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Chapter 3 Chemical Equations and Stoichiometry

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1 Chapter 3 Chemical Equations and Stoichiometry
3.1 Formulae of Compounds 3.2 Derivation of Empirical Formulae 3.3 Derivation of Molecular Formulae Chemical Equations Calculations Based on Equations Simple Titrations

2 1st way = by chemical formula
Formulae of Compounds (SB p.54) Formulae of Compounds How can you describe the composition of compound X? 1st way = by chemical formula C?H? ratio of no. of atoms

3 2nd way = by percentage by mass
Formulae of Compounds (SB p.54) How can you describe the composition of compound X? Compound X 2nd way = by percentage by mass Mass of carbon atoms inside = …. g Mass of hydrogen atoms inside = …. g carbon atoms hydrogen atoms

4 3.1 Formulae of Compounds (SB p.54)
Check Point 3-1 Give the empirical molecular and structural formula for the following compounds C6H12O6 (d) Glucose C2H5OH C2H6O (c) Ethanol HNO3 (b) Nitric acid C3H6 CH2 (a) Propene Structural formula Molecular formula Empirical formula Compound Answer

5 The different types of formulae of some compounds
Formulae of Compounds (SB p.55) The 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-

6 Example 3-1 Solution: The relative molecular mass of CO2
3.2 Derivation of Empirical Formulae (SB p.56) Solution: The relative molecular mass of CO2 = x 16.0 = 44.0 Mass of carbon in 2.93 g of CO2 = 2.93 g x 12.0/44.0 = 0.80 g The relative molecular mass of H2O = 2 x = 18.0 Mass of hydrogen in 1.80 g of H2O = 1.80 g x 2.0/18.0 = 0.20 g Example 3-1 A hydrogen was burnt completely in excess oxygen. It was found that 1.00 g of the hydrocarbon gives 2.93 g of carbon dioxide and 1.8 g of water. Find the empirical formula of the hydrocarbon.? (R.a.m.* : H = 1.0, C = 12.0, O = 16.0) Answer

7 Let the empirical formula of the hydrocarbon be CxHy.
3.2 Derivation of Empirical Formulae (SB p.57) Solution: (cont’d) 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 The simplest whole number ratio of x and y can be determined by the following the steps in the below table.

8 Relative number of moles
3.2 Derivation of Empirical Formulae (SB p.57) Carbon Hydrogen Mass (g) 0.80 0.20 Number of moles (mol) 0.80/12.0 = 0.2/ = 3 Relative number of moles / =1 0.20/ =3 Simplest mole ratio 1 3

9 Example 3-2 Solution: Mass of compound X = 0.46g
3.2 Derivation of Empirical Formulae (SB p.57) Solution: Mass of compound X = 0.46g Mass of carbon in compound X = 0.88 g x 12.0/44.0 = 0.24 g Mass of hydrogen in compound X = 0.54 g x 2.0/18.0 = 0..06g Mass of oxygen in compound X = 0.46 g – 0.24 g – 0.06 g = 0.16 g Example 3-2 Compound X is known to contain carbon, hydrogen 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. (R.a.m.* : H = 1.0, C = 12.0, O = 16.0) Answer

10 Let the empirical formula of compound X be CxHyOz.
3.2 Derivation of Empirical Formulae (SB p.57) Solution: (cont’d) Let the empirical formula of compound X be CxHyOz. Therefore, the empirical formula of compound X is C2H6O. Carbon Hydrogen Oxygen Mass (g) 0.24 0.06 0.16 Number of moles (mol) 0.02 0.01 Relative number of moles 2 6 1 Simplest mole ratio

11 Relative Number of moles
(a)Mass of sulphur = 5 g Mass of oxygen = (10 – 5) g The empirical formula of the sulphur oxide is SO2. Formulae of Compounds (SB p.58) Check Point 3-2 (a) 5 g of sulphur forms 10 g of an oxide on burning.What is the empirical formula of the oxide? (R.a.m. : O = 16.0, S = 32.1) (b) f of element M combines with g of oxygen to form an oxide. If the relative atomic mass of M is 331.0, find the empirical formula of the oxide. (R.a.m. : O =16.0) (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 (R.a.m. : Cu = 63.5, O = 16.0) Sulphur Oxygen Mass (g) 5 Number of moles (mol) 5/ 32.1 = 0.156 5/16.0 = 0.313 Relative Number of moles 0.156/0.156=1 0.313/0.156 =2 Simplest mole ratio 1 2 Answer

