Revision of Important Concepts 1. Types of Bonding
Electronegativity (EN) often molecular often ionic compounds
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 400700KJ/mol NaCl, CsF, MgO (salts) Covalent 100400KJ/mol H 2, HF, O 2, S 8 (molecules) Metallic 100400KJ/mol Ag, Cu 3 Au, Pb (metals)
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)
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.
Counting Objects of Fixed Relative Mass Oxygen 32.00 g 55.85g Fe = 6.022 x 10 23 atoms Fe 32.07g S = 6.022 x 10 23 atoms S CaCO 3 100.09 g Water 18.02 g Copper 63.55 g
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
Mass % Mass % of element A in a compound = mass of A mass of compound x 100
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
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
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 0.460 mol 1 L = 0.805 mol Na 2 HPO 4 0.805mol 141.96g/mol = 114g Na 2 HPO 4
Revision of Important Concepts 3. Reactions Limiting Reagents and Yields Precipitation Reactions Acid Base Reactions Redox reactions
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
. 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 55.85 g/mol = 256.35 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%
Important classes of chemical reactions a) Precipitation Reactions AgCl PbI 2
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 2+ + 2NO 3 + 2Na + + 2I PbI 2 (s) + 2Na + + 2NO 3 net ionic equation Pb 2+ + 2I PbI 2 (s) Na + and NO 3 are spectator ions
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 2 2+. 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 4+. 3. All common sulfides are insoluble except those of group 1A(1), group 2A(2) and NH 4+.
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 4 + + H 2 O H 3 O + + NH 3 HCO 3 + H 2 O H 3 O + + CO 3 2
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
The redox process in compound formation
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
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
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: +7 +4 MnO 4 +3e MnO 2 Reduction 1 +5 Br BrO 3 +6e Oxidation Balance the equations and add the two half reaction together.
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
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)
Revision of Important Concepts 3. Past Questions
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
5e +5 0 Reduction: IO 3 I 2 2IO 3 +10e I 2 Charge 12 0 2IO 3 +10e + 12H + I 2 2IO 3 +10e + 12H + I 2 + 6H 2 O Oxidation: 1e 1 0 2I I 2 + 2e
Overall: 2IO 3 +10e + 12H + I 2 + 6H 2 O 2I I 2 + 2e x5 2IO 3 + 10I + 12H + 6 I 2 + 6H 2 O IO 3 + 5I + 6H + 3 I 2 + 3H 2 O
+2 +3 S 2 O 2 3 S 2 O 2 4 +2e Charge 2 4 S 2 O 3 2 S 2 O 4 2 +2e + 2H + S 2 O 3 2 + H 2 O S 2 O 4 2 +2e + 2H + I 2 + 2e 2I S 2 O 3 2 + H 2 O + I 2 S 2 O 4 2 + 2I + 2H +
IO 3 + 5I + 6H + 3 I 2 + 3H 2 O S 2 O 3 2 + H 2 O + I 2 S 2 O 4 2 + 2I + 2H + x3 IO 3 + 5I 6H + + 3S 2 O 2 3 + 3H 2 O + 3 I 2 3 I 2 + 3H 2 O + 3S 2 O 4 2 + 6I + 6H + Overall: IO 3 + 3S 2 O 3 2 3S 2 O 4 2 + I
KIO 3 M= 214.00 g/mol 2.075g/214gmol 1 = 9.696x10 3 mol in 1l /20 4.848x10 4 mol in 50 ml From the reaction equation: ratio: 3 (S 2 O 3 2 ) : 1 (IO 3 ) 1.45x10 3 mol in 36.26 ml n = m/m c = n/v c = n / V = 0.04 M
Sulfur S and sulfate, SO 4 2
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
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: +7 +2 MnO 4 + 5e Mn 2+ 2(+3) 2(+4) C 2 O 2 4 2CO 2 + 2e
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: +7 +2 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 2+ + 4H 2 O Oxidation: C 2 O 4 2 2CO 2 + 2e
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 2+ + 4H 2 O x 2 Oxidation: H 2 C 2 O 4 2CO 2 + 2e x 5 Reduction: Oxidation: 2MnO 4 + 10e + 16H + 2 Mn 2 + + 8 H 2 O 5C 2 O 2 4 10CO 2 + 10e Redox: 2 MnO 4 + 5C 2 O 4 2 + 16H + 2Mn 2+ + 10CO 2 + 8H 2 O
A redox titration known concentration unknown concentration all the oxalic acid used for the reduction
25ml 0.02M KMnO 4 C= n/v n=c x V n= 0.0005 mol 2:5 ratio according to the reaction equation Redox: 2 MnO 4 + 5C 2 O 4 2 + 16H + 2Mn 2+ + 10CO 2 + 8H 2 O n= 0.0005 mol MnO 4 n= 0.00125 mol C 2 O 4 2 M(C 2 O 4 2 ) = 88.02 g/mol n=m/m m= nxm m = 0.11g Weight% of oxalate ligand in complex: 0.11g/0.18g = 61.1%
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 2.40.960.16= 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) = 12 + 2+ 16 = 30 g/mol n=2 C 2 H 4 O 2 or CH 3 COOH