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Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen. Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l) Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 2. Calculate the volume of air required to completely reacts with 50cm3 of Hydrogen.(assume Oxygen is 21% by volume of air) Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l)
21 21 Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 21% = 25cm3 100% = 100 x 25 = 21 3.If 5cm3 of a hydrocarbon CxHy burn in 15cm3 of Oxygen to form 10cm3 of Carbon(IV)oxide and 10cm3 of water vapour/steam, obtain the equation for the reaction and hence find the value of x and y in CxHy. Chemical equation: CxHy (g) + O2 (g) -> H2O(g) + CO2(g) Volumes 5cm3 : 15cm3 : 10cm3 : 10cm3 Volume ratios 5cm3 : 15cm3 : 10cm3 : 10cm3 (divide by lowest volume) 5 5 5 5 Reacting volume ratios 1volume 3 volume 2 volume 2 volume Balanced chemical equation: CxHy (g) + 3O2 (g) -> 2H2O(g) + 2CO2(g) If “4H” are in 2H2O(g) the y=4 If “2C” are in 2CO2 (g) the x=2 Thus(i) chemical formula of hydrocarbon = C2H4 (ii) chemical name of hydrocarbon = Ethene 4.100cm3 of nitrogen (II)oxide NO combine with 50cm3 of Oxygen to form 100cm3 of a single gaseous compound of nitrogen. All volumes measured at the same temperature and pressure. Obtain the equation for the reaction and name the gaseous product.
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Chemical equation: CxHy (g) + O2 (g) -> H2O(g) + CO2(g) Volumes 5cm3 : 15cm3 : 10cm3 : 10cm3 Volume ratios 5cm3 : 15cm3 : 10cm3 : 10cm3 (divide by lowest volume) 5 5 5 5 Reacting volume ratios 1volume 3 volume 2 volume 2 volume Balanced chemical equation: CxHy (g) + 3O2 (g) -> 2H2O(g) + 2CO2(g) If “4H” are in 2H2O(g) the y=4 If “2C” are in 2CO2 (g) the x=2 Thus(i) chemical formula of hydrocarbon = C2H4 (ii) chemical name of hydrocarbon = Ethene 4.100cm3 of nitrogen (II)oxide NO combine with 50cm3 of Oxygen to form 100cm3 of a single gaseous compound of nitrogen. All volumes measured at the same temperature and pressure. Obtain the equation for the reaction and name the gaseous product. Chemical equation: NO (g) + O2 (g) -> NOx Volumes 100cm3 : 50cm3 : 100 Volume ratios 100cm3 : 50cm3 : 100cm3 (divide by lowest volume) 50 50 50 Reacting volume ratios 2volume 1 volume 2 volume Balanced chemical equation: 2 NO (g) + O2 (g) -> 2NO x(g) Thus(i) chemical formula of the nitrogen compound = 2 NO2 (ii) chemical name of compound = Nitrogen(IV)oxide 5.When 15cm3 of a gaseous hydrocarbon was burnt in 100cm3 of Oxygen ,the resulting gaseous mixture occupied70cm3 at room temperature and pressure. 22 22 When the gaseous mixture was passed through, potassium hydroxide its volume decreased to 25cm3.
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Obtain the equation for the reaction and name the gaseous product. Chemical equation: NO (g) + O2 (g) -> NOx Volumes 100cm3 : 50cm3 : 100 Volume ratios 100cm3 : 50cm3 : 100cm3 (divide by lowest volume) 50 50 50 Reacting volume ratios 2volume 1 volume 2 volume Balanced chemical equation: 2 NO (g) + O2 (g) -> 2NO x(g) Thus(i) chemical formula of the nitrogen compound = 2 NO2 (ii) chemical name of compound = Nitrogen(IV)oxide 5.When 15cm3 of a gaseous hydrocarbon was burnt in 100cm3 of Oxygen ,the resulting gaseous mixture occupied70cm3 at room temperature and pressure. 22 22 When the gaseous mixture was passed through, potassium hydroxide its volume decreased to 25cm3. (a)What volume of Oxygen was used during the reaction.(1mk) Volume of Oxygen used =100-25 =75cm3√ (P was completely burnt) (b)Determine the molecular formula of the hydrocarbon(2mk) CxHy + O2 -> xCO2 + yH2O 15cm3 : 75cm3 15 15 1 : 3√ => 1 atom of C react with 6 (3x2)atoms of Oxygen Thus x = 1 and y = 2 => P has molecula formula CH4√ (g) Ionic equations An ionic equation is a chemical statement showing the movement of ions (cations and anions ) from reactants to products. Solids, gases and liquids do not ionize/dissociate into free ions. Only ionic compounds in aqueous/solution or molten state ionize/dissociate into free cations and anions (ions) An ionic equation is usually derived from a stoichiometric equation by using the following guidelines Guidelines for writing ionic equations 1.Write the balanced stoichiometric equation 2.Indicate the state symbols of the reactants and products 3.Split into cations and anions all the reactants and products that exist in aqueous state. 4.Cancel out any cation and anion that appear on both the product and reactant side. 5. Rewrite the chemical equation. It is an ionic equation.
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5. Rewrite the chemical equation. It is an ionic equation. Practice (a)Precipitation of an insoluble salt
23 23 All insoluble salts are prepared in the laboratory from double decomposition /precipitation. This involves mixing two soluble salts to form one soluble and one insoluble salt 1. When silver nitrate(V) solution is added to sodium chloride solution,sodium nitrate(V) solution and a white precipitate of silver chloride are formed. Balanced stoichiometric equation AgNO3(aq) + NaCl(aq) -> AgCl (s) + NaNO3 (aq) Split reactants product existing in aqueous state as cation/anion Ag+(aq) + NO3- (aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq)+ NO3- (aq) Cancel out ions appearing on reactant and product side Ag+(aq) + NO3- (aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq)+ NO3- (aq) Rewrite the equation Ag+(aq) + Cl-(aq) -> AgCl(s) (ionic equation) 2. When barium nitrate(V) solution is added to copper(II)sulphate(VI) solution, copper(II) nitrate(V) solution and a white precipitate of barium sulphate(VI) are formed. Balanced stoichiometric equation Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq) Split reactants product existing in aqueous state as cation/anion Ba2+(aq) + 2NO3- (aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3- (aq)+ Cu2+(aq) Cancel out ions appearing on reactant and product side Ba2+(aq) + 2NO3- (aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3- (aq) + Cu2+(aq) Rewrite the equation Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ionic equation) 3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide.
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Balanced stoichiometric equation AgNO3(aq) + NaCl(aq) -> AgCl (s) + NaNO3 (aq) Split reactants product existing in aqueous state as cation/anion Ag+(aq) + NO3- (aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq)+ NO3- (aq) Cancel out ions appearing on reactant and product side Ag+(aq) + NO3- (aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq)+ NO3- (aq) Rewrite the equation Ag+(aq) + Cl-(aq) -> AgCl(s) (ionic equation) 2. When barium nitrate(V) solution is added to copper(II)sulphate(VI) solution, copper(II) nitrate(V) solution and a white precipitate of barium sulphate(VI) are formed. Balanced stoichiometric equation Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq) Split reactants product existing in aqueous state as cation/anion Ba2+(aq) + 2NO3- (aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3- (aq)+ Cu2+(aq) Cancel out ions appearing on reactant and product side Ba2+(aq) + 2NO3- (aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3- (aq) + Cu2+(aq) Rewrite the equation Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ionic equation) 3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide. Balanced stoichiometric equation Pb(NO3)2(aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)
24 24 Split reactants product existing in aqueous state as cation/anion Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Cancel out ions appearing on reactant and product side Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Rewrite the equation Pb2+(aq) + 2I - (aq) -> PbI2 (s) (ionic equation) (b)Neutralization Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base.
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When barium nitrate(V) solution is added to copper(II)sulphate(VI) solution, copper(II) nitrate(V) solution and a white precipitate of barium sulphate(VI) are formed. Balanced stoichiometric equation Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq) Split reactants product existing in aqueous state as cation/anion Ba2+(aq) + 2NO3- (aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3- (aq)+ Cu2+(aq) Cancel out ions appearing on reactant and product side Ba2+(aq) + 2NO3- (aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3- (aq) + Cu2+(aq) Rewrite the equation Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ionic equation) 3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide. Balanced stoichiometric equation Pb(NO3)2(aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)
24 24 Split reactants product existing in aqueous state as cation/anion Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Cancel out ions appearing on reactant and product side Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Rewrite the equation Pb2+(aq) + 2I - (aq) -> PbI2 (s) (ionic equation) (b)Neutralization Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base. (i)Reaction of alkalis with acids 1.Reaction of nitric(V)acid with potassium hydroxide Balanced stoichiometric equation HNO3(aq) + KOH (aq) -> H2O (l) + KNO3 (aq) Split reactants product existing in aqueous state as cation/anion H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Cancel out ions appearing on reactant and product side H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Rewrite the equation H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 2.Reaction of sulphuric(VI)acid with ammonia solution Balanced stoichiometric equation H2SO4(aq) + 2NH4OH (aq) -> H2O (l) + (NH4) 2SO4 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)
25 25 Rewrite the equation 2H+ (aq) + 2OH - (aq) -> 2H2O (l) H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 3.Reaction of hydrochloric acid with Zinc hydroxide Balanced stoichiometric equation 2HCl(aq) + Zn(OH)2 (s) -> 2H2O (l) + ZnCl 2 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Rewrite the equation 2H+(aq) + Zn(OH)2 (s) ->2H2O (l) + Zn 2+ (aq) (ionic equation) (h)Molar solutions A molar solution is one whose concentration is known.
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Balanced stoichiometric equation Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq) Split reactants product existing in aqueous state as cation/anion Ba2+(aq) + 2NO3- (aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3- (aq)+ Cu2+(aq) Cancel out ions appearing on reactant and product side Ba2+(aq) + 2NO3- (aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3- (aq) + Cu2+(aq) Rewrite the equation Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ionic equation) 3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide. Balanced stoichiometric equation Pb(NO3)2(aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)
24 24 Split reactants product existing in aqueous state as cation/anion Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Cancel out ions appearing on reactant and product side Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Rewrite the equation Pb2+(aq) + 2I - (aq) -> PbI2 (s) (ionic equation) (b)Neutralization Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base. (i)Reaction of alkalis with acids 1.Reaction of nitric(V)acid with potassium hydroxide Balanced stoichiometric equation HNO3(aq) + KOH (aq) -> H2O (l) + KNO3 (aq) Split reactants product existing in aqueous state as cation/anion H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Cancel out ions appearing on reactant and product side H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Rewrite the equation H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 2.Reaction of sulphuric(VI)acid with ammonia solution Balanced stoichiometric equation H2SO4(aq) + 2NH4OH (aq) -> H2O (l) + (NH4) 2SO4 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)
25 25 Rewrite the equation 2H+ (aq) + 2OH - (aq) -> 2H2O (l) H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 3.Reaction of hydrochloric acid with Zinc hydroxide Balanced stoichiometric equation 2HCl(aq) + Zn(OH)2 (s) -> 2H2O (l) + ZnCl 2 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Rewrite the equation 2H+(aq) + Zn(OH)2 (s) ->2H2O (l) + Zn 2+ (aq) (ionic equation) (h)Molar solutions A molar solution is one whose concentration is known. The SI unit of concentration is Molarity denoted M.
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Balanced stoichiometric equation Pb(NO3)2(aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)
24 24 Split reactants product existing in aqueous state as cation/anion Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Cancel out ions appearing on reactant and product side Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Rewrite the equation Pb2+(aq) + 2I - (aq) -> PbI2 (s) (ionic equation) (b)Neutralization Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base. (i)Reaction of alkalis with acids 1.Reaction of nitric(V)acid with potassium hydroxide Balanced stoichiometric equation HNO3(aq) + KOH (aq) -> H2O (l) + KNO3 (aq) Split reactants product existing in aqueous state as cation/anion H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Cancel out ions appearing on reactant and product side H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Rewrite the equation H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 2.Reaction of sulphuric(VI)acid with ammonia solution Balanced stoichiometric equation H2SO4(aq) + 2NH4OH (aq) -> H2O (l) + (NH4) 2SO4 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)
25 25 Rewrite the equation 2H+ (aq) + 2OH - (aq) -> 2H2O (l) H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 3.Reaction of hydrochloric acid with Zinc hydroxide Balanced stoichiometric equation 2HCl(aq) + Zn(OH)2 (s) -> 2H2O (l) + ZnCl 2 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Rewrite the equation 2H+(aq) + Zn(OH)2 (s) ->2H2O (l) + Zn 2+ (aq) (ionic equation) (h)Molar solutions A molar solution is one whose concentration is known. The SI unit of concentration is Molarity denoted M. Molarity may be defined as the number of moles of solute present in one cubic decimeter of solution.
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(i)Reaction of alkalis with acids 1.Reaction of nitric(V)acid with potassium hydroxide Balanced stoichiometric equation HNO3(aq) + KOH (aq) -> H2O (l) + KNO3 (aq) Split reactants product existing in aqueous state as cation/anion H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Cancel out ions appearing on reactant and product side H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Rewrite the equation H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 2.Reaction of sulphuric(VI)acid with ammonia solution Balanced stoichiometric equation H2SO4(aq) + 2NH4OH (aq) -> H2O (l) + (NH4) 2SO4 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)
25 25 Rewrite the equation 2H+ (aq) + 2OH - (aq) -> 2H2O (l) H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 3.Reaction of hydrochloric acid with Zinc hydroxide Balanced stoichiometric equation 2HCl(aq) + Zn(OH)2 (s) -> 2H2O (l) + ZnCl 2 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Rewrite the equation 2H+(aq) + Zn(OH)2 (s) ->2H2O (l) + Zn 2+ (aq) (ionic equation) (h)Molar solutions A molar solution is one whose concentration is known. The SI unit of concentration is Molarity denoted M. Molarity may be defined as the number of moles of solute present in one cubic decimeter of solution. One cubic decimeter is equal to one litre and also equal to 1000cm3.
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The SI unit of concentration is Molarity denoted M. Molarity may be defined as the number of moles of solute present in one cubic decimeter of solution. One cubic decimeter is equal to one litre and also equal to 1000cm3. The higher the molarity the higher the concentration and the higher/more solute has been dissolved in the solvent to make one cubic decimeter/ litre/1000cm3 solution. Examples 2M sodium hydroxide means 2 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter/ litre/1000cm3 uniform solution mixture of sodium hydroxide and water. 0.02M sodium hydroxide means 0.02 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter/ litre/1000cm3 uniform solution mixture of sodium hydroxide and water. “2M” is more concentrated than“0.02M”. Preparation of molar solution
26 26 Procedure Weigh accurately 4.0 g of sodium hydroxide pellets into a 250cm3 volumetric flask. Using a wash bottle add about 200cm3 of distilled water. Stopper the flask. Shake vigorously for three minutes. Remove the stopper for a second then continue to shake for about another two minutes until all the solid has dissolved. Add more water slowly upto exactly the 250 cm3 mark. Sample questions 1.Calculate the number of moles of sodium hydroxide pellets present in: (i) 4.0 g. Molar mass of NaOH = (23 + 16 + 1) = 40g Moles = Mass => 4.0 = 0.1 / 1.0 x 10 -1 moles Molar mass 40 (ii) 250 cm3 solution in the volumetric flask.
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x 10-3moles 1000 1000 Mass of HNO3 =Moles x molar mass => 2.5 x 10-3 x 40 = 0.1 g 3. Calculate the volume required to dissolve : (a)(i) 0.25moles of sodium hydroxide solution to form a 0.8M solution Volume (in cm3) = moles x 1000 => 0.25 x 1000 = 312.5cm3 Molarity 0.8 (ii) 100cm3 was added to the sodium hydroxide solution above. Calculate the concentration of the solution. C1 x V1 = C2 x V2 where: C1 = molarity/concentration before diluting/adding water C2 = molarity/concentration after diluting/adding water V1 = volume before diluting/adding water V2 = volume after diluting/adding water => 0.8M x 312.5cm3 = C2 x (312.5 + 100) C2 = 0.8M x 312.5cm3 = 0.6061M
31 31 412.5 (b)(ii) 0.01M solution containing 0.01moles of sodium hydroxide solution . Volume (in cm3) = moles x 1000 => 0.01 x 1000 = 1000 cm3 Molarity 0.01 (ii) Determine the quantity of water which must be added to the sodium hydroxide solution above to form a 0.008M solution. C1 x V1 = C2 x V2 where: C1 = molarity/concentration before diluting/adding water C2 = molarity/concentration after diluting/adding water V1 = volume before diluting/adding water V2 = volume after diluting/adding water => 0.01M x 1000 cm3 = 0.008 x V2 V2 = 0.01M x 1000cm3 = 1250cm3 0.008 Volume added = 1250 - 1000 = 250cm3 (c)Volumetric analysis/Titration Volumetric analysis/Titration is the process of determining unknown concentration of one reactant from a known concentration and volume of another. Reactions take place in simple mole ratio of reactants and products.
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Volume (in cm3) = moles x 1000 => 0.01 x 1000 = 1000 cm3 Molarity 0.01 (ii) Determine the quantity of water which must be added to the sodium hydroxide solution above to form a 0.008M solution. C1 x V1 = C2 x V2 where: C1 = molarity/concentration before diluting/adding water C2 = molarity/concentration after diluting/adding water V1 = volume before diluting/adding water V2 = volume after diluting/adding water => 0.01M x 1000 cm3 = 0.008 x V2 V2 = 0.01M x 1000cm3 = 1250cm3 0.008 Volume added = 1250 - 1000 = 250cm3 (c)Volumetric analysis/Titration Volumetric analysis/Titration is the process of determining unknown concentration of one reactant from a known concentration and volume of another. Reactions take place in simple mole ratio of reactants and products. Knowing the concentration/ volume of one reactant, the other can be determined from the relationship: M1V1 = M2V2 where: n1 n2 M1 = Molarity of 1st reactant M2 = Molarity of 2nd reactant V1 = Volume of 1st reactant V1 = Volume of 2nd reactant n1 = number of moles of 1st reactant from stoichiometric equation n2 = number of moles of 2nd reactant from stoichiometric equation Examples 1.Calculate the molarity of MCO3 if 5.0cm3 of MCO3 react with 25.0cm3 of 0.5M hydrochloric acid.(C=12.0 ,O =16.0) Stoichiometric equation: MCO3(s) + 2HCl(aq) -> MCl2(aq) + CO2(g) + H2O(l) Method 1
32 32 M1V1 = M2V2 -> M1 x 5.0cm3 = 0.5M x 25.0cm3 n1 n2 1 2 => M1 = 0.5 x 25.0 x1 = 1.25M / 1.25 moledm-3 5.0 x 2 Method 2 Moles of HCl used = molarity x volume 1000 => 0.5 x 25.0 = 0.0125 /1.25 x 10-2moles 1000 Mole ratio MCO3 : HCl = 1:2 Moles MCO3 = 0.0125 /1.25 x 10-2moles = 0.00625 / 6.25 x 10-3 moles 2 Molarity MCO3 = moles x 1000 => 0.00625 / 6.25 x 10-3 x 1000 Volume 5 = 1.25M / 1.25 moledm-3 2.
