Presentation is loading. Please wait.

Presentation is loading. Please wait.

1 Chemical Equations and Stoichiometry 3.1Formulae of Compounds 3.2Derivation of Empirical Formulae 3.3Derivation of Molecular Formulae 3.4Chemical Equations.

Similar presentations


Presentation on theme: "1 Chemical Equations and Stoichiometry 3.1Formulae of Compounds 3.2Derivation of Empirical Formulae 3.3Derivation of Molecular Formulae 3.4Chemical Equations."— Presentation transcript:

1

2 1 Chemical Equations and Stoichiometry 3.1Formulae of Compounds 3.2Derivation of Empirical Formulae 3.3Derivation of Molecular Formulae 3.4Chemical Equations 3.5Calculations Based on Chemical Equations 3.6Simple Titrations 3

3 2 Deals with quantitative relationships (a)among atoms, molecules and ions RAM / RMM / Formula Mass Stoichiometry ( 化學計量學 )p.19

4 3 Deals with quantitative relationships (b)among the constituent elements of a compound Empirical / Molecular Formulae Stoichiometry ( 化學計量學 )p.19

5 4 Deals with quantitative relationships (c)among the substances participating in chemical reaction Calculations involving chemical equation Stoichiometry ( 化學計量學 )p.19

6 5 3.1 Formulae of Compounds

7 6 3.1 Formulae of compounds (SB p.43) Empirical formula Shows the simplest whole number ratio of the atoms or ions present E.g.Methane,CH 4 Sodium chloride,NaCl

8 7 3.1 Formulae of compounds (SB p.43) Molecular formula Shows the actual number of each kind of atoms present in one molecule E.g. CH 4 methane Ionic compounds do not have molecular formulae

9 8 3.1 Formulae of compounds (SB p.43) Structural formula Shows the bonding order of atoms in one molecule E.g. CH 4 methane

10 9 3.1 Formulae of compounds (SB p.44) CompoundEmpirical formula Molecular formula Structural formula Carbon dioxide CO 2 O = C =O WaterH2OH2OH2OH2O MethaneCH 4 GlucoseCH 2 OC 6 H 12 O 6 Sodium fluoride NaFNot applicable Na + F - Different types of formulae of some compounds

11 Derivation of Empirical Formulae

12 11 From composition by mass 3.2 Derivation of empirical formulae (SB p.45) Example 1 Mg + N 2 Mg x N y g0.623 gexcess Mass of N used = ( )g = g

13 12 From composition by mass 3.2 Derivation of empirical formulae (SB p.45) Mg 3 N 2 Example 1 Mg + N 2 Mg x N y g0.623 gexcess

14 Derivation of empirical formulae (SB p.45) Example 2 Q.14 Water of Crystallization Derived from Composition by Mass Example 3-3C Example 3-3C Check Point 3-3A Check Point 3-3A

15 14 Q.14 CuSO 4.xH 2 O CuSO 4 + xH 2 O heat 10.0 g6.4 g (10.0–6.4) g = 3.6 g CuSO 4 : H 2 O = x = 5

16 15 heat CxHyOzCxHyOz C x H y O z + O 2  CO 2 (g) + H 2 O(g) g g g From combustion data Vitamin C excess

17 16 Mass of C in sample = mass of C in CO 2 formed C x H y O z + O 2  CO 2 (g) + H 2 O(g) Mass of H in sample = mass of H in H 2 O formed Mass of O in sample = total mass of sample – mass of C – mass of H

18 17 Mass of O in sample = ( ) g = g

19 Simplest whole no. ratio Number of moles (mol) Mass (g) OHC C3H4O3C3H4O3

20 Derivation of Molecular Formulae

21 20 What is molecular formulae? Molecular formula = (Empirical formula) n 3.3 Derivation of molecular formulae (SB p.49)

22 21 Empirical formulaMolecular mass Molecular formula 3.3 Derivation of Molecular Formulae (SB p.49) From empirical formula and known relative molecular mass Example 3-3A Example 3-3A Example 3-3B Example 3-3B

23 22 Q.15 = 142 g mol  1 Relative molecular mass = 142 C n H 2n+2 (l) C n H 2n+2 (g) 12.0 g 3.28 dm K, 1.00 atm

24 23 RMM = 12n + 2n+2 = 142 n = 10 The molecular formula is C 10 H 22 Q.15 C n H 2n+2 (l) C n H 2n+2 (g) 12.0 g 3.28 dm K, 1.00 atm

25 24 Ethanoic acid C, H, O are present Qualitative Analysis Quantitative Analysis Empirical formula CH 2 O Structural formula CH 3 COOH IR, NMR, MS Determination of Chemical Formulae RMM Molecular formula C 2 H 4 O 2 = 60.0

26 25 Structural formula : bond-line structure

27 26 Na 2 SO 4 ·Al 2 (SO 4 ) 3 ·24H 2 O Calculate the % by mass of the constituent elements of soda alum. Na :S :Al :

28 27 Na 2 SO 4 ·Al 2 (SO 4 ) 3 ·24H 2 O H :O : Calculate the % by mass of the constituent elements of soda alum.

