3 Chemical Equations and Stoichiometry 3.1 Formulae of Compounds

3 Chemical Equations and Stoichiometry 3.1 Formulae of Compounds
3.2 Derivation of Empirical Formulae 3.3 Derivation of Molecular Formulae 3.4 Chemical Equations 3.5 Calculations Based on Chemical Equations 3.6 Simple Titrations

3.1 Formulae of Compounds

1st way = by chemical formula
3.1 Formulae of compounds (SB p.43) Formulae of compounds How can you describe the composition of compound X? 1st way = by chemical formula C?H? ratio of no. of atoms

2nd way = by percentage by mass
3.1 Formulae of compounds (SB p.43) How can you describe the composition of compound X? 2nd way = by percentage by mass Compound X Mass of carbon atoms inside = …. g Mass of hydrogen atoms inside = …. g carbon atoms hydrogen atoms Check Point 3-1

Different types of formulae of some compounds
3.1 Formulae of compounds (SB p.44) Different types of formulae of some compounds Compound Empirical formula Molecular formula Structural formula Carbon dioxide CO2 O = C =O Water H2O Methane CH4 Glucose CH2O C6H12O6 Sodium fluoride NaF Not applicable Na+F-

Derivation of Empirical Formulae
3.2 Derivation of Empirical Formulae

3.2 Derivation of empirical formulae (SB p.45)
From combustion data During complete combustion, elements in a compound are oxidized. e.g. carbon to carbon dioxide, hydrogen to water, sulphur to sulphur dioxide From the masses of the products formed, the number of moles of these atoms originally present can be found

Example 3-2A Example 3-2B Check Point 3-2A
3.2 Derivation of empirical formulae (SB p.45) The laboratory set-up used for determining the empirical formula of a gaseous hydrocarbon Example 3-2A Example 3-2B Check Point 3-2A

From combustion by mass
3.2 Derivation of Empirical Formulae (SB p.48) From combustion by mass Composition by mass Empirical formula Example 3-2C Example 3-2D Check Point 3-2B

Derivation of Molecular Formulae
3.3 Derivation of Molecular Formulae

What is molecular formulae?
3.3 Derivation of molecular formulae (SB p.49) What is molecular formulae? Molecular formula ? = (Empirical formula)n

From empirical formula and known relative molecular mass
3.3 Derivation of Molecular Formulae (SB p.49) From empirical formula and known relative molecular mass Empirical formula Molecular mass Example 3-3A Molecular formula Example 3-3B

(Colourless crystals)
3.3 Derivation of Molecular Formulae (SB p.51) Water of Crystallization Derived from Composition by Mass Example 3-3C Hydrated salt Anhydrous salt CuSO45H2O (Blue crystals) Anhydrous CuSO4 (White powder) Na2CO310H2O (Colourless crystals) Anhydrous Na2CO3 CoCl2 2H2O (Pink crystals) Anhydrous CoCl2 Check Point 3-3A

Find composition by mass from formula
3.3 Derivation of Molecular Formulae (SB p.52) Find composition by mass from formula Formula of a compound Example 3-3D Example 3-3E Composition by mass Check Point 3-3B

3.4 Chemical Equations

Chemical equations a A + b B  c C + d D mole ratios
3.4 Chemical equations (SB p.53) Chemical equations a A + b B  c C + d D mole ratios (can also be volume ratios for gases) Stoichiometry = relative no. of moles of substances involved in a chemical reaction Check Point 3-4

Calculations Based on Chemical Equations
3.5 Calculations Based on Chemical Equations

Calculations involving reacting masses
3.5 Calculations Based on Equations (SB p.65) Calculations based on equations Calculations involving reacting masses Example 3-5A Example 3-5B

Calculations based on equations
3.5 Calculations Based on Equations (SB p.66) Calculations based on equations Calculations involving volumes of gases Example 3-5C Check Point 3-5 Example 3-5D

3.6 Simple Titrations

Simple titrations Acid-base titrations
3.6 Simple titrations (SB p.58) Simple titrations Acid-base titrations Acid-base titrations with indicators Acid-base titrations without indicators (to be discussed in later chapters)

Finding the concentration of a solution
3.6 Simple titrations (SB p.59) Finding the concentration of a solution Copper(II) sulphate + solute Water solvent Copper(II) sulphate solution solution