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0.388g of a monobasic organic acid B required 46.5 cm3 of 0.095M sodium hydroxide for complete neutralization. Name and draw the structural formula of B Moles of NaOH used = molarity x volume 1000 => 0.095 x 46.5 = 0.0044175 /4.4175 x 10-3moles 1000 Mole ratio B : NaOH = 1:1
34 34 Moles B = 0.0044175 /4.4175 x 10-3moles Molar mass B = mass => 0.388 moles 0.0044175 /4.4175 x 10-3moles = 87.8324 gmole-1 X-COOH = 87.8324 where X is an alkyl group X =87.8324- 42 = 42.8324=43 By elimination: CH3 = 15 CH3CH2 = 29 CH3CH2 CH2 = 43 Molecula formula : CH3CH2 CH2COOH Molecule name : Butan-1-oic acid Molecular structure H H H O H C C C C O H H H H H 5. 10.5 g of an impure sample containing ammonium sulphate (VI) fertilizer was warmed with 250cm3 of o.8M sodium hydroxide solution.The excess of the alkali was neutralized by 85cm3 of 0.5M hydrochloric acid. Calculate the % of impurities in the ammonium sulphate (VI)fertilizer.
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10.5 g of an impure sample containing ammonium sulphate (VI) fertilizer was warmed with 250cm3 of o.8M sodium hydroxide solution.The excess of the alkali was neutralized by 85cm3 of 0.5M hydrochloric acid. Calculate the % of impurities in the ammonium sulphate (VI)fertilizer. (N=14.0,S=32.0,O=16.0, H=1.0) Equation for neutralization NaOH(aq) + HCl(aq) -> NaOH(aq) + H2O(l) Mole ratio NaOH(aq):HCl(aq)= 1:1 Moles of HCl = Molarity x volume => 0.5 x 85 = 0.0425 moles 1000 1000 Excess moles of NaOH(aq)= 0.0425 moles Equation for reaction with ammonium salt 2NaOH(aq) + (NH4) 2SO4(aq) -> Na 2SO4(aq) + 2NH3 (g)+ 2H2O(l) Mole ratio NaOH(aq): (NH4) 2SO4(aq)= 2:1 Total moles of NaOH = Molarity x volume => 0.8 x 250 = 0.2 moles 1000 1000 Moles of NaOH that reacted with(NH4) 2SO4 = 0.2 - 0.0425 = 0.1575moles Moles (NH4) 2SO4 = ½ x 0.1575moles = 0. 07875moles Molar mass (NH4) 2SO4= 132 gmole-1
35 35 Mass of in impure sample = moles x molar mass =>0. 07875 x 132 = 10.395 g Mass of impurities = 10.5 -10.395 = 0.105 g % impurities = 0.105 x 100 = 1.0 % 10.5 Practically volumetric analysis involves titration. Titration generally involves filling a burette with known/unknown concentration of a solution then adding the solution to unknown/known concentration of another solution in a conical flask until there is complete reaction.
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07875moles Molar mass (NH4) 2SO4= 132 gmole-1
35 35 Mass of in impure sample = moles x molar mass =>0. 07875 x 132 = 10.395 g Mass of impurities = 10.5 -10.395 = 0.105 g % impurities = 0.105 x 100 = 1.0 % 10.5 Practically volumetric analysis involves titration. Titration generally involves filling a burette with known/unknown concentration of a solution then adding the solution to unknown/known concentration of another solution in a conical flask until there is complete reaction. If the solutions used are both colourless, an indicator is added to the conical flask. When the reaction is over, a slight/little excess of burette contents change the colour of the indicator. This is called the end point. Set up of titration apparatus The titration process involve involves determination of titre. The titre is the volume of burette contents/reading before and after the end point. Burette contents/reading before titration is usually called the Initial burette reading. Burette contents/reading after titration is usually called the Final burette reading. The titre value is thus a sum of the Final less Initial burette readings. To reduce errors, titration process should be repeated at least once more. The results of titration are recorded in a titration table as below
36 36 Sample titration table Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution used(cm3) 20.0 20.0 20.0 As evidence of a titration actually done examining body requires the candidate to record their burette readings before and after the titration. For KCSE candidates burette readings must be recorded in a titration table in the format provided by the Kenya National Examination Council. As evidence of all titration actually done Kenya National Examination Council require the candidate to record their burette readings before and after the titration to complete the titration table in the format provided.
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The results of titration are recorded in a titration table as below
36 36 Sample titration table Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution used(cm3) 20.0 20.0 20.0 As evidence of a titration actually done examining body requires the candidate to record their burette readings before and after the titration. For KCSE candidates burette readings must be recorded in a titration table in the format provided by the Kenya National Examination Council. As evidence of all titration actually done Kenya National Examination Council require the candidate to record their burette readings before and after the titration to complete the titration table in the format provided. Calculate the average volume of solution used 24.0 + 24.0 + 24.0 = 24.0 cm3 3 As evidence of understanding the degree of accuracy of burettes , all readings must be recorded to a decimal point. As evidence of accuracy in carrying the out the titration , candidates value should be within 0.2 of the school value . The school value is the teachers readings presented to the examining body/council based on the concentrations of the solutions s/he presented to her/his candidates. Bonus mark is awarded for averaged reading within 0.1 school value as Final answer. Calculations involved after the titration require candidates thorough practical and theoretical practice mastery on the: (i)relationship among the mole, molar mass, mole ratios, concentration, molarity. (ii) mathematical application of 1st principles. Very useful information which candidates forget appears usually in the beginning of the question paper as: “You are provided with…” All calculation must be to the 4th decimal point unless they divide fully to a lesser decimal point. Candidates are expected to use a non programmable scientific calculator. 37 37 (a)Sample Titration Practice 1 (Simple Titration) You are provided with: 0.1M sodium hydroxide solution A Hydrochloric acid solution B You are required to determine the concentration of solution B in moles per litre. Procedure Fill the burette with solution B. Pipette 25.0cm3 of solution A into a conical flask.
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37 37 (a)Sample Titration Practice 1 (Simple Titration) You are provided with: 0.1M sodium hydroxide solution A Hydrochloric acid solution B You are required to determine the concentration of solution B in moles per litre. Procedure Fill the burette with solution B. Pipette 25.0cm3 of solution A into a conical flask. Titrate solution A with solution B using phenolphthalein indicator to complete the titration table 1 Sample results Titration table 1 Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution B used(cm3) 20.0 20.0 20.0 Sample worked questions 1. Calculate the average volume of solution B used Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3 3 3 2. How many moles of: (i)solution A were present in 25cm3 solution. Moles of solution A = Molarity x volume = 0.1 x 25 = 2.5 x 10-3 moles 1000 1000 (ii)solution B were present in the average volume. Chemical equation: NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l) Mole ratio 1:1 => Moles of A = Moles of B = 2.5 x 10-3 moles (iii) solution B in moles per litre.
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How many moles of: (i)solution A were present in 25cm3 solution. Moles of solution A = Molarity x volume = 0.1 x 25 = 2.5 x 10-3 moles 1000 1000 (ii)solution B were present in the average volume. Chemical equation: NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l) Mole ratio 1:1 => Moles of A = Moles of B = 2.5 x 10-3 moles (iii) solution B in moles per litre. Moles of B per litre = moles x 1000 = 2.5 x 10-3 x 1000 = 0.1M Volume 20
38 38 (b)Sample Titration Practice 2 (Redox Titration) You are provided with: Acidified Potassium manganate(VII) solution A 0.1M of an iron (II)salt solution B 8.5g of ammonium iron(II)sulphate(VI) crystals(NH4)2 SO4FeSO4.xH2O solid C You are required to (i)standardize acidified potassium manganate(VII) (ii)determine the value of x in the formula (NH4)2 SO4FeSO4.xH2O. Procedure 1 Fill the burette with solution A. Pipette 25.0cm3 of solution B into a conical flask. Titrate solution A with solution B until a pink colour just appears. Record your results to complete table 1. Table 1:Sample results Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution A used(cm3) 20.0 20.0 20.0 Sample worked questions 1. Calculate the average volume of solution A used Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3 3 3 2. How many moles of: (i)solution B were present in 25cm3 solution.
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Table 1:Sample results Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution A used(cm3) 20.0 20.0 20.0 Sample worked questions 1. Calculate the average volume of solution A used Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3 3 3 2. How many moles of: (i)solution B were present in 25cm3 solution. Moles of solution A = Molarity x volume = 0.1 x 25 = 2.5 x 10-3 moles 1000 1000 (ii)solution A were present in the average volume. Assume one mole of B react with five moles of B Mole ratio A : B = 1:5 => Moles of A = Moles of B = 2.5 x 10-3 moles = 5.0 x 10 -4 moles 5 5
39 39 (iii) solution B in moles per litre. Moles of B per litre = moles x 1000 = 2.5 x 10-3 x 1000 Volume 20 = 0.025 M /moles per litre /moles l-1 Procedure 2 Place all the solid C into the 250cm3 volumetric flask carefully. Add about 200cm3 of distilled water. Shake to dissolve. Make up to the 250cm3 of solution by adding more distilled water. Label this solution C. Pipette 25cm3 of solution C into a conical flask, Titrate solution C with solution A until a permanent pink colour just appears. Complete table 2. Table 2:Sample results Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution A used(cm3) 20.0 20.0 20.0 Sample worked questions 1.
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Pipette 25cm3 of solution C into a conical flask, Titrate solution C with solution A until a permanent pink colour just appears. Complete table 2. Table 2:Sample results Titration number 1 2 3 Final burette reading (cm3) 20.0 20.0 20.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of solution A used(cm3) 20.0 20.0 20.0 Sample worked questions 1. Calculate the average volume of solution A used Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3 3 3 2. How many moles of: (i)solution A were present inin the average titre. Moles of solution A = Molarity x volume = 0.025 x 20 = 5.0 x 10-4 moles 1000 1000 (ii)solution C in 25cm3 solution given the equation for the reaction: MnO4- (aq) + 8H+(aq) + 5Fe2+ (aq) -> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) Mole ratio MnO4- (aq): 5Fe2+ (aq) = 1:5 => Moles of 5Fe2+ (aq) = Moles of MnO4- (aq) = 5.0 x 10-4 moles = 1.0 x 10 -4 moles 5 5 (iii) solution B in 250cm3. Moles of B per litre = moles x 250 = 1.0 x 10 -4 x 250 = 1.0 x 10 -3 moles Volume 25 3. Calculate the molar mass of solid C and hence the value of x in the chemical formula (NH4)2SO4FeSO4.xH2O.
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Moles of solution A = Molarity x volume = 0.025 x 20 = 5.0 x 10-4 moles 1000 1000 (ii)solution C in 25cm3 solution given the equation for the reaction: MnO4- (aq) + 8H+(aq) + 5Fe2+ (aq) -> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) Mole ratio MnO4- (aq): 5Fe2+ (aq) = 1:5 => Moles of 5Fe2+ (aq) = Moles of MnO4- (aq) = 5.0 x 10-4 moles = 1.0 x 10 -4 moles 5 5 (iii) solution B in 250cm3. Moles of B per litre = moles x 250 = 1.0 x 10 -4 x 250 = 1.0 x 10 -3 moles Volume 25 3. Calculate the molar mass of solid C and hence the value of x in the chemical formula (NH4)2SO4FeSO4.xH2O. 40 40 (N=14.0, S=32.0, Fe=56.0, H=1.0 O=16.0) Molar mass = mass perlitre = 8.5 = 8500 g Moles per litre 1.0 x 10 -3 moles NH4)2SO4FeSO4.xH2O = 8500 284 + 18x =8500 8500 - 284 = 8216 = 18x = 454.4444 18 18 x = 454 (whole number) (c)Sample Titration Practice 3 (Back titration) You are provided with: (i)an impure calcium carbonate labeled M (ii)Hydrochloric acid labeled solution N (iii)solution L containing 20g per litre sodium hydroxide. You are required to determine the concentration of N in moles per litre and the % of calcium carbonate in mixture M. Procedure 1 Pipette 25.0cm3 of solution L into a conical flask. Add 2-3 drops of phenolphthalein indicator.
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You are required to determine the concentration of N in moles per litre and the % of calcium carbonate in mixture M. Procedure 1 Pipette 25.0cm3 of solution L into a conical flask. Add 2-3 drops of phenolphthalein indicator. Titrate with dilute hydrochloric acid solution N and record your results in table 1(4mark) Sample Table 1 1 2 3 Final burette reading (cm3) 6.5 6.5 6.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 6.5 6.5 6.5 Sample questions (a) Calculate the average volume of solution N used 6.5 + 6.5 + 6.5 = 6.5 cm3 3 (b) How many moles of sodium hydroxide are contained in 25cm3of solution L Molar mass NaOH =40g Molarity of L = mass per litre => 20 = 0.5M
41 41 Molar mass NaOH 40 Moles NaOH in 25cm3 = molarity x volume => 0.5M x 25cm3 = 0.0125 moles 1000 1000 (c)Calculate: (i)the number of moles of hydrochloric acidthat react with sodium hydroxide in (b)above. Mole ratio NaOH : HCl from stoichiometric equation= 1:1 Moles HCl =Moles NaOH => 0.0125 moles (ii)the molarity of hydrochloric acid solution N. Molarity = moles x 1000 => 0.0125 moles x 1000 =1.9231M/moledm-3 6.5 6.5 Procedure 2 Place the 4.0 g of M provided into a conical flask and add 25.0cm3 of the dilute hydrochloric acid to it using a clean pipette.
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Titrate with dilute hydrochloric acid solution N and record your results in table 1(4mark) Sample Table 1 1 2 3 Final burette reading (cm3) 6.5 6.5 6.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 6.5 6.5 6.5 Sample questions (a) Calculate the average volume of solution N used 6.5 + 6.5 + 6.5 = 6.5 cm3 3 (b) How many moles of sodium hydroxide are contained in 25cm3of solution L Molar mass NaOH =40g Molarity of L = mass per litre => 20 = 0.5M
41 41 Molar mass NaOH 40 Moles NaOH in 25cm3 = molarity x volume => 0.5M x 25cm3 = 0.0125 moles 1000 1000 (c)Calculate: (i)the number of moles of hydrochloric acidthat react with sodium hydroxide in (b)above. Mole ratio NaOH : HCl from stoichiometric equation= 1:1 Moles HCl =Moles NaOH => 0.0125 moles (ii)the molarity of hydrochloric acid solution N. Molarity = moles x 1000 => 0.0125 moles x 1000 =1.9231M/moledm-3 6.5 6.5 Procedure 2 Place the 4.0 g of M provided into a conical flask and add 25.0cm3 of the dilute hydrochloric acid to it using a clean pipette. Swirl the contents of the flask vigorously until effervescence stop.Using a 100ml measuring cylinder add 175cm3 distilled waterto make up the solution up to 200cm3.Label this solution K.Using a clean pipettetransfer 25.0cm3 of the solution into a clean conical flask and titrate with solution L from the burette using 2-3 drops of methyl orange indicator.Record your observations in table 2.
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Mole ratio NaOH : HCl from stoichiometric equation= 1:1 Moles HCl =Moles NaOH => 0.0125 moles (ii)the molarity of hydrochloric acid solution N. Molarity = moles x 1000 => 0.0125 moles x 1000 =1.9231M/moledm-3 6.5 6.5 Procedure 2 Place the 4.0 g of M provided into a conical flask and add 25.0cm3 of the dilute hydrochloric acid to it using a clean pipette. Swirl the contents of the flask vigorously until effervescence stop.Using a 100ml measuring cylinder add 175cm3 distilled waterto make up the solution up to 200cm3.Label this solution K.Using a clean pipettetransfer 25.0cm3 of the solution into a clean conical flask and titrate with solution L from the burette using 2-3 drops of methyl orange indicator.Record your observations in table 2. Sample Table 2 1 2 3 Final burette reading (cm3) 24.5 24.5 24.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.5 24.5 24.5 Sample calculations (a)Calculate the average volume of solution L used(1mk) 24.5 + 24.5 + 24.5 = 24.5cm3 3 (b)How many moles of sodium hydroxide are present in the average volume of solution L used? Moles = molarity x average burette volume => 0.5 x 24.5 1000 1000 = 0.01225 /1.225 x 10-2 moles
42 42 (c) How many moles of hydrochloric acid are present in the original 200cm3 of solution K?
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Swirl the contents of the flask vigorously until effervescence stop.Using a 100ml measuring cylinder add 175cm3 distilled waterto make up the solution up to 200cm3.Label this solution K.Using a clean pipettetransfer 25.0cm3 of the solution into a clean conical flask and titrate with solution L from the burette using 2-3 drops of methyl orange indicator.Record your observations in table 2. Sample Table 2 1 2 3 Final burette reading (cm3) 24.5 24.5 24.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.5 24.5 24.5 Sample calculations (a)Calculate the average volume of solution L used(1mk) 24.5 + 24.5 + 24.5 = 24.5cm3 3 (b)How many moles of sodium hydroxide are present in the average volume of solution L used? Moles = molarity x average burette volume => 0.5 x 24.5 1000 1000 = 0.01225 /1.225 x 10-2 moles
42 42 (c) How many moles of hydrochloric acid are present in the original 200cm3 of solution K? Mole ratio NaOH: HCl = 1:1 => moles of HCl = 0.01225 /1.225 x 10-2 moles Moles in 200cm3 = 200cm3 x 0.01225 /1.225 x 10-2moles 25cm3(volume pipetted) =0.49 /4.9 x 10-1moles (d)How many moles of hydrochloric acid were contained in original 25 cm3 solution N used Original moles = Original molarity x pipetted volume => 1000cm3 1.9231M/moledm-3 x 25 = 0.04807/4.807 x 10-2 moles 1000 (e)How many moles of hydrochloric acid were used to react with calcium carbonate present?