29 28 A certain compound was known to have a formula which could be represented as [PdC x H y Nz](ClO 4 ) 2. Analysis showed that the compound contained 30.15% carbon and 5.06% hydrogen. When converted to the corresponding thiocyanate, [PdC x H y N z ](SCN) 2, the analysis was 40.46% carbon and 5.94% hydrogen. Calculate the values of x, y and z. (Relative atomic masses : C = 12.0, H = 1.0, N = 14.0, O = 16.0, Cl = 35.5, S = 32.0, Pd = 106.0)

30 29 Let M be the formula mass of [PdC x H y N z ](ClO 4 ) 2 % by mass of C in [PdC x H y N z ](SCN) 2 Then, the formula mass of [PdC x H y N z ](SCN) 2 % by mass of C in [PdC x H y N z ](ClO 4 ) 2 = M + 2( ) – 2(  16.0) = M – 83.0

31 30 Solving by simultaneous equations, x = 14, M = 557 % by mass of H in [PdC x H y N z ](ClO 4 ) 2 y = =   z + 2(  16.0) z = 4

32 Chemical Equations

33 32 Chemical equations a A + b B  c C + d D a, b, c, d are stoichiometric coefficients 3.4 Chemical equations (SB p.53)

34 33 Calculations based on equations 3.5 Calculations Based on Equations (SB p.65) Calculations involving reacting masses Example 3-5A Example 3-5A Example 3-5B Example 3-5B Check Point 3-4 Check Point 3-4

35 Chemical equations (SB p.53) Example(a) Excess oxygen 2Mg(s) + O 2 (g)  2MgO(s) 2.43 gexcess?

36 Chemical equations (SB p.53) Example(a) Excess oxygen 2Mg(s) + O 2 (g)  2MgO(s) 2.43 gexcess? = 4.03 g

37 Chemical equations (SB p.53) Example(b) limiting reagent to be determined 2Mg(s) + O 2 (g)  2MgO(s) 2.43 g1.28g?

38 Chemical equations (SB p.53) Example(b) limiting reagent to be determined 2Mg(s) + O 2 (g)  2MgO(s) mol0.040 mol? Mg is in excess, O 2 is the limiting reagent

39 Chemical equations (SB p.53) Example(b) limiting reagent to be determined 2Mg(s) + O 2 (g)  2MgO(s) mol0.080 mol Check Point 3-4 Check Point 3-4 Q.16, 17 = 3.22 g

40 39 Q.16 P 4 (s) + 5O 2 (g) 2P 2 O 5 (s) 4.00 g6.00 g

41 40 P 4 (s) + 5O 2 (g) 2P 2 O 5 (s) excess limiting reactant mol0.188 mol O 2 is in excess

42 41 Q.16 P 4 (s) + 5O 2 (g) 2P 2 O 5 (s) mol = 9.17 g 2  mol

43 42 Q.17 2Al(s) + 3Cu 2+ (aq)  2Al 3+ (aq) + 3Cu(s) = 2 : 3

44 Calculations Based on Equations (SB p.66) Calculations involving volumes of gases 1.Gases not at the same conditions 2.Gases at the same conditions - Gay Lussac’s Law

45 44 CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) 2.8 dm 3 25  C 1.65 atm 35.0 dm 3 31  C 1.25 atm O 2 is in excess and CH 4 is the limiting reactant ? dm  C 2.50 atm

46 45 CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) mol ? dm  C 2.50 atm = 2.47 dm 3

47 46 Gay Lussac’s law : When gases reacts, they do so in volumes which bears a simple whole number ratio to one another, and to the volumes of gaseous products, all volumes being measured under the same conditions of temperature and pressure. Watch video

48 47 Ammonia nitrogen + hydrogen Fe as catalyst H 2 (g) + CuO Cu(s) + H 2 O(l) heat 20 cm 3 10 cm cm cm 3 2NH 3 (g)  1N 2 (g) + 3H 2 (g)

49 48 Gay Lussac’s law is an application of the Avogadro’s law. a A(g) + b B(g)  c C(g) + d D(g) At fixed T & P,n  V

50 49 Gay Lussac’s law a A(g) + b B(g)  c C(g) + d D(g) At fixed T & P

51 50 Determination of Molecular Formula from Reacting Volumes of Gases For C x H y C x H y (g) + O 2 (g)  xCO 2 (g) + H 2 O(?)

52 51 Determination of Molecular Formula from Reacting Volumes of Gases For C x H y O z C x H y O z (g) + O 2 (g)  xCO 2 (g) + H 2 O(?)