12 Relative number of moles
Formulae of Compounds (SB p.58) (b) The empirical formula of the oxide is M2O5. M O Mass (g) 19.85 25.61 Number of moles (mol) 19.85/31.0 = 0.64 25.61/16.0 = 1.6 Relative number of moles 0.64/0.64 = 1 1.6/0.64 = 2.5 Simplest mole ratio 2 5

13 Relative number of moles
Formulae of Compounds (SB p.58) (c) Mass of Cu = ( ) g = 1.51g Mass of O = ( ) = g The empirical formula of the oxide is CuO. Cu O Mass (g) 1.51 0.381 Number of moles (mol) 1.51/63.5 = 0.381/16.0 Relative number of moles 0.0238/ =1 Simplest mole ratio 1

14 Determination of Empirical Formula
3.2 Derivation of Empirical Formulae (SB p.58) Determination of Empirical Formula From Combustion by Mass Composition by mass Empirical formula

15 3.2 Derivation of Empirical Formulae (SB p.58)
Solution: Let the empirical formula of the hydrocarbon be CxHy, and the mass of the compound be 100 g. Mass of carbon in the compound = 75 g Mass of hydrogen in the compound=(100 –75) g = 25 g Therefore, the empirical formula of the hydrocarbon is CH4. Example3-3 Compound A contains carbon and hydrogen only. It is found that the compound contains 75% carbon by mass. Determine its empirical formula. (Relative atomic masses: C=12, H=1 ) Carbon Hydrogen Mass (g) 75 25 Number of moles (mol) 75/12.0 = 6.25 25/1.0 = 25 Relative number of moles 6.25/6.25 = 1 25/6.25 = 4 Simplest mole ratio 1 4 Answer

16 3.2 Derivation of Empirical Formulae (SB p.59)
Solution: Let the mass of phosphorus chloride be 100g. Then, Mass of phosphorus in the compound = 22.55g Mass of chloride in the compound = 77.45g Therefore, the empirical formula of the phosphorus chloride is PCl3. Example 3-4 The percentage by mass of phosphorus and chlorine in a sample of a phosphorus chloride are 22.55% and 77.45% respectively. Find the empirical formula of the chloride. (R.a.m. : P = 31.0, Cl = 35.5) Phosphorus Chloride Mass (g) 22.55 77.45 Number of mole (mol) 22.55/31.0 = 0.727 77.45/35.5 =2.182 Relative number of moles 0.727/0.727 = 1 2.182/0.727 = 3 Simplest mole ratio 1 3 Answer

17 Relative number of moles
3.2 Derivation of Empirical Formulae (SB p.59) Let the mass of vitamin C analyzed be 100g. The empirical formula of vitamin C is C3H4O3. Check Point 3-3 Find the empirical formula of vitamin C if it consists of 40.9% caarbon, 54.5% oxygen and 4.6% hydrogen by mass. ( R.a.m.: C = 12.0, H = 1.0, O = 16.0) Each 325 mg tablet of aspirin consists of mg carbon 14.6 mg hydrogen and 115.4mg oxygen. Determine the empirical formula of aspirin. (R.a.m. : C= 12.0, H = 1.0, O = 16.0) Carbon Hydrogen Oxygen Mass (g) 40.9 4.6 54.5 Number of moles (mol) 40.9/12.0 = 3.41 4.6/1.0 =4.60 54.5/16.0 =3.41 Relative number of moles 3.41/3.41 =1 4.61/3.41 =1.35 Simplest mole ratio 3 4 Answer

18 Relative number of moles
3.2 Derivation of Empirical Formulae (SB p.59) (b) In order to facilitate calculation, the masses of the elements are multiplied by 1000 first. The empirical formula of aspirin is C9H8O4. Carbon Hydrogen Oxygen Mass (g) 195.0 14.6 115.4 Number of moles (mol) 195.0/12.0 =16.25 14.6/7.21 =2.02 7.21/7.21 =1 Relative number of moles 16.25/7.21 = 2.25 = 2.02 Simplest mole ratio 9 8 4

19 What is Molecular Formulae?
3.3 Derivation of Molecular Formulae (SB p.60) What is Molecular Formulae? Molecular formula ? = (Empirical formula)n

