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Tambaya 1 Rahoto
(a)(i) Name a suitable drying agent for the preparation of carbon (IV) oxide in the laboratory.
(ii) Using one chemical test, distinguish between carbon (II) oxide and carbon (IV) oxide.
(b)(i) Describe briefly how oxygen and nitrogen could be obtained separately from air on an industrial scale
(ii) State how a lighted splint can be used to distinguish between samples of oxygen and nitrogen.
(c)(i) Give one reason why bauxite is usually preferred as the ore for the extraction of aluminium.
(ii). List two main impurities. usually present in bauxite.
(iii) State the function of sodium hydroxide solution in the extraction of aluminium from its ore.
(iv) Explain briefly why it is difficult to extract aluminium by chemical reduction of aluminium oxide
(v) Write an equation for the reaction of aluminium oxide with aqueous sodium hydroxide.
(d) (i) The melting and boiling points of sodium chloride are 801 °C and 141.3 °C respectively. Explain briefly why sodium chloride does not conduct electricity at 25°C but does so between 801 °C and 1413 °C.
(ii) State the reason why sodium metal is stored under paraffin oil in the laboratory.
(e)(i) State what would be observed when aqueous sodium trioxocarbonate(IV) is added to a solution containing iron (III) ions
(ii) Write a balanced equation for the reaction in (e)(i).
(a)(i) A suitable drying agent for carbon(IV) oxide is concentrated tetraoxosulphate(VI) acid, \(H_2SO_4\). (Fused calcium chloride is also acceptable; solid NaOH or CaO must not be used because they react with \(CO_2\).)
(ii) To distinguish carbon(II) oxide from carbon(IV) oxide, bubble each gas through limewater (calcium hydroxide solution): carbon(IV) oxide turns limewater milky, whereas carbon(II) oxide does not. (Alternatively, carbon(II) oxide burns with a blue flame; carbon(IV) oxide does not burn.)
(b)(i) Oxygen and nitrogen are obtained industrially by the fractional distillation of liquid air: air is filtered, and its carbon(IV) oxide and water vapour removed; it is then compressed and cooled repeatedly until it liquefies; the liquid air is allowed to warm in a fractionating column, and because nitrogen has a lower boiling point (\(-196^{\circ}C\)) it boils off first, leaving oxygen (b.p. \(-183^{\circ}C\)) behind.
(ii) Using a lighted splint: oxygen relights (rekindles) a glowing splint and makes a lighted splint burn more brightly, whereas nitrogen puts out (extinguishes) a lighted splint.
(c)(i) Bauxite is preferred as the ore for aluminium extraction because it is the most abundant and richest ore of aluminium (high proportion of aluminium oxide) and is cheap and readily available.
(ii) Two main impurities in bauxite: iron(III) oxide (\(Fe_2O_3\)) and silicon(IV) oxide/silica (\(SiO_2\)).
(iii) Sodium hydroxide solution dissolves the amphoteric aluminium oxide (forming soluble sodium aluminate), leaving the insoluble impurities behind to be filtered off; that is, it purifies the ore.
(iv) It is difficult to extract aluminium by chemical reduction of its oxide because aluminium is very reactive and high in the electrochemical series; it has a strong affinity for oxygen, so ordinary reducing agents (carbon, carbon(II) oxide, hydrogen) cannot reduce aluminium oxide. Electrolysis is used instead.
(v) \[Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O\]
(d)(i) Sodium chloride does not conduct electricity at 25\(^{\circ}\)C because in the solid state the ions are held firmly in fixed positions in the crystal lattice and are not free to move. Between 801\(^{\circ}\)C and 1413\(^{\circ}\)C it is molten, so the ions become free (mobile) and can move to carry the current.
(ii) Sodium is stored under paraffin oil because it is very reactive and would otherwise react rapidly with the oxygen and moisture of the air; the paraffin excludes air and water from the metal.
(e)(i) When aqueous sodium trioxocarbonate(IV) is added to a solution containing iron(III) ions, a reddish-brown precipitate (of iron(III) hydroxide) is formed, with effervescence (bubbles) of carbon(IV) oxide.
(ii) \[2FeCl_3 + 3Na_2CO_3 + 3H_2O \rightarrow 2Fe(OH)_3 + 6NaCl + 3CO_2\]
Bayanin Amsa
(a)(i) A suitable drying agent for carbon(IV) oxide is concentrated tetraoxosulphate(VI) acid, \(H_2SO_4\). (Fused calcium chloride is also acceptable; solid NaOH or CaO must not be used because they react with \(CO_2\).)
