Revision of Important Concepts. 1. Types of Bonding

Transcription

1 Revision of Important Concepts 1. Types of Bonding

2 Electronegativity (EN) often molecular often ionic compounds

3 Bonding in chemical substances Bond energy: Is the energy that is released when a bond is formed Dissociation energy: Is the energy necessary to break a bond Bond Type Bond Energy Examples Ionic bonds KJ/mol NaCl, CsF, MgO (salts) Covalent KJ/mol H 2, HF, O 2, S 8 (molecules) Metallic KJ/mol Ag, Cu 3 Au, Pb (metals)

4 Bonding in chemical substances Weaker Bonds Type Bond Energy (KJ/mol) Example Hydrogen bonding < 30 HF; H 2 O; NH 3 ; CH 3 CO 2 H van der Waals < 3 He, Ne, hydrocarbons Interactions between positively and negatively polarised centres Permanent dipoles (Hbond) Interactions between temporary positively and negatively polarised centres Induced dipoles (van der Waals bond)

5 Revision of Important Concepts 2. Stoichiometry Derived from the Greek stoicheion ( element ) and metron ( measure ) The total mass of all substances present after a chemical reaction is the same as the total mass before the reaction.

6 Counting Objects of Fixed Relative Mass Oxygen g 55.85g Fe = x atoms Fe 32.07g S = x atoms S CaCO g Water g Copper g

7 Number of moles (n) = Mass (g) Molar Mass (g mol1) M = m/n n = m M m = n M No. of entities = no. of moles x 6.022x1023 entities 1 mol 1 mole of an ideal gas: 22.4l

8 Mass % Mass % of element A in a compound = mass of A mass of compound x 100

9 Converting a Concentrated Solution to a Dilute Solution molarity or concentration Fundamentals of Solution Stoichiometry Solute: The substance that dissolves in the solvent Solvent: The substance in which the solute dissolves

10 Molarity is the number of moles of solute per litre of solution. c = Number of moles Volume mol L = M also called: concentration: Symbol: c or [ ] (sometimes M) c(h + ) = 5 mol/l = 5M [H + ] = 5 mol/l = 5M c = n V

11 Sample Problem Calculating Mass of Solute in a Given Volume of Solution PROBLEM: How many grams of solute are in 1.75L of 0.460M sodium monohydrogen phosphate? c = n/v n n = m/m m SOLUTION: 1.75L mol 1 L = mol Na 2 HPO mol g/mol = 114g Na 2 HPO 4

12 Revision of Important Concepts 3. Reactions Limiting Reagents and Yields Precipitation Reactions Acid Base Reactions Redox reactions

13 Yield of a Reaction Theoretical is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield x 100

14 . yield of our previous experiment 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe 4.59 mol Maximum (theoretical) yield of Fe: 4.59 mol In this experiment we obtain: 241g Fe Theoretical yield: Actual yield: 4.59 mol x g/mol = g [m = n x M(Fe)] 241g Fe % Yield of this Experiment: (241g/256g)x100 = 94.1% 94% Same result if you calculate the yield from moles: 241g Fe: 241g/55.85 gmol 1 = 4.32 mol % yield: (4.32 mol/ 4.59 mol) x 100 = 94.1% 94%

15 Important classes of chemical reactions a) Precipitation Reactions AgCl PbI 2

16 Precipitation Reactions Precipitate insoluble solid that separates from solution molecular equation ionic equation Pb(NO 3 ) 2 (aq) + 2NaI (aq) PbI 2 (s) + 2NaNO 3 (aq) precipitate Pb NO 3 + 2Na + + 2I PbI 2 (s) + 2Na + + 2NO 3 net ionic equation Pb I PbI 2 (s) Na + and NO 3 are spectator ions

17 Solubility Rules For Ionic Compounds in Water Soluble Ionic Compounds 1. All common compounds of group 1A(1) ions (Li +, Na +, K +, etc.) and ammonium ion (NH 4+ ) are soluble. 2. All common nitrates (NO 3 ), acetates (CH 3 COO ) and most perchlorates (ClO 4 ) are soluble. 3. All common chlorides (Cl ), bromides (Br ) and iodides (I ) are soluble, except those of Ag +, Pb 2+, Cu +, and Hg Insoluble Ionic Compounds 1. All common metal hydroxides are insoluble, except those of group 1A(1) and the larger members of group 2A(2)(beginning with Ca 2+ ). 2. All common carbonates (CO 3 2 ) and phosphates (PO 4 3 ) are insoluble, except those of group 1A(1) and NH All common sulfides are insoluble except those of group 1A(1), group 2A(2) and NH 4+.