3.6 Simple titrations (SB p.59) Finding the concentration of a solution ~50 cm3

3.6 Simple titrations (SB p.59) Finding the concentration of a solution 50 cm3 Solution A

3.6 Simple titrations (SB p.59) Finding the concentration of a solution 50 cm3 Solution B

3.6 Simple titrations (SB p.59) Finding the concentration of a solution 100 cm3 Solution C

Comment on the concentrations of solutions A, B and C !
3.6 Simple titrations (SB p.59) Comment on the concentrations of solutions A, B and C ! Concentration of solution B is 2 times that of the concentrations of solutions A & B. 2 x the amount of solute contain the same amount of solute (same concentration) Concentration is the amount of solute in a unit volume of solution.

Comment on the concentrations of solutions A, B and C !
3.6 Simple titrations (SB p.59) Comment on the concentrations of solutions A, B and C ! no. of spoons no. of moles mass Concentration is the amount of solute in a unit volume of solution.

Molarity (M) A way of expressing concentrations
3.6 Simple titrations (SB p.59) Molarity A way of expressing concentrations Molarity is the number of moles of solute dissolved in 1 dm3 (1000 cm3) of solution. Unit: mol dm-3 (M)

“In every 1 dm3 of the solution, 2 moles of HCl is dissolved.”
3.6 Simple titrations (SB p.59) What does this mean? 1 dm3 contains 2 moles of HCl “In every 1 dm3 of the solution, 2 moles of HCl is dissolved.” Example 3-6A Example 3-6B

Titration without an indicator
3.6 Simple titrations (SB p.62) Titration without an indicator By change in pH value Example 3-6C

By change in temperature
3.6 Simple titrations (SB p.62) Titration without an indicator By change in temperature Example 3-6D

Redox titrations 1. Iodometric titration in burette in conical flask
3.6 Simple Titrations (SB p.65) Redox titrations 1. Iodometric titration in conical flask in burette I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) brown colourless

During titration : brown  yellow
3.6 Simple Titrations (SB p.65) Redox titrations 1. Iodometric titration Add starch in burette in conical flask I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) brown colourless During titration : brown  yellow

Redox titrations 1. Iodometric titration
3.6 Simple Titrations (SB p.65) Redox titrations 1. Iodometric titration I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) brown colourless During titration : brown  yellow

Redox titrations Example 3-6E 1. Iodometric titration
3.6 Simple Titrations (SB p.65) Redox titrations Example 3-6E 1. Iodometric titration I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) brown colourless During titration : brown  yellow End point : blue black  colourless (after addition of starch indicator)

Redox titrations 2. Titrations involving potassium permanganate
3.6 Simple Titrations (SB p.66) Redox titrations 2. Titrations involving potassium permanganate In conical flask In burette MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + H2O(aq) + 5Fe3+(aq) purple pale green Example 3-6F Check Point 3-6

The END

Check Point 3-1 Answer Back
3.1 Formulae of compounds (SB p.45) Back Check Point 3-1 Give the empirical, molecular and structural formulae for the following compounds: (a) Propene (b) Nitric acid (c) Ethanol (d) Glucose C6H12O6 (d) Glucose C2H5OH C2H6O (c) Ethanol HNO3 (b) Nitric acid C3H6 CH2 (a) Propene Structural formula Molecular formula Empirical formula Compound Answer

Example 3-2A A hydrocarbon was burnt completely in excess oxygen. It was found that 1.00 g of the hydrocarbon gives 2.93 g of carbon dioxide and 1.80 g of water. Find the empirical formula of the hydrocarbon. Answer

Example 3-2A 3.2 Derivation of empirical formulae (SB p.46)
The relative molecular mass of CO2 =  2 = 44.0 Mass of carbon in 2.93 g of CO2 = 2.93 g  = 0.80 g The relative molecular mass of H2O = 1.0  = 18.0 Mass of hydrogen in 1.80 g of H2O = 1.80 g  = 0.20 g Let the empirical formula of the hydrocarbon be CxHy. Mass of carbon in CxHy = Mass of carbon in CO2 Mass of hydrogen in CxHy = Mass of hydrogen in H2O