53 52 Q cm 3 75 cm 3 45 cm 3 Volume of CO 2 formed = volume of gas absorbed by NaOH = (70 – 25) cm 3 = 45 cm 3 Before the reaction,After the reaction, used = 75 cm 3 C x H y (g) + O 2 (g)  xCO 2 (g) + H 2 O(l)

54 53 Q cm 3 75 cm 3 45 cm 3 C3H8C3H8 C x H y (g) + O 2 (g)  xCO 2 (g) + H 2 O(l)

55 54 Q cm cm 3 C2H4O2C2H4O2 C x H y O z (g) + O 2 (g)  xCO 2 (g) + H 2 O(g)

56 Simple Titrations

57 56 Simple titrations Acid-base titrations Redox titrations 3.6 Simple titrations (SB p.58)

58 57 Simple titrations Acid-base titrations Acid-base titrations with indicators Acid-base titrations without indicators 3.6 Simple titrations (SB p.58) Refer to Notes on ‘Detection of Equivalence Point in Acid-base Titration

59 58 End point : - The point at which the indicator undergoes a sharp colour change in a titration Equivalence point : - The point at which the reaction is just complete without excess of any reactant.

60 59 Titration without an indicator 3.6 Simple titrations (SB p.62) By following the change (a) in pH value (pH titration) (b) in temperature (thermometric titration) (c) in electrical conductivity (conductometric titration) in the course of the reaction

61 60 pH Titration Curves Equivalence point, pH 7.00 Phenolphthalein Methyl orange

62 61 pH Titration Curves Equivalence point, pH 5.28 Methyl orange

63 62 pH Titration Curves Equivalence point, pH 8.72 Phenolphthalein

64 63 pH Titration Curves Slow change in pH around equivalence point Most indicators are not suitable Equivalence point, ~ pH 7

65 64 Thermometric Titration Volume of acid added / cm 3 Temperature / K Equivalence point

66 65 Volume of acid added / cm 3 Temperature / K Equivalence point H + (aq) + OH  (aq)  H 2 O(l) + heat cooling by excess acid

67 66 Electrical Conductivity Volume of acid added /cm 3 Conductometric Titration – NaOH vs HCl Equivalence point

68 67 Electrical Conductivity Volume of acid added /cm 3 Conductometric Titration – NaOH vs HCl H + (aq) + Cl  (aq) + Na + (aq) + OH  (aq) H 2 O(l) + Na + (aq) + Cl  (aq) more mobile less mobile

69 68 Electrical Conductivity Volume of acid added /cm 3 Conductometric Titration – NaOH vs HCl Equivalence point Beyond the equivalence point, conductivity  sharply due to excess H + (most mobile) & Cl  Steeper slope

70 69 Electrical Conductivity Volume of acid added /cm 3 At the equivalence point H + (aq) + Cl  (aq) + Na + (aq) + OH  (aq) H 2 O(l) + Na + (aq) + Cl  (aq) conductivity > 0

71 70 Volume of acid added /cm 3 Electrical Conductivity Conductometric Titration – NH 3 vs HCl Equivalence point

72 71 Volume of acid added /cm 3 Electrical Conductivity Equivalence point Initial conductivity is low NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH  (aq)

73 72 Volume of acid added /cm 3 Electrical Conductivity H + (aq) + Cl  (aq) + NH 3 (aq) NH 4 + (aq) + Cl  (aq) more conducting less conducting

74 73 Volume of acid added /cm 3 Electrical Conductivity Equivalence point Beyond the equivalence point, conductivity  sharply due to excess H + (most mobile) & Cl  steeper slope

75 74 Electrical Conductivity Volume of acid added /cm 3 Conductometric Titration – NaOH vs CH 3 COOH Equivalence point

76 75 Electrical Conductivity Volume of acid added /cm 3 CH 3 COOH(aq) + Na + (aq) + OH  (aq) H 2 O(l) + Na + (aq) + CH 3 COO  (aq) more conducting less conducting

77 76 Electrical Conductivity Volume of acid added /cm 3 Equivalence point Beyond the equivalence point, conductivity  slowly because the excess acid is weak smaller slope

78 77 Electrical Conductivity Volume of acid added /cm 3 Conductometric Titration – NH 3 vs CH 3 COOH Equivalence point

79 78 Electrical Conductivity Volume of acid added /cm 3 Equivalence point Initial conductivity is low NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH  (aq)

80 79 Electrical Conductivity Volume of acid added /cm 3 CH 3 COOH (aq) + NH 3 (aq) CH 3 COO  (aq) + NH 4 + (aq) more conducting

81 80 Electrical Conductivity Volume of acid added /cm 3 Equivalence point Beyond the equivalence point, conductivity  slowly because the excess acid is weak smaller slope

82 81 Electrical Conductivity Volume of acid added / cm 3 Conductometric Titration – Ba(OH) 2 vs H 2 SO 4 Equivalence point At equivalence point, conductivity  0 because BaSO 4 is insoluble in water

83 82 Simple titrations Acid-base titrations Redox titrations 3.6 Simple titrations (SB p.58) 6B

84 83 1. Iodometric titration I 2 (aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq) browncolourless 3.6 Simple Titrations (SB p.65) Redox titrations I2I2 S 2 O 3 2- (aq) Very few I 2 brown  yellow During the course of the titration