(ii) To distinguish carbon(II) oxide from carbon(IV) oxide, bubble each gas through limewater (calcium hydroxide solution): carbon(IV) oxide turns limewater milky, whereas carbon(II) oxide does not. (Alternatively, carbon(II) oxide burns with a blue flame; carbon(IV) oxide does not burn.)
(b)(i) Oxygen and nitrogen are obtained industrially by the fractional distillation of liquid air: air is filtered, and its carbon(IV) oxide and water vapour removed; it is then compressed and cooled repeatedly until it liquefies; the liquid air is allowed to warm in a fractionating column, and because nitrogen has a lower boiling point (\(-196^{\circ}C\)) it boils off first, leaving oxygen (b.p. \(-183^{\circ}C\)) behind.
(ii) Using a lighted splint: oxygen relights (rekindles) a glowing splint and makes a lighted splint burn more brightly, whereas nitrogen puts out (extinguishes) a lighted splint.
(c)(i) Bauxite is preferred as the ore for aluminium extraction because it is the most abundant and richest ore of aluminium (high proportion of aluminium oxide) and is cheap and readily available.
(ii) Two main impurities in bauxite: iron(III) oxide (\(Fe_2O_3\)) and silicon(IV) oxide/silica (\(SiO_2\)).
(iii) Sodium hydroxide solution dissolves the amphoteric aluminium oxide (forming soluble sodium aluminate), leaving the insoluble impurities behind to be filtered off; that is, it purifies the ore.
(iv) It is difficult to extract aluminium by chemical reduction of its oxide because aluminium is very reactive and high in the electrochemical series; it has a strong affinity for oxygen, so ordinary reducing agents (carbon, carbon(II) oxide, hydrogen) cannot reduce aluminium oxide. Electrolysis is used instead.
(v) \[Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O\]
(d)(i) Sodium chloride does not conduct electricity at 25\(^{\circ}\)C because in the solid state the ions are held firmly in fixed positions in the crystal lattice and are not free to move. Between 801\(^{\circ}\)C and 1413\(^{\circ}\)C it is molten, so the ions become free (mobile) and can move to carry the current.
(ii) Sodium is stored under paraffin oil because it is very reactive and would otherwise react rapidly with the oxygen and moisture of the air; the paraffin excludes air and water from the metal.
(e)(i) When aqueous sodium trioxocarbonate(IV) is added to a solution containing iron(III) ions, a reddish-brown precipitate (of iron(III) hydroxide) is formed, with effervescence (bubbles) of carbon(IV) oxide.
(ii) \[2FeCl_3 + 3Na_2CO_3 + 3H_2O \rightarrow 2Fe(OH)_3 + 6NaCl + 3CO_2\]
Tambaya 2 Rahoto
(a) What are nucleons?
(b) State Graham's law of diffusion.
(c) Explain briefly why aluminium does not corrode easily.
(d) State three examples of periodic properties.
(e) State two reasons why real gases deviate from ideal gas behaviour.
(f) List three uses of fractional distillation in industry.
(g) What factors determine the selective discharge of ions at the electrodes during electrolysis?
(h) State the type of reaction represented by each of the following equations:
(i) C\(_2\)H\(_6\) + Br\(_2\) ---> C\(_2\)H\(_5\)Br + HBr;
(ii) C\(_2\)H\(_4\) + .Br\(_2\) ---> CH\(_4\)Br\(_2\)
(i) Name the products formed when butane burns in limited supply of air.
(j) List three methods of separating a solid from a liquid.
(a) Nucleons are the particles found in the nucleus of an atom, that is, the protons and neutrons.
(b) Graham's law of diffusion: at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or of its molar mass).
\[\text{rate} \propto \frac{1}{\sqrt{\rho}}\]
(c) Aluminium does not corrode easily because it forms a thin, tough, coherent and impervious layer of aluminium oxide on its surface. This oxide layer sticks firmly to the metal and prevents air and moisture from reaching the metal underneath, protecting it from further attack.
(d) Three periodic properties: atomic radius, ionization energy and electronegativity (electron affinity and metallic/non-metallic character are also acceptable).
(e) Two reasons real gases deviate from ideal behaviour:
(f) Three industrial uses of fractional distillation: refining of crude petroleum into its various fractions; separation of liquid air into oxygen and nitrogen; separation of ethanol from a water-ethanol mixture (rectification of spirits).