Example 3-2A Back 3.2 Derivation of empirical formulae (SB p.46)
Therefore, the empirical formula of the hydrocarbon is CH3. Carbon Hydrogen Mass (g) 0.80 0.20 No. of moles (mol) Relative no. of moles Simplest mole ratio 1 3

Example 3-2B Compound X is known to contain carbon, hdyrogen and oxygen only. When it is burnt completely in excess oxygen, carbon dioxide and water are given out as the only products. It is found that 0.46 g of compound X gives 0.88 g of carbon dioxide and 0.54 g of water. Find the empirical formula of compound X. Answer

Example 3-2B 3.2 Derivation of empirical formulae (SB p.47)
Mass of compound X = 0.46 g Mass of carbon in compound X = 0.88 g  = 0.24 g Mass of hydrogen in compound X = 0.54 g  = 0.06 g Mass of oxygen in compound X = 0.46 g – 0.24 g – 0.06 g = 0.16 g Let the empirical formula of compound X be CxHyOz.

Example 3-2B Back 3.2 Derivation of empirical formulae (SB p.47)
Therefore, the empirical formula of compound X is C2H6O. Carbon Hydrogen Oxygen Mass (g) 0.24 0.06 0.16 No. of moles (mol) Relative no. of moles Simplest mole ratio 2 6 1

Check Point 3-2A 5 g of sulphur forms 10 g of an oxide on complete combustion. What is the empirical formula of the oxide? Answer Mass of sulphur = 5 g Mass of oxygen = (10 – 5) g = 5 g 2 1 Simplest mole ratio Relative no. of moles No. of moles (mol) 5 Mass (g) Oxygen Sulphur

Check Point 3-2A (b) g of element M combines with g of oxygen to form an oxide. If the relative atomic mass of M is 31.0, find the empirical formula of the oxide. Answer (b) The empirical formula of the oxide is M2O5. 5 2 Simplest mole ratio Relative no. of moles No. of moles (mol) 25.61 19.85 Mass (g) O M

Check Point 3-2A (c) Determine the empirical formula of copper(II) oxide using the following results. Experimental results: Mass of test tube = g Mass of test tube + Mass of copper(II) oxide = g Mass of test tube + Mass of copper = g Answer

Check Point 3-2A Back 3.2 Derivation of empirical formulae (SB p.47)
Mass of Cu = ( – ) g = 1.51 g Mass of O = ( – ) g = g Therefore, the empirical formula of copper(II) oxide is CuO. 1 Simplest mole ratio Relative no. of moles No. of moles (mol) 0.381 1.51 Mass (g) Oxygen Copper

Example 3-2C Compound A contains carbon and hydrogen atoms only. It is found that the compound contains 75 % carbon by mass. Determine its empirical formula. Answer

Example 3-2C Back 3.2 Derivation of empirical formulae (SB p.48)
Let the empirical formula of compound A be CxHy, and the mass of the compound be 100 g. Then, mass of carbon in the compound = 75 g Mass of hydrogen in the compound = (100 – 75) g = 25 g Therefore, the empirical formula of compound A is CH4. 4 1 Simplest mole ratio Relative no. of moles No. of moles (mol) 25 75 Mass (g) Hydrogen Carbon

Example 3-2D The percentages by mass of phosphorus and chlorine in a sample of phosphorus chloride are % and % respectively. Find the empirical formula of the phosphorus chloride. Answer

Example 3-2D Back 3.2 Derivation of empirical formulae (SB p.48)
Let the mass of phosphorus chloride be 100 g. Then, Mass of phosphorus in the compound = g Mass of chlorine in the compound = g Therefore, the empirical formula of the phosphorus chloride is PCl3. 3 1 Simplest mole ratio Relative no. of moles No. of moles (mol) 77.45 22.55 Mass (g) Chlorine Phosphorus

Check Point 3-2B Find the empirical formula of vitamin C if it consists of 40.9 % carbon, 54.5 % oxygen and 4.6 % hydrogen by mass. Answer (a) Let the mass of vitamin C analyzed be 100 g. The empirical formula of vitamin C is C3H4O3. 3 4 Simplest mole ratio Relative no. of moles No. of moles (mol) 54.5 4.6 40.9 Mass (g) Oxygen Hydrogen Carbon