18 Acid Strength Base Strength AcidBase Theory of Brönsted (1923) Acid: donate protons (H + ) Base: accept protons (H + ) Corresponding acid and base Acid 1 Base 1 Acid 2 Base 2 HCl + H 2 O H 3 O + + Cl H 2 SO 4 + H 2 O H 3 O + + HSO 4 HSO 4 + H 2 O H 3 O + + SO 4 2 NH H 2 O H 3 O + + NH 3 HCO 3 + H 2 O H 3 O + + CO 3 2

19 An acidbase titration Solution with known concentration Solution with an unknown concentration HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O neutral point addition of base

20 The redox process in compound formation

21 A summary of terminology for oxidationreduction (redox) reactions e X transfer or shift of electrons Y X loses electron(s) X is oxidized X is the reducing agent X increases its oxidation number Y gains electron(s) Y is reduced Y is the oxidizing agent Y decreases its oxidation number

22 Rules for Assigning an Oxidation Number (O.N.) General rules 1. For an atom in its elemental form (Na, O 2, Cl 2, etc.): O.N. = 0 2. For a monoatomic ion: O.N. = ion charge 3. The sum of O.N. values for the atoms in a compound equals zero. The sum of O.N. values for the atoms in a polyatomic ion equals the ion s charge. Rules for specific atoms or periodic table groups 1. For Group 1A(1): O.N. = +1 in all compounds 2. For Group 2A(2): O.N. = +2 in all compounds 3. For hydrogen: O.N. = +1 in combination with nonmetals 4. For fluorine: O.N. = 1 in combination with metals and boron 5. For oxygen: O.N. = 1 in peroxides O.N. = 2 in all other compounds(except with F) 6. For Group 7A(17): O.N. = 1 in combination with metals, nonmetals (except O), and other halogens lower in the group

23 Exercise Permanganate ions react with bromide ions in basic solution to form MnO 2 and BrO 3. Write the balanced equation for the reaction. MnO 4 + Br MnO 2 + BrO 3 O.N. Br Br = 1 BrO 3 Br = +5 O.N. MnO 4 Mn = +7 MnO 2 Mn = +4 Half reactions: MnO 4 +3e MnO 2 Reduction 1 +5 Br BrO 3 +6e Oxidation Balance the equations and add the two half reaction together.

24 continued Reduction: Oxidation: MnO 4 +3e MnO 2 Br BrO 3 +6e Charge: 4 0 Charge: 1 7 Reduction: MnO 4 +3e MnO 2 + 4OH Oxidation: Br + 6OH BrO 3 +6e Reduction: MnO 4 +3e + 2H 2 O MnO 2 + 4OH Oxidation: Br + 6OH BrO 3 +6e + 3H 2 O

25 continued Reduction: MnO 4 +3e + 2H 2 O MnO 2 + 4OH x2 Oxidation: Br + 6OH BrO 3 +6e + 3H 2 O Reduction: Oxidation: 2MnO 4 +6e + 4H 2MnO 2 + 8OH 2 O 2 Br + 6OH BrO 3 +6e + 3H 2 O 2MnO 4 (aq) + Br (aq) + H 2 O(l) 2MnO 2 (s) + BrO 3 (aq) + 2OH (aq)

26 Revision of Important Concepts 3. Past Questions

27 Oxidation States: KI 1+ I = 0 I=1 I 2 I =0 KIO 3 3x(2) + 1+ I = 0 I=+5 S 2 O 2 3 3x(2) + 2xS = 2 S=+2 S 2 O 2 4 4x(2) + 2xS = 2 S=+3

28 5e +5 0 Reduction: IO 3 I 2 2IO 3 +10e I 2 Charge IO 3 +10e + 12H + I 2 2IO 3 +10e + 12H + I 2 + 6H 2 O Oxidation: 1e 1 0 2I I 2 + 2e