(g) Factors that determine selective discharge of ions during electrolysis: the position of the ion in the electrochemical (discharge) series, the concentration of the ion in the electrolyte, and the nature of the electrode used.
(h) Types of reaction:
(i) When butane burns in a limited supply of air, incomplete combustion occurs, giving carbon(II) oxide (and carbon/soot) and water vapour.
(j) Three methods of separating a solid from a liquid: filtration, evaporation and decantation (centrifugation and crystallization are also acceptable).
Bayanin Amsa
(a) Nucleons are the particles found in the nucleus of an atom, that is, the protons and neutrons.
(b) Graham's law of diffusion: at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or of its molar mass).
\[\text{rate} \propto \frac{1}{\sqrt{\rho}}\]
(c) Aluminium does not corrode easily because it forms a thin, tough, coherent and impervious layer of aluminium oxide on its surface. This oxide layer sticks firmly to the metal and prevents air and moisture from reaching the metal underneath, protecting it from further attack.
(d) Three periodic properties: atomic radius, ionization energy and electronegativity (electron affinity and metallic/non-metallic character are also acceptable).
(e) Two reasons real gases deviate from ideal behaviour:
(f) Three industrial uses of fractional distillation: refining of crude petroleum into its various fractions; separation of liquid air into oxygen and nitrogen; separation of ethanol from a water-ethanol mixture (rectification of spirits).
(g) Factors that determine selective discharge of ions during electrolysis: the position of the ion in the electrochemical (discharge) series, the concentration of the ion in the electrolyte, and the nature of the electrode used.
(h) Types of reaction:
(i) When butane burns in a limited supply of air, incomplete combustion occurs, giving carbon(II) oxide (and carbon/soot) and water vapour.
(j) Three methods of separating a solid from a liquid: filtration, evaporation and decantation (centrifugation and crystallization are also acceptable).
Tambaya 3 Rahoto
(a)(i) Determine the oxidation number of sulphur in Na\(_2\)S\(_2\)O\(_3\)
(ii) Name the allotropes of sulphur.
(iii) State two ways in which the structure of graphite and diamond are similar.
(b)(i) Name two green-house gases.
(ii) State one effect of an increased level of green-house gases on the environment.
(iii) State one source from which nitrogen (I) oxide is released into the environment.
(iv) Write a chemical equation to show the effect of heat on each of the following compounds: I. KNO\(_{3(s)}\) II. AgNO\(_{3(s)}\)
(C)(i)Describe briefly how pure crystals of calcium chloride could be obtained from a solution of calcium chloride
ii) Explain briefly each of the following observations:
I. ammonia gas is highly soluble in water;
II. boiling ploint of chlorine is lower than that of iodine
(d) Consider the reaction represented by the following equation:- 2NaCl + H\(_2\)SO\(_{4(s)}\) \(\to\) Na\(_2\)SO\(_{4(s)}\) + 2HCl\(_{(g)}\)
Calculate the volume of HCl gas that can be obtained at s.t.p. from 5.85 g of sodium. chloride. [ Na = 23.0, Cl = 35.5, Molar volume of gas at s.t.p. = 22.4 dm\(^3\)]
(a)(i) Oxidation number of sulphur in \(Na_2S_2O_3\). Let it be \(x\); Na is \(+1\), O is \(-2\), and the compound is neutral:
\[2(+1) + 2x + 3(-2) = 0\]
\[2 + 2x - 6 = 0 \;\Rightarrow\; 2x = 4 \;\Rightarrow\; x = +2\]
Sulphur has an average oxidation number of +2.
(ii) The allotropes of sulphur are rhombic sulphur and monoclinic sulphur (plastic sulphur is also sometimes listed).
(iii) Two ways in which graphite and diamond are similar: both are allotropes of carbon (made up of carbon atoms only), and both are giant covalent (macromolecular) structures with very high melting points.
(b)(i) Two greenhouse gases: carbon(IV) oxide (\(CO_2\)) and methane (\(CH_4\)). (Water vapour and CFCs are also acceptable.)
(ii) One effect of an increased level of greenhouse gases: global warming (a rise in the average atmospheric temperature, leading to melting of ice caps and rising sea levels).
(iii) One source of nitrogen(I) oxide, \(N_2O\): bacterial action on nitrogenous fertilizers in the soil (or from vehicle exhausts).