Check Point 3-2B Answer Back
3.2 Derivation of empirical formulae (SB p.49) Back Check Point 3-2B (b) Each 325 mg tablet of aspirin consists of mg carbon, 14.6 mg hydrogen and mg oxygen. Determine the empirical formula of aspirin. Answer (b) The masses of the elements are multiplied by 1000 first. The empirical formula of aspirin is C9H8O4. 4 8 9 Simplest mole ratio Relative no. of moles No. of moles (mol) 115.4 14.6 195.0 Mass (g) Oxygen Hydrogen Carbon

3.3 Derivation of molecular formulae (SB p.50)
Example 3-3A A hydrocarbon was burnt completely in excess oxygen. It was found that 5.0 g of the hydrocarbon gave 14.6 g of carbon dioxide and 9.0 g of water. Given that the relative molecular mass of the hydrocarbon is 30.0, determine its molecular formula. Answer

Example 3-3A 3.3 Derivation of molecular formulae (SB p.50)
Let the empirical formula of the hydrocarbon be CxHy. Mass of carbon in the hydrocarbon = 14.6 g  = 4.0 g Mass of hydrogen in the hydrocarbon = 9.0 g  = 1.0 g 3 1 Simplest mole ratio Relative no. of moles No. of moles (mol) 1.0 4.0 Mass (g) Hydrogen Carbon

Example 3-3A Back 3.3 Derivation of molecular formulae (SB p.50)
Therefore, the empirical formula of the hydrocarbon is CH3. Let the molecular formula of the hydrocarbon be (CH3)n. Relative molecular mass of (CH3)n = 30.0 n  (  3) = 30.0 n = 2 Therefore, the molecular formula of the hydrocarbon is C2H6.

Example 3-3B Compound X is known to contain % carbon, 6.18 % hydrogen and % oxygen by mass. A typical analysis shows that it has a relative molecular mass of Find its molecular formula. Answer

Example 3-3B 3.3 Derivation of molecular formulae (SB p.50)
Let the empirical formula of compound X is CxHyOz, and the mass of the compound be 100 g. Then, Mass of carbon in the compound = g Mass of hydrogen in the compound = 6.18 g Mass of oxygen in the compound = g The empirical formula of compound X is C6H10O5. 5 10 6 Simplest mole ratio Relative no. of moles No. of moles (mol) 49.38 6.18 44.44 Mass (g) Oxygen Hydrogen Carbon

Example 3-3B Back 3.3 Derivation of molecular formulae (SB p.50)
Let the molecular formula of compound X be (C6H10O5)n. Relative molecular mass of (C6H10O5)n = 162.0 n  (12.0    5) = 162.0 n = 1 Therefore, the molecular formula of compound X is C6H10O5.

Example 3-3C The chemical formula of hydrated copper(II) sulphate is known to be CuSO4 · xH2O. It is found that the percentage of water of crystallization by mass in the compound is 36 %. Find the value of x. Answer

Example 3-3C Back 3.3 Derivation of molecular formulae (SB p.51)
Relative formula mass of CuSO4 · xH2O =  4 + (1.0  )x = x Relative molecular mass of water of crystallization = 18x 1800x = x 1152x = x = 4.99  5 Therefore, the chemical formula of the hydrated copper(II) sulphate is CuSO4 · 5H2O.

Check Point 3-3A Compound Z is the major ingredient of a healthy drink. It contains % carbon, 6.67 % hydrogen and % oxygen. (i) Find the empirical formula of compound Z. (ii) If the relative molecular mass of compound Z is 180, find its molecular formula. Answer

Check Point 3-3A 3.3 Derivation of molecular formulae (SB p.52)
(i) Let the mass of compound Z be 100 g. Therefore, the empirical formula of compound Z is CH2O. Carbon Hydrogen Oxygen Mass (g) 40.00 6.67 53.33 No. of moles (mol) Relative no. of moles Simplest mole ratio 1 2

Check Point 3-3A 3.3 Derivation of molecular formulae (SB p.52)
(ii) Let the empirical formula of compound Z be (CH2O)n. n  (  ) = 180 30n = 180 n = 6 Therefore, the molecular formula of compound Z is C6H12O6.