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Sample Table 2 1 2 3 Final burette reading (cm3) 24.5 24.5 24.5 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.5 24.5 24.5 Sample calculations (a)Calculate the average volume of solution L used(1mk) 24.5 + 24.5 + 24.5 = 24.5cm3 3 (b)How many moles of sodium hydroxide are present in the average volume of solution L used? Moles = molarity x average burette volume => 0.5 x 24.5 1000 1000 = 0.01225 /1.225 x 10-2 moles
42 42 (c) How many moles of hydrochloric acid are present in the original 200cm3 of solution K? Mole ratio NaOH: HCl = 1:1 => moles of HCl = 0.01225 /1.225 x 10-2 moles Moles in 200cm3 = 200cm3 x 0.01225 /1.225 x 10-2moles 25cm3(volume pipetted) =0.49 /4.9 x 10-1moles (d)How many moles of hydrochloric acid were contained in original 25 cm3 solution N used Original moles = Original molarity x pipetted volume => 1000cm3 1.9231M/moledm-3 x 25 = 0.04807/4.807 x 10-2 moles 1000 (e)How many moles of hydrochloric acid were used to react with calcium carbonate present? Moles that reacted = original moles –moles in average titre => 0.04807/4.807 x 10-2moles - 0.01225 /1.225 x 10-2moles = 0.03582/3.582 x 10 -2 moles (f)Write the equation for the reaction between calcium carbonate and hydrochloric acid.
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Moles = molarity x average burette volume => 0.5 x 24.5 1000 1000 = 0.01225 /1.225 x 10-2 moles
42 42 (c) How many moles of hydrochloric acid are present in the original 200cm3 of solution K? Mole ratio NaOH: HCl = 1:1 => moles of HCl = 0.01225 /1.225 x 10-2 moles Moles in 200cm3 = 200cm3 x 0.01225 /1.225 x 10-2moles 25cm3(volume pipetted) =0.49 /4.9 x 10-1moles (d)How many moles of hydrochloric acid were contained in original 25 cm3 solution N used Original moles = Original molarity x pipetted volume => 1000cm3 1.9231M/moledm-3 x 25 = 0.04807/4.807 x 10-2 moles 1000 (e)How many moles of hydrochloric acid were used to react with calcium carbonate present? Moles that reacted = original moles –moles in average titre => 0.04807/4.807 x 10-2moles - 0.01225 /1.225 x 10-2moles = 0.03582/3.582 x 10 -2 moles (f)Write the equation for the reaction between calcium carbonate and hydrochloric acid. CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l) (g)Calculate the number of moles of calcium carbonate that reacted with hydrochloric acid.
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Mole ratio NaOH: HCl = 1:1 => moles of HCl = 0.01225 /1.225 x 10-2 moles Moles in 200cm3 = 200cm3 x 0.01225 /1.225 x 10-2moles 25cm3(volume pipetted) =0.49 /4.9 x 10-1moles (d)How many moles of hydrochloric acid were contained in original 25 cm3 solution N used Original moles = Original molarity x pipetted volume => 1000cm3 1.9231M/moledm-3 x 25 = 0.04807/4.807 x 10-2 moles 1000 (e)How many moles of hydrochloric acid were used to react with calcium carbonate present? Moles that reacted = original moles –moles in average titre => 0.04807/4.807 x 10-2moles - 0.01225 /1.225 x 10-2moles = 0.03582/3.582 x 10 -2 moles (f)Write the equation for the reaction between calcium carbonate and hydrochloric acid. CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l) (g)Calculate the number of moles of calcium carbonate that reacted with hydrochloric acid. From the equation CaCO3(s):2HCl(aq) = 1:2 => Moles CaCO3(s) = 1/2moles HCl = 1/2 x 0.03582/3.582 x 10 -2 moles = 0.01791 /1.791 x 10-2moles
43 43 (h)Calculate the mass of calcium carbonate in 4.0g of mixture M (Ca=40.0,O = 16.0,C=12.0) Molar mass CaCO3 = 100g Mass CaCO3 = moles x molar mass => 0.01791 /1.791 x 10-2moles x 100g = 1.791g (i)Determine the % of calcium carbonate present in the mixture % CaCO3 = mass of pure x 100% => 1.791g x 100% = 44.775% Mass of impure 4.0 (d)Sample titration practice 4 (Multiple titration) You are provided with: (i)sodium L containing 5.0g per litre of a dibasic organic acid H2X.2H2O.
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Moles that reacted = original moles –moles in average titre => 0.04807/4.807 x 10-2moles - 0.01225 /1.225 x 10-2moles = 0.03582/3.582 x 10 -2 moles (f)Write the equation for the reaction between calcium carbonate and hydrochloric acid. CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l) (g)Calculate the number of moles of calcium carbonate that reacted with hydrochloric acid. From the equation CaCO3(s):2HCl(aq) = 1:2 => Moles CaCO3(s) = 1/2moles HCl = 1/2 x 0.03582/3.582 x 10 -2 moles = 0.01791 /1.791 x 10-2moles
43 43 (h)Calculate the mass of calcium carbonate in 4.0g of mixture M (Ca=40.0,O = 16.0,C=12.0) Molar mass CaCO3 = 100g Mass CaCO3 = moles x molar mass => 0.01791 /1.791 x 10-2moles x 100g = 1.791g (i)Determine the % of calcium carbonate present in the mixture % CaCO3 = mass of pure x 100% => 1.791g x 100% = 44.775% Mass of impure 4.0 (d)Sample titration practice 4 (Multiple titration) You are provided with: (i)sodium L containing 5.0g per litre of a dibasic organic acid H2X.2H2O. (ii)solution M which is acidified potassium manganate(VII) (iii)solution N a mixture of sodium ethanedioate and ethanedioic acid (iv)0.1M sodium hydroxide solution P (v)1.0M sulphuric(VI) You are required to: (i)standardize solution M using solution L (ii)use standardized solution M and solution P to determine the % of sodium ethanedioate in the mixture. Procedure 1 Fill the burette with solution M.
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From the equation CaCO3(s):2HCl(aq) = 1:2 => Moles CaCO3(s) = 1/2moles HCl = 1/2 x 0.03582/3.582 x 10 -2 moles = 0.01791 /1.791 x 10-2moles
43 43 (h)Calculate the mass of calcium carbonate in 4.0g of mixture M (Ca=40.0,O = 16.0,C=12.0) Molar mass CaCO3 = 100g Mass CaCO3 = moles x molar mass => 0.01791 /1.791 x 10-2moles x 100g = 1.791g (i)Determine the % of calcium carbonate present in the mixture % CaCO3 = mass of pure x 100% => 1.791g x 100% = 44.775% Mass of impure 4.0 (d)Sample titration practice 4 (Multiple titration) You are provided with: (i)sodium L containing 5.0g per litre of a dibasic organic acid H2X.2H2O. (ii)solution M which is acidified potassium manganate(VII) (iii)solution N a mixture of sodium ethanedioate and ethanedioic acid (iv)0.1M sodium hydroxide solution P (v)1.0M sulphuric(VI) You are required to: (i)standardize solution M using solution L (ii)use standardized solution M and solution P to determine the % of sodium ethanedioate in the mixture. Procedure 1 Fill the burette with solution M. Pipette 25.0cm3 of solution L into a conical flask. Heat this solution to about 70oC(but not to boil).Titrate the hot solution L with solution M until a permanent pink colour just appears .Shake thoroughly during the titration. Repeat this procedure to complete table 1.
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Pipette 25.0cm3 of solution L into a conical flask. Heat this solution to about 70oC(but not to boil).Titrate the hot solution L with solution M until a permanent pink colour just appears .Shake thoroughly during the titration. Repeat this procedure to complete table 1. Sample Table 1 1 2 3 Final burette reading (cm3) 24.0 24.0 24.0 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.0 24.0 24.0 Sample calculations
44 44 (a)Calculate the average volume of solution L used (1mk) 24.0 + 24.0 + 24.0 = 24.0cm3 3 (b)Given that the concentration of the dibasic acid is 0.05molesdm-3.determine the value of x in the formula H2X.2H2O (H=1.0,O=16.0) Molar mass H2X.2H2O= mass per litre => 5.0g/litre = 100g Moles/litre 0.05molesdm-3 H2X.2H2O =100 X = 100 – ((2 x1) + 2 x (2 x1) + (2 x 16) => 100 – 34 = 66 (c) Calculate the number of moles of the dibasic acid H2X.2H2O. Moles = molarity x pipette volume => 0.5 x 25 = 0.0125/1.25 x10 -2 moles 1000 1000 (d)Given the mole ratio manganate(VII)(MnO4-): acid H2X is 2:5, calculate the number of moles of manganate(VII) (MnO4-) in the average titre. Moles H2X = 2/5 moles of MnO4- => 2/5 x 0.0125/1.25 x10 -2 moles = 0.005/5.0 x 10 -3moles (e)Calculate the concentration of the manganate(VII)(MnO4-) in moles per litre.
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Moles = molarity x pipette volume => 0.5 x 25 = 0.0125/1.25 x10 -2 moles 1000 1000 (d)Given the mole ratio manganate(VII)(MnO4-): acid H2X is 2:5, calculate the number of moles of manganate(VII) (MnO4-) in the average titre. Moles H2X = 2/5 moles of MnO4- => 2/5 x 0.0125/1.25 x10 -2 moles = 0.005/5.0 x 10 -3moles (e)Calculate the concentration of the manganate(VII)(MnO4-) in moles per litre. Moles per litre/molarity = moles x 1000 average burette volume =>0.005/5.0 x 10 -3moles x 1000 = 0.2083 molesl-1/M 24.0 Procedure 2 With solution M still in the burette ,pipette 25.0cm3 of solution N into a conical flask. Heat the conical flask containing solution N to about 70oC.Titrate while hot with solution M.Repeat the experiment to complete table 2. Sample Table 2 1 2 3 Final burette reading (cm3) 12.5 12.5 12.5
45 45 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 12.5 12.5 12.5 Sample calculations (a)Calculate the average volume of solution L used (1mk) 12.5 + 12.5 + 12.5 =12.5cm3 3 (b)Calculations: (i)How many moles of manganate(VII)ions are contained in the average volume of solution M used?
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Moles per litre/molarity = moles x 1000 average burette volume =>0.005/5.0 x 10 -3moles x 1000 = 0.2083 molesl-1/M 24.0 Procedure 2 With solution M still in the burette ,pipette 25.0cm3 of solution N into a conical flask. Heat the conical flask containing solution N to about 70oC.Titrate while hot with solution M.Repeat the experiment to complete table 2. Sample Table 2 1 2 3 Final burette reading (cm3) 12.5 12.5 12.5
45 45 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 12.5 12.5 12.5 Sample calculations (a)Calculate the average volume of solution L used (1mk) 12.5 + 12.5 + 12.5 =12.5cm3 3 (b)Calculations: (i)How many moles of manganate(VII)ions are contained in the average volume of solution M used? Moles = molarity of solution M x average burette volume 1000 => 0.2083 molesl-1/ M x 12.5 = 0.0026 / 2.5 x 10-3 moles 1000 (ii)The reaction between manganate(VII)ions and ethanedioate ions that reacted with is as in the equation: 2MnO4- (aq) + 5C2O42- (aq) + 16H+ (aq) -> 2Mn2+(aq) + 10CO2(g) + 8H2O(l) Calculate the number of moles of ethanedioate ions that reacted with manganate (VII) ions in the average volume of solution M.
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Heat the conical flask containing solution N to about 70oC.Titrate while hot with solution M.Repeat the experiment to complete table 2. Sample Table 2 1 2 3 Final burette reading (cm3) 12.5 12.5 12.5
45 45 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 12.5 12.5 12.5 Sample calculations (a)Calculate the average volume of solution L used (1mk) 12.5 + 12.5 + 12.5 =12.5cm3 3 (b)Calculations: (i)How many moles of manganate(VII)ions are contained in the average volume of solution M used? Moles = molarity of solution M x average burette volume 1000 => 0.2083 molesl-1/ M x 12.5 = 0.0026 / 2.5 x 10-3 moles 1000 (ii)The reaction between manganate(VII)ions and ethanedioate ions that reacted with is as in the equation: 2MnO4- (aq) + 5C2O42- (aq) + 16H+ (aq) -> 2Mn2+(aq) + 10CO2(g) + 8H2O(l) Calculate the number of moles of ethanedioate ions that reacted with manganate (VII) ions in the average volume of solution M. From the stoichiometric equation,mole ratio MnO4- (aq): C2O42- (aq) = 2:5 => moles C2O42- = 5/2 moles MnO4- => 5/2 x 0.0026 / 2.5 x 10-3 moles = 0.0065 /6.5 x10-3 moles (iii)Calculate the number of moles of ethanedioate ions contained in 250cm3 solution N.
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Moles = molarity of solution M x average burette volume 1000 => 0.2083 molesl-1/ M x 12.5 = 0.0026 / 2.5 x 10-3 moles 1000 (ii)The reaction between manganate(VII)ions and ethanedioate ions that reacted with is as in the equation: 2MnO4- (aq) + 5C2O42- (aq) + 16H+ (aq) -> 2Mn2+(aq) + 10CO2(g) + 8H2O(l) Calculate the number of moles of ethanedioate ions that reacted with manganate (VII) ions in the average volume of solution M. From the stoichiometric equation,mole ratio MnO4- (aq): C2O42- (aq) = 2:5 => moles C2O42- = 5/2 moles MnO4- => 5/2 x 0.0026 / 2.5 x 10-3 moles = 0.0065 /6.5 x10-3 moles (iii)Calculate the number of moles of ethanedioate ions contained in 250cm3 solution N. 25cm3 pipette volume -> 0.0065 /6.5 x10-3 moles 250cm3 -> 0.0065 /6.5 x10-3 moles x 250 = 0.065 / 6.5 x10-2 moles 25 Procedure 3
46 46 Remove solution M from the burette and rinse it with distilled water. Fill the burette with sodium hydroxide solution P. Pipette 25cm3 of solution N into a conical flask and add 2-3 drops of phenolphthalein indicator. Titrate this solution N with solution P from the burette. Repeat the procedure to complete table 3.
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Pipette 25cm3 of solution N into a conical flask and add 2-3 drops of phenolphthalein indicator. Titrate this solution N with solution P from the burette. Repeat the procedure to complete table 3. Sample Table 2 1 2 3 Final burette reading (cm3) 24.9 24.9 24.9 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.9 24.9 24.9 Sample calculations (a)Calculate the average volume of solution L used (1mk) 24.9 + 24.9 + 24.9 = 24.9 cm3 3 (b)Calculations: (i)How many moles of sodium hydroxide solution P were contained in the average volume? Moles = molarity of solution P x average burette volume 1000 => 0.1 molesl-1 x 24.9 = 0.00249 / 2.49 x 10-3 moles 1000 (ii)Given that NaOH solution P reacted with the ethanedioate ions from the acid only and the equation for the reaction is: 2NaOH (aq) + H2C2O4 (aq) -> Na2C2O4(g) + 2H2O(l) Calculate the number of moles of ethanedioic acid that were used in the reaction From the stoichiometric equation,mole ratio NaOH(aq): H2C2O4 (aq) = 2:1 => moles H2C2O4 = 1/2 moles NaOH => 1/2 x 0.00249 / 2.49 x 10-3 moles = 0.001245/1.245 x10-3 moles. (iii)How many moles of ethanedioic acid were contained in 250cm3 of solution N?
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Sample Table 2 1 2 3 Final burette reading (cm3) 24.9 24.9 24.9 Initial burette reading (cm3) 0.0 0.0 0.0 Volume of N used (cm3) 24.9 24.9 24.9 Sample calculations (a)Calculate the average volume of solution L used (1mk) 24.9 + 24.9 + 24.9 = 24.9 cm3 3 (b)Calculations: (i)How many moles of sodium hydroxide solution P were contained in the average volume? Moles = molarity of solution P x average burette volume 1000 => 0.1 molesl-1 x 24.9 = 0.00249 / 2.49 x 10-3 moles 1000 (ii)Given that NaOH solution P reacted with the ethanedioate ions from the acid only and the equation for the reaction is: 2NaOH (aq) + H2C2O4 (aq) -> Na2C2O4(g) + 2H2O(l) Calculate the number of moles of ethanedioic acid that were used in the reaction From the stoichiometric equation,mole ratio NaOH(aq): H2C2O4 (aq) = 2:1 => moles H2C2O4 = 1/2 moles NaOH => 1/2 x 0.00249 / 2.49 x 10-3 moles = 0.001245/1.245 x10-3 moles. (iii)How many moles of ethanedioic acid were contained in 250cm3 of solution N? 47 47 25cm3 pipette volume -> 0.001245/1.245 x10-3 moles 250cm3 -> 0.001245/1.245 x10-3 moles x 250 = 0.01245/1.245 x10-2 moles 25 (iii)Determine the % by mass of sodium ethanedioate in the micture (H= 1.0,O=16.0,C=12.0 and total mass of mixture =2.0 g in 250cm3 solution) Molar mass H2C2O4 = 90.0g Mass of H2C2O4 in 250cm3 = moles in 250cm3 x molar mass H2C2O4 =>0.01245/1.245 x10-2 moles x 90.0 = 1.1205g % by mass of sodium ethanedioate =(Mass of mixture - mass of H2C2O4) x 100% Mass of mixture => 2.0 - 1.1205 g = 43.975% 2.0 Note (i) L is 0.05M Oxalic acid (ii) M is 0.01M KMnO4 (iii) N is 0.03M oxalic acid(without sodium oxalate) Practice example 5.(Determining equation for a reaction) You are provided with -0.1M hydrochloric acid solution A -0.5M sodium hydroxide solution B You are to determine the equation for thereaction between solution A and B Procedure Fill the burette with solution A.Using a pipette and pipette filler transfer 25.0cm3 of solution B into a conical flask.Add 2-3 drops of phenolphthalein indicator.Run solution A into solution B until a permanent pink colour just appears.Record your results in Table 1.Repeat the experiment to obtain three concordant results to complete Table 1 Table 1(Sample results) Titration 1 2 3 Final volume(cm3) 12.5 25.0 37.5 Initial volume(cm3) 0.0 12.5 25.0 Volume of solution A used(cm3) 12.5 12.5 12.5
48 48 Sample questions Calculate the average volume of solution A used.