20 Empirical formula Molecular mass Molecular formula
3.3 Derivation of Molecular Formulae (SB p.60) Empirical formula Molecular mass Molecular formula

21 3.3 Derivation of Molecular Formulae (SB p.60)
Solution: Let the empirical formula of the hydrocarbon be CxHy. Mass of carbon in the hydrocarbon = 14.6g x 12.0/44.0 = 4.0g Mass of hydrogen in the hydrocarbon = 9.0g x 2.0/18.0 = 1.0g Example 3-5 A hydrogen was burnt completely in excess oxygen. It was found that 5.00 g of the hydrocarbon gives 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. hydrocarbon.? (R.a.m.* : H = 1.0, C = 12.0, O = 16.0) Carbon Hydrogen Mass (g) 4.0 1.0 Number of moles (mol) 4.0/12.0 = 0.333 1.0/1.0 = 1 Relative number of moles 0.333/0.333 = 1 1/0.333 = 3 Simplest mole ratio 1 3 Answer

22 Relative molecular mass of (CH3)n = 30.0 n x (12.0 + 1.0 x 3) = 30.0
3.3 Derivation of Molecular Formulae (SB p.60) Solution: (cont’d) Therefore, the empirical formula of the hydrocarbon is CH3. The molecular formula of the hydrocarbon is (CH3)n. Relative molecular mass of (CH3)n = 30.0 n x ( x 3) = 30.0 n= 2 Therefore, the molecular formula of the hydrocarbon is C2H6.

23 3.3 Derivation of Molecular Formulae (SB p.61)
Solution: Let the empirical formula of the hydrocarbon be CxHyOz. Mass of carbon in the compound = 44.44g Mass of hydrogen in the compound = 6.18g Mass of oxygen in the compound = 49.38g Example 3-6 Compound X is known to contain 44.44% carbon, 6.18% hydrogen and 49.38% oxygen by mass. A typical analysis shows that it has a relative molecular mass of Find its molecular formula(R.a.m.* : H = 1.0, C = 12.0, O = 16.0) Carbon Hydrogen Oxygen Mass (g) 44.44 6.18 49.38 Number of moles (mol) 44.44/12.0 = 3.70 6.18/1.0 = 6.18 49.38/16.0 = 3.09 Relative number of moles 3.70/3.09 = 1.2 6.18/3.09 = 2 3.09/3.09 = 1 Simplest mole ratio 6 10 5 Answer

24 The empirical formula of compound X is C6H10O5.
3.3 Derivation of Molecular Formulae (SB p.61) Solution(cont’d) The empirical formula of compound X is C6H10O5. The molecular formula of compound X is (C6H10O5)n. Relative molecular mass of (C6H10O5)n = 162.0 n x (12.0 x x x 5) = 162.0 n = 1 Therefore, the molecular formula of compound is C6H10O5.

25 Water of Crystallization Derived from Composition by Mass
3.3 Derivation of Molecular Formulae (SB p.61) Water of Crystallization Derived from Composition by Mass Hydrated salt Anhydrous salt CuSO45H2O Blue crystals Anhydrous CuSO4 White powder Na2CO310H2O Colourless crystals Anhydrous Na2CO3 CoCl2 2H2O Pink crystals Anhydrous CoCl2

26 3.3 Derivation of Molecular Formulae (SB p.61)
Example 3-7 The chemical formula of hydrated copper(II) sulphate is known to be CuSO4.xH2O. It is found that the percentage of water by mass in the compound is 36%. Find x. (R.a.m. : H=1.0, O=16.0, S=32.1, Cu=63.5) Solution: Let Relative molecular mass of CuSO4xH2O = x 4 + (1.0x2 = 16.0)x = x Relative molecular mass of water of crystallization =18x 18x/( x) = 36/100 1800x= x 1152x= x = 4.99  5 Therefore, the chemical formula of hydrated copper(II) sulphate is CuSO4 5H2O Answer

27 Relative number of moles
(i) Let the mass of compound Z be 100g. The empirical formula of compound Z is CH2O. 3.3 Derivation of Molecular Formulae (SB p.63) Check Point 3-4 Find Compound Z is the major component of a healthy drink. It contains 40.00% carbon, 6.67% hydrogen and 53.33% oxygen. (i) Find the empirical formula of compound Z. (ii) If the relative molecular mass of compound Z is 180, finds its molecular formula.(R.a.m. : C= 12.0, H = 1.0, O = 16.0) Carbon Hydrogen Oxygen Mass (g) 40.00 6.67 53.33 Number of moles (mol) 40.00/12.0 = 3.33 6.67/1.0 = 6.67 53.33/16.0=3.33 Relative number of moles 3.33/3.33 =1 6.67/3.33 =2 Simplest mole ratio 1 2 Answer