Check Point 3-3A (b) (NH4)2Sx contains % sulphur by mass. Find the value of x. Answer (b) Since the chemical formula of (NH4)Sx is (NH4)2S3, the value of x is 3. 3 2 Simplest mole ratio Relative no. of moles No. of moles (mol) 72.72 27.28 Mass (g) S (NH4) unit

Check Point 3-3A Answer Back
3.3 Derivation of molecular formulae (SB p.52) Back Check Point 3-3A (c) In the compound MgSO4 · nH2O, % by mass is water. Find the value of n. Answer (c) Since the chemical formula of MgSO4 · nH2O is MgSO4 · 7H2O, the value of n is 7. 7 1 Simplest mole ratio Relative no. of moles No. of moles (mol) 51.22 48.78 Mass (g) H2O MgSO4

Example 3-3D The chemical formula of ethanoic acid is CH3COOH. Calculate the percentage of mass of carbon, hydrogen and oxygen respectively. Answer

Example 3-3D Back 3.3 Derivation of molecular formulae (SB p.52)
Relative molecular mass of CH3COOH = 12.0    2 = 60.0 % by mass of C = = % % by mass of H = = 6.67 % % by mass of O = = % The percentage by mass of carbon, hydrogen and oxygen are %, 6.67 % and % respectively.

Back Example 3-3E Calculate the mass of iron in a sample of 20 g of hydrated iron(II) sulphate, FeSO4 · 7H2O. Answer Relative formula mass of FeSO4 · 7H2O =  4 + (1.0  )  7 = 277.9 % by mass of Fe = = % Mass of Fe = 20 g  % = 4.02 g

Check Point 3-3B Calculate the percentages by mass of potassium , chromium and oxygen in potassium chromate(VI), K2Cr2O7. Answer Molar mass of K2Cr2O7 = (39.1    7) g mol-1 = g mol-1 % by mass of K = = % % by mass of Cr = = % % by mass of O = = %

Check Point 3-3B Find the mass of metal and water of crystallization in (i) 100 g of Na2SO4 · 10H2O (ii) 70 g of Fe2O3 · 8H2O Answer

Check Point 3-3B Back 3.3 Derivation of molecular formulae (SB p.53)
(b) (i) Molar mass of Na2SO4 · 10H2O = g mol-1 Mass of Na = = g Mass of H2O = = g (ii) Molar mass of Fe2O3 · 8H2O = g mol-1 Mass of Fe = = g = g

3.4 Chemical equations (SB p.54) Check Point 3-4 Back Give the chemical equations for the following reactions: Zinc + steam  zinc oxide + hydrogen (b) Magnesium + silver nitrate  silver + magnesium nitrate (c) Butane + oxygen  carbon dioxide + water Zn(s) + H2O(g)  ZnO(s) + H2(g) (b) Mg(s) + 2AgNO3(aq)  2Ag(s) + Mg(NO3)2(aq) (c) 2C4H10(g) + 13O2(g)  8CO2(g) + 10H2O(l) Answer

3.5 Calculations based on chemical equations (SB p.55)
Example 3-5A Calculate the mass of copper formed when g of copper(II) oxide is completely reduced by hydrogen. Answer

Back Example 3-5A CuO(s) + H2(g)  Cu(s) + H2O(l) As the mole ratio of CuO : Cu is 1 : 1, the number of moles of Cu formed is the same as the number of moles of CuO reduced. Number of moles of CuO reduced = = mol Number of moles of Cu formed = mol Mass of Cu = mol  63.5 g mol-1 = 9.97 g Therefore, the mass of copper formed in the reaction is 9.97 g.

Example 3-5B Sodium hydrogencarbonate decomposes according to the following chemial equation: 2NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(l) In order to obtain 240 cm3 of CO2 at room temperature and pressure, what is the minimum amount of sodium hydrogencarbonate required? (Molar volume of gas at R.T.P. = 24.0 dm3 mol-1) Answer

Back Example 3-5B Number of moles of CO2 required = = 0.01 mol From the chemical equation, 2 moles of NaHCO3(s) give 1 mole of CO2(g). Number of moles of NaHCO3 required = 0.01  2 = 0.02 mol Mass of NaHCO3 required = 0.02 mol  (  3) g mol-1 = 0.02 mol  84.0 g mol-1 = 1.68 g Therefore, the minimum amount of sodium hydrogencarbonate required is 1.68 g.