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Moles = molarity of solution P x average burette volume 1000 => 0.1 molesl-1 x 24.9 = 0.00249 / 2.49 x 10-3 moles 1000 (ii)Given that NaOH solution P reacted with the ethanedioate ions from the acid only and the equation for the reaction is: 2NaOH (aq) + H2C2O4 (aq) -> Na2C2O4(g) + 2H2O(l) Calculate the number of moles of ethanedioic acid that were used in the reaction From the stoichiometric equation,mole ratio NaOH(aq): H2C2O4 (aq) = 2:1 => moles H2C2O4 = 1/2 moles NaOH => 1/2 x 0.00249 / 2.49 x 10-3 moles = 0.001245/1.245 x10-3 moles. (iii)How many moles of ethanedioic acid were contained in 250cm3 of solution N? 47 47 25cm3 pipette volume -> 0.001245/1.245 x10-3 moles 250cm3 -> 0.001245/1.245 x10-3 moles x 250 = 0.01245/1.245 x10-2 moles 25 (iii)Determine the % by mass of sodium ethanedioate in the micture (H= 1.0,O=16.0,C=12.0 and total mass of mixture =2.0 g in 250cm3 solution) Molar mass H2C2O4 = 90.0g Mass of H2C2O4 in 250cm3 = moles in 250cm3 x molar mass H2C2O4 =>0.01245/1.245 x10-2 moles x 90.0 = 1.1205g % by mass of sodium ethanedioate =(Mass of mixture - mass of H2C2O4) x 100% Mass of mixture => 2.0 - 1.1205 g = 43.975% 2.0 Note (i) L is 0.05M Oxalic acid (ii) M is 0.01M KMnO4 (iii) N is 0.03M oxalic acid(without sodium oxalate) Practice example 5.(Determining equation for a reaction) You are provided with -0.1M hydrochloric acid solution A -0.5M sodium hydroxide solution B You are to determine the equation for thereaction between solution A and B Procedure Fill the burette with solution A.Using a pipette and pipette filler transfer 25.0cm3 of solution B into a conical flask.Add 2-3 drops of phenolphthalein indicator.Run solution A into solution B until a permanent pink colour just appears.Record your results in Table 1.Repeat the experiment to obtain three concordant results to complete Table 1 Table 1(Sample results) Titration 1 2 3 Final volume(cm3) 12.5 25.0 37.5 Initial volume(cm3) 0.0 12.5 25.0 Volume of solution A used(cm3) 12.5 12.5 12.5
48 48 Sample questions Calculate the average volume of solution A used. 12.5+12.5+12.5 = 12.5cm3 3 Theoretical Practice examples 1.
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(iii)How many moles of ethanedioic acid were contained in 250cm3 of solution N? 47 47 25cm3 pipette volume -> 0.001245/1.245 x10-3 moles 250cm3 -> 0.001245/1.245 x10-3 moles x 250 = 0.01245/1.245 x10-2 moles 25 (iii)Determine the % by mass of sodium ethanedioate in the micture (H= 1.0,O=16.0,C=12.0 and total mass of mixture =2.0 g in 250cm3 solution) Molar mass H2C2O4 = 90.0g Mass of H2C2O4 in 250cm3 = moles in 250cm3 x molar mass H2C2O4 =>0.01245/1.245 x10-2 moles x 90.0 = 1.1205g % by mass of sodium ethanedioate =(Mass of mixture - mass of H2C2O4) x 100% Mass of mixture => 2.0 - 1.1205 g = 43.975% 2.0 Note (i) L is 0.05M Oxalic acid (ii) M is 0.01M KMnO4 (iii) N is 0.03M oxalic acid(without sodium oxalate) Practice example 5.(Determining equation for a reaction) You are provided with -0.1M hydrochloric acid solution A -0.5M sodium hydroxide solution B You are to determine the equation for thereaction between solution A and B Procedure Fill the burette with solution A.Using a pipette and pipette filler transfer 25.0cm3 of solution B into a conical flask.Add 2-3 drops of phenolphthalein indicator.Run solution A into solution B until a permanent pink colour just appears.Record your results in Table 1.Repeat the experiment to obtain three concordant results to complete Table 1 Table 1(Sample results) Titration 1 2 3 Final volume(cm3) 12.5 25.0 37.5 Initial volume(cm3) 0.0 12.5 25.0 Volume of solution A used(cm3) 12.5 12.5 12.5
48 48 Sample questions Calculate the average volume of solution A used. 12.5+12.5+12.5 = 12.5cm3 3 Theoretical Practice examples 1. 1.0g of dibasic acid HOOC(CH2)xCOOH was dissolved in 250cm3 solution.
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47 47 25cm3 pipette volume -> 0.001245/1.245 x10-3 moles 250cm3 -> 0.001245/1.245 x10-3 moles x 250 = 0.01245/1.245 x10-2 moles 25 (iii)Determine the % by mass of sodium ethanedioate in the micture (H= 1.0,O=16.0,C=12.0 and total mass of mixture =2.0 g in 250cm3 solution) Molar mass H2C2O4 = 90.0g Mass of H2C2O4 in 250cm3 = moles in 250cm3 x molar mass H2C2O4 =>0.01245/1.245 x10-2 moles x 90.0 = 1.1205g % by mass of sodium ethanedioate =(Mass of mixture - mass of H2C2O4) x 100% Mass of mixture => 2.0 - 1.1205 g = 43.975% 2.0 Note (i) L is 0.05M Oxalic acid (ii) M is 0.01M KMnO4 (iii) N is 0.03M oxalic acid(without sodium oxalate) Practice example 5.(Determining equation for a reaction) You are provided with -0.1M hydrochloric acid solution A -0.5M sodium hydroxide solution B You are to determine the equation for thereaction between solution A and B Procedure Fill the burette with solution A.Using a pipette and pipette filler transfer 25.0cm3 of solution B into a conical flask.Add 2-3 drops of phenolphthalein indicator.Run solution A into solution B until a permanent pink colour just appears.Record your results in Table 1.Repeat the experiment to obtain three concordant results to complete Table 1 Table 1(Sample results) Titration 1 2 3 Final volume(cm3) 12.5 25.0 37.5 Initial volume(cm3) 0.0 12.5 25.0 Volume of solution A used(cm3) 12.5 12.5 12.5
48 48 Sample questions Calculate the average volume of solution A used. 12.5+12.5+12.5 = 12.5cm3 3 Theoretical Practice examples 1. 1.0g of dibasic acid HOOC(CH2)xCOOH was dissolved in 250cm3 solution. 25.0 cm3 of this solution reacted with 30.0cm3 of 0.06M sodium hydroxide solution.
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12.5+12.5+12.5 = 12.5cm3 3 Theoretical Practice examples 1. 1.0g of dibasic acid HOOC(CH2)xCOOH was dissolved in 250cm3 solution. 25.0 cm3 of this solution reacted with 30.0cm3 of 0.06M sodium hydroxide solution. Calculate the value of x in HOOC(CH2)xCOOH.
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46.0g of a metal carbonate MCO3 was dissolved 160cm3 of 0.1M excess hydrochloric acid and the resultant solution diluted to one litre.25.0cm3 of this solution required 20.0cm3 of 0.1M sodium hydroxide solution for complete neutralization. Calculate the atomic mass of ‘M’ Equation Chemical equation NaOH(aq) + HCl(aq) -> KCl(aq) + H2O(l) Moles of NaOH = Molarity x volume=> 0.1 x20 = 0.002 moles 1000 1000 Mole ratio HCl; NaOH = 1:1 Excess moles of HCl = 0.002 moles 25cm3 -> 0.002 moles 1000cm3 -> 1000 x 0.002 = 0.08moles 25cm3 Original moles of HCl = Molarity x volume => 1M x 1litre = 1.0 moles Moles of HCl reacted with MCO3 = 1.0 - 0.08 moles = 0.92moles Chemical equation MCO3(s) + 2HCl(aq) -> MCl2 (aq) + CO2 (g) + H2O(l) Mole ratio MCO3(s) : HCl(aq) =1:2 Moles of MCO3 = 0.92moles => 0.46moles 2 Molar mass of MCO3= mass => 46g = 100 g moles 0.46moles M= MCO3 - CO3 =>100g – (12+ 16 x3 = 60) = 40
52 52 6. 25.0cm3 of a mixture of Fe2+ and Fe3+ ions in an aqueous salt was acidified with sulphuric(VI)acid then titrated against potassium manganate(VI).The salt required 15cm3 ofe0.02M potassium manganate(VI) for complete reaction. A second 25cm3 portion of the Fe2+ and Fe3+ ion salt was reduced by Zinc then titrated against the same concentration of potassium manganate(VI).19.0cm3 of potassium manganate(VI)solution was used for complete reaction.
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Calculate the atomic mass of ‘M’ Equation Chemical equation NaOH(aq) + HCl(aq) -> KCl(aq) + H2O(l) Moles of NaOH = Molarity x volume=> 0.1 x20 = 0.002 moles 1000 1000 Mole ratio HCl; NaOH = 1:1 Excess moles of HCl = 0.002 moles 25cm3 -> 0.002 moles 1000cm3 -> 1000 x 0.002 = 0.08moles 25cm3 Original moles of HCl = Molarity x volume => 1M x 1litre = 1.0 moles Moles of HCl reacted with MCO3 = 1.0 - 0.08 moles = 0.92moles Chemical equation MCO3(s) + 2HCl(aq) -> MCl2 (aq) + CO2 (g) + H2O(l) Mole ratio MCO3(s) : HCl(aq) =1:2 Moles of MCO3 = 0.92moles => 0.46moles 2 Molar mass of MCO3= mass => 46g = 100 g moles 0.46moles M= MCO3 - CO3 =>100g – (12+ 16 x3 = 60) = 40
52 52 6. 25.0cm3 of a mixture of Fe2+ and Fe3+ ions in an aqueous salt was acidified with sulphuric(VI)acid then titrated against potassium manganate(VI).The salt required 15cm3 ofe0.02M potassium manganate(VI) for complete reaction. A second 25cm3 portion of the Fe2+ and Fe3+ ion salt was reduced by Zinc then titrated against the same concentration of potassium manganate(VI).19.0cm3 of potassium manganate(VI)solution was used for complete reaction. Calculate the concentration of Fe2+ and Fe3+ ion in the solution on moles per litre.
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25.0cm3 of a mixture of Fe2+ and Fe3+ ions in an aqueous salt was acidified with sulphuric(VI)acid then titrated against potassium manganate(VI).The salt required 15cm3 ofe0.02M potassium manganate(VI) for complete reaction. A second 25cm3 portion of the Fe2+ and Fe3+ ion salt was reduced by Zinc then titrated against the same concentration of potassium manganate(VI).19.0cm3 of potassium manganate(VI)solution was used for complete reaction. Calculate the concentration of Fe2+ and Fe3+ ion in the solution on moles per litre. Mole ratio Fe2+ :Mn04- = 5:1 Moles Mn04- used = 0.02 x 15 = 3.0 x 10-4 moles 1000 Moles Fe2+ = 3.0 x 10-4 moles = 6.0 x 10-5 moles 5 Molarity of Fe2+ = 6.0 x 10-4 moles x 1000 = 2.4 x 10-3 moles l-1 25 Since Zinc reduces Fe3+ to Fe2+ in the mixture: Moles Mn04- that reacted with all Fe2+= 0.02 x 19 = 3.8 x 10-4 moles 1000 Moles of all Fe2+ = 3.8 x 10-4 moles = 7.6 x 10-5 moles 5 Moles of Fe3+ = 3.8 x 10-4 - 6.0 x 10-5 = 1.6 x 10-5 moles Molarity of Fe3+ = 1.6 x 10-5 moles x 1000 = 4.0 x 10-4 moles l-1 25
14.0.0 ORGANIC CHEMISTRY I (HYDROCARBONS) (25 LESSONS) Introduction to Organic chemistry Organic chemistry is the branch of chemistry that studies carbon compounds present in living things, once living things or synthetic/man-made. Compounds that makes up living things whether alive or dead mainly contain carbon. Carbon is tetravalent.
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Mole ratio Fe2+ :Mn04- = 5:1 Moles Mn04- used = 0.02 x 15 = 3.0 x 10-4 moles 1000 Moles Fe2+ = 3.0 x 10-4 moles = 6.0 x 10-5 moles 5 Molarity of Fe2+ = 6.0 x 10-4 moles x 1000 = 2.4 x 10-3 moles l-1 25 Since Zinc reduces Fe3+ to Fe2+ in the mixture: Moles Mn04- that reacted with all Fe2+= 0.02 x 19 = 3.8 x 10-4 moles 1000 Moles of all Fe2+ = 3.8 x 10-4 moles = 7.6 x 10-5 moles 5 Moles of Fe3+ = 3.8 x 10-4 - 6.0 x 10-5 = 1.6 x 10-5 moles Molarity of Fe3+ = 1.6 x 10-5 moles x 1000 = 4.0 x 10-4 moles l-1 25
14.0.0 ORGANIC CHEMISTRY I (HYDROCARBONS) (25 LESSONS) Introduction to Organic chemistry Organic chemistry is the branch of chemistry that studies carbon compounds present in living things, once living things or synthetic/man-made. Compounds that makes up living things whether alive or dead mainly contain carbon. Carbon is tetravalent. It is able to form stable covalent bonds with itself and many non-metals like hydrogen, nitrogen ,oxygen and halogens to form a variety of compounds. This is because: (i) carbon uses all the four valence electrons to form four strong covalent bond. (ii)carbon can covalently bond to form a single, double or triple covalent bond with itself. (iii)carbon atoms can covalently bond to form a very long chain or ring. When carbon covalently bond with Hydrogen, it forms a group of organic compounds called Hydrocarbons A.HYDROCARBONS (HCs) Hydrocarbons are a group of organic compounds containing /made up of hydrogen and carbon atoms only.
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(ii)carbon can covalently bond to form a single, double or triple covalent bond with itself. (iii)carbon atoms can covalently bond to form a very long chain or ring. When carbon covalently bond with Hydrogen, it forms a group of organic compounds called Hydrocarbons A.HYDROCARBONS (HCs) Hydrocarbons are a group of organic compounds containing /made up of hydrogen and carbon atoms only. Depending on the type of bond that exist between the individual carbon atoms, hydrocarbon are classified as: (i) Alkanes (ii) Alkenes (iii) Alkynes (i) Alkanes (a)Nomenclature/Naming These are hydrocarbons with a general formula CnH2n+2 where n is the number of Carbon atoms in a molecule. The carbon atoms are linked by single bond to each other and to hydrogen atoms.
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When carbon covalently bond with Hydrogen, it forms a group of organic compounds called Hydrocarbons A.HYDROCARBONS (HCs) Hydrocarbons are a group of organic compounds containing /made up of hydrogen and carbon atoms only. Depending on the type of bond that exist between the individual carbon atoms, hydrocarbon are classified as: (i) Alkanes (ii) Alkenes (iii) Alkynes (i) Alkanes (a)Nomenclature/Naming These are hydrocarbons with a general formula CnH2n+2 where n is the number of Carbon atoms in a molecule. The carbon atoms are linked by single bond to each other and to hydrogen atoms. 2 They include: n General/ Molecular formula Structural formula Name 1 CH4 H H C H H Methane 2 C2H6 H H H C C H H H Ethane 3 C3H8 H H H H C C C H H H H Propane 4 C4H10 H H H H H C C C C H H H H H Butane 5 C5H12 H H H H H H C C C C C H CH3 (CH2) 6CH3 H H H H H Pentane 6 C6H14 H H H H H H Hexane
3 H C C C C C C H CH3 (CH2) 6CH3 H H H H H H 7 C7H16 H H H H H H H H C C C C C C C H H H H H H H H Heptane 8 C8H18 H H H H H H H H H C C C C C C C C H H H H H H H H H Octane 9 C9H20 H H H H H H H H H H C C C C C C C C C H H H H H H H H H H Nonane 10 C10H22 H H H H H H H H H H H C C C C C C C C C C H H H H H H H H H H H decane Note 1.The general formula/molecular formular of a compound shows the number of each atoms of elements making the compound e.g. Decane has a general/molecular formula C10H22 ;this means there are 10 carbon atoms and 22 hydrogen atoms in a molecule of decane.
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The carbon atoms are linked by single bond to each other and to hydrogen atoms. 2 They include: n General/ Molecular formula Structural formula Name 1 CH4 H H C H H Methane 2 C2H6 H H H C C H H H Ethane 3 C3H8 H H H H C C C H H H H Propane 4 C4H10 H H H H H C C C C H H H H H Butane 5 C5H12 H H H H H H C C C C C H CH3 (CH2) 6CH3 H H H H H Pentane 6 C6H14 H H H H H H Hexane
3 H C C C C C C H CH3 (CH2) 6CH3 H H H H H H 7 C7H16 H H H H H H H H C C C C C C C H H H H H H H H Heptane 8 C8H18 H H H H H H H H H C C C C C C C C H H H H H H H H H Octane 9 C9H20 H H H H H H H H H H C C C C C C C C C H H H H H H H H H H Nonane 10 C10H22 H H H H H H H H H H H C C C C C C C C C C H H H H H H H H H H H decane Note 1.The general formula/molecular formular of a compound shows the number of each atoms of elements making the compound e.g. Decane has a general/molecular formula C10H22 ;this means there are 10 carbon atoms and 22 hydrogen atoms in a molecule of decane. 2.The structural formula shows the arrangement/bonding of atoms of each element making the compound e.g Decane has the structural formula as in the table above ;this means the 1st carbon from left to right is bonded to three hydrogen atoms and one carbon atom. The 2nd carbon atom is joined/bonded to two other carbon atoms and two Hydrogen atoms. 4 3.Since carbon is tetravalent ,each atom of carbon in the alkane MUST always be bonded using four covalent bond /four shared pairs of electrons.