28 (a) (ii) Let the molecular formula of compound Z be (CH2O)n.
3.3 Derivation of Molecular Formulae (SB p.63) (a) (ii) Let the molecular formula of compound Z be (CH2O)n. n x (12.0 = 1.0 x 2 = 16.0) = 180 30n = 180 n = 6 The molecular formula of Z is C6H12O6.

29 (b) Since the chemical formula of (NH4)2Sx is (NH4)2S3, the value of x is 3. 3.3 Derivation of Molecular Formulae (SB p.63) Check Point 3-4 (b) (NH4)2Sx contains 72.72% sulphur by mass is water. Find the value of x. (R.a.m.: H = 1.0, N = 14.0, O = 16.0) (c) In the compound MgSO4nH2O, 51.22% by mass is water. Find the value of n. (R.a.m.: H = 1.0, O = 16.0, Mg = 24.3, S = 32.1) (NH4) unit S Mass (g) 27.28 72.72 Number of moles (mol) 27.28/18.0 = 1.52 72.72/32.1 =2.27 Relative number of moles 1.52/1.52 =1 2.27/1.52 =1.49 Simplest mole ratio 2 3 Answer

30 Since the chemical formula of MgSO4nH2O
is MgSO47H2O , the value of x is 7. 3.3 Derivation of Molecular Formulae (SB p.63) MgSO4 H2O Mass (g) 48.78 51.22 Number of moles (mol) 48.78/120.4 =0.405 51.22/18.0 =2.846 Relative number of moles 0.405/0.405 =1 2.846/0.405=7 Simplest mole ratio 1 7

31 3.3 Derivation of Molecular Formulae (SB p.63)
Example 3-8 The chemical formula of ethanoic acid is CH3COOH. Calculate the percentages by mass of carbon, hydrogen and oxygen by mass respectively. (R.a.m. : C=12.0, H=1.0, O=16.0 ) Solution: Relative molecular mass of CH3COOH = 12.0 x x x 2 = 60.0 % by mass of C = 12.0 x 2/ 60.0 x 100%= 40.00% % by mass of H = 1.0 x 4 /60.0 x 100% = 6.67% % by mass of O = 16.0 x 2/60.0 x 100% = 53.33% The percentage by mass of carbon, hydrogen and oxygen are 40.00%, 6.67% and 53.33% respectively. Answer

32 3.3 Derivation of Molecular Formulae (SB p.63)
Example 3-9 Calculate the mass of iron metal in a sample of 20g of hydrated iron (II) sulphate, FeSO47H2O. (R.a.m. : Fe = 55.8 , H=1.0, O=16.0 ) Solution: Relative molecular mass of FeSO4·7H2O = x 4 + (1.0x2+16.0) x 7=277.9 % by mass of Fe = 55.8/277.9 x 100% = 20.08% Mass of Fe = 20g x 20.08% = 4.02g Answer

33 3.3 Derivation of Molecular Formulae (SB p.63)
Molar mass of K2Cr2O7 = (39.1x x7) g mol-1 = g mol-1 % by mass of K = 39.1 x 2 g mol-1/294.2 g mol-1 x 100% = 26.58% % by mass of Cr = 52.0 x 2 g mol-1/294.2g mol-1 x 100% =35.25% % by mass of O = 16.0 x 7 g mol-1/294.2g mol-1 x 100% = 38.07% Check Point 3-5 (a) Calculate percentages by mass of potassium, chromium and oxygen in potassium chromate (VI), K2Cr2O7.(R.a.m. : K = Cr = 52.0, O = 16.0) (b) Find the mass of metal and water of crystallization in 100 g of Na2SO4·10H2O; 70g of Fe2O3·8H2O. (R.a.m.: H = 1.0, O = 16.0, Na = 23, S = 32.1, Fe = 55.8) Answer

34 (b)( i) Molar mass of Na2SO4·10H2O= 322.1 g mol-1
3.3 Derivation of Molecular Formulae (SB p.63) (b)( i) Molar mass of Na2SO4·10H2O= g mol-1 Mass of Na = 23.0 x 2 g mol-1/ g mol-1 x 100g = g Mass of H2O = 18.0 x 10 g mol-1/ g mol-1 x 100g = g (ii) Molar mass of Fe2O3·8H2O= g mol-1 Mass of Fe = 55.8 x 2 g mol-1/303.6g mol-1 x 70g = g Mass of H2O = 18.0 x 8 g mol-1/303.6g mol-1 x 70g = g

35 Chemical Equations a A + b B  c C + d D mole ratios
3.4 Chemical Equations (SB p.64) 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.