Example 3-5C Calculate the volume of carbon dioxide formed when 20 cm3 of ethane and 70 cm3 of oxygen are exploded, assuming all volumes of gases are measured at room temperature and pressure. Answer

Back Example 3-5C 2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(l) 2 mol : 7 mol : 4 mol : 6 mol (from equation) 2 volumes : 7 volumes : 4 volumes : - (by Avogadro’s law) It can be judged from the chemical equation that the mole ratio of CO2 : C2H6 is 4 : 2, and the volume ratio of CO2 : C2H6 should also be 4 : 2 according to the Avogadro’s law. Let x be the volume of CO2(g) formed. Number of moles of CO2(g) formed : number of moles of C2H6(g) used = 4 : 2 Volume of CO2(g) : volume of C2H6(g) = 4 : 2 x : 20 cm3 = 4 : 2 x = 40 cm3 Therefore, the volume of CO2(g) formed is 40 cm3.

Example 3-5D 10 cm3 of a gaseous hydrocarbon was mixed with 80 cm3 of oxygen which was in excess. The mixture was exploded and then cooled. The volume left was 70 cm3. Upon passing the resulting gaseous mixture through concentrated sodium hydroxide solution (to absorb carbon dioxide), the volume of the residual gas became 50 cm3. Find the molecular formula of the hydrocarbon. Answer

Example 3-5D 3.5 Calculations based on chemical equations (SB p.57)
Let the molecular formula of the hydrocarbon be CxHy. Volume of hydrocarbon reacted = 10 cm3 Volume of O2(g) unreacted = 50 cm3 (the residual gas after reaction) Volume of O2(g) reacted = (80 – 50) cm3 = 30 cm3 Volume of CO2(g) formed = (70 – 50) cm3 = 20 cm3 CxHy + O2  xCO H2O 1 mol : mol : x mol 1 volume : volumes : x volumes

Back Example 3-5D = x = 2 = 3 As x = 2, = 3 y = 4 Therefore, the molecular formula of the hydrocarbon is C2H4.

Check Point 3-5 Find the volume of hydrogen produced at R.T.P. when 2.43 g of magnesium reacts with excess hydrochloric acid. (Molar volume of gas at R.T.P. = 24.0 dm3 mol-1) Answer Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g) No. of moles of H2 = No. of moles of Mg = Volume of H2 = 2.4 dm3

Check Point 3-5 (b) Find the minimum mass of chlorine required to produce 100 g of phosphorus trichloride (PCl3). Answer (b) 2P(s) + 3Cl2(g)  2PCl3(l)  No. of moles of Cl2 =  No. of moles of PCl3 = Mass of Cl2 = g The minimum mass of chlorine required is g.

Check Point 3-5 (c) 20 cm3 of a gaseous hydrocarbon and 150 cm3 of oxygen (which was in excess) were exploded in a closed vessel. After cooling, 110 cm3 of gases remained. After passing the resulting gaseous mixture through concentrated sodium hydroxide solution, the volume of the residual gas became 50 cm3. Determine the molecular formula of the hydrocarbon. Answer

Check Point 3-5 3.5 Calculations based on chemical equations (SB p.58)
CxHy(g) O2(g)  xCO2(g) + H2O(l) Volume of CxHy used = 20 cm3 Volume of CO2 formed = (110 – 50) cm3 = 60 cm3 Volume of O2 used = (150 – 50) cm3 = 100 cm3 Volume of CxHy : Volume of CO2 = 1 : x = 20 : 60 x = 3 Volume of CxHy : Volume of O2 = 1 : = 20 : 100 = 5

Check Point 3-5 3.5 Calculations based on chemical equations (SB p.58)
As x = 3, = 5 = 2 y = 8 The molecular formula of the hydrocarbon is C3H8.

3.5 Calculations based on chemical equations (SB p.58) Check Point 3-5 Back Calculate the volume of carbon dioxide formed when 5 cm3 of methane was burnt completely in excess oxygen, assuming all volumes of gases are measured at room temperature and pressure. Answer CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) 1 mol : 2 mol : 1 mol : 2 mol (from equation) 1 volume : 2 volumes : 1 volume : - (from Avogadro’s law) 5 cm3 x cm3 It can be judged from the equation that the mole ratio of CO2 : CH4 is 1 : 1, the volume ratio of CO2 : CH4 should also be 1 : 1. = x = 5 cm3 The volume of CO2(g) formed is 5 cm3.

