Revision of Important Concepts

1. Types of Bonding

Electronegativity (EN)

often ionic compounds

often molecular

Bond energy: Is the energy that is released when a bond is formed

Dissociation energy: Is the energy necessary to break a bond

Ionic bonds 400-700KJ/mol NaCl, CsF, MgO (salts)

Covalent 100-400KJ/mol H2, HF, O2, S8 (molecules)

Metallic 100-400KJ/mol Ag, Cu3Au, Pb (metals)

Bond Energy

Bonding in chemical substances

Bond Type Examples

Weaker Bonds

Hydrogen bonding < 30 HF; H2O; NH3; CH3CO2H

van der Waals < 3 He, Ne, hydrocarbons

Interactions between positively and negatively polarised centres

Permanent dipoles (H-bond)

Interactions between temporary positively and

negatively polarised centres

Induced dipoles (van der Waals bond)

Bonding in chemical substances

Type Bond Energy (KJ/mol) Example

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

55.85g Fe = 6.022 x 1023 atoms Fe

32.07g S = 6.022 x 1023 atoms S

Oxygen

32.00 g

Water

18.02 g

Copper

63.55 g

CaCO3

100.09 g

n = M

m m = n · M M = m/n

Number of moles (n) =

Molar Mass (g mol-1)

Mass (g)

No. of entities = no. of moles x 6.022x1023 entities

1 mol

1 mole of an ideal gas: 22.4l

Mass % of element A in a compound =

mass of A

mass of compound x 100

Mass - %

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

c or [ ] (sometimes M)

c(H+) = 5 mol/L = 5M

[H+] = 5 mol/L = 5M

also called: concentration: Symbol:

c = n

V

= M

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?

SOLUTION: 1.75L 0.460 mol

1 L

0.805mol 141.96g/mol

= 0.805 mol Na2HPO4

= 114g Na2HPO4

n = m/M c = n/V n m

Revision of Important Concepts

3. Reactions

Limiting Reagents and Yields

Precipitation Reactions

Acid – Base Reactions

Redox reactions

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 a Reaction

…. yield of our previous experiment

4.59 mol x 55.85 g/mol = 256.35 g [m = n x M(Fe)] Theoretical yield:

241g Fe Actual yield:

% 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%

2Al + Fe2O3 Al2O3 + 2Fe

4.59 mol Maximum (theoretical)

yield of Fe: 4.59 mol

In this experiment we obtain: 241g Fe

a) Precipitation Reactions

Important classes of chemical reactions

PbI2

AgCl

Precipitation Reactions

Precipitate – insoluble solid that separates

from solution

molecular

equation

ionic

equation

net

ionic equation

Pb2+ + 2NO3- + 2Na+ + 2I- PbI2 (s) + 2Na+ + 2NO3

-

Na+ and NO3- are

spectator ions

Pb(NO3)2 (aq) + 2NaI (aq) PbI2 (s) + 2NaNO3 (aq)

precipitate

Pb2+ + 2I- PbI2 (s)

Solubility Rules For Ionic Compounds in Water

1. All common compounds of group 1A(1) ions (Li+, Na+, K+, etc.) and

ammonium ion (NH4+) are soluble.

2. All common nitrates (NO3-), acetates (CH3COO–) and most

perchlorates (ClO4-) are soluble.

3. All common chlorides (Cl-), bromides (Br-) and iodides (I-) are soluble,

except those of Ag+, Pb2+, Cu+, and Hg22+.

1. All common metal hydroxides are insoluble, except those of group 1A(1)

and the larger members of group 2A(2)(beginning with Ca2+).

2. All common carbonates (CO32-) and phosphates (PO4

3-) are insoluble,

except those of group 1A(1) and NH4+.

3. All common sulfides are insoluble except those of group 1A(1), group

2A(2) and NH4+.