85 84 1. Iodometric titration I 2 (aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq) browncolourless 3.6 Simple Titrations (SB p.65) Redox titrations Starch solution Very few I 2 yellow  dark blue Complex of starch & I 2 During the course of the titration

86 85 1. Iodometric titration I 2 (aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq) browncolourless 3.6 Simple Titrations (SB p.65) Redox titrations dark blue  colourless Complex of starch & I 2 S 2 O 3 2- (aq) 2I - (aq) + S 4 O 6 2- (aq) Example 3-6E Example 3-6E At the end point

87 86 2. Titrations involving potassium manganate(VII) MnO 4 - (aq) + 8H + (aq) + 5Fe 2+ (aq)  Mn 2+ (aq) + H 2 O(aq) + 5Fe 3+ (aq) purplepale green 3.6 Simple Titrations (SB p.66) Fe 2+ H + (aq)/MnO 4  (aq) Fe 2+ & Fe 3+ yellowcolourless During the course of the titration

88 87 2. Titrations involving potassium manganate(VII) MnO 4 - (aq) + 8H + (aq) + 5Fe 2+ (aq)  Mn 2+ (aq) + H 2 O(aq) + 5Fe 3+ (aq) purplepale green 3.6 Simple Titrations (SB p.66) H + (aq)/MnO 4  (aq) Fe 3+ & MnO 4  yellowcolourless At the end point Example 3-6F Example 3-6F Check Point 3-6 Check Point 3-6 Fe 2+ & Fe 3+

89 88 Double indicator titration Na 2 CO 3 + HCl  NaHCO 3 + NaCl NaOH + HCl  NaCl + H 2 O Red in phenolphthalein colourless in phenolphthalein

90 89 Double indicator titration NaHCO 3 + HCl  NaCl + H 2 O + CO 2 yellow in methyl orange red in methyl orange

91 90 The END

92 91 Give the empirical, molecular and structural formulae for the following compounds: (a) Propene (b) Nitric acid (c) Ethanol (d) Glucose Back Answer 3.1 Formulae of compounds (SB p.45) C 6 H 12 O 6 (d) Glucose C 2 H 5 OHC2H6OC2H6O(c) Ethanol HNO 3 (b) Nitric acid C3H6C3H6 CH 2 (a) Propene Structural formula Molecular formula Empirical formula Compound

93 92 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. 3.2 Derivation of empirical formulae (SB p.46) Answer

94 Derivation of empirical formulae (SB p.46) The relative molecular mass of CO 2 =  2 = 44.0 Mass of carbon in 2.93 g of CO 2 = 2.93 g  = 0.80 g The relative molecular mass of H 2 O = 1.0  = 18.0 Mass of hydrogen in 1.80 g of H 2 O = 1.80 g  = 0.20 g Let the empirical formula of the hydrocarbon be C x H y. Mass of carbon in C x H y = Mass of carbon in CO 2 Mass of hydrogen in C x H y = Mass of hydrogen in H 2 O

95 94 Therefore, the empirical formula of the hydrocarbon is CH Derivation of empirical formulae (SB p.46) Back CarbonHydrogen Mass (g) No. of moles (mol) Relative no. of moles Simplest mole ratio 13

96 95 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. 3.2 Derivation of empirical formulae (SB p.46) Answer

97 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 C x H y O z.

98 97 Therefore, the empirical formula of compound X is C 2 H 6 O. 3.2 Derivation of empirical formulae (SB p.47) Back CarbonHydrogenOxygen Mass (g) No. of moles (mol) Relative no. of moles Simplest mole ratio 261

99 98 (a)5 g of sulphur forms 10 g of an oxide on complete combustion. What is the empirical formula of the oxide? 3.2 Derivation of empirical formulae (SB p.47) Answer (a)Mass of sulphur = 5 g Mass of oxygen = (10 – 5) g = 5 g 21Simplest mole ratio Relative no. of moles No. of moles (mol) 55Mass (g) OxygenSulphur

100 99 (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. 3.2 Derivation of empirical formulae (SB p.47) Answer (b) The empirical formula of the oxide is M 2 O 5. 52Simplest mole ratio Relative no. of moles No. of moles (mol) Mass (g) OM

101 100 (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 3.2 Derivation of empirical formulae (SB p.47) Answer

102 Derivation of empirical formulae (SB p.47) (c)Mass of Cu = ( – ) g = 1.51 g Mass of O = ( – ) g = g Therefore, the empirical formula of copper(II) oxide is CuO. 11Simplest mole ratio Relative no. of moles No. of moles (mol) Mass (g) OxygenCopper Back

103 102 Compound A contains carbon and hydrogen atoms only. It is found that the compound contains 75 % carbon by mass. Determine its empirical formula. 3.2 Derivation of empirical formulae (SB p.48) Answer