3.6 Simple titrations (SB p.61)
Example 3-6A 25.0 cm3 of sodium hydroxide solution was titrated against M sulphuric(VI) acid using methyl orange as an indicator. The indicator changed colour from yellow to red when 22.5 cm3 of sulphuric(VI) acid had been added. Calculate the molarity of the sodium hydroxide solution. Answer

Example 3-6A Back 3.6 Simple titrations (SB p.61)
2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(l) =  Number of moles of NaOH(aq) = Number of moles of H2SO4(aq) Number of moles of H2SO4(aq) = mol dm-3  22.5  10-3 dm3 =  10-3 mol Number of moles of NaOH(aq) = 2   10-3 mol =  10-3 mol Molarity of NaOH(aq) = = mol dm-3 Therefore, the molarity of the sodium hydroxide solution was M.

Example 3-6B 2.52 g of a pure dibasic acid with formula mass of was dissolved in water and made up to cm3 in a volumetric flask cm3 of this solution was found to neutralize cm3 of sodium hydroxide solution. Calculate the molarity of the acid solution. Answer Number of moles of acid = = 0.02 mol Molarity of acid solution = = 0.08 M

Example 3-6B (b) If the dibasic acid is represented by H2X, write an equation for the reaction between the acid and sodium hydroxide. Answer (b) H2X(aq) + 2NaOH(aq)  Na2X(aq) + 2H2O(l)

Back Example 3-6B (c) Calculate the molarity of the sodium hydroxide solution. Answer Number of moles of H2X =  Number of moles of NaOH 0.08 mol dm-3  25.0  10-3 dm3 =  Molarity of NaOH  28.5  10-3 dm3 Molarity of NaOH = 0.14 M Therefore, the molarity of the sodium hydroxide solution was 0.14 M.

Example 3-6C 0.186 g of a sample of hydrated sodium carbonate, Na2CO3 · nH2O, was dissolved in 100 cm3 of distilled water in a conical flask M hydrochloric acid was added from a burette, 2 cm3 at a time. The pH value of the reaction mixture was measured with a pH meter. The results were recorded and shown in the following figure. Calculate the value of n in Na2CO3 · nH2O. Answer

Example 3-6C Back 3.6 Simple titrations (SB p.63)
There is a sudden drop in the pH value of the solution (from pH 8 to pH 3) with the equivalence point at 30.0 cm3. Na2CO3 · nH2O(s) + 2HCl(aq)  2NaCl(aq) + CO2(g) + (n + 1)H2O(l) Number of moles of Na2CO3 · nH2O =  Number of moles of HCl =  0.10 mol dm-3  30.0  10-3 dm3 n = 124.0 n = 1 Therefore, the chemical formula of the hydrated sodium carbonate is Na2CO3 · H2O.

Volume of H2SO4 added (cm3)
3.6 Simple titrations (SB p.63) Example 3-6D 5 cm3 of 0.5 M sulphuric(VI) acid was added to 25.0 cm3 of potassium hydroxide solution. The mixture was then stirred and the highest temperature was recorded. The experiment was repeated with different volumes of the sulphuric(VI) acid. The laboratory set-up and the results were as follows: Volume of H2SO4 added (cm3) Temperature (oC) 20.0 5 21.8 10 23.4 15 25.0 20 26.5 25 25.2 30 24.0

Example 3-6D Plot a graph of temperature against volume of sulphuric(VI) acid added. Answer

Example 3-6D (b) Calculate the molarity of the potassium hydroxide solution. Answer From the graph, it is found that the equivalence point of the titration is reached when 20 cm3 of H2SO4 is added. Number of moles of H2SO4 = 0.5 mol dm-3  20  10-3 dm3 = 0.01 mol 2KOH(aq) + H2SO4(aq)  K2SO4(aq) + 2H2O(l) 2 mol : 1 mol From the equation, mole ratio of KOH(aq) : H2SO4(aq) = 2 : 1 Number of moles of KOH(aq) = 2  0.01 mol = 0.02 mol Molarity of KOH(aq) = = 0.8 M Therefore, the molarity of potassium hydroxide solution was 0.8 M.