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2.The structural formula shows the arrangement/bonding of atoms of each element making the compound e.g Decane has the structural formula as in the table above ;this means the 1st carbon from left to right is bonded to three hydrogen atoms and one carbon atom. The 2nd carbon atom is joined/bonded to two other carbon atoms and two Hydrogen atoms. 4 3.Since carbon is tetravalent ,each atom of carbon in the alkane MUST always be bonded using four covalent bond /four shared pairs of electrons. 4.Since Hydrogen is monovalent ,each atom of hydrogen in the alkane MUST always be bonded using one covalent bond/one shared pair of electrons. 5.One member of the alkane differ from the next/previous by a CH2 group. e.g Propane differ from ethane by one carbon and two Hydrogen atoms form ethane. Ethane differ from methane also by one carbon and two Hydrogen atoms 6.A group of compounds that differ by a CH2 group from the next /previous consecutively is called a homologous series. 7.A homologous series: (i) differ by a CH2 group from the next /previous consecutively (ii)have similar chemical properties (iii)have similar chemical formula that can be represented by a general formula e.g alkanes have the general formula CnH2n+2. (iv)the physical properties (e.g.melting/boiling points)show steady gradual change) 8.The 1st four alkanes have the prefix meth_,eth_,prop_ and but_ to represent 1,2,3 and 4 carbons in the compound. All other use the numeral prefix pent_,Hex_,hept_ , etc to show also the number of carbon atoms. 9.If one hydrogen atom in an alkane is removed, an alkyl group is formed.e.g Alkane name molecular structure CnH2n+2 Alkyl name Molecula structure CnH2n+1 methane CH4 methyl CH3 ethane CH3CH3 ethyl CH3 CH2 propane CH3 CH2 CH3 propyl CH3 CH2 CH2 butane CH3 CH2 CH2 CH3 butyl CH3 CH2 CH2 CH2 (b)Isomers of alkanes Isomers are compounds with the same molecular general formula but different molecular structural formula.
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(iv)the physical properties (e.g.melting/boiling points)show steady gradual change) 8.The 1st four alkanes have the prefix meth_,eth_,prop_ and but_ to represent 1,2,3 and 4 carbons in the compound. All other use the numeral prefix pent_,Hex_,hept_ , etc to show also the number of carbon atoms. 9.If one hydrogen atom in an alkane is removed, an alkyl group is formed.e.g Alkane name molecular structure CnH2n+2 Alkyl name Molecula structure CnH2n+1 methane CH4 methyl CH3 ethane CH3CH3 ethyl CH3 CH2 propane CH3 CH2 CH3 propyl CH3 CH2 CH2 butane CH3 CH2 CH2 CH3 butyl CH3 CH2 CH2 CH2 (b)Isomers of alkanes Isomers are compounds with the same molecular general formula but different molecular structural formula. Isomerism is the existence of a compounds having the same general/molecular formula but different structural formula. The 1st three alkanes do not form isomers.Isomers are named by using the IUPAC(International Union of Pure and Applied Chemistry) system of nomenclature/naming. The IUPAC system of nomenclature uses the following basic rules/guidelines: 1.Identify the longest continuous carbon chain to get/determine the parent alkane. 5 2.Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible 3. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane. Name them fluoro-,chloro-,bromo-,iodo- if they are halogens 4.Use prefix di-,tri-,tetra-,penta-,hexa- to show the number of branches attached to the parent alkane. Practice on IUPAC nomenclature of alkanes (a)Draw the structure of: (i)2-methylpentane Procedure 1. Identify the longest continuous carbon chain to get/determine the parent alkane. Butane is the parent name CH3 CH2 CH2 CH3 2.
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Practice on IUPAC nomenclature of alkanes (a)Draw the structure of: (i)2-methylpentane Procedure 1. Identify the longest continuous carbon chain to get/determine the parent alkane. Butane is the parent name CH3 CH2 CH2 CH3 2. Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible The methyl group is attached to Carbon “2” 3. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane i.e Position of the branch at carbon “2” Number of branches at carbon “1” Type of the branch “methyl” hence Molecular formula CH3 CH3 CH CH2 CH3 // CH3 CH (CH3 ) CH2CH3 Structural formula H H H H H C C C C H H H H H C H
6 H (ii)2,2-dimethylpentane Procedure 1. Identify the longest continuous carbon chain to get/determine the parent alkane. Butane is the parent name CH3 CH2 CH2 CH3 2. Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible The methyl group is attached to Carbon “2” 3. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane i.e Position of the branch at carbon “2” Number of branches at carbon “2” Type of the branch two“methyl” hence Molecular formular CH3 CH3 C CH2 CH3 // CH3 C (CH3 )2 CH2CH3 CH3 Structural formula H H C H H H H H C C C C H H H H H C H H
7 (iii) 2,2,3-trimethylbutane Procedure 1. Identify the longest continuous carbon chain to get/determine the parent alkane. Butane is the parent name CH3 CH2 CH2 CH3 2. Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible The methyl group is attached to Carbon “2 and 3” 3.
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Identify the longest continuous carbon chain to get/determine the parent alkane. Butane is the parent name CH3 CH2 CH2 CH3 2. Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible The methyl group is attached to Carbon “2 and 3” 3. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane i.e Position of the branch at carbon “2 and 3” Number of branches at carbon “3” Type of the branch three “methyl” hence Molecular formular CH3 CH3 C CH CH3 // CH3 C (CH3 )3 CH2CH3 CH3 CH3 Structural formula H H C H H H H C C C H H H H H C C H H H C H
8 H (iv) 1,1,1,2,2,2-hexabromoethane Molecular formula CBr3 CBr3 Structural formula Br Br Br C C Br Br Br (v) 1,1,1-tetrachloro-2,2-dimethylbutane CH3 CCl 3 C CH3 // C Cl 3 C (CH3 )2 CH3 CH3 Structural formula Cl Cl C Cl H H H C C C H H H H C H H (c)Occurrence and extraction Crude oil ,natural gas and biogas are the main sources of alkanes: (i)Natural gas is found on top of crude oil deposits and consists mainly of methane. 9 (ii)Biogas is formed from the decay of waste organic products like animal dung and cellulose. When the decay takes place in absence of oxygen , 60-75% by volume of the gaseous mixture of methane gas is produced. (iii)Crude oil is a mixture of many flammable hydrocarbons/substances. Using fractional distillation, each hydrocarbon fraction can be separated from the other. The hydrocarbon with lower /smaller number of carbon atoms in the chain have lower boiling point and thus collected first. As the carbon chain increase, the boiling point, viscosity (ease of flow) and colour intensity increase as flammability decrease. Hydrocarbons in crude oil are not pure. They thus have no sharp fixed boiling point.
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As the carbon chain increase, the boiling point, viscosity (ease of flow) and colour intensity increase as flammability decrease. Hydrocarbons in crude oil are not pure. They thus have no sharp fixed boiling point. Uses of different crude oil fractions Carbon atoms in a molecule Common name of fraction Uses of fraction 1-4 Gas L.P.G gas for domestic use 5-12 Petrol Fuel for petrol engines 9-16 Kerosene/Paraffin Jet fuel and domestic lighting/cooking 15-18 Light diesel Heavy diesel engine fuel 18-25 Diesel oil Light diesel engine fuel 20-70 Lubricating oil Lubricating oil to reduce friction. Over 70 Bitumen/Asphalt Tarmacking roads (d)School laboratory preparation of alkanes In a school laboratory, alkanes may be prepared from the reaction of a sodium alkanoate with solid sodium hydroxide/soda lime. Chemical equation: Sodium alkanoate + soda lime -> alkane + Sodium carbonate CnH2n+1COONa(s) + NaOH(s) -> C n H2n+2 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CnH2n+1 in CnH2n+1COONa(s) to form C n H2n+2. 10 Examples 1. Methane is prepared from the heating of a mixture of sodium ethanoate and soda lime and collecting over water Sodium ethanoate + soda lime -> methane + Sodium carbonate CH3COONa(s) + NaOH(s) -> C H4 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CH3 in CH3COONa(s) to form CH4. 2. Ethane is prepared from the heating of a mixture of sodium propanoate and soda lime and collecting over water Sodium propanoate + soda lime -> ethane + Sodium carbonate CH3 CH2COONa(s) + NaOH(s) -> CH3 CH3 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CH3 CH2 in CH3 CH2COONa (s) to form CH3 CH3 3.
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Methane is prepared from the heating of a mixture of sodium ethanoate and soda lime and collecting over water Sodium ethanoate + soda lime -> methane + Sodium carbonate CH3COONa(s) + NaOH(s) -> C H4 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CH3 in CH3COONa(s) to form CH4. 2. Ethane is prepared from the heating of a mixture of sodium propanoate and soda lime and collecting over water Sodium propanoate + soda lime -> ethane + Sodium carbonate CH3 CH2COONa(s) + NaOH(s) -> CH3 CH3 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CH3 CH2 in CH3 CH2COONa (s) to form CH3 CH3 3. Propane is prepared from the heating of a mixture of sodium butanoate and soda lime and collecting over water Sodium butanoate + soda lime -> propane + Sodium carbonate CH3 CH2CH2COONa(s) + NaOH(s) -> CH3 CH2CH3 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CH3 CH2 CH2 in CH3 CH2CH2COONa (s) to form CH3 CH2CH3 4. Butane is prepared from the heating of a mixture of sodium pentanoate and soda lime and collecting over water Sodium pentanoate + soda lime -> butane + Sodium carbonate CH3 CH2 CH2CH2COONa(s)+NaOH(s) -> CH3 CH2CH2CH3 + Na2CO3(s) The “H” in NaOH is transferred/moves to the CH3CH2 CH2 CH2 in CH3 CH2CH2 CH2COONa (s) to form CH3 CH2 CH2CH3 Laboratory set up for the preparation of alkanes
11 (d)Properties of alkanes I. Physical properties Alkanes are colourless gases, solids and liquids that are not poisonous. They are slightly soluble in water. The solubility decrease as the carbon chain and thus the molar mass increase The melting and boiling point increase as the carbon chain increase. This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase.
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They are slightly soluble in water. The solubility decrease as the carbon chain and thus the molar mass increase The melting and boiling point increase as the carbon chain increase. This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase. The 1st four straight chain alkanes (methane,ethane,propane and butane)are therefore gases ,the nect six(pentane ,hexane, heptane,octane,nonane, and decane) are liquids while the rest from unidecane(11 carbon atoms) are solids . The density of straight chain alkanes increase with increasing carbon chain as the intermolecular forces increases. This reduces the volume occupied by a given mass of the compound. Summary of physical properties of alkanes Alkane General formula Melting point(K) Boiling point(K) Density gcm-3 State at room(298K) temperature and pressure atmosphere (101300Pa) Methane CH4 90 112 0.424 gas
12 Ethane CH3CH3 91 184 0.546 gas Propane CH3CH2CH3 105 231 0.501 gas Butane CH3(CH2)2CH3 138 275 0.579 gas Pentane CH3(CH2)3CH3 143 309 0.626 liquid Hexane CH3(CH2)4CH3 178 342 0.657 liquid Heptane CH3(CH2)5CH3 182 372 0.684 liquid Octane CH3(CH2)6CH3 216 399 0.703 liquid Nonane CH3(CH2)7CH3 219 424 0.708 liquid Octane CH3(CH2)8CH3 243 447 0.730 liquid II.Chemical properties (i)Burning. Alkanes burn with a blue/non-luminous non-sooty/non-smoky flame in excess air to form carbon(IV) oxide and water. Alkane + Air -> carbon(IV) oxide + water (excess air/oxygen) Alkanes burn with a blue/non-luminous no-sooty/non-smoky flame in limited air to form carbon(II) oxide and water.
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Summary of physical properties of alkanes Alkane General formula Melting point(K) Boiling point(K) Density gcm-3 State at room(298K) temperature and pressure atmosphere (101300Pa) Methane CH4 90 112 0.424 gas
12 Ethane CH3CH3 91 184 0.546 gas Propane CH3CH2CH3 105 231 0.501 gas Butane CH3(CH2)2CH3 138 275 0.579 gas Pentane CH3(CH2)3CH3 143 309 0.626 liquid Hexane CH3(CH2)4CH3 178 342 0.657 liquid Heptane CH3(CH2)5CH3 182 372 0.684 liquid Octane CH3(CH2)6CH3 216 399 0.703 liquid Nonane CH3(CH2)7CH3 219 424 0.708 liquid Octane CH3(CH2)8CH3 243 447 0.730 liquid II.Chemical properties (i)Burning. Alkanes burn with a blue/non-luminous non-sooty/non-smoky flame in excess air to form carbon(IV) oxide and water. Alkane + Air -> carbon(IV) oxide + water (excess air/oxygen) Alkanes burn with a blue/non-luminous no-sooty/non-smoky flame in limited air to form carbon(II) oxide and water. Alkane + Air -> carbon(II) oxide + water (limited air) Examples 1.(a) Methane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water. Methane + Air -> carbon(IV) oxide + water (excess air/oxygen) CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l/g) (b) Methane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water.
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Alkane + Air -> carbon(IV) oxide + water (excess air/oxygen) Alkanes burn with a blue/non-luminous no-sooty/non-smoky flame in limited air to form carbon(II) oxide and water. Alkane + Air -> carbon(II) oxide + water (limited air) Examples 1.(a) Methane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water. Methane + Air -> carbon(IV) oxide + water (excess air/oxygen) CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l/g) (b) Methane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water. Methane + Air -> carbon(II) oxide + water (excess air/oxygen) 2CH4(g) + 3O2(g) -> 2CO(g) + 4H2O(l/g) 2.(a) Ethane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water. Ethane + Air -> carbon(IV) oxide + water (excess air/oxygen) 2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l/g) (b) Ethane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water. 13 Ethane + Air -> carbon(II) oxide + water (excess air/oxygen) 2C2H6(g) + 5O2(g) -> 4CO(g) + 6H2O(l/g) 3.(a) Propane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water. Propane + Air -> carbon(IV) oxide + water (excess air/oxygen) C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l/g) (b) Ethane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water.
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Ethane + Air -> carbon(IV) oxide + water (excess air/oxygen) 2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l/g) (b) Ethane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water. 13 Ethane + Air -> carbon(II) oxide + water (excess air/oxygen) 2C2H6(g) + 5O2(g) -> 4CO(g) + 6H2O(l/g) 3.(a) Propane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water. Propane + Air -> carbon(IV) oxide + water (excess air/oxygen) C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l/g) (b) Ethane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water. Ethane + Air -> carbon(II) oxide + water (excess air/oxygen) 2C3H8(g) + 7O2(g) -> 6CO(g) + 8H2O(l/g) ii)Substitution Substitution reaction is one in which a hydrogen atom is replaced by a halogen in presence of ultraviolet light. Alkanes react with halogens in presence of ultraviolet light to form halogenoalkanes. During substitution: (i)the halogen molecule is split into free atom/radicals. (ii)one free halogen radical/atoms knock /remove one hydrogen from the alkane leaving an alkyl radical. (iii) the alkyl radical combine with the other free halogen atom/radical to form halogenoalkane. (iv)the chlorine atoms substitute repeatedly in the alkane. Each substitution removes a hydrogen atom from the alkane and form hydrogen halide. (v)substitution stops when all the hydrogen in alkanes are replaced with halogens. Substitution reaction is a highly explosive reaction in presence of sunlight / ultraviolet light that act as catalyst. Examples of substitution reactions Methane has no effect on bromine or chlorine in diffused light/dark.
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(v)substitution stops when all the hydrogen in alkanes are replaced with halogens. Substitution reaction is a highly explosive reaction in presence of sunlight / ultraviolet light that act as catalyst. Examples of substitution reactions Methane has no effect on bromine or chlorine in diffused light/dark. In sunlight , a mixture of chlorine and methane explode to form colourless mixture of chloromethane and hydrogen chloride gas. The pale green colour of chlorine gas fades. Chemical equation 1.(a)Methane + chlorine -> Chloromethane + Hydrogen chloride
14 CH4(g) + Cl2(g) -> CH3Cl (g) + HCl (g) H H H C H + Cl Cl -> H C Cl + H Cl H H (b) Chloromethane + chlorine -> dichloromethane + Hydrogen chloride CH3Cl (g) + Cl2(g) -> CH2Cl2 (g) + HCl (g) H H H C Cl + Cl Cl -> H C Cl + H Cl H Cl (c) dichloromethane + chlorine -> trichloromethane + Hydrogen chloride CH2Cl2 (g) + Cl2(g) -> CHCl3 (g) + HCl (g) Cl H H C Cl + Cl Cl -> Cl C Cl + H Cl H Cl (c) trichloromethane + chlorine -> tetrachloromethane + Hydrogen chloride CHCl3 (g) + Cl2(g) -> CCl4 (g) + HCl (g) H Cl Cl C Cl + Cl Cl -> Cl C Cl + H Cl Cl Cl
15 Ethane has no effect on bromine or chlorine in diffused light/dark. In sunlight , a mixture of bromine and ethane explode to form colourless mixture of bromoethane and hydrogen chloride gas. The red/brown colour of bromine gas fades.
