3RD TERM SS2 CHEMISTRY SCHEME OF WORK AND NOTE

THIRD TERM E-LEARNING NOTE

SUBJECT: CHEMISTRY  CLASS: SSS 2

SCHEME OF WORK

WEEK TOPIC

1. Water

• Sources, Types, Uses and Structure of Water.
• Laboratory Preparation of Water.
• Test for Water
• Causes/ Removal of Hardness of Water.
• Purification of Water for Municipal Supply.

1. Solubility and Solutions

• Definition of Terms.
• Calculations based on Solubility.
• Solubility Curves.
• Uses of Solubility.

1. Mass/Volume Relationship

• Mole and Molar Quantities
• Relative Atomic Mass and Relative Molecular Mass.
• Calculations involving Mass and Volume.

1. Acid/ Base Reactions

• Preparation of Standard Solutions.
• Indicators
• Calculations based on Acid-Base Titration.

1. Hydrocarbons

• Unique Nature of Carbon.
• Characteristics Features of Organic Compounds
• Classification of Hydrocarbons.
• Definition of Terms used in Organic Chemistry
  

1. Saturated Hydrocarbon (Alkanes)

• Nomenclature
• Preparation, Properties and Uses

1. Unsaturated Hydrocarbon (Alkenes)

• Nomenclature
• Preparation, Properties and Uses

1. Unsaturated Hydrocarbon (Alkynes)

• Nomenclature
• Preparation, Properties and Uses

  Aromatic Hydrocarbons

  Also Read:

  • 3RD TERM SS2 AGRICULTURAL SCIENCE SCHEME OF WORK AND NOTE
  • 3RD TERM SS2 BIOLOGY SCHEME OF WORK AND NOTE
• Benzene Structure
• Preparation, Properties and Uses

1. Alkanols

• Types and Classes
• Industrial Production by Fermentation
• Properties and Uses

REFERENCE MATERIALS

• New School Chemistry for Senior Secondary Schools by O. Y. Ababio
• New System Chemistry for Senior Secondary Schools by T. Y. Toon et al
• S.S.C.E Past Questions and Answers on Chemistry
• U.T.M.E Past Questions and Answers on Chemistry

WEEK ONE

TOPIC: WATER

CONTENT

• Sources, Types, Uses and Structure of Water.
• Laboratory Preparation of Water.
• Test for Water
• Causes/ Removal of Hardness of Water.
• Purification of Water for Municipal Supply.

   

  WATER
  

Water is regarded as the universal solvent. It is a good solvent for many substances.

SOURCES OF WATER

The following are the sources of water:

1. Natural water:Rainwater, Well water, Spring water and Sea water
   1. Treated water: Distilled water, Pipe – borne water and chlorinated water.

       

      TYPES OF WATER
      

Water is of two types namely: soft water and hard water. Soft water forms lather with soap easily while hard water does not form lather readily with soap since it contains some dissolved salt in it.

STRUCTURE OF WATER

In a molecule of water, H₂O, the central atom is Oxygen. Oxygen has the following electronic configuration: 1s² 2s² 2p⁴.

The valence shell of oxygen has two lone pairs of electrons (2s²2p²) and two unpaired electrons (2p_y¹2p_z¹). Each unpaired electron forms a covalent bond with an electron from a hydrogen atom. The water molecule has two lone pairs and two bond pairs of electrons in the valence shell of its central atom, thereby satisfying the octet rule for stability.

Ideally, the four electron pairs should be directed towards corners of a tetrahedron. However, when lone pairs of electrons is located near another lone pair, the repulsion between them is so great that they squeeze the other two bond pairs of electrons closer together. As a result, the bond angle in water is compressed to approximately 105^o, such that the structure of the water molecule is V-shaped or angular shape.

O

H H

LABORATORY PREPARATION OF WATER  

To prepare water in laboratory, dry hydrogen gas is ignited in air. It burns with a faint blue flame to give steam, which will condense on contact with any cold surface to form water.