36 Zn(s) + H2O(g) ZnO(s) + H2(g)
(b) Mg(s) + 2 AgNO3(aq) Ag(s) + Mg(NO3)2(aq) (c) 2C4H10(g) + 13O2(g) CO2(g) + 10H2O(l) 3.4 Chemical Equations (SB p.64) Check Point 3-6 Give the chemical equations for the following reactions: (a) Zinc + steam zinc oxide + hydrogen (b) Magnesium + silver nitrate silver + magnesium nitrate (c) Butane + oxygen carbon dioxide + water Answer

37 Calculations Based on Equations
3.5 Calculations Based on Equations (SB p.65) Calculations Based on Equations Calculations involving Reacting Masses

38 Example 3-10 Solution: CuO(s) + H2(g) Cu(s) + H2O(l)
3.5 Calculations Based on Equations (SB p.65) Solution: CuO(s) + H2(g) Cu(s) + H2O(l) As the mole ratio of Cu : CuO 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 = 12.45/ ( ) g mol-1 = mol Number of mole of Cu formed = mol Mass of Cu / 63.5 g mol-1 = 0.157 Mass of Cu = mol x 63.5 g mol-1= 9.97g Therefore, the mass of copper formed in the reaction is 9.97g. Example 3-10 Calculate the mass of copper formed when 12.45g of copper(II) oxide is completely reduced by hydrogen. (R.a.m. : H=1.0, O=16.0, Cu = 63.5 ) Answer

39 3.5 Calculations Based on Equations (SB p.65)
Example 3-11 Sodium hydrogencarbonate decomposes according to the following 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? (R.a.m. : H = 1.0, C =12.0, O = 16.0, Na = 23.0; molar volume of gas at R.T.P. = 24.0 dm3mol-1) Solution: Number of moles of CO2 formed = 240cm3/ 24000cm3 mol-1 = 0.01 mol From the equation, 2 moles of NaHCO3(s) will form 1 mole of CO2(g). Number of moles of NaHCO3 required = 0.01 x 2 = 0.02 mol Mass of NaHCO3 required = 0.02 mol x( x 3) g mol-1 = 0.02 mol x 84.0g mol-1 = 1.68 g Therefore, the minimum amount of sodium hydrogencarbonate required is 1.68g. Answer

40 Calculations Based on Equations
3.5 Calculations Based on Equations (SB p.66) Calculations Based on Equations Calculations involving Volumes of Gases

41 3.5 Calculations Based on Equations (SB p.66)
Example 3-12 Calculate the volume of carbon dioxide formed when 20 cm3 of ethane and 70 cm3 of oxygen are exploded, assuming all volumes are measured at room temperature and pressure. Solution: Number of moles of CO2 formed C2H6(g) + 7O2(g) CO2(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 equation that the mole ratio of CO2 : C2H6 is 4 :2, and the volume ratio of CO2 : C2H6 should also be 4:2. Let x be the volume of CO2(g) formed x /20cm3 = 4/2 x = 40 cm3 Therefore, the volume of CO2 formed is 40 cm3. Answer

42 3.5 Calculations Based on Equations (SB p.67)
Solution: Let the molecular formula of the hydrocarbon be CxHy. Volume of hydrogen reacted = 10 cm3 Volume of O2(g) unreacted = 50 cm3 Volume of O2(g) reacted = 30 cm3 Volume of CO2(g) formed = 20 cm3 CxHy (x + y/4) O CO y/2 H2O 1 volume : (x + y/4) volumes : x volumes Volume of CO2 (g)/ volume of CxHy(g) = 20 cm3/ 10cm3 = X =2 Volume of O2(g) / volume of CxHy(g) =(x + y/4) / 1= 30/ 10 (x + y/4)= Y = 2 Molecular formula is C2H4 Example 3-13 10 cm3 of a gaseous hydrocarbon was mixed with 80cm3 of oxygen which was in excess. The mixture was exploded and then cooled. The volume left was 70cm3. 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