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Chemical equation 1.(a)Methane + chlorine -> Chloromethane + Hydrogen chloride
14 CH4(g) + Cl2(g) -> CH3Cl (g) + HCl (g) H H H C H + Cl Cl -> H C Cl + H Cl H H (b) Chloromethane + chlorine -> dichloromethane + Hydrogen chloride CH3Cl (g) + Cl2(g) -> CH2Cl2 (g) + HCl (g) H H H C Cl + Cl Cl -> H C Cl + H Cl H Cl (c) dichloromethane + chlorine -> trichloromethane + Hydrogen chloride CH2Cl2 (g) + Cl2(g) -> CHCl3 (g) + HCl (g) Cl H H C Cl + Cl Cl -> Cl C Cl + H Cl H Cl (c) trichloromethane + chlorine -> tetrachloromethane + Hydrogen chloride CHCl3 (g) + Cl2(g) -> CCl4 (g) + HCl (g) H Cl Cl C Cl + Cl Cl -> Cl C Cl + H Cl Cl Cl
15 Ethane has no effect on bromine or chlorine in diffused light/dark. In sunlight , a mixture of bromine and ethane explode to form colourless mixture of bromoethane and hydrogen chloride gas. The red/brown colour of bromine gas fades. Chemical equation (a)Ethane + chlorine -> Chloroethane + Hydrogen chloride CH3CH3(g) + Br2(g) -> CH3CH2Br (g) + HBr (g) H H H H H C C H + Br Br -> H C C H + H Br H H H Br Bromoethane H H H Br H C C H + Br Br -> H C C H + H Br H Br H Br 1,1-dibromoethane H Br H Br H C C H + Br Br -> H C C Br + H Br H Br H Br 1,1,1-tribromoethane H Br H Br H C C Br + Br Br -> H C C Br + H Br H Br Br Br 1,1,1,2-tetrabromoethane H Br H Br H C C Br + Br Br -> Br C C Br + H Br
16 Br Br Br Br 1,1,1,2,2-pentabromoethane H Br Br Br Br C C Br + Br Br -> Br C C Br + H Br Br Br Br Br 1,1,1,2,2,2-hexabromoethane Uses of alkanes 1.Most alkanes are used as fuel e.g.
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In sunlight , a mixture of bromine and ethane explode to form colourless mixture of bromoethane and hydrogen chloride gas. The red/brown colour of bromine gas fades. Chemical equation (a)Ethane + chlorine -> Chloroethane + Hydrogen chloride CH3CH3(g) + Br2(g) -> CH3CH2Br (g) + HBr (g) H H H H H C C H + Br Br -> H C C H + H Br H H H Br Bromoethane H H H Br H C C H + Br Br -> H C C H + H Br H Br H Br 1,1-dibromoethane H Br H Br H C C H + Br Br -> H C C Br + H Br H Br H Br 1,1,1-tribromoethane H Br H Br H C C Br + Br Br -> H C C Br + H Br H Br Br Br 1,1,1,2-tetrabromoethane H Br H Br H C C Br + Br Br -> Br C C Br + H Br
16 Br Br Br Br 1,1,1,2,2-pentabromoethane H Br Br Br Br C C Br + Br Br -> Br C C Br + H Br Br Br Br Br 1,1,1,2,2,2-hexabromoethane Uses of alkanes 1.Most alkanes are used as fuel e.g. Methane is used as biogas in homes.Butane is used as the Laboratory gas. 2.On cracking ,alkanes are a major source of Hydrogen for the manufacture of ammonia/Haber process. 3.In manufacture of Carbon black which is a component in printers ink. 4.In manufacture of useful industrial chemicals like methanol, methanol, and chloromethane. (ii) Alkenes (a)Nomenclature/Naming These are hydrocarbons with a general formula CnH2n and C C double bond as the functional group . n is the number of Carbon atoms in the molecule. The carbon atoms are linked by at least one double bond to each other and single bonds to hydrogen atoms.
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(ii) Alkenes (a)Nomenclature/Naming These are hydrocarbons with a general formula CnH2n and C C double bond as the functional group . n is the number of Carbon atoms in the molecule. The carbon atoms are linked by at least one double bond to each other and single bonds to hydrogen atoms. They include: n General/ Molecular formula Structural formula Name 1 Does not exist 2 C2H6 H H H C C H Ethene
17 CH2 CH2 3 C3H8 H H H H C C C H H CH2 CH CH3 Propene 4 C4H10 H H H H H C C C C H H H CH2 CH CH2CH3 Butene 5 C5H12 H H H H H H C C C C C H H H H CH2 CH (CH2)2CH3 Pentene 6 C6H14 H H H H H H H C C C C C C H H H H H CH2 CH (CH2)3CH3 Hexene 7 C7H16 H H H H H H H H C C C C C C C H H H H H H H H Heptene
18 CH2 CH (CH2)4CH3 8 C8H18 H H H H H H H H H C C C C C C C C H H H H H H H CH2 CH (CH2)5CH3 Octene 9 C9H20 H H H H H H H H H H C C C C C C C C C H H H H H H H H CH2 CH (CH2)6CH3 Nonene 10 C10H22 H H H H H H H H H H H C C C C C C C C C C H H H H H H H H H CH2 CH (CH2)7CH3 decene Note 1.Since carbon is tetravalent ,each atom of carbon in the alkene MUST always be bonded using four covalent bond /four shared pairs of electrons including at the double bond. 2.Since Hydrogen is monovalent ,each atom of hydrogen in the alkene MUST always be bonded using one covalent bond/one shared pair of electrons. 3.One member of the alkene ,like alkanes,differ from the next/previous by a CH2 group.They also form a homologous series.
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They include: n General/ Molecular formula Structural formula Name 1 Does not exist 2 C2H6 H H H C C H Ethene
17 CH2 CH2 3 C3H8 H H H H C C C H H CH2 CH CH3 Propene 4 C4H10 H H H H H C C C C H H H CH2 CH CH2CH3 Butene 5 C5H12 H H H H H H C C C C C H H H H CH2 CH (CH2)2CH3 Pentene 6 C6H14 H H H H H H H C C C C C C H H H H H CH2 CH (CH2)3CH3 Hexene 7 C7H16 H H H H H H H H C C C C C C C H H H H H H H H Heptene
18 CH2 CH (CH2)4CH3 8 C8H18 H H H H H H H H H C C C C C C C C H H H H H H H CH2 CH (CH2)5CH3 Octene 9 C9H20 H H H H H H H H H H C C C C C C C C C H H H H H H H H CH2 CH (CH2)6CH3 Nonene 10 C10H22 H H H H H H H H H H H C C C C C C C C C C H H H H H H H H H CH2 CH (CH2)7CH3 decene Note 1.Since carbon is tetravalent ,each atom of carbon in the alkene MUST always be bonded using four covalent bond /four shared pairs of electrons including at the double bond. 2.Since Hydrogen is monovalent ,each atom of hydrogen in the alkene MUST always be bonded using one covalent bond/one shared pair of electrons. 3.One member of the alkene ,like alkanes,differ from the next/previous by a CH2 group.They also form a homologous series. e.g Propene differ from ethene by one carbon and two Hydrogen atoms from ethene.
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2.Since Hydrogen is monovalent ,each atom of hydrogen in the alkene MUST always be bonded using one covalent bond/one shared pair of electrons. 3.One member of the alkene ,like alkanes,differ from the next/previous by a CH2 group.They also form a homologous series. e.g Propene differ from ethene by one carbon and two Hydrogen atoms from ethene. 4.A homologous series of alkenes like that of alkanes: (i) differ by a CH2 group from the next /previous consecutively (ii)have similar chemical properties
19 (iii)have similar chemical formula represented by the general formula CnH2n (iv)the physical properties also show steady gradual change 5.The = C= C = double bond in alkene is the functional group. A functional group is the reacting site of a molecule/compound. 6. The = C= C = double bond in alkene can easily be broken to accommodate more two more monovalent atoms. The = C= C = double bond in alkenes make it thus unsaturated. 7. An unsaturated hydrocarbon is one with a double =C=C= or triple – C C – carbon bonds in their molecular structure. Unsaturated hydrocarbon easily reacts to be saturated. 8.A saturated hydrocarbon is one without a double =C=C= or triple – C C – carbon bonds in their molecular structure. Most of the reactions of alkenes take place at the = C = C =bond. (b)Isomers of alkenes Isomers are alkenes lie alkanes have the same molecular general formula but different molecular structural formula. Ethene and propene do not form isomers. Isomers of alkenes are also named by using the IUPAC(International Union of Pure and Applied Chemistry) system of nomenclature/naming. The IUPAC system of nomenclature of naming alkenes uses the following basic rules/guidelines: 1.Identify the longest continuous/straight carbon chain which contains the =C = C= double bond get/determine the parent alkene. 2.Number the longest chain form the end of the chain which contains the =C = C= double bond so he =C = C= double bond lowest number possible. 3 Indicate the positions by splitting “alk-positions-ene” e.g. but-2-ene, pent-1,3diene.
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2.Number the longest chain form the end of the chain which contains the =C = C= double bond so he =C = C= double bond lowest number possible. 3 Indicate the positions by splitting “alk-positions-ene” e.g. but-2-ene, pent-1,3diene. 4.The position indicated must be for the carbon atom at the lower position in the =C = C= double bond.i.e But-2-ene means the double =C = C= is between Carbon “2”and “3” Pent-1,3-diene means there are two double bond one between carbon “1” and “2”and another between carbon “3” and “4” 5. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of alkyl carbon chains attached to the alkene. Name them fluoro-,chloro-,bromo-,iodo- if they are halogens 6.Use prefix di-,tri-,tetra-,penta-,hexa- to show the number of double C = C bonds and branches attached to the alkene. 20 7.Position isomers can be formed when the=C = C= double bond is shifted between carbon atoms e.g. But-2-ene means the double =C = C= is between Carbon “2”and “3” But-1-ene means the double =C = C= is between Carbon “1”and “2” Both But-1-ene and But-2-ene are position isomers of Butene 8.Position isomers are molecules/compounds having the same general formular but different position of the functional group.i.e. Butene has the molecular/general formular C4H8 position but can form both But-1ene and But-2-ene as position isomers. 9. Like alkanes ,an alkyl group can be attached to the alkene. Chain/branch isomers are thus formed. 10.Chain/branch isomers are molecules/compounds having the same general formula but different structural formula e.g Butene and 2-methyl propene both have the same general formualr but different branching chain.
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Like alkanes ,an alkyl group can be attached to the alkene. Chain/branch isomers are thus formed. 10.Chain/branch isomers are molecules/compounds having the same general formula but different structural formula e.g Butene and 2-methyl propene both have the same general formualr but different branching chain. Practice on IUPAC nomenclature of alkenes Name the following isomers of alkene H H H H H C C C C H But-1-ene H H H H H H H C C C C H But-2-ene H H H H H H H H H C C C C C C H 4-methylhex-1-ene H H H H C H
21 H H H C H H H H H H H C C C C C C H 4,4-dimethylhex-1-ene H H H H C H H 3. H H C H H H H H H C C C C C H 4,4-dimethylpent -1- ene H H H C H H 4. H H C H H H H H H C C C C C H 5,5-dimethylhex-2- ene H C H H H H C H H H
22 5. H H C H H H H H C C C C H 2,2-dimethylbut -2- ene H H H C H H 8.H2C CHCH2 CH2 CH3 pent -1- ene 9.H2C C(CH3)CH2 CH2 CH3 2-methylpent -1- ene 10.H2C C(CH3)C(CH3)2 CH2 CH3 2,3,3-trimethylpent -1- ene 11.H2C C(CH3)C(CH3)2 C(CH3)2 CH3 2,3,3,4,4-pentamethylpent -1- ene 12.H3C C(CH3)C(CH3) C(CH3)2 CH3 2,3,4,4-tetramethylpent -2- ene 13. H2C C(CH3)C(CH3) C(CH3) CH3 2,3,4-trimethylpent -1,3- diene 14.
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I2C CICI CI2 1,1,2,3,4,4-hexaiodobut -1,3- diene 18. H2C C(CH3)C(CH3) CH2 2,3-dimethylbut -1,3- diene (c)Occurrence and extraction
23 At indusrial level,alkenes are obtained from the cracking of alkanes.Cracking is the process of breaking long chain alkanes to smaller/shorter alkanes, an alkene and hydrogen gas at high temperatures. Cracking is a major source of useful hydrogen gas for manufacture of ammonia/nitric(V)acid/HCl i.e. Long chain alkane -> smaller/shorter alkane + Alkene + Hydrogen gas Examples 1.When irradiated with high energy radiation,Propane undergo cracking to form methane gas, ethene and hydrogen gas. Chemical equation CH3CH2CH3 (g) -> CH4(g) + CH2=CH2(g) + H2(g) 2.Octane undergo cracking to form hydrogen gas, butene and butane gases Chemical equation CH3(CH2) 6 CH3 (g) -> CH3CH2CH2CH3(g) + CH3 CH2CH=CH2(g) + H2(g) (d)School laboratory preparation of alkenes In a school laboratory, alkenes may be prepared from dehydration of alkanols using: (i) concentrated sulphuric(VI)acid(H2SO4). (a) aluminium(III)oxide(Al2O3) i.e Alkanol --Conc. H2SO4 --> Alkene + Water Alkanol --Al2O3 --> Alkene + Water e.g. 1.(a)At about 180oC,concentrated sulphuric(VI)acid dehydrates/removes water from ethanol to form ethene. The gas produced contain traces of carbon(IV)oxide and sulphur(IV)oxide gas as impurities. It is thus passed through concentrated sodium/potassium hydroxide solution to remove the impurities.
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1.(a)At about 180oC,concentrated sulphuric(VI)acid dehydrates/removes water from ethanol to form ethene. The gas produced contain traces of carbon(IV)oxide and sulphur(IV)oxide gas as impurities. It is thus passed through concentrated sodium/potassium hydroxide solution to remove the impurities. Chemical equation CH3CH2OH (l) --conc H2SO4/180oC--> CH2=CH2(g) + H2O(l) (b)On heating strongly aluminium(III)oxide(Al2O3),it dehydrates/removes water from ethanol to form ethene. 24 Ethanol vapour passes through the hot aluminium (III) oxide which catalyses the dehydration. Activated aluminium(III)oxide has a very high affinity for water molecules/elements of water and thus dehydrates/ removes water from ethanol to form ethene. Chemical equation CH3CH2OH (l) --(Al2O3/strong heat--> CH2=CH2(g) + H2O(l) 2(a) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by conc H2SO4 at about 180oC to propene(propene has no position isomers).
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24 Ethanol vapour passes through the hot aluminium (III) oxide which catalyses the dehydration. Activated aluminium(III)oxide has a very high affinity for water molecules/elements of water and thus dehydrates/ removes water from ethanol to form ethene. Chemical equation CH3CH2OH (l) --(Al2O3/strong heat--> CH2=CH2(g) + H2O(l) 2(a) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by conc H2SO4 at about 180oC to propene(propene has no position isomers). Chemical equation CH3CH2 CH2OH (l) -- conc H2SO4/180oC --> CH3CH2=CH2(g) + H2O(l) Propan-1-ol Prop-1-ene CH3CHOH CH3 (l) -- conc H2SO4/180oC --> CH3CH2=CH2(g) + H2O(l) Propan-2-ol Prop-1-ene (b) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by heating strongly aluminium(III)oxide(Al2O3) form propene Chemical equation CH3CH2 CH2OH (l) -- Heat/Al2O3 --> CH3CH2=CH2(g) + H2O(l) Propan-1-ol Prop-1-ene CH3CHOH CH3 (l) -- Heat/Al2O3 --> CH3CH2=CH2(g) + H2O(l) Propan-2-ol Prop-1-ene 3(a) Butan-1-ol and Butan-2-ol(position isomers of butanol) are dehydrated by conc H2SO4 at about 180oC to But-1-ene and But-2-ene respectively Chemical equation CH3CH2 CH2 CH2OH (l) -- conc H2SO4/180oC -->CH3 CH2CH2=CH2(g) + H2O(l) Butan-1-ol But-1-ene CH3CHOH CH2CH3 (l)-- conc H2SO4/180oC -->CH3CH=CH CH2(g) + H2O(l) Butan-2-ol But-2-ene (b) Butan-1-ol and Butan-2-ol are dehydrated by heating strongly aluminium (III) oxide (Al2O3) form But-1-ene and But-2-ene respectively.
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Activated aluminium(III)oxide has a very high affinity for water molecules/elements of water and thus dehydrates/ removes water from ethanol to form ethene. Chemical equation CH3CH2OH (l) --(Al2O3/strong heat--> CH2=CH2(g) + H2O(l) 2(a) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by conc H2SO4 at about 180oC to propene(propene has no position isomers). Chemical equation CH3CH2 CH2OH (l) -- conc H2SO4/180oC --> CH3CH2=CH2(g) + H2O(l) Propan-1-ol Prop-1-ene CH3CHOH CH3 (l) -- conc H2SO4/180oC --> CH3CH2=CH2(g) + H2O(l) Propan-2-ol Prop-1-ene (b) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by heating strongly aluminium(III)oxide(Al2O3) form propene Chemical equation CH3CH2 CH2OH (l) -- Heat/Al2O3 --> CH3CH2=CH2(g) + H2O(l) Propan-1-ol Prop-1-ene CH3CHOH CH3 (l) -- Heat/Al2O3 --> CH3CH2=CH2(g) + H2O(l) Propan-2-ol Prop-1-ene 3(a) Butan-1-ol and Butan-2-ol(position isomers of butanol) are dehydrated by conc H2SO4 at about 180oC to But-1-ene and But-2-ene respectively Chemical equation CH3CH2 CH2 CH2OH (l) -- conc H2SO4/180oC -->CH3 CH2CH2=CH2(g) + H2O(l) Butan-1-ol But-1-ene CH3CHOH CH2CH3 (l)-- conc H2SO4/180oC -->CH3CH=CH CH2(g) + H2O(l) Butan-2-ol But-2-ene (b) Butan-1-ol and Butan-2-ol are dehydrated by heating strongly aluminium (III) oxide (Al2O3) form But-1-ene and But-2-ene respectively. Chemical equation CH3CH2 CH2 CH2OH (l) -- Heat/Al2O3 --> CH3 CH2CH2=CH2(g) + H2O(l) Butan-1-ol But-1-ene
25 CH3CHOH CH2CH3 (l) -- Heat/Al2O3 --> CH3CH=CH CH2(g) + H2O(l) Butan-2-ol But-2-ene Laboratory set up for the preparation of alkenes/ethene Caution (i)Ethanol is highly inflammable (ii)Conc H2SO4 is highly corrosive on skin contact.
