = = [CH C] c = Ka c = = [H ] mol dm 2. ph = - log [H + ] = - log 0.2 = 0.7. HS + C S + HC A 1 B 2 B 1 A 2. By comparison of the ionisation constants we get: K a (HCl) > K a ( HS ) > K a ( C) - Thus, the strength of the acids in relation to water decreases in the above given order. C - is the strongest conjugate base, whereas Cl - is the weakest one HS + HP H + S 2 P K (HS ) >> K (HP ) a - 2- a Equilibrium is shifted to the formation of H P and S C + C - + C C - + HC HC C - + H 2 C K a ( C) > K a (H 2 C ) > K a ( HC ) Equilibrium is shifted to the formation of C - a H 2 C. 7. n(h 2 S ) = c V = 0.2 mol dm dm = 0.00 mol n mol V ( 0.2 molar Na) = = = 0.0 dm c 0.2 mol dm 71

11 n R T 0.2 mol 8.1 J mol K 10 K V ( HCl) = = = p kpa 2.5 dm 72

12 PRBLEM A mixture contains two organic compounds, A and B. Both of them have in their molecules oxygen and they can be mixed together in arbitrary ratios. xidation of this mixture on cooling yields the only substance C that combines with NaHS. The ratio of the molar mass of the substance being formed in the reaction with NaHS to that of substance C, is equal to The mixture of substances A and B is burned in the presence of a stoichiometric amount of air (20 % 2 and 80 % of N 2 by volume) in an eudiometer to produce a mixture of gases with a total volume of 5.2 dm at STP. After the gaseous mixture is bubbled through a Ba() 2 solution, its volume is decreased by 15.6 %. Problems:.1 Write structural formulas of substance A and B..2 Calculate the molar ratio of substances A and B in the mixture. A r (C) = 12; A r () = 16; A r (S) = 2; A r (Na) = 2. SLUTIN.1 S Na R C H (R) + NaHS R C H (R) M r (C) M r (NaHS ) = 10 M r (C) + 10 Mr ( C) + 10 M ( C) r = M r (C) = 58 A... CH C... C 7

13 .2 At STP conditions the gaseous mixture can only contain C 2 and N 2. Carbon dioxide is absorbed in a barium hydroxide solution and therefore: B... C (a) V(C 2 ) = 5.2 dm = 0.8 dm (b) V(N 2 ) = 5.2 dm 0.8 dm =.592 dm (c) -CH- + 9/2 ( 2 + N 2 ) = C 2 + H N 2 (d) -C- + ( 2 + N 2 ) = C 2 + H N 2 Let us mark the amounts of substances as: n( -CH- ) = x n( -C- ) = y From equations (a), (c) and (d): (e) (x 22.) + (y 22.) = 0.8 From equations (b), (c) and (d): (f) (18x 22.) + (16y 22.) =.592 In solving equations (e) and (f) we get: x = mol y = 0.01 mol x y = 1 7

14 PRBLEM 5 A mixture of two metals found in Mendelejev's periodical table in different groups, reacted with 56 cm of hydrogen on heating (measured at STP conditions) to produce two ionic compounds. These compounds were allowed to react with 270 mg of water but only one third of water reacted. A basic solution was formed in which the content of hydroxides was 0 % by mass and at the same time deposited a precipitate with a mass that represented % of a total mass of the products formed by the reaction. After filtration the precipitate was heated and its mass decreased by 27 mg. When a stoichiometric amount of ammonium carbonate was added to the basic solution, a slightly soluble precipitate was obtained, at the same time ammonia was liberated and the content of hydroxides in the solution decreased to %. Problem: 5.1 Determine the metals in the starting mixture and their masses. SLUTIN Ionic hydrides are formed by combining of alkali metals or alkaline earth metals with hydrogen. In relation to the conditions in the task, there will be an alkali metal (M I ) as well as an alkaline earth metal (M II ) in the mixture. Equations: (1) M I + 1/2 H 2 M I H (2) M II + H 2 M II H 2 () M I H + H 2 M I + H 2 () M II H H 2 M II () H 2 reacted: 0.09 g H 2, i. e mol unreacted: 0.18 g H 2, i. e mol Since all hydroxides of alkali metals are readily soluble in water, the undissolved precipitate is M II () 2, however, it is slightly soluble in water, too. Thus, the mass of hydroxides dissolved in the solution: (5) m'(m I + M II () 2 ) = Z 75

15 Therefore: 0 = Z 100 Z Z = g (6) m'(m I + M II () 2 ) = g It represents 0.95 % of the total mass of the hydroxides, i. e. the total mass of hydroxides is as follows: (7) m'(m I + M II g 100 () 2 ) = 0.95 = g The mass of solid M II () 2 : (8) g g = g Heating: (9) M II () 2 M II + H 2 Decrease of the mass: g (H 2 ) (10) Mass of M II : 0.08 g In relation to (8), (9), and (10): II Mr ( M ) 0.08 = II M ( M ) r M r (M II ) = 56 g mol -1 M r (M II ) = M r (M II ) M r () = = 0 M II = Ca Precipitation with (NH C ): (11) Ca() 2 + (NH ) 2 C CaC + 2 NH + 2 H 2 According to (5) and (6) the mass of the solution was: 0.18 g g = g After precipitation with (NH ) 2 C : I m(m ) = 100 m (solution) Let us mark as n' the amount of substance of Ca() 2 being present in the solution. M(Ca() 2 ) = 7 g mol -1 Taking into account the condition in the task as well as equation (11), we get: ( n' ) = n' + 2n' 18 76

16 n' = mol The total amount of substance of Ca() 2 (both in the precipitate and in the solution): g (12) n(ca() 2) = mol = mol (i. e g) 1 7 g mol According to equations () and (): n(h 2 ) = 0.00 mol (for M II H 2 ) n(h 2 ) = mol (for M I H) n(m I ) = mol According to equations (7) and (11): m(m I ) = g 0.18 g = 0.0 g M I (M ) = n m I (M ) I (M ) = 0.0 g mol = 0 g mol 1 M I = Na Composition of the mixture: mol Ca mol Na or g Ca g Na 77

17 PRACTICAL PRBLEMS PRBLEM 1 (practical) Test tubes with unknown samples contain: - a salt of carboxylic acid, - a phenol, - a carbohydrate, - an amide. Determine the content of each test tube using reagents that are available on the laboratory desk. PRBLEM 2 (practical) Determine cations in solutions No 5, 6, 8 and 9 using the solution in test tube 7. Without using any indicator find out whether the solution in test tube 7 is an acid or a hydroxide. SLUTIN + Test tube: No 5 - NH ; No 6 - Hg 2+ ; No ; No 8 Fe + ; No 9 Cu 2+ PRBLEM (practical) The solution in test tube No 10 contains two cations and two anions. Prove those ions by means of reagents that are available on the laboratory desk. SLUTIN The solution in test tube No 10 contained: Ba 2+, Al +, Cl -, C 78

5 theoretical problems 3 practical problems

6 th 5 theoretical problems 3 practical problems THE 6 TH INTERNATINAL CHEMISTRY LYMPIAD, 1974 THE SIXTH INTERNATINAL CHEMISTRY LYMPIAD 1 10 JULY 1974, BUCURESTI, RMANIA THERETICAL PRBLEMS PRBLEM 1 By