43 Check Point 3-7 No. of moles of H2 = No. of moles of Mg
3.5 Calculations Based on Equations (SB p.68) No. of moles of H2 = No. of moles of Mg Volume of H2/ 24.0 dm3 mol-1 = 2.43 g / 24.3 g mol-1 Volume of H2 = 2.4 dm3 (b) 1/3 x no. of moles of Cl2 = 1/2x no. of moles of PCl3 1/3 x mass of Cl2 / (35.5 x 2) g mol-1 = 1/2 x 100g / ( ) g mol-1 Mass of Cl2 = 77.45g Check Point 3-7 Find the volume of hydrogen produced at R.T.P. when 2.43g of magnesium reacts with excess hydrochloric acid. (R.a.m. : Mg = 24.3; molar volume of gas at R.T.P. = 24.0 dm3mol-1. Find the minimum mass of chlorine required to produced 100 g of phosphorus trichloride ( PCl3). 20 cm3 of a gaseous hydrocarbon and 150 cm3 of oxygen were exploded in a closed vessel. After cooling, 110 cm3 of gases remained. After passing through a solution of concentrated sodium hydroxide, the volume left was 50 cm3 . Determine the molecular of the hydrocarbon. Calculate the volume of carbon dioxide formed when cm3 of methane burns in excess oxygen, assuming all volumes are measured at room temperature and pressure. Answer

44 (c)Volume of CxHy used = 20 cm3 Volume of CO2 formed = 60 cm3
3.5 Calculations Based on Equations (SB p.68) (c)Volume of CxHy used = 20 cm3 Volume of CO2 formed = 60 cm3 Volume of O2 used = 100 cm3 Volume of CxHy : volume of CO2 = 1 : x = 20 : 60 x = 3 Volume of CxHy : volume of O2 = 1 : x + y/4 = 20 : 100 x + y/4 = 5 3 + y/4 = 5 y = 8

45 (d)Volume of CxHy used = 20 cm3
3.5 Calculations Based on Equations (SB p.68) (d)Volume of CxHy used = 20 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 = 1/1 x = 5 The volume of carbon dioxide gas is 5 cm3.

46 Simple Titrations Acid-Base Titrations
3.6 Simple Titrations (SB p.68) Simple Titrations Acid-Base Titrations Acid-Base Titrations with Indicators Acid-Base Titrations without Indicators (to be discussed in later chapters)

47 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution Copper(II) sulphate + solute Water solvent Copper(II) sulphate solution solution

48 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~50 cm3

49 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~50 cm3

50 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~50 cm3

51 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~50 cm3

52 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~50 cm3

53 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~50 cm3

54 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution 50 cm3 Solution A

55 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~50 cm3

56 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~50 cm3

57 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~50 cm3

58 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~50 cm3

59 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~50 cm3

60 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~50 cm3

61 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution 50 cm3 Solution B

62 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~100 cm3

63 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~100 cm3

64 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~100 cm3

65 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~100 cm3

66 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~100 cm3

67 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution ~100 cm3

68 Finding the concentration of a solution
3.6 Simple Titrations (SB p.69) Finding the concentration of a solution 100 cm3 Solution C

69 Comment on the Concentrations of Solutions A, B and C !
3.6 Simple Titrations (SB p.69) 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.

70 3.6 Simple Titrations (SB p.69)
Class Practice Suppose the right-handed side figure shows the number of solute particles in solution D. Draw similar particle models for Solutions A, B and C.

71 Class Practice Answers
3.6 Simple Titrations (SB p.69) Class Practice Answers

72 Comment on the Concentrations of Solutions A, B and C !
3.6 Simple Titrations (SB p.69) 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.

73 Molarity Unit: moles/dm3 (M) A way of expressing concentrations
3.6 Simple Titrations (SB p.69) Molarity A way of expressing concentrations Molarity is the number of moles of solute dissolved in 1 dm3 (1000 cm3) of solution. Unit: moles/dm3 (M)

74 “In every 1 dm3 of the solution, 2 moles of HCl is dissolved.”
3.6 Simple Titrations (SB p.69) What does this mean? 1 dm3 contains 2 moles of HCl “In every 1 dm3 of the solution, 2 moles of HCl is dissolved.”