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Chemical equation CH3CH2OH (l) --(Al2O3/strong heat--> CH2=CH2(g) + H2O(l) 2(a) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by conc H2SO4 at about 180oC to propene(propene has no position isomers). Chemical equation CH3CH2 CH2OH (l) -- conc H2SO4/180oC --> CH3CH2=CH2(g) + H2O(l) Propan-1-ol Prop-1-ene CH3CHOH CH3 (l) -- conc H2SO4/180oC --> CH3CH2=CH2(g) + H2O(l) Propan-2-ol Prop-1-ene (b) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by heating strongly aluminium(III)oxide(Al2O3) form propene Chemical equation CH3CH2 CH2OH (l) -- Heat/Al2O3 --> CH3CH2=CH2(g) + H2O(l) Propan-1-ol Prop-1-ene CH3CHOH CH3 (l) -- Heat/Al2O3 --> CH3CH2=CH2(g) + H2O(l) Propan-2-ol Prop-1-ene 3(a) Butan-1-ol and Butan-2-ol(position isomers of butanol) are dehydrated by conc H2SO4 at about 180oC to But-1-ene and But-2-ene respectively Chemical equation CH3CH2 CH2 CH2OH (l) -- conc H2SO4/180oC -->CH3 CH2CH2=CH2(g) + H2O(l) Butan-1-ol But-1-ene CH3CHOH CH2CH3 (l)-- conc H2SO4/180oC -->CH3CH=CH CH2(g) + H2O(l) Butan-2-ol But-2-ene (b) Butan-1-ol and Butan-2-ol are dehydrated by heating strongly aluminium (III) oxide (Al2O3) form But-1-ene and But-2-ene respectively. Chemical equation CH3CH2 CH2 CH2OH (l) -- Heat/Al2O3 --> CH3 CH2CH2=CH2(g) + H2O(l) Butan-1-ol But-1-ene
25 CH3CHOH CH2CH3 (l) -- Heat/Al2O3 --> CH3CH=CH CH2(g) + H2O(l) Butan-2-ol But-2-ene Laboratory set up for the preparation of alkenes/ethene Caution (i)Ethanol is highly inflammable (ii)Conc H2SO4 is highly corrosive on skin contact. (iii)Common school thermometer has maximum calibration of 110oC and thus cannot be used.
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Chemical equation CH3CH2 CH2OH (l) -- conc H2SO4/180oC --> CH3CH2=CH2(g) + H2O(l) Propan-1-ol Prop-1-ene CH3CHOH CH3 (l) -- conc H2SO4/180oC --> CH3CH2=CH2(g) + H2O(l) Propan-2-ol Prop-1-ene (b) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by heating strongly aluminium(III)oxide(Al2O3) form propene Chemical equation CH3CH2 CH2OH (l) -- Heat/Al2O3 --> CH3CH2=CH2(g) + H2O(l) Propan-1-ol Prop-1-ene CH3CHOH CH3 (l) -- Heat/Al2O3 --> CH3CH2=CH2(g) + H2O(l) Propan-2-ol Prop-1-ene 3(a) Butan-1-ol and Butan-2-ol(position isomers of butanol) are dehydrated by conc H2SO4 at about 180oC to But-1-ene and But-2-ene respectively Chemical equation CH3CH2 CH2 CH2OH (l) -- conc H2SO4/180oC -->CH3 CH2CH2=CH2(g) + H2O(l) Butan-1-ol But-1-ene CH3CHOH CH2CH3 (l)-- conc H2SO4/180oC -->CH3CH=CH CH2(g) + H2O(l) Butan-2-ol But-2-ene (b) Butan-1-ol and Butan-2-ol are dehydrated by heating strongly aluminium (III) oxide (Al2O3) form But-1-ene and But-2-ene respectively. Chemical equation CH3CH2 CH2 CH2OH (l) -- Heat/Al2O3 --> CH3 CH2CH2=CH2(g) + H2O(l) Butan-1-ol But-1-ene
25 CH3CHOH CH2CH3 (l) -- Heat/Al2O3 --> CH3CH=CH CH2(g) + H2O(l) Butan-2-ol But-2-ene Laboratory set up for the preparation of alkenes/ethene Caution (i)Ethanol is highly inflammable (ii)Conc H2SO4 is highly corrosive on skin contact. (iii)Common school thermometer has maximum calibration of 110oC and thus cannot be used. It breaks/cracks.
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Chemical equation CH3CH2 CH2 CH2OH (l) -- Heat/Al2O3 --> CH3 CH2CH2=CH2(g) + H2O(l) Butan-1-ol But-1-ene
25 CH3CHOH CH2CH3 (l) -- Heat/Al2O3 --> CH3CH=CH CH2(g) + H2O(l) Butan-2-ol But-2-ene Laboratory set up for the preparation of alkenes/ethene Caution (i)Ethanol is highly inflammable (ii)Conc H2SO4 is highly corrosive on skin contact. (iii)Common school thermometer has maximum calibration of 110oC and thus cannot be used. It breaks/cracks. (i)Using conentrated sulphuric(VI)acid Some broken porcelain or sand should be put in the flask when heating to: (i)prevent bumping which may break the flask. (ii)ensure uniform and smooth boiling of the mixture The temperatures should be maintained at above160oC. At lower temperatures another compound -ether is predominantly formed instead of ethene gas. (ii)Using aluminium(III)oxide
26 (e)Properties of alkenes I. Physical properties Like alkanes, alkenes are colourles gases, solids and liquids that are not poisonous. They are slightly soluble in water. The solubility in water decrease as the carbon chain and as the molar mass increase but very soluble in organic solvents like tetrachloromethane and methylbenzene. The melting and boiling point increase as the carbon chain increase. This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase. The 1st four straight chain alkenes (ethene,propane,but-1-ene and pent-1-ene)are gases at room temperature and pressure. The density of straight chain alkenes,like alkanes, increase with increasing carbon chain as the intermolecular forces increases reducing the volume occupied by a given mass of the alkene.
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This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase. The 1st four straight chain alkenes (ethene,propane,but-1-ene and pent-1-ene)are gases at room temperature and pressure. The density of straight chain alkenes,like alkanes, increase with increasing carbon chain as the intermolecular forces increases reducing the volume occupied by a given mass of the alkene. Summary of physical properties of the 1st five alkenes Alkene General formula Melting point(oC) Boiling point(K) State at room(298K) temperature and pressure atmosphere (101300Pa)
27 Ethene CH2CH2 -169 -104 gas Propene CH3 CHCH2 -145 -47 gas Butene CH3CH2 CHCH2 -141 -26 gas Pent-1ene CH3(CH2 CHCH2 -138 30 liquid Hex-1ene CH3(CH2) CHCH2 -98 64 liquid II. Chemical properties (a)Burning/combustion Alkenes burn with a yellow/ luminous sooty/ smoky flame in excess air to form carbon(IV) oxide and water. Alkene + Air -> carbon(IV) oxide + water (excess air/oxygen) Alkenes burn with a yellow/ luminous sooty/ smoky flame in limited air to form carbon(II) oxide and water. Alkene + Air -> carbon(II) oxide + water (limited air) Burning of alkenes with a yellow/ luminous sooty/ smoky flame is a confirmatory test for the presence of the =C=C= double bond because they have higher C:H ratio. A homologous series with C = C double or C C triple bond is said to be unsaturated. A homologous series with C C single bond is said to be saturated.Most of the reactions of the unsaturated compound involve trying to be saturated to form a C C single bond . Examples of burning alkenes 1.(a) Ethene when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water.
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A homologous series with C = C double or C C triple bond is said to be unsaturated. A homologous series with C C single bond is said to be saturated.Most of the reactions of the unsaturated compound involve trying to be saturated to form a C C single bond . Examples of burning alkenes 1.(a) Ethene when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water. Ethene + Air -> carbon(IV) oxide + water (excess air/oxygen) C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l/g) (b) Ethene when ignited burns with a yellow sooty flame in limited air to form carbon(II) oxide and water. 28 Ethene + Air -> carbon(II) oxide + water (limited air ) C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l/g) 2.(a) Propene when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water. Propene + Air -> carbon(IV) oxide + water (excess air/oxygen) 2C3H6(g) + 9O2(g) -> 6CO2(g) + 6H2O(l/g) (a) Propene when ignited burns with a yellow sooty flame in limited air to form carbon(II) oxide and water. Propene + Air -> carbon(IV) oxide + water (excess air/oxygen) C3H6(g) + 3O2(g) -> 3CO(g) + 3H2O(l/g) (b)Addition reactions An addition reaction is one which an unsaturated compound reacts to form a saturated compound.Addition reactions of alkenes are named from the reagent used to cause the addtion/convert the double =C=C= to single C-C bond. (i)Hydrogenation Hydrogenation is an addition reaction in which hydrogen in presence of Palladium/Nickel catalyst at high temperatures react with alkenes to form alkanes. Examples 1.When Hydrogen gas is passed through liquid vegetable and animal oil at about 180oC in presence of Nickel catalyst,solid fat is formed.
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Propene + Air -> carbon(IV) oxide + water (excess air/oxygen) C3H6(g) + 3O2(g) -> 3CO(g) + 3H2O(l/g) (b)Addition reactions An addition reaction is one which an unsaturated compound reacts to form a saturated compound.Addition reactions of alkenes are named from the reagent used to cause the addtion/convert the double =C=C= to single C-C bond. (i)Hydrogenation Hydrogenation is an addition reaction in which hydrogen in presence of Palladium/Nickel catalyst at high temperatures react with alkenes to form alkanes. Examples 1.When Hydrogen gas is passed through liquid vegetable and animal oil at about 180oC in presence of Nickel catalyst,solid fat is formed. Hydrogenation is thus used to harden oils to solid fat especially margarine. During hydrogenation, one hydrogen atom in the hydrogen molecule attach itself to one carbon and the other hydrogen to the second carbon breaking the double bond to single bond.
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Examples 1.When Hydrogen gas is passed through liquid vegetable and animal oil at about 180oC in presence of Nickel catalyst,solid fat is formed. Hydrogenation is thus used to harden oils to solid fat especially margarine. During hydrogenation, one hydrogen atom in the hydrogen molecule attach itself to one carbon and the other hydrogen to the second carbon breaking the double bond to single bond. Chemical equation H2C=CH2 + H2 -Ni/Pa-> H3C - CH3 H H H H C = C + H – H - Ni/Pa -> H - C – C - H H H H H 2.Propene undergo hydrogenation to form Propane
29 Chemical equation H3C CH=CH2 + H2 -Ni/Pa-> H3C CH - CH3 H H H H H H H C C = C + H – H - Ni/Pa-> H - C – C - C- H H H H H H 3.Both But-1-ene and But-2-ene undergo hydrogenation to form Butane Chemical equation But-1-ene + Hydrogen –Ni/Pa-> Butane H3C CH2 CH=CH2 + H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H H C C - C = C + H – H - Ni/Pa-> H - C- C – C - C- H H H H H H H H But-2-ene + Hydrogen –Ni/Pa-> Butane H3C CH2 =CH CH2 + H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H H C C = C - C -H + H – H - Ni/Pa-> H - C- C – C - C- H H H H H H H 4. But-1,3-diene should undergo hydrogenation to form Butane. The reaction uses two moles of hydrogen molecules/four hydrogen atoms to break the two double bonds.
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Chemical equation H2C=CH2 + H2 -Ni/Pa-> H3C - CH3 H H H H C = C + H – H - Ni/Pa -> H - C – C - H H H H H 2.Propene undergo hydrogenation to form Propane
29 Chemical equation H3C CH=CH2 + H2 -Ni/Pa-> H3C CH - CH3 H H H H H H H C C = C + H – H - Ni/Pa-> H - C – C - C- H H H H H H 3.Both But-1-ene and But-2-ene undergo hydrogenation to form Butane Chemical equation But-1-ene + Hydrogen –Ni/Pa-> Butane H3C CH2 CH=CH2 + H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H H C C - C = C + H – H - Ni/Pa-> H - C- C – C - C- H H H H H H H H But-2-ene + Hydrogen –Ni/Pa-> Butane H3C CH2 =CH CH2 + H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H H C C = C - C -H + H – H - Ni/Pa-> H - C- C – C - C- H H H H H H H 4. But-1,3-diene should undergo hydrogenation to form Butane. The reaction uses two moles of hydrogen molecules/four hydrogen atoms to break the two double bonds. But-1,3-diene + Hydrogen –Ni/Pa-> Butane H2C CH CH=CH2 + 2H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H H C C - C = C -H + 2(H – H) - Ni/Pa-> H - C- C – C - C- H
30 H H H H (ii) Halogenation. Halogenation is an addition reaction in which a halogen (Fluorine, chlorine, bromine, iodine) reacts with an alkene to form an alkane. The double bond in the alkene break and form a single bond.
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But-1,3-diene + Hydrogen –Ni/Pa-> Butane H2C CH CH=CH2 + 2H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H H C C - C = C -H + 2(H – H) - Ni/Pa-> H - C- C – C - C- H
30 H H H H (ii) Halogenation. Halogenation is an addition reaction in which a halogen (Fluorine, chlorine, bromine, iodine) reacts with an alkene to form an alkane. The double bond in the alkene break and form a single bond. The colour of the halogen fades as the number of moles of the halogens remaining unreacted decreases/reduces. One bromine atom bond at the 1st carbon in the double bond while the other goes to the 2nd carbon. Examples 1Ethene reacts with bromine to form 1,2-dibromoethane. Chemical equation H2C=CH2 + Br2 H2 Br C - CH2 Br H H H H C = C + Br – Br Br - C – C - Br H H H H Ethene + Bromine 1,2-dibromoethane 2.Propene reacts with chlorine to form 1,2-dichloropropane.
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One bromine atom bond at the 1st carbon in the double bond while the other goes to the 2nd carbon. Examples 1Ethene reacts with bromine to form 1,2-dibromoethane. Chemical equation H2C=CH2 + Br2 H2 Br C - CH2 Br H H H H C = C + Br – Br Br - C – C - Br H H H H Ethene + Bromine 1,2-dibromoethane 2.Propene reacts with chlorine to form 1,2-dichloropropane. Chemical equation H3C CH=CH2 + Cl2 H3C CHCl - CH2Cl Propene + Chlorine 1,2-dichloropropane H H H H H H H C C = C + Cl – Cl H - C – C - C- Cl H H H Cl H H H H H H H H H H C C - C = C + I – I H - C- C – C - C- I
31 H H H H H H H H 3.Both But-1-ene and But-2-ene undergo halogenation with iodine to form 1,2diiodobutane and 2,3-diiodobutane Chemical equation But-1-ene + iodine 1,2 diiodobutane H3C CH2 CH=CH2 + I2 H3C CH2CH I - CH2I But-2-ene + Iodine 2,3-diiodobutane H3C CH= CH-CH2 + F2 H3C CHICHI - CH3 H H H H H H H H H C C = C - C -H + I – I H - C- C – C - C- H H H H I I H 4. But-1,3-diene should undergo halogenation to form Butane. The reaction uses two moles of iodine molecules/four iodine atoms to break the two double bonds.
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Chemical equation H3C CH=CH2 + Cl2 H3C CHCl - CH2Cl Propene + Chlorine 1,2-dichloropropane H H H H H H H C C = C + Cl – Cl H - C – C - C- Cl H H H Cl H H H H H H H H H H C C - C = C + I – I H - C- C – C - C- I
31 H H H H H H H H 3.Both But-1-ene and But-2-ene undergo halogenation with iodine to form 1,2diiodobutane and 2,3-diiodobutane Chemical equation But-1-ene + iodine 1,2 diiodobutane H3C CH2 CH=CH2 + I2 H3C CH2CH I - CH2I But-2-ene + Iodine 2,3-diiodobutane H3C CH= CH-CH2 + F2 H3C CHICHI - CH3 H H H H H H H H H C C = C - C -H + I – I H - C- C – C - C- H H H H I I H 4. But-1,3-diene should undergo halogenation to form Butane. The reaction uses two moles of iodine molecules/four iodine atoms to break the two double bonds. But-1,3-diene + iodine 1,2,3,4-tetraiodobutane H2C= CH CH=CH2 + 2I2 H2CI CHICHI - CHI H H H H H H H H H C C - C = C -H + 2(I – I) H - C- C – C - C- H I I I I (iii) Reaction with hydrogen halides. Hydrogen halides reacts with alkene to form a halogenoalkane. The double bond in the alkene break and form a single bond. The main compound is one which the hydrogen atom bond at the carbon with more hydrogen . Examples 1. Ethene reacts with hydrogen bromide to form bromoethane.
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The main compound is one which the hydrogen atom bond at the carbon with more hydrogen . Examples 1. Ethene reacts with hydrogen bromide to form bromoethane. Chemical equation H2C=CH2 + HBr H3 C - CH2 Br H H H H
32 C = C + H – Br H - C – C - Br H H H H Ethene + Bromine bromoethane 2. Propene reacts with hydrogen iodide to form 2-iodopropane. Chemical equation H3C CH=CH2 + HI H3C CHI - CH3 Propene + Chlorine 2-chloropropane H H H H H H H C C = C + H – Cl H - C – C - C- H H H H Cl H 3. Both But-1-ene and But-2-ene reacts with hydrogen bromide to form 2- bromobutane Chemical equation But-1-ene + hydrogen bromide 2-bromobutane H3C CH2 CH=CH2 + HBr H3C CH2CHBr -CH3 H H H H H H H H H C C - C = C + H – Br H - C- C – C - C- H H H H H H Br H But-2-ene + Hydrogen bromide 2-bromobutane H3C CH= CH-CH2 + HBr H3C CHBrCH2 - CH3 H H H H H H H H H C C = C - C -H + Br – H H - C- C – C - C- H H H H Br H H 4. But-1,3-diene react with hydrogen iodide to form 2,3- diiodobutane. The reaction uses two moles of hydrogen iodide molecules/two iodine atoms and two hydrogen atoms to break the two double bonds. But-1,3-diene + iodine 2,3-diiodobutane Carbon atom with more Hydrogen atoms gets extra hydrogen
33 H2C= CH CH=CH2 + 2HI2 H3CCHICHI - CH3 H H H H H H H H H C C - C = C -H + 2(H – I) H - C- C – C - C- H H I I H (iv) Reaction with bromine/chlorine water.