Soluble Ionic Compounds

Insoluble Ionic Compounds

Acid-Base Theory of Brönsted (1923)

Acid: donate protons (H+)

Base: accept protons (H+)

HCl + H2O H3O+ + Cl–

H2SO4 + H2O H3O+ + HSO4

–

HSO4– + H2O H3O

+ + SO42–

NH4+ + H2O H3O

+ + NH3

HCO3– + H2O H3O

+ + CO32–

Acid

Str

en

gth

Ba

se

Str

en

gth

Acid 1 Base 1 Base 2 Acid 2

Corresponding acid and base

An acid-base titration

Solution with

an unknown

concentration

Solution with

known

concentration

neutral point

addition of base

HCl (aq) + NaOH (aq) NaCl (aq) + H2O

The redox process in compound formation

A summary of terminology for oxidation-

reduction (redox) reactions

X Y

e–

transfer

or shift of

electrons

X loses electron(s) Y gains electron(s)

X is oxidized Y is reduced

X is the reducing agent Y is the oxidizing agent

X increases its

oxidation number

Y decreases its

oxidation number

Rules for Assigning an Oxidation Number (O.N.)

1. For an atom in its elemental form (Na, O2, Cl2, 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.

General rules

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

6. For Group 7A(17): O.N. = -1 in combination with metals, nonmetals

(except O), and other halogens lower in the group

5. For oxygen: O.N. = -1 in peroxides

O.N. = -2 in all other compounds(except with F)

Permanganate ions react with bromide ions in basic solution to

form MnO2 and BrO3-. Write the balanced equation for the reaction.

Exercise

O.N.

MnO4- Mn = +7

MnO2 Mn = +4

MnO4- MnO2 +3e-

Br- BrO3-

O.N.

Br- Br = -1

BrO3- Br = +5

+6e-

Balance the equations and add the two half reaction together.

Half reactions:

Reduction

Oxidation

MnO4- MnO2 + BrO3

– + Br-

+7 +4

-1 +5

MnO4- MnO2 +3e-

Br- BrO3- +6e-

Charge: –4 0

Charge: –1 –7

MnO4- MnO2 +3e-

Br– BrO3- +6e-

+ 4OH–

+ 6OH–

MnO4- MnO2 +3e-

Br– BrO3- +6e-

+ 4OH–

+ 6OH– + 3H2O

+ 2H2O

Reduction:

Reduction:

Reduction:

Oxidation:

Oxidation:

Oxidation:

continued

MnO4- MnO2 +3e-

Br– BrO3- +6e-

+ 4OH–

+ 6OH– + 3H2O

+ 2H2O Reduction:

Oxidation:

x2

2MnO4- 2MnO2 +6e-

Br– BrO3- +6e-

+ 8OH–

+ 6OH– + 3H2O

+ 4H2O Reduction:

Oxidation:

2

2MnO4-(aq) + Br-(aq) + H2O(l) 2MnO2(s) + BrO3

- (aq) + 2OH-(aq)

continued

Revision of Important Concepts

3. Past Questions

Oxidation States:

S2O32-

3x(-2) + 2xS = -2 S=+2

KIO3 3x(-2) + 1+ I = 0 I=+5

KI 1+ I = 0 I=-1

I2 I =0

S2O42-

4x(-2) + 2xS = -2 S=+3

IO3- I2

Reduction: +5 0

5e-

2IO3- I2 +10e- Charge -12 ↔ 0

2IO3- + 12H+ I2 +10e-

2IO3- + 12H+ I2 + 6H2O +10e-

2I- I2 + 2e-

Oxidation:

-1 0

-1e-

2IO3- + 12H+ I2 + 6H2O +10e-

2I- I2 + 2e- x5

2IO3- + 12H+ 6 I2 + 6H2O + 10I-

IO3- + 6H+ 3 I2 + 3H2O + 5I-

Overall:

S2O32- S2O4

2- +2e- + 2H+

I2 + 2e- 2I-

S2O32- S2O4

2- +2e- +2 +3

Charge -2 ↔ -4

S2O32- + H2O S2O4

2- +2e- + 2H+

S2O32- + H2O + I2 S2O4

2- + 2I- + 2H+

IO3- + 6H+ 3 I2 + 3H2O + 5I-

S2O32- + H2O + I2 S2O4

2- + 2I- + 2H+ x3

IO3- 6H+ + 3S2O3

2- + 3H2O + 3 I2

3 I2 + 3H2O + 3S2O42- + 6I- + 6H+

+ 5I-

Overall: IO3- + 3S2O3

2- 3S2O42- + I-

KIO3 M= 214.00 g/mol /20 4.848x10-4 mol in 50 mL From the reaction equation: ratio: 3 (S2O3

2-) : 1 (IO3-)

1.45x10-3 mol in 36.26 mL c = n / V = 0.04 M

2.075g/214gmol-1 = 9.696x10-3 mol in 1l

c = n/V

n = m/M

Sulfur S and sulfate, SO42-

Balancing Redox Equations

Oxalate is oxidised by the permanganate ion MnO4- in acidic solution. During

the reaction Mn2+ and CO2 are formed.

MnO4- (aq) + C2O4

2-(aq) Mn2+(aq) + CO2(g)

Calculate the oxidation numbers:

MnO4- Mn + 4(-2) = -1

C2O42-

C = +3

Mn = +7

CO2 C + 2(-2) = 0 C = +4

C O O

2C + 4(-2) = -2

Balancing Redox Equations – Half Equations

MnO4- (aq) Mn2+(aq)

H2C2O4(aq) 2CO2(g)

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)

+7 +2 Reduction

Oxidation +3 +4

1st step:

MnO4– Mn2+

+7 +2

+ 5e–

C2O4 2- 2CO2 + 2e– 2(+3) 2(+4)

Reduction:

Oxidation:

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).

MnO4– Mn2+

+7 +2

+ 5e–

C2O4 2- 2CO2 + 2e–

2(+3) +4

charge: -6 charge: +2

+ 8H+

Balance with water, so that you obtain “proper” half equations.

MnO4– Mn2+ + 5e– + 8H+ + 4H2O

2nd step:

3rd step:

Reduction:

Oxidation:

Reduction:

Oxidation: C2O4 2- 2CO2 + 2e–

2 MnO4– + 5C2O4

2- + 16H+ 2Mn2+ + 10CO2 + 8H2O

MnO4– Mn2+ + 5e–

H2C2O4 2CO2 + 2e–

+ 8H+ + 4H2O x 2

x 5

2MnO4– 2 Mn2 + + 10e–

5C2O4 2- 10CO2 + 10e–

+ 16H+ + 8 H2O

Oxidation:

Reduction:

Oxidation:

Reduction:

Redox:

4th step: Multiply the equations to have the same number of electrons on each side. Simplify and add the equations.

A redox titration

known

concentration

unknown

concentration

all the

oxalic acid

used for the

reduction

25ml 0.02M KMnO4

C= n/V n=c x V n= 0.0005 mol

2:5 ratio according to the reaction equation

n= 0.0005 mol MnO4- n= 0.00125 mol C2O4

2-

M(C2O42-) = 88.02 g/mol n=m/M m= nxM

m = 0.11g

2 MnO4– + 5C2O4

2- + 16H+ 2Mn2+ + 10CO2 + 8H2O Redox:

Weight-% of oxalate

ligand in complex:

0.11g/0.18g = 61.1%

n(CO2) = 3.52g/44g mol-1 = 0.08 mol

n(H2O) = 1.44g/18g mol-1 = 0.08 mol

2.4-0.96-0.16= 1.28g O n = m/M = 1.28g/ 16 = 0.08 mol

M(CH2O) = 12 + 2+ 16 = 30 g/mol n=2

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

Empirical Formula: CH2O

Molecular Formula: (CH2O)n M= 60 g/mol

C2H4O2 or

CH3COOH