(iv) Effect of heat:
I. \[2KNO_3 \rightarrow 2KNO_2 + O_2\]
II. \[2AgNO_3 \rightarrow 2Ag + 2NO_2 + O_2\]
(c)(i) To obtain pure crystals of calcium chloride from its solution: evaporate the solution carefully to concentrate it to the point of crystallization, then allow it to cool so that crystals form; filter off the crystals and dry them (calcium chloride is deliquescent, so drying should be done quickly in a desiccator).
(ii) I. Ammonia is highly soluble in water because ammonia molecules are polar and form hydrogen bonds with water (and react with water to form ammonium hydroxide).
II. The boiling point of chlorine is lower than that of iodine because chlorine molecules are smaller and lighter, so the van der Waals (intermolecular) forces between them are weaker than the stronger van der Waals forces between the larger iodine molecules; less energy is needed to separate chlorine molecules.
(d) \(2NaCl + H_2SO_4 \rightarrow Na_2SO_4 + 2HCl\). From 5.85 g of NaCl:
Molar mass of \(NaCl = 23 + 35.5 = 58.5\,g\,mol^{-1}\)
\[\text{moles of } NaCl = \frac{5.85}{58.5} = 0.1\,mol\]
From the equation, 2 moles NaCl give 2 moles HCl (1:1), so moles of \(HCl = 0.1\,mol\).
\[\text{volume at s.t.p.} = 0.1 \times 22.4 = 2.24\,dm^3\]
Bayanin Amsa
(a)(i) Oxidation number of sulphur in \(Na_2S_2O_3\). Let it be \(x\); Na is \(+1\), O is \(-2\), and the compound is neutral:
\[2(+1) + 2x + 3(-2) = 0\]
\[2 + 2x - 6 = 0 \;\Rightarrow\; 2x = 4 \;\Rightarrow\; x = +2\]
Sulphur has an average oxidation number of +2.
(ii) The allotropes of sulphur are rhombic sulphur and monoclinic sulphur (plastic sulphur is also sometimes listed).
(iii) Two ways in which graphite and diamond are similar: both are allotropes of carbon (made up of carbon atoms only), and both are giant covalent (macromolecular) structures with very high melting points.
(b)(i) Two greenhouse gases: carbon(IV) oxide (\(CO_2\)) and methane (\(CH_4\)). (Water vapour and CFCs are also acceptable.)
(ii) One effect of an increased level of greenhouse gases: global warming (a rise in the average atmospheric temperature, leading to melting of ice caps and rising sea levels).
(iii) One source of nitrogen(I) oxide, \(N_2O\): bacterial action on nitrogenous fertilizers in the soil (or from vehicle exhausts).
(iv) Effect of heat:
I. \[2KNO_3 \rightarrow 2KNO_2 + O_2\]
II. \[2AgNO_3 \rightarrow 2Ag + 2NO_2 + O_2\]
(c)(i) To obtain pure crystals of calcium chloride from its solution: evaporate the solution carefully to concentrate it to the point of crystallization, then allow it to cool so that crystals form; filter off the crystals and dry them (calcium chloride is deliquescent, so drying should be done quickly in a desiccator).
(ii) I. Ammonia is highly soluble in water because ammonia molecules are polar and form hydrogen bonds with water (and react with water to form ammonium hydroxide).
II. The boiling point of chlorine is lower than that of iodine because chlorine molecules are smaller and lighter, so the van der Waals (intermolecular) forces between them are weaker than the stronger van der Waals forces between the larger iodine molecules; less energy is needed to separate chlorine molecules.
(d) \(2NaCl + H_2SO_4 \rightarrow Na_2SO_4 + 2HCl\). From 5.85 g of NaCl:
Molar mass of \(NaCl = 23 + 35.5 = 58.5\,g\,mol^{-1}\)
\[\text{moles of } NaCl = \frac{5.85}{58.5} = 0.1\,mol\]
From the equation, 2 moles NaCl give 2 moles HCl (1:1), so moles of \(HCl = 0.1\,mol\).
\[\text{volume at s.t.p.} = 0.1 \times 22.4 = 2.24\,dm^3\]
Tambaya 4 Rahoto
(a) Consider the following atoms: \(^R_T\)X; \(^S_T\)X.
(i) State the phenomenon exhibited by the two atoms.
(ii) What is the difference between the atoms?
(iii) Give two examples of elements that exhibit the phenomenon stated in (ai)
(iv) lf T is 17, write the electron configuration of the element
(b)(i) State two differences between metals and-non-metals with respect to their:
I. physical properties;
2. chemical properties.