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But-1,3-diene react with hydrogen iodide to form 2,3- diiodobutane. The reaction uses two moles of hydrogen iodide molecules/two iodine atoms and two hydrogen atoms to break the two double bonds. But-1,3-diene + iodine 2,3-diiodobutane Carbon atom with more Hydrogen atoms gets extra hydrogen
33 H2C= CH CH=CH2 + 2HI2 H3CCHICHI - CH3 H H H H H H H H H C C - C = C -H + 2(H – I) H - C- C – C - C- H H I I H (iv) Reaction with bromine/chlorine water. Chlorine and bromine water is formed when the halogen is dissolved in distilled water.Chlorine water has the formular HOCl(hypochlorous/chloric(I)acid) .Bromine water has the formular HOBr(hydrobromic(I)acid). During the addition reaction .the halogen move to one carbon and the OH to the other carbon in the alkene at the =C=C= double bond to form a halogenoalkanol. Bromine water + Alkene -> bromoalkanol Chlorine water + Alkene -> bromoalkanol Examples 1Ethene reacts with bromine water to form bromoethanol. Chemical equation H2C=CH2 + HOBr H2 Br C - CH2 OH H H H H C = C + Br – OH Br - C – C - OH H H H H Ethene + Bromine water bromoethanol 2.Propene reacts with chlorine water to form chloropropan-2-ol / 2-chloropropan-1ol.
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During the addition reaction .the halogen move to one carbon and the OH to the other carbon in the alkene at the =C=C= double bond to form a halogenoalkanol. Bromine water + Alkene -> bromoalkanol Chlorine water + Alkene -> bromoalkanol Examples 1Ethene reacts with bromine water to form bromoethanol. Chemical equation H2C=CH2 + HOBr H2 Br C - CH2 OH H H H H C = C + Br – OH Br - C – C - OH H H H H Ethene + Bromine water bromoethanol 2.Propene reacts with chlorine water to form chloropropan-2-ol / 2-chloropropan-1ol. Chemical equation I.H3C CH=CH2 + HOCl H3C CHCl - CH2OH Propene + Chlorine water 2-chloropropane H H H H H H H C C = C + HO – Cl H - C – C - C- OH
34 H H H Cl H II.H3C CH=CH2 + HOCl H3C CHOH - CH2Cl Propene + Chlorine chloropropan-2-ol H H H H H H H C C = C + HO – Cl H - C – C - C- Cl H H H OH H 3.Both But-1-ene and But-2-ene react with bromine water to form 2-bromobutan-1ol /3-bromobutan-2-ol respectively Chemical equation I.But-1-ene + bromine water 2-bromobutan-1-ol H3C CH2 CH=CH2 + HOBr H3C CH2CH Br - CH2OH H H H H H H H H H C C - C = C + HO– Br H - C- C – C - C- OH H H H H H Br H II.But-2-ene + bromine water 3-bromobutan-2-ol H3C CH= CHCH3 + HOBr H3C CH2OHCH Br CH3 H H H H H H H H H C C - C = C + HO– Br H - C- C – C - C- OH H H H H H Br H 4.
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Bromine water + Alkene -> bromoalkanol Chlorine water + Alkene -> bromoalkanol Examples 1Ethene reacts with bromine water to form bromoethanol. Chemical equation H2C=CH2 + HOBr H2 Br C - CH2 OH H H H H C = C + Br – OH Br - C – C - OH H H H H Ethene + Bromine water bromoethanol 2.Propene reacts with chlorine water to form chloropropan-2-ol / 2-chloropropan-1ol. Chemical equation I.H3C CH=CH2 + HOCl H3C CHCl - CH2OH Propene + Chlorine water 2-chloropropane H H H H H H H C C = C + HO – Cl H - C – C - C- OH
34 H H H Cl H II.H3C CH=CH2 + HOCl H3C CHOH - CH2Cl Propene + Chlorine chloropropan-2-ol H H H H H H H C C = C + HO – Cl H - C – C - C- Cl H H H OH H 3.Both But-1-ene and But-2-ene react with bromine water to form 2-bromobutan-1ol /3-bromobutan-2-ol respectively Chemical equation I.But-1-ene + bromine water 2-bromobutan-1-ol H3C CH2 CH=CH2 + HOBr H3C CH2CH Br - CH2OH H H H H H H H H H C C - C = C + HO– Br H - C- C – C - C- OH H H H H H Br H II.But-2-ene + bromine water 3-bromobutan-2-ol H3C CH= CHCH3 + HOBr H3C CH2OHCH Br CH3 H H H H H H H H H C C - C = C + HO– Br H - C- C – C - C- OH H H H H H Br H 4. But-1,3-diene reacts with bromine water to form Butan-1,3-diol. The reaction uses two moles of bromine water molecules to break the two double bonds.
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Chemical equation I.H3C CH=CH2 + HOCl H3C CHCl - CH2OH Propene + Chlorine water 2-chloropropane H H H H H H H C C = C + HO – Cl H - C – C - C- OH
34 H H H Cl H II.H3C CH=CH2 + HOCl H3C CHOH - CH2Cl Propene + Chlorine chloropropan-2-ol H H H H H H H C C = C + HO – Cl H - C – C - C- Cl H H H OH H 3.Both But-1-ene and But-2-ene react with bromine water to form 2-bromobutan-1ol /3-bromobutan-2-ol respectively Chemical equation I.But-1-ene + bromine water 2-bromobutan-1-ol H3C CH2 CH=CH2 + HOBr H3C CH2CH Br - CH2OH H H H H H H H H H C C - C = C + HO– Br H - C- C – C - C- OH H H H H H Br H II.But-2-ene + bromine water 3-bromobutan-2-ol H3C CH= CHCH3 + HOBr H3C CH2OHCH Br CH3 H H H H H H H H H C C - C = C + HO– Br H - C- C – C - C- OH H H H H H Br H 4. But-1,3-diene reacts with bromine water to form Butan-1,3-diol. The reaction uses two moles of bromine water molecules to break the two double bonds. But-1,3-diene + bromine water 2,4-dibromobutan-1,3-diol H2C= CH CH=CH2 + 2HOBr H2COH CHBrCHOH CHBr H H H H H H H H
35 H C C - C = C -H + 2(HO – Br) H - C- C – C - C- H HO Br HO Br (v) Oxidation. Alkenes are oxidized to alkanols with duo/double functional groups by oxidizing agents.
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The reaction uses two moles of bromine water molecules to break the two double bonds. But-1,3-diene + bromine water 2,4-dibromobutan-1,3-diol H2C= CH CH=CH2 + 2HOBr H2COH CHBrCHOH CHBr H H H H H H H H
35 H C C - C = C -H + 2(HO – Br) H - C- C – C - C- H HO Br HO Br (v) Oxidation. Alkenes are oxidized to alkanols with duo/double functional groups by oxidizing agents. When an alkene is bubbled into orange acidified potassium/sodium dichromate (VI) solution,the colour of the oxidizing agent changes to green. When an alkene is bubbled into purple acidified potassium/sodium manganate(VII) solution, the oxidizing agent is decolorized. Examples 1Ethene is oxidized to ethan-1,2-diol by acidified potassium/sodium manganate(VII) solution/ acidified potassium/sodium dichromate(VI) solution. The purple acidified potassium/sodium manganate(VII) solution is decolorized. The orange acidified potassium/sodium dichromate(VI) solution turns to green. Chemical equation H2C=CH2 [O] in H+/K2Cr2O7 HO CH2 - CH2 OH H H H H C = C+ [O] in H+/KMnO4 H - C – C - H H H OH OH Ethene + [O] in H+/KMnO4 ethan-1,2-diol 2. Propene is oxidized to propan-1,2-diol by acidified potassium/sodium manganate(VII) solution/ acidified potassium/sodium dichromate(VI) solution. The purple acidified potassium/sodium manganate(VII) solution is decolorized. The orange acidified potassium/sodium dichromate(VI) solution turns to green.
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Propene is oxidized to propan-1,2-diol by acidified potassium/sodium manganate(VII) solution/ acidified potassium/sodium dichromate(VI) solution. The purple acidified potassium/sodium manganate(VII) solution is decolorized. The orange acidified potassium/sodium dichromate(VI) solution turns to green. Chemical equation H3C CH=CH2 [O] in H+/KMnO4 H3C CHOH - CH2OH Propene [O] in H+/KMnO4 propan-1,2-diol H H H H H H H C C = C [O] in H+/KMnO4 H - C – C - C- OH
36 H H H OH H 3.Both But-1-ene and But-2-ene react with bromine water to form butan-1,2-diol and butan-2,3-diol Chemical equation I.But-1-ene + [O] in H+/KMnO4 butan-1,2-diol H3C CH2 CH=CH2 + [O] H3C CH2CHOH - CH2OH H H H H H H H H H C C - C = C + [O] H - C- C – C - C- OH H H H H H OH H (v) Hydrolysis. Hydrolysis is the reaction of a compound with water/addition of H-OH to a compound. Alkenes undergo hydrolysis to form alkanols . This takes place in two steps: (i)Alkenes react with concentrated sulphuric(VI)acid at room temperature and pressure to form alkylhydrogen sulphate(VI). Alkenes + concentrated sulphuric(VI)acid -> alkylhydrogen sulphate(VI) (ii)On adding water to alkylhydrogen sulphate(VI) then warming, an alkanol is formed. alkylhydrogen sulphate(VI) + water -warm-> Alkanol.
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This takes place in two steps: (i)Alkenes react with concentrated sulphuric(VI)acid at room temperature and pressure to form alkylhydrogen sulphate(VI). Alkenes + concentrated sulphuric(VI)acid -> alkylhydrogen sulphate(VI) (ii)On adding water to alkylhydrogen sulphate(VI) then warming, an alkanol is formed. alkylhydrogen sulphate(VI) + water -warm-> Alkanol. Examples (i)Ethene reacts with cold concentrated sulphuric(VI)acid to form ethyl hydrogen sulphate(VII) Chemical equation H2C=CH2 + H2SO4 CH3 - CH2OSO3H H H H O-SO3H C = C + H2SO4 H - C – C - H
37 H H H H Ethene + H2SO4 ethylhydrogen sulphate(VI) (ii) Ethylhydrogen sulphate(VI) is hydrolysed by water to ethanol Chemical equation CH3 - CH2OSO3H + H2O CH3 - CH2OH + H2SO4 H OSO3H H OH H - C - C - H + H2O H - C – C - H + H2SO4 H H H H ethylhydrogen sulphate(VI) + H2O Ethanol 2.
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Alkenes + concentrated sulphuric(VI)acid -> alkylhydrogen sulphate(VI) (ii)On adding water to alkylhydrogen sulphate(VI) then warming, an alkanol is formed. alkylhydrogen sulphate(VI) + water -warm-> Alkanol. Examples (i)Ethene reacts with cold concentrated sulphuric(VI)acid to form ethyl hydrogen sulphate(VII) Chemical equation H2C=CH2 + H2SO4 CH3 - CH2OSO3H H H H O-SO3H C = C + H2SO4 H - C – C - H
37 H H H H Ethene + H2SO4 ethylhydrogen sulphate(VI) (ii) Ethylhydrogen sulphate(VI) is hydrolysed by water to ethanol Chemical equation CH3 - CH2OSO3H + H2O CH3 - CH2OH + H2SO4 H OSO3H H OH H - C - C - H + H2O H - C – C - H + H2SO4 H H H H ethylhydrogen sulphate(VI) + H2O Ethanol 2. Propene reacts with cold concentrated sulphuric(VI)acid to form propyl hydrogen sulphate(VII) Chemical equation CH3H2C=CH2 + H2SO4 CH3CH2 - CH2OSO3H H H H H H O-SO3H C = C - C - H + H2SO4 H - C - C – C - H H H H H H H Propene + H2SO4 propylhydrogen sulphate(VI) (ii) Propylhydrogen sulphate(VI) is hydrolysed by water to propanol Chemical equation CH3 - CH2OSO3H + H2O CH3 - CH2OH + H2SO4 H H OSO3H H H OH H - C - C - C - H + H2O H - C - C – C - H + H2SO4 H H H H H H propylhydrogen sulphate(VI) + H2O propanol (vi) Polymerization/self addition
38 Addition polymerization is the process where a small unsaturated monomer (alkene ) molecule join together to form a large saturated molecule. Only alkenes undergo addition polymerization.
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Examples (i)Ethene reacts with cold concentrated sulphuric(VI)acid to form ethyl hydrogen sulphate(VII) Chemical equation H2C=CH2 + H2SO4 CH3 - CH2OSO3H H H H O-SO3H C = C + H2SO4 H - C – C - H
37 H H H H Ethene + H2SO4 ethylhydrogen sulphate(VI) (ii) Ethylhydrogen sulphate(VI) is hydrolysed by water to ethanol Chemical equation CH3 - CH2OSO3H + H2O CH3 - CH2OH + H2SO4 H OSO3H H OH H - C - C - H + H2O H - C – C - H + H2SO4 H H H H ethylhydrogen sulphate(VI) + H2O Ethanol 2. Propene reacts with cold concentrated sulphuric(VI)acid to form propyl hydrogen sulphate(VII) Chemical equation CH3H2C=CH2 + H2SO4 CH3CH2 - CH2OSO3H H H H H H O-SO3H C = C - C - H + H2SO4 H - C - C – C - H H H H H H H Propene + H2SO4 propylhydrogen sulphate(VI) (ii) Propylhydrogen sulphate(VI) is hydrolysed by water to propanol Chemical equation CH3 - CH2OSO3H + H2O CH3 - CH2OH + H2SO4 H H OSO3H H H OH H - C - C - C - H + H2O H - C - C – C - H + H2SO4 H H H H H H propylhydrogen sulphate(VI) + H2O propanol (vi) Polymerization/self addition
38 Addition polymerization is the process where a small unsaturated monomer (alkene ) molecule join together to form a large saturated molecule. Only alkenes undergo addition polymerization. Addition polymers are named from the alkene/monomer making the polymer and adding the prefix “poly” before the name of monomer to form a polyalkene During addition polymerization (i)the double bond in alkenes break (ii)free radicals are formed (iii)the free radicals collide with each other and join to form a larger molecule.
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Propene reacts with cold concentrated sulphuric(VI)acid to form propyl hydrogen sulphate(VII) Chemical equation CH3H2C=CH2 + H2SO4 CH3CH2 - CH2OSO3H H H H H H O-SO3H C = C - C - H + H2SO4 H - C - C – C - H H H H H H H Propene + H2SO4 propylhydrogen sulphate(VI) (ii) Propylhydrogen sulphate(VI) is hydrolysed by water to propanol Chemical equation CH3 - CH2OSO3H + H2O CH3 - CH2OH + H2SO4 H H OSO3H H H OH H - C - C - C - H + H2O H - C - C – C - H + H2SO4 H H H H H H propylhydrogen sulphate(VI) + H2O propanol (vi) Polymerization/self addition
38 Addition polymerization is the process where a small unsaturated monomer (alkene ) molecule join together to form a large saturated molecule. Only alkenes undergo addition polymerization. Addition polymers are named from the alkene/monomer making the polymer and adding the prefix “poly” before the name of monomer to form a polyalkene During addition polymerization (i)the double bond in alkenes break (ii)free radicals are formed (iii)the free radicals collide with each other and join to form a larger molecule. The more collisions the larger the molecule. Examples of addition polymerization 1.Formation of Polyethene Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
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Addition polymers are named from the alkene/monomer making the polymer and adding the prefix “poly” before the name of monomer to form a polyalkene During addition polymerization (i)the double bond in alkenes break (ii)free radicals are formed (iii)the free radicals collide with each other and join to form a larger molecule. The more collisions the larger the molecule. Examples of addition polymerization 1.Formation of Polyethene Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles) H H H H H H H H C = C + C = C + C = C + C = C + … H H H H H H H H Ethene + Ethene + Ethene + Ethene + … (ii)the double bond joining the ethane molecule break to free readicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H H H H H H H H Ethene radical + Ethene radical + Ethene radical + Ethene radical + … (iii)the free radicals collide with each other and join to form a larger molecule
39 H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H H H H H H H H Lone pair of electrons can be used to join more monomers to form longer polyethene. Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H H Where n is the number of monomers in the polymer.
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Examples of addition polymerization 1.Formation of Polyethene Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles) H H H H H H H H C = C + C = C + C = C + C = C + … H H H H H H H H Ethene + Ethene + Ethene + Ethene + … (ii)the double bond joining the ethane molecule break to free readicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H H H H H H H H Ethene radical + Ethene radical + Ethene radical + Ethene radical + … (iii)the free radicals collide with each other and join to form a larger molecule
39 H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H H H H H H H H Lone pair of electrons can be used to join more monomers to form longer polyethene. Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H H Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene.
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During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles) H H H H H H H H C = C + C = C + C = C + C = C + … H H H H H H H H Ethene + Ethene + Ethene + Ethene + … (ii)the double bond joining the ethane molecule break to free readicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H H H H H H H H Ethene radical + Ethene radical + Ethene radical + Ethene radical + … (iii)the free radicals collide with each other and join to form a larger molecule
39 H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H H H H H H H H Lone pair of electrons can be used to join more monomers to form longer polyethene. Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H H Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene. 40 It is an elastic, tough, transparent and durable plastic.
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Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H H Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene. 40 It is an elastic, tough, transparent and durable plastic. Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
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The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene. 40 It is an elastic, tough, transparent and durable plastic. Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene.
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40 It is an elastic, tough, transparent and durable plastic. Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene. Polychloroethene molecule can be represented as:
41 H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + … H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC).
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Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene. Polychloroethene molecule can be represented as:
41 H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + … H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic.
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During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene. Polychloroethene molecule can be represented as:
41 H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + … H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
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