29 Overall: 2IO 3 +10e + 12H + I 2 + 6H 2 O 2I I 2 + 2e x5 2IO I + 12H + 6 I 2 + 6H 2 O IO 3 + 5I + 6H + 3 I 2 + 3H 2 O

30 +2 +3 S 2 O 2 3 S 2 O e Charge 2 4 S 2 O 3 2 S 2 O e + 2H + S 2 O H 2 O S 2 O e + 2H + I 2 + 2e 2I S 2 O H 2 O + I 2 S 2 O I + 2H +

31 IO 3 + 5I + 6H + 3 I 2 + 3H 2 O S 2 O H 2 O + I 2 S 2 O I + 2H + x3 IO 3 + 5I 6H + + 3S 2 O H 2 O + 3 I 2 3 I 2 + 3H 2 O + 3S 2 O I + 6H + Overall: IO 3 + 3S 2 O 3 2 3S 2 O I

32 KIO 3 M= g/mol 2.075g/214gmol 1 = 9.696x10 3 mol in 1l / x10 4 mol in 50 ml From the reaction equation: ratio: 3 (S 2 O 3 2 ) : 1 (IO 3 ) 1.45x10 3 mol in ml n = m/m c = n/v c = n / V = 0.04 M

33 Sulfur S and sulfate, SO 4 2

34

35 Balancing Redox Equations Oxalate is oxidised by the permanganate ion MnO 4 in acidic solution. During the reaction Mn 2+ and CO 2 are formed. MnO 4 (aq) + C 2 O 4 2 (aq) Mn 2+ (aq) + CO 2 (g) Calculate the oxidation numbers: MnO 4 Mn + 4(2) = 1 Mn = +7 C 2 O 2 2C + 4(2) = 2 4 C = +3 CO 2 C + 2(2) = 0 C = +4 O C O

36 Balancing Redox Equations Half Equations +7 MnO 4 (aq) +2 Mn 2+ (aq) +3 H 2 C 2 O 4 (aq) +4 2CO 2 (g) Reduction Oxidation 1 st step: Identify the oxidation and the reduction and then balance the total oxidation numbers with electrons (Red: electrons on the left side, Ox: electrons on the right side) Reduction: Oxidation: MnO 4 + 5e Mn 2+ 2(+3) 2(+4) C 2 O 2 4 2CO 2 + 2e

37 2 nd step: Here the reaction is performed under acidic conditions. So balance the charge of the half equations with protons. (If a reaction occurs under basic conditions OH ions are used to balance the equation). Reduction: Oxidation: MnO 4 + 5e + 8H + Mn 2+ charge: 6 charge: +2 2(+3) +4 C 2 O 2 4 2CO 2 + 2e 3 rd step: Balance with water, so that you obtain proper half equations. Reduction: MnO 4 + 5e + 8H + Mn H 2 O Oxidation: C 2 O 4 2 2CO 2 + 2e

38 4 th step: Multiply the equations to have the same number of electrons on each side. Simplify and add the equations. Reduction: MnO 4 + 5e + 8H + Mn H 2 O x 2 Oxidation: H 2 C 2 O 4 2CO 2 + 2e x 5 Reduction: Oxidation: 2MnO e + 16H + 2 Mn H 2 O 5C 2 O CO e Redox: 2 MnO 4 + 5C 2 O H + 2Mn CO 2 + 8H 2 O

39 A redox titration known concentration unknown concentration all the oxalic acid used for the reduction

40 25ml 0.02M KMnO 4 C= n/v n=c x V n= mol 2:5 ratio according to the reaction equation Redox: 2 MnO 4 + 5C 2 O H + 2Mn CO 2 + 8H 2 O n= mol MnO 4 n= mol C 2 O 4 2 M(C 2 O 4 2 ) = g/mol n=m/m m= nxm m = 0.11g Weight% of oxalate ligand in complex: 0.11g/0.18g = 61.1%

41 n(co 2 ) = 3.52g/44g mol 1 = 0.08 mol n(h 2 O) = 1.44g/18g mol 1 = 0.08 mol m = nxm = 0.08 mol x 12g mol 1 = 0.96g C m = nxm = 2x 0.08 mol x 1g mol 1 = 0.16g H = 1.28g O n = m/m = 1.28g/ 16 = 0.08 mol Empirical Formula: CH 2 O Molecular Formula: (CH 2 O) n M= 60 g/mol M(CH 2 O) = = 30 g/mol n=2 C 2 H 4 O 2 or CH 3 COOH