75 Example 3-14 Solution: Number of moles of NaOH(aq) 2
3.6 Simple Titrations (SB p.71) Solution: Number of moles of NaOH(aq) Number of moles of H2SO4(aq) = 1 ½ x Number of moles of NaOH(aq) = Number of moles of H2SO4 (aq) = mol dm-3 x 22.5 x 10-3 dm3 = x 10-3 mol Number of moles of NaOH(aq) = 2 x x 10-3 mol = x 10-3 mol Molarity of NaOH(aq) = x 10-3 mol / 25.0 x 10-3 mol = mol dm-3 The molarity of NaOH is 0.121M Example 3-14 25.0cm3 of sodium hydroxide solution was titrated against M of sulphuric(VI) acid using methyl orange as indicator. The indicator changed colour from yellow to red when 22.5 cm3 of sulphuric(VI) acid had benn added. Calculate the molarity of the sodium hydroxide solution. Answer

76 3.6 Simple Titrations (SB p.71)
Example 3-15 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 25.0 cm3 of this solution was found to neutralize 28.5 cm3 of sodium hydroxide solution. Calculate the molarity of the acid solution. If the dibasic acid is represented by H2X, write an equation for the reaction between the acid and sodium hydroxide. Calculate the molarity of the sodium hydroxide solution. Solution: Number of moles of acid = 2.52 g/ g mol-1 = 0.02 mol Molarity of acid solution = 0.02 mol / 250 x 10-3 dm3 = 0.08M (b) H2X(aq) + 2NaOH(aq) Na2X(aq) + 2NaOH(l) (c) Number of moles of H2X =½ x number of moles of NaOH Molarity of NaOH = 0.14M Answer

77 3.6 Simple Titrations (SB p.72)
Example 3-16 0.186g of sample of hydrate sodium carbonate, NaCO2·nH2O, was dissolved in 100 cm3 of distilled water in conical flask M by hydrochloric acid was added from a burette, 2 cm3 at a time. The pH value of the solution was measured by a pH meter. The result was recorded and shown in the following figure. Calculate the value of n in NaCO2·nH2O. Solution: Number of moles of acid = 2.52 g / g mol-1 There is a sudden drop in the pH value of the solution (from pH 3 to pH 8) with the end point at 30.0 cm3. Na2CO3·nH2O(s) + 2 HCl(aq) 2NaCl(aq) + CO2(g) + (n+1)H2O(l) Number of moles of Na2CO3·nH2O = ½ x 0.1 mol dm-3 x 30 x 10-3 dm3 n = 124.0 n = 1 The formula is Na2CO3·H2O Answer

78 Example 3-17 Solution: (a)
3.6 Simple Titrations (SB p.73) Example 3-17 5 cm3 of 0.5M 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: Plot the graph of temperature against volume of sulphuric(VI) acid added. Calculate the molarity of the potassium hydroxide solution. Explain why the temperature rose to a maximum and the fell. Solution: (a) Answer

79 Number of moles of = 0.5 mol dm-3 x 20/1000 dm3 = 0.01 mol
3.6 Simple Titrations (SB p.74) Solution: (cont’d) (b) From the graph, it is found that the end point of the titration is reached when 20 cm3 of H2SO4 is added. Number of moles of = 0.5 mol dm-3 x 20/1000 dm3 = 0.01 mol 2KOH(aq) + H2SO4(aq) K2SO4(aq) + 2H2O(l) 2 mol mol Mole of KOH(aq) : H2SO4 = 2 : 1 Number of moles of KOH(aq) = 2 x 0.01 mol = 0.02 mol Molarity of KOH(aq) = 0.02 mol / (25 x 10-3 dm3) = 0.8M

80 3.6 Simple Titrations (SB p.74)
Solution: (cont’d) (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 cooled down the reacting solution, causing the temperature to drop.