Back Example 3-6D (c) Explain why the temperature rose to a maximum and then fell. Answer (c) Neutralization is an exothermic reaction. When more and more sulphuric(VI) acid was added and reacted with potassium hydroxide, the temperature rose. The temperature rose to a maximum value at which the equivalence point of the reaction was reached. After that, any excess sulphuric(VI) acid added would cool down the reacting solution, causing the temperature to drop.

Example 3-6E When excess potassium iodide solution (KI) is added to 25.0 cm3 of acidified potassium iodate solution (KIO3) of unknown concentration, the solution turns brown. This brown solution requires 22.0 cm3 of 0.05 M sodium thiosulphate solution to react completely with the iodine formed, using starch solution as an indicator. Find the molarity of the acidified potassium iodate solution. Answer

Example 3-6E Back 3.6 Simple titrations (SB p.66)
IO3-(aq) + 5I-(aq) + 6H+(aq)  3I2(aq) + 3H2O(l) … … (1) I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) … … (2) From (1), Number of moles of IO3-(aq) =  Number of moles of I2(aq) From (2), Number of moles of I2(aq) =  Number of moles of S2O32-(aq) Number of moles of IO3-(aq) =  Number of moles of S2O32-(aq) Molarity of IO3-(aq)  25.0  10-3 dm3 =  0.05 mol dm-3  22.0  10-3 dm3 Molarity of IO3-(aq) = 7.33  10-3 M Therefore, the molarity of the acidified potassium iodate solution is 7.33  10-3 M.

Example 3-6F A piece of impure iron wire weighs 0.22 g. When it is dissolved in hydrochloric acid, it is oxidized to iron(II) ions. The solution requires 36.5 cm3 of 0.02 M acidified potassium manganate(VII) solution for complete reaction to form iron(III) ions. What is the percentage purity of the iron wire? Answer

Example 3-6F Back 3.6 Simple titrations (SB p.67)
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)  Mn2+(aq) + 4H2O(l) + 5Fe3+(aq) Number of moles of MnO4-(aq) : Number of moles of Fe2+(aq) = 1 : 5 Number of moles of Fe2+(aq) = 5  Number of moles of MnO4-(aq) = 5  0.02 mol dm-3  36.5  10-3 dm3 = 3.65  10-3 mol Number of moles of Fe dissolved = Number of moles of Fe2+ formed Mass of Fe = 3.65  10-3 mol  55.8 g mol-1 = g Percentage purity of Fe =  100 % = % Therefore, the percentage purity of the iron wire is %.

Check Point 3-6 5 g of anhydrous sodium carbonate is added to 100 cm3 of 2 M hydrochloric acid. What is the volume of gas evolved at room temperature and pressure? (Molar volume of gas at R.T.P. = 24.0 dm3 mol-1) Answer

Check Point 3-6 3.6 Simple titrations (SB p.67)
Na2CO3(s) + 2HCl(aq)  2NaCl(aq) + H2O(l) + CO2(g) No. of moles of Na2CO3 used = = mol No. of moles of HCl used = 2 M  = 0.2 mol Since HCl is in excess, Na2CO3 is the limiting agent. No. of moles of CO2 produced = No. of moles of Na2CO3 used Volume of CO2 produced = mol  24.0 dm3 mol-1 = dm3

Check Point 3-6 (b) 8.54 g of impure hydrated iron(II) sulphate (formula mass of ) was dissolved in water and made up to cm cm3 of this solution required cm3 of M acidified potassium manganate(VII) solution for complete reaction. Determine the percentage purity of the hydrated iron(II) sulphate. Answer

Check Point 3-6 Back 3.6 Simple titrations (SB p.67)
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)  Mn2+(aq) + 4H2O(l) + 5Fe3+(aq) No. of moles of MnO4- ions = M  =  10-4 mol No. of moles of Fe2+ ions = 5  No. of moles of MnO4- ions =  10-3 mol No. of moles of Fe2+ ions in 25.0 cm3 solution =  10-3 mol No. of moles of Fe2+ ions in cm3 solution = mol Molar mass of hydrated FeSO4 = g mol-1 Mass of hydrated FeSO4 = mol  g mol-1 = 8.26 g % purity of FeSO4 = = %

3 Chemical Equations and Stoichiometry 3.1 Formulae of Compounds

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3 Chemical Equations and Stoichiometry 3.1 Formulae of Compounds

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