(ii) Give one example of each for the following compounds:
I. an amphoteric oxide;
II. a hydride which evolves hydrogen when reacted with water;
Ill. a trioxocarbonate (IV) salt which is readily decomposed on heating;
IV. a chloride salt which is readily hydrolyzed in water.
(c)(i) State three characteristic properties of transition metals.
(ii) Write the electron configuration of \(_{30}Zn\)
(iii.) Explain briefly why zinc is not considered as a typical transition element.
(d) Consider the reaction represented by the following equation:- Na\(_2\)CO\(_{3(aq)}\) + MgCl\(_{2(aq)}\) ----> 2NaCl\(_{(aq)}\) + MgCO\(_{3(aq)}\). Calculate the mass of sodium trioxocarbonate (IV) needed to produce 3.36 of magnesium trioxocarbonate (IV). [C = 12.0, O = 16.0, Na = 23.0, Mg = 24:0 ]
(a) The atoms \(^R_TX\) and \(^S_TX\) have the same atomic number T but different mass numbers R and S.
(b)(i) Metals compared with non-metals:
| Metals | Non-metals | |
|---|---|---|
| Physical | Shiny/lustrous, malleable and ductile, good conductors of heat and electricity, mostly solids with high melting points. | Dull, brittle (if solid), poor conductors (except graphite), many are gases or low-melting solids. |
| Chemical | Form basic oxides; are electropositive, lose electrons to form positive ions (reducing agents). | Form acidic oxides; are electronegative, gain electrons to form negative ions (oxidizing agents). |
(ii) Examples:
(c)(i) Three characteristic properties of transition metals: they show variable oxidation states, form coloured ions/compounds, and act as catalysts (they also form complex ions).
(ii) Electron configuration of \(_{30}Zn\): 2, 8, 18, 2 (\(1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^2\)).
(iii) Zinc is not regarded as a typical transition element because its 3d subshell is completely filled (3d^{10}); it shows only one oxidation state (+2) and forms mainly white/colourless compounds, so it lacks the characteristic transition-metal properties.
(d) \(Na_2CO_3 + MgCl_2 \rightarrow 2NaCl + MgCO_3\). To produce 3.36 g of \(MgCO_3\):
Molar mass of \(MgCO_3 = 24 + 12 + (3 \times 16) = 84\,g\,mol^{-1}\)
\[\text{moles of } MgCO_3 = \frac{3.36}{84} = 0.04\,mol\]
From the equation, 1 mole \(Na_2CO_3\) gives 1 mole \(MgCO_3\), so moles of \(Na_2CO_3 = 0.04\,mol\).
Molar mass of \(Na_2CO_3 = (2 \times 23) + 12 + (3 \times 16) = 106\,g\,mol^{-1}\)
\[\text{mass of } Na_2CO_3 = 0.04 \times 106 = 4.24\,g\]
Bayanin Amsa
(a) The atoms \(^R_TX\) and \(^S_TX\) have the same atomic number T but different mass numbers R and S.
(b)(i) Metals compared with non-metals:
| Metals | Non-metals | |
|---|---|---|
| Physical | Shiny/lustrous, malleable and ductile, good conductors of heat and electricity, mostly solids with high melting points. | Dull, brittle (if solid), poor conductors (except graphite), many are gases or low-melting solids. |
| Chemical | Form basic oxides; are electropositive, lose electrons to form positive ions (reducing agents). | Form acidic oxides; are electronegative, gain electrons to form negative ions (oxidizing agents). |
(ii) Examples:
(c)(i) Three characteristic properties of transition metals: they show variable oxidation states, form coloured ions/compounds, and act as catalysts (they also form complex ions).
(ii) Electron configuration of \(_{30}Zn\): 2, 8, 18, 2 (\(1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^2\)).
(iii) Zinc is not regarded as a typical transition element because its 3d subshell is completely filled (3d^{10}); it shows only one oxidation state (+2) and forms mainly white/colourless compounds, so it lacks the characteristic transition-metal properties.
(d) \(Na_2CO_3 + MgCl_2 \rightarrow 2NaCl + MgCO_3\). To produce 3.36 g of \(MgCO_3\):
Molar mass of \(MgCO_3 = 24 + 12 + (3 \times 16) = 84\,g\,mol^{-1}\)
\[\text{moles of } MgCO_3 = \frac{3.36}{84} = 0.04\,mol\]
From the equation, 1 mole \(Na_2CO_3\) gives 1 mole \(MgCO_3\), so moles of \(Na_2CO_3 = 0.04\,mol\).