104 Derivation of empirical formulae (SB p.48) Back Let the empirical formula of compound A be C x H y, 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 CH 4. 41Simplest mole ratio Relative no. of moles No. of moles (mol) 2575Mass (g) HydrogenCarbon

105 104 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. 3.2 Derivation of empirical formulae (SB p.48) Answer

106 Derivation of empirical formulae (SB p.48) Back 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 PCl 3. 31Simplest mole ratio Relative no. of moles No. of moles (mol) Mass (g) ChlorinePhosphorus

107 106 (a)Find the empirical formula of vitamin C if it consists of 40.9 % carbon, 54.5 % oxygen and 4.6 % hydrogen by mass. 3.2 Derivation of empirical formulae (SB p.49) Answer (a) Let the mass of vitamin C analyzed be 100 g. The empirical formula of vitamin C is C 3 H 4 O Simplest mole ratio Relative no. of moles No. of moles (mol) Mass (g) OxygenHydrogenCarbon

108 107 (b) Each 325 mg tablet of aspirin consists of mg carbon, 14.6 mg hydrogen and mg oxygen. Determine the empirical formula of aspirin. 3.2 Derivation of empirical formulae (SB p.49) Answer (b) The masses of the elements are multiplied by 1000 first. The empirical formula of aspirin is C 9 H 8 O Simplest mole ratio Relative no. of moles No. of moles (mol) Mass (g) OxygenHydrogenCarbon Back

109 108 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. 3.3 Derivation of molecular formulae (SB p.50) Answer

110 Derivation of molecular formulae (SB p.50) Let the empirical formula of the hydrocarbon be C x H y. Mass of carbon in the hydrocarbon = 14.6 g  = 4.0 g Mass of hydrogen in the hydrocarbon = 9.0 g  = 1.0 g 31Simplest mole ratio Relative no. of moles No. of moles (mol) Mass (g) HydrogenCarbon

111 Derivation of molecular formulae (SB p.50) Therefore, the empirical formula of the hydrocarbon is CH 3. Let the molecular formula of the hydrocarbon be (CH 3 ) n. Relative molecular mass of (CH 3 ) n = 30.0 n  (  3) = 30.0 n = 2 Therefore, the molecular formula of the hydrocarbon is C 2 H 6. Back

112 111 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. 3.3 Derivation of molecular formulae (SB p.50) Answer

113 Derivation of molecular formulae (SB p.50) Let the empirical formula of compound X is C x H y O z, 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 C 6 H 10 O Simplest mole ratio Relative no. of moles No. of moles (mol) Mass (g) OxygenHydrogenCarbon

114 Derivation of molecular formulae (SB p.50) Let the molecular formula of compound X be (C 6 H 10 O 5 )n. Relative molecular mass of (C 6 H 10 O 5 )n = n  (12.0    5) = n = 1 Therefore, the molecular formula of compound X is C 6 H 10 O 5. Back

115 114 The chemical formula of hydrated copper(II) sulphate is known to be CuSO 4 · xH 2 O. It is found that the percentage of water of crystallization by mass in the compound is 36 %. Find the value of x. 3.3 Derivation of molecular formulae (SB p.51) Answer

116 Derivation of molecular formulae (SB p.51) Relative formula mass of CuSO 4 · 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 CuSO 4 · 5H 2 O. Back

117 116 (a)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 3.3 Derivation of molecular formulae (SB p.52)

118 117 (a)(i)Let the mass of compound Z be 100 g. Therefore, the empirical formula of compound Z is CH 2 O. 3.3 Derivation of molecular formulae (SB p.52) CarbonHydrogenOxygen Mass (g) No. of moles (mol) Relative no. of moles Simplest mole ratio 121

119 118 (ii)Let the empirical formula of compound Z be (CH 2 O) n. n  (  )= n= 180 n= 6 Therefore, the molecular formula of compound Z is C 6 H 12 O Derivation of molecular formulae (SB p.52)

120 119 (b)(NH 4 ) 2 S x contains % sulphur by mass. Find the value of x. Answer 3.3 Derivation of molecular formulae (SB p.52) (b) Since the chemical formula of (NH 4 )S x is (NH 4 ) 2 S 3, the value of x is 3. 32Simplest mole ratio Relative no. of moles No. of moles (mol) Mass (g) S(NH 4 ) unit

121 120 (c)In the compound MgSO 4 · nH 2 O, % by mass is water. Find the value of n. Answer 3.3 Derivation of molecular formulae (SB p.52) (c) Since the chemical formula of MgSO 4 · nH 2 O is MgSO 4 · 7H 2 O, the value of n is 7. 71Simplest mole ratio Relative no. of moles No. of moles (mol) Mass (g) H2OH2OMgSO 4 Back

122 121 The chemical formula of ethanoic acid is CH 3 COOH. Calculate the percentage of mass of carbon, hydrogen and oxygen respectively. 3.3 Derivation of molecular formulae (SB p.52) Answer

123 Derivation of molecular formulae (SB p.52) Relative molecular mass of CH 3 COOH = 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