81 Redox Titrations Some Examples in burette in conical flask
3.6 Simple Titrations (SB p.76) Redox Titrations Some Examples in conical flask in burette Iodometric Titrations I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) brown colourless

82 Redox Titrations Some Examples Add starch Iodometric Titrations
3.6 Simple Titrations (SB p.76) Redox Titrations Some Examples Add starch Iodometric Titrations in burette in conical flask I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) brown colourless During titration : brown  yellow

83 Redox Titrations Some Examples Iodometric Titrations
3.6 Simple Titrations (SB p.76) Redox Titrations Some Examples Iodometric Titrations I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) brown colourless During titration : brown  yellow

84 Redox Titrations Some Examples Iodometric Titrations
3.6 Simple Titrations (SB p.76) Redox Titrations Some Examples Iodometric Titrations I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) brown colourless During titration : brown  yellow End point : blue black  colourless (after addition of starch indicator)

85 3.6 Simple Titrations (SB p.76)
Example 3-18 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 indicator. Find the molarity of the potassium iodate solution. Solution: IO3-(aq) + 5I- + 6H+(aq) I2(aq) + 3H2O(l) …(1) I2(aq) + 2S2O3 2-(aq) I-(aq) + S4O6 2-(aq) …(2) From (1), Number of moles of IO3-(aq) =1/3 x number of moles of I2(aq) From(2), Number of moles of I2(aq) =1/2 x number of moles of S2O32-(aq) Number of moles of IO3-(aq) = 1/6 x number of moles of S2O3 2-(aq) Molority of IO3-(aq) x 25.0/1000 dm3 =1/6 x 0.05 mol dm-3 x 22.0/1000 dm3 Molarity of IO3-(aq) = 7.33 x 10-3 M Answer

86 Redox Titrations In conical flask In burette Some Examples
3.6 Simple Titrations (SB p.76) Redox Titrations In conical flask In burette Some Examples Titrations Involving Potassium Permanganate MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + H2O(aq) + 5Fe3+(aq) purple colourless

87 Redox Titrations Some Examples
3.6 Simple Titrations (SB p.76) Redox Titrations In burette In conical flask Some Examples Titrations Involving Potassium Permanganate MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)  Mn2+(aq) + H2O(aq) + 5Fe3+(aq) purple colourless During titration : pale green  yellow

88 Redox Titrations Some Examples
3.6 Simple Titrations (SB p.76) Redox Titrations In burette In conical flask Some Examples Titrations Involving Potassium Permanganate MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)  Mn2+(aq) + H2O(aq) + 5Fe3+(aq) purple colourless During titration : pale green  yellow End point : yellow  light purple

89 3.6 Simple Titrations (SB p.77)
Solution: MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + 4H2O(l) + 5Fe3+(aq) Number of moles of Fe2+(aq) = 5 x number of moles of MnO4-(aq) = 5 x 0.02 mol dm-3 x 36.5 x 10-3 dm3 = 3.65 x 10-3 mol Number of moles of Fe dissolved = number of mole of Fe 2+ formed = x 10-3 mol Mass of Fe = 3.65 x 10-3 mol x 55.8 g mol-1= 0.204g Percentage purity of Fe = 0.204g/0.22g x 100% = 92.73% Example 3-19 A piece of impure iron wire weighs 0.22g. 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) for complete reaction to form iron(III) ions. What is the percentage purity of the iron wire? Answer

90 3.6 Simple Titrations (SB p.78)
Na2CO3 (s) HCl (aq) 2NaCl(aq) + H2O(l) + CO2(g) No. of moles of Na2CO3 used = 5g / (23.0x x 3)g mol-1 = mol No. of moles of HCl used = 2M x 100/ 1000 dm3 = 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 = mol Volume of CO2 produced = mol x dm3 mol-1= dm3 Check Point 3-8 5g 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? (R.a.m. : C = 12.0, O = 16.0, Na = 23.0; molar volume of gas at R.T.P. = 24.0 dm3 mol-1) (b) 8.54g of impure hydrated iron(II) sulphate (formula mass of ) was dissolved in water and made up to 250 cm3. 25cm3 of this solution required 20.76cm3 of M acidified potassium manganate(VII) solution for complete reaction. Determine the percentage purity of the hydrated iron(II) sulphate. Answer

91 = 5 x no. of moles of MnO4- = 2.107 x 10-3 mol
3.6 Simple Titrations (SB p.78) (b)No. of moles of MnO4- = M x 20.76/1000 dm3 = x 10-4 mol No. of moles of Fe2+ = 5 x no. of moles of MnO4- = x 10-3 mol No. of mole of Fe2+ in 25.0 cm3 solution = x 10-3 mol No. of mole of Fe2+ in cm3 solution = mol Molar mass of hydrated FeSO4 = g mol –1 Mass of hydrated FeSO4 = mol x g mol –1= 8.26g 5 purity of FeSO4 = 8.26g/8.54g x 100% = 96.72%

92 The END


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