Molar mass of \(Na_2CO_3 = (2 \times 23) + 12 + (3 \times 16) = 106\,g\,mol^{-1}\)
\[\text{mass of } Na_2CO_3 = 0.04 \times 106 = 4.24\,g\]
Tambaya 5 Rahoto
(a) (i) Draw the structures of the isomers of the alkene with molecular formurat C\(_4\)H\(_8\)
(ii) State the class of alkanols to which each of the following compounds belongs:
I. CH\(_3\)C(CH\(_3\))\(_2\)OH;
II. CH\(_3\)CH(CH\(_3\))CH\(_2\)OH;
III. CH\(_3\)CH\(_2\)CH(CH\(_3\))OH.
(b) (i) Write the formulae of the products formed in the following reactions:
I. CH\(_3\)CH\(_2\)COOH \(\frac{K_{(s)}}{}\)
II. CH\(_3\)CH\(_2\)COOH. \(\frac{C_4H_6OH, heat}{Conc.H_2SO_4}\)
III. CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)OH \(\frac{H^+/KMnO_4}{(excess)}\)
(ii) Name the major product(s) of each of the reactions in (b)(i).
(c) A gaseous hydrocarbon R of mass 7.0 g occupies a volume of 2.24 dm\(^3\) at s. t.p. If the percentage composition by mass of hydrogen is 14.3, determine its:
(i) empirical formula;
(ii) molecular formula. [ H = 1.00, C = 12.0, Molar volume of gas at s.t.p, = 22.4 dm\(^3\) ]
(d) Define structural isomerism.
(a)(i) Isomers of the alkene \(C_4H_8\):
(But-2-ene also shows cis and trans geometric forms.)
(ii) Class of alkanols:
(b)(i) Products:
(ii) Major products: I. potassium propanoate (\(CH_3CH_2COOK\)); II. ethyl propanoate (\(CH_3CH_2COOC_2H_5\)); III. butanoic acid (\(CH_3CH_2CH_2COOH\)).
(c) Hydrocarbon R: mass 7.0 g, volume 2.24 dm\(^3\) at s.t.p.
\[\text{moles of R} = \frac{2.24}{22.4} = 0.1\,mol\]
\[\text{molar mass} = \frac{7.0}{0.1} = 70\,g\,mol^{-1}\]
Percentage of H = 14.3, so percentage of C = 85.7.
(i) Empirical formula:
| C | H | |
|---|---|---|
| Mole ratio | \(\frac{85.7}{12}=7.14\) | \(\frac{14.3}{1}=14.3\) |
| Divide by 7.14 | 1 | 2 |
Empirical formula = \(CH_2\).
(ii) Molecular formula: \((CH_2)_n = 70\), so \(14n = 70\), \(n = 5\). Molecular formula = \(C_5H_{10}\).
(d) Structural isomerism is the existence of two or more compounds that have the same molecular formula but different structural formulae (different arrangement of their atoms).
Bayanin Amsa
(a)(i) Isomers of the alkene \(C_4H_8\):
(But-2-ene also shows cis and trans geometric forms.)
(ii) Class of alkanols:
(b)(i) Products:
(ii) Major products: I. potassium propanoate (\(CH_3CH_2COOK\)); II. ethyl propanoate (\(CH_3CH_2COOC_2H_5\)); III. butanoic acid (\(CH_3CH_2CH_2COOH\)).
(c) Hydrocarbon R: mass 7.0 g, volume 2.24 dm\(^3\) at s.t.p.
\[\text{moles of R} = \frac{2.24}{22.4} = 0.1\,mol\]
\[\text{molar mass} = \frac{7.0}{0.1} = 70\,g\,mol^{-1}\]
Percentage of H = 14.3, so percentage of C = 85.7.
(i) Empirical formula:
| C | H | |
|---|---|---|
| Mole ratio | \(\frac{85.7}{12}=7.14\) | \(\frac{14.3}{1}=14.3\) |
| Divide by 7.14 | 1 | 2 |
Empirical formula = \(CH_2\).
(ii) Molecular formula: \((CH_2)_n = 70\), so \(14n = 70\), \(n = 5\). Molecular formula = \(C_5H_{10}\).
(d) Structural isomerism is the existence of two or more compounds that have the same molecular formula but different structural formulae (different arrangement of their atoms).
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