124 123 Calculate the mass of iron in a sample of 20 g of hydrated iron(II) sulphate, FeSO 4 · 7H 2 O. 3.3 Derivation of molecular formulae (SB p.53) Answer Relative formula mass of FeSO 4 · 7H 2 O =  4 + (1.0  )  7 = % by mass of Fe = = % Mass of Fe = 20 g  % = 4.02 g Back

125 124 (a)Calculate the percentages by mass of potassium, chromium and oxygen in potassium dichromate(VI), K 2 Cr 2 O 7. Answer 3.3 Derivation of molecular formulae (SB p.53) (a)Molar mass of K 2 Cr 2 O 7 = (39.1    7) g mol -1 = g mol -1 % by mass of K = = % % by mass of Cr = = % % by mass of O = = %

126 125 (b)Find the mass of metal and water of crystallization in (i)100 g of Na 2 SO 4 · 10H 2 O (ii)70 g of Fe 2 O 3 · 8H 2 O Answer 3.3 Derivation of molecular formulae (SB p.53)

127 Derivation of molecular formulae (SB p.53) (b)(i)Molar mass of Na 2 SO 4 · 10H 2 O = g mol -1 Mass of Na = = g Mass of H 2 O = = g (ii)Molar mass of Fe 2 O 3 · 8H 2 O = g mol-1 Mass of Fe = = g Mass of H 2 O = = g Back

128 127 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 Answer 3.4 Chemical equations (SB p.54) (a)Zn(s) + H 2 O(g)  ZnO(s) + H 2 (g) (b)Mg(s) + 2AgNO 3 (aq)  2Ag(s) + Mg(NO 3 ) 2 (aq) (c)2C 4 H 10 (g) + 13O 2 (g)  8CO 2 (g) + 10H 2 O(l) Back

129 128 Calculate the mass of copper formed when g of copper(II) oxide is completely reduced by hydrogen. 3.5 Calculations based on chemical equations (SB p.55) Answer

130 Calculations based on chemical equations (SB p.55) CuO(s) + H 2 (g)  Cu(s) + H 2 O(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 = 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. Back

131 130 Sodium hydrogencarbonate decomposes according to the following chemial equation: 2NaHCO 3 (s)  Na 2 CO 3 (s) + CO 2 (g) + H 2 O(l) In order to obtain 240 cm 3 of CO 2 at room temperature and pressure, what is the minimum amount of sodium hydrogencarbonate required? (Molar volume of gas at R.T.P. = 24.0 dm 3 mol -1 ) 3.5 Calculations based on chemical equations (SB p.55) Answer

132 Calculations based on chemical equations (SB p.55) Number of moles of CO 2 required = = 0.01 mol From the chemical equation, 2 moles of NaHCO 3 (s) give 1 mole of CO 2 (g). Number of moles of NaHCO 3 required = 0.01  2 = 0.02 mol Mass of NaHCO 3 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. Back

133 132 Calculate the volume of carbon dioxide formed when 20 cm 3 of ethane and 70 cm 3 of oxygen are exploded, assuming all volumes of gases are measured at room temperature and pressure. 3.5 Calculations based on chemical equations (SB p.56) Answer

134 Calculations based on chemical equations (SB p.56) 2C 2 H 6 (g)+7O 2 (g)  4CO 2 (g)+6H 2 O(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 CO 2 : C 2 H 6 is 4 : 2, and the volume ratio of CO 2 : C 2 H 6 should also be 4 : 2 according to the Avogadro’s law. Let x be the volume of CO 2 (g) formed. Number of moles of CO 2 (g) formed : number of moles of C 2 H 6 (g) used = 4 : 2 Volume of CO 2 (g) : volume of C 2 H 6 (g) = 4 : 2 x : 20 cm 3 = 4 : 2 x = 40 cm 3 Therefore, the volume of CO 2 (g) formed is 40 cm 3. Back

135 cm 3 of a gaseous hydrocarbon was mixed with 80 cm 3 of oxygen which was in excess. The mixture was exploded and then cooled. The volume left was 70 cm 3. Upon passing the resulting gaseous mixture through concentrated sodium hydroxide solution (to absorb carbon dioxide), the volume of the residual gas became 50 cm 3. Find the molecular formula of the hydrocarbon. 3.5 Calculations based on chemical equations (SB p.57) Answer

136 Calculations based on chemical equations (SB p.57) Let the molecular formula of the hydrocarbon be C x H y. Volume of hydrocarbon reacted = 10 cm 3 Volume of O 2 (g) unreacted = 50 cm 3 (the residual gas after reaction) Volume of O 2 (g) reacted = (80 – 50) cm 3 = 30 cm 3 Volume of CO 2 (g) formed = (70 – 50) cm 3 = 20 cm 3 C x H y +O 2  xCO 2 + H 2 O 1 mol:mol:x mol 1 volume :volumes:x volumes

137 Calculations based on chemical equations (SB p.57) = x = 2 = = 3 As x = 2, = 3 y = 4 Therefore, the molecular formula of the hydrocarbon is C 2 H 4. Back

138 137 (a)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 dm 3 mol -1 ) Answer 3.5 Calculations based on chemical equations (SB p.58) (a)Mg(s) + 2HCl(aq)  MgCl 2 (aq) + H 2 (g) No. of moles of H 2 = No. of moles of Mg = Volume of H 2 = 2.4 dm 3

139 138 (b)Find the minimum mass of chlorine required to produce 100 g of phosphorus trichloride (PCl 3 ). Answer 3.5 Calculations based on chemical equations (SB p.58)

140 139 (c)20 cm 3 of a gaseous hydrocarbon and 150 cm 3 of oxygen (which was in excess) were exploded in a closed vessel. After cooling, 110 cm 3 of gases remained. After passing the resulting gaseous mixture through concentrated sodium hydroxide solution, the volume of the residual gas became 50 cm 3. Determine the molecular formula of the hydrocarbon. Answer 3.5 Calculations based on chemical equations (SB p.58)

141 Calculations based on chemical equations (SB p.58) (c)C x H y (g) + O 2 (g)  xCO 2 (g) + H 2 O(l) Volume of C x H y used = 20 cm 3 Volume of CO 2 formed = (110 – 50) cm 3 = 60 cm 3 Volume of O 2 used = (150 – 50) cm 3 = 100 cm 3 Volume of C x H y : Volume of CO 2 = 1 : x = 20 : 60 x = 3 Volume of C x H y : Volume of O 2 = 1 : = 20 : 100 = 5

142 Calculations based on chemical equations (SB p.58) (c)As x = 3, = 5 = 2 y = 8 The molecular formula of the hydrocarbon is C 3 H 8.

143 142 (d)Calculate the volume of carbon dioxide formed when 5 cm 3 of methane was burnt completely in excess oxygen, assuming all volumes of gases are measured at room temperature and pressure. Answer 3.5 Calculations based on chemical equations (SB p.58) (d)CH 4 (g)+ 2O 2 (g)  CO 2 (g)+ 2H 2 O(l) 1 mol:2 mol:1 mol:2 mol (from equation) 1 volume:2 volumes:1 volume:- (from Avogadro’s law) 5 cm 3 x cm 3 It can be judged from the equation that the mole ratio of CO 2 : CH 4 is 1 : 1, the volume ratio of CO 2 : CH 4 should also be 1 : 1. = x = 5 cm 3 The volume of CO 2 (g) formed is 5 cm 3. Back

144 cm 3 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 cm 3 of sulphuric(VI) acid had been added. Calculate the molarity of the sodium hydroxide solution. 3.6 Simple titrations (SB p.61) Answer

145 Simple titrations (SB p.61) 2NaOH(aq) + H 2 SO 4 (aq)  Na 2 SO 4 (aq) + 2H 2 O(l) =  Number of moles of NaOH(aq) = Number of moles of H 2 SO 4 (aq) Number of moles of H 2 SO 4 (aq) = mol dm -3  22.5  dm 3 =  mol Number of moles of NaOH(aq) = 2   mol =  mol Molarity of NaOH(aq) = = mol dm -3 Therefore, the molarity of the sodium hydroxide solution was M. Back

146 g of a pure dibasic acid with formula mass of was dissolved in water and made up to cm 3 in a volumetric flask cm 3 of this solution was found to neutralize 28.5 cm 3 of sodium hydroxide solution. (a)Calculate the molarity of the acid solution. 3.6 Simple titrations (SB p.61) Answer (a)Number of moles of acid = = 0.02 mol Molarity of acid solution = = 0.08 M

147 146 (b) If the dibasic acid is represented by H 2 X, write an equation for the reaction between the acid and sodium hydroxide. 3.6 Simple titrations (SB p.61) Answer (b) H 2 X(aq) + 2NaOH(aq)  Na 2 X(aq) + 2H 2 O(l)

148 147 (c) Calculate the molarity of the sodium hydroxide solution. 3.6 Simple titrations (SB p.61) Answer (c)Number of moles of H 2 X =  Number of moles of NaOH 0.08 mol dm -3  25.0  dm 3 =  Molarity of NaOH  28.5  dm 3 Molarity of NaOH = 0.14 M Therefore, the molarity of the sodium hydroxide solution was 0.14 M. Back

149 g of a sample of hydrated sodium carbonate, Na 2 CO 3 · nH 2 O, was dissolved in 100 cm 3 of distilled water in a conical flask M hydrochloric acid was added from a burette, 2 cm 3 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 Na 2 CO 3 · nH 2 O. 3.6 Simple titrations (SB p.62) Answer

150 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 cm 3. Na 2 CO 3 · nH 2 O(s) + 2HCl(aq)  2NaCl(aq) + CO 2 (g) + (n + 1)H 2 O(l) Number of moles of Na 2 CO 3 · nH 2 O =  Number of moles of HCl =  0.10 mol dm -3  30.0  dm n = n = 1 Therefore, the chemical formula of the hydrated sodium carbonate is Na 2 CO 3 · H 2 O. Back

151 150 5 cm 3 of 0.5 M sulphuric(VI) acid was added to 25.0 cm 3 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: 3.6 Simple titrations (SB p.63) Volume of H 2 SO 4 added (cm 3 ) Temperature ( o C)

152 151 (a)Plot a graph of temperature against volume of sulphuric(VI) acid added. 3.6 Simple titrations (SB p.63) Answer

153 152 (b) Calculate the molarity of the potassium hydroxide solution. 3.6 Simple titrations (SB p.63) Answer (b)From the graph, it is found that the equivalence point of the titration is reached when 20 cm 3 of H 2 SO 4 is added. Number of moles of H 2 SO 4 = 0.5 mol dm -3  20  dm 3 = 0.01 mol 2KOH(aq) + H 2 SO 4 (aq)  K 2 SO 4 (aq) + 2H 2 O(l) 2 mol : 1 mol From the equation, mole ratio of KOH(aq) : H 2 SO 4 (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.

154 153 (c) Explain why the temperature rose to a maximum and then fell. 3.6 Simple titrations (SB p.63) 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. Back

155 154 When excess potassium iodide solution (KI) is added to 25.0 cm 3 of acidified potassium iodate solution (KIO 3 ) of unknown concentration, the solution turns brown. This brown solution requires 22.0 cm 3 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. 3.6 Simple titrations (SB p.66) Answer

156 Simple titrations (SB p.66) IO 3 - (aq) + 5I - (aq) + 6H + (aq)  3I 2 (aq) + 3H 2 O(l) … … (1) I 2 (aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq) … … (2) Back From (1), Number of moles of IO 3 - (aq) =  Number of moles of I 2 (aq) From (2), Number of moles of I 2 (aq) =  Number of moles of S 2 O 3 2- (aq) Number of moles of IO 3 - (aq) =  Number of moles of S 2 O 3 2- (aq) Molarity of IO 3 - (aq)  25.0  dm 3 =  0.05 mol dm -3  22.0  dm 3 Molarity of IO 3 - (aq) = 7.33  M Therefore, the molarity of the acidified potassium iodate solution is 7.33  M.

157 Simple titrations (SB p.67) 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 cm 3 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

158 Simple titrations (SB p.67) MnO 4 - (aq) + 8H + (aq) + 5Fe 2+ (aq)  Mn 2+ (aq) + 4H 2 O(l) + 5Fe 3+ (aq) Number of moles of MnO 4 - (aq) : Number of moles of Fe 2+ (aq) = 1 : 5 Number of moles of Fe 2+ (aq) = 5  Number of moles of MnO 4 - (aq) = 5  0.02 mol dm -3  36.5  dm 3 = 3.65  mol Number of moles of Fe dissolved = Number of moles of Fe 2+ formed = 3.65  mol Mass of Fe = 3.65  mol  55.8 g mol -1 = g Back Percentage purity of Fe =  100 % = % Therefore, the percentage purity of the iron wire is %.

159 Simple titrations (SB p.67) (a)5 g of anhydrous sodium carbonate is added to 100 cm 3 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 dm 3 mol -1 ) Answer

160 Simple titrations (SB p.67) No. of moles of Na 2 CO 3 used = = mol No. of moles of HCl used = 2 M  = 0.2 mol Since HCl is in excess, Na 2 CO 3 is the limiting agent. No. of moles of CO 2 produced = No. of moles of Na 2 CO 3 used = mol Volume of CO 2 produced = mol  24.0 dm 3 mol -1 = dm 3 Na 2 CO 3 (s) + 2HCl(aq)  2NaCl(aq) + H 2 O(l) + CO 2 (g)

161 Simple titrations (SB p.67) (b) 8.54 g of impure hydrated iron(II) sulphate (formula mass of ) was dissolved in water and made up to cm cm 3 of this solution required cm 3 of M acidified potassium manganate(VII) solution for complete reaction. Determine the percentage purity of the hydrated iron(II) sulphate. Answer

162 Simple titrations (SB p.67) Back No. of moles of MnO 4 - ions = M  =  mol (b) MnO 4 - (aq) + 8H + (aq) + 5Fe 2+ (aq)  Mn 2+ (aq) + 4H 2 O(l) + 5Fe 3+ (aq) % purity of FeSO 4 = = % No. of moles of Fe 2+ ions = 5  No. of moles of MnO 4 - ions =  mol No. of moles of Fe 2+ ions in 25.0 cm 3 solution =  mol No. of moles of Fe 2+ ions in cm 3 solution = mol Molar mass of hydrated FeSO 4 = g mol -1 Mass of hydrated FeSO 4 = mol  g mol -1 = 8.26 g


Download ppt "1 Chemical Equations and Stoichiometry 3.1Formulae of Compounds 3.2Derivation of Empirical Formulae 3.3Derivation of Molecular Formulae 3.4Chemical Equations."

Similar presentations


Ads by Google