PHYSICAL PROPERTIES OF WATER

1. Water boils at 100^oC and freezes at 0^oC
2. It has a maximum density of 1gcm^-3 at 4^oC
3. It is neutral to litmus.

    

CHEMICAL PROPERTIES

1. Water reacts with electropositive metals to form alkali and liberate hydrogen gas. E.g

   Na_(s) + H₂O_(aq)  NaOH_(aq) + H_2(g)

 Mg & Zn react with steam

 Cu, Au, Ag, Hg do not react with water to form alkaline solution

1. Non-metal like chlorine reacts with water to form acid solution.

   H₂0_(aq) + Cl_2(g) HCl_(aq) + HOCl_(aq)

TEST FOR WATER

When few drops of water are added to

1. White anhydrous copper (II) tetraoxosulphate (VI), it turns blue.
2. Blue cobalt (II) chloride, it turns pink.

NOTE: These two tests are not specific for water. They only indicate the presence of water. Any aqueous solution or substance containing water will give a positive test for water.

EVALUATION

1. Describe the structure of water.
2. How will you identify a give solution to be water?

    

HARDNESS OF WATER

Hard water is the water that does not form lather readily with soap.

Water acquired hardness when insoluble salts of CaSO₄, MgSO₄ and Ca(HCO₃)₂ dissolves in it from the soil which it flows through.

TYPES OF HARDNESS OF WATER

1. Temporary hard water
2. Permanent hard water

TEMPORARY HARDNESS: This is caused by the presences of Ca²⁺ and Mg²⁺ in the form of hydrogen trioxocarbonate IV i.e. Ca(HCO₃)₂

REMOVAL OF TEMPORARY HARDNESS

1. Physical method: By boiling

   Ca(HCO₃)_2(aq)^heat   CaCO_3(s) + H₂O_(l) + CO_2(g)

2. Chemical method: By using of slaked lime (calcium hydroxide solution)

   Ca(HCO₃)_2(aq) + Ca(OH)_2(aq) 2CaCO_3(s)+  2H₂O_(l)

EFFECTS OF TEMPORARY HARDNESS: It causes

1. Furring of kettles and boilers.
2. Stalagmite and stalactites in caves.

    

PERMANENT HARDNESS

Permanent hardness in water is caused by the presence of Calcium and Magnesium ions in the form of soluble tetraoxosulphate (VI) and chlorides (i.e. CaSO₄, MgSO₄, MgCl₂, CaCl₂)

Removal of permanent hardness: By chemical method only

1.  Addition of washing soda

 Na₂CO_3(aq)  + CaSO_4(aq) CaCO_3(s) + Na₂SO_4(aq)

1. Addition of caustic soda

    2NaOH_(aq)  + CaSO_4(aq) Ca(OH)_2(s)  + Na₂SO_4(aq)

   3.  Ion exchange resin

    CaSO_4(aq) + Sodium zeolite Calcium zeolite + NaSO_4(aq)
   

   (insoluble)

ADVANTAGES OF HARD WATER

1. It has better taste than soft water.
2. Calcium salts in it helps to build strong teeth and bones.
3. It provides CaCO₃, that crab and snail use to build their shells.
4. It does not dissolvelead, hence it can be supplied in lead pipes.

    

DISADVANTAGES OF HARD WATER

1. It causes furring of kettles and boilers.
2. It wastes soap.
3. It cannot be used in dying and tanning.

EVALUATION

1. Mention TWO compounds that can cause permanent hardness of water.
2. Write two equations to show the removal of permanent hardness of water.

TREATMENT OF WATER FOR MUNICIPAL SUPPLY

The following are the processes of treating river water for town supply

1. Coagulation:Chemicals like potash alum, KAl(SO₄)₂, or sodium aluminate III, NaAlO₂ is added to water in a large settling tank.
2. Sedimentation: The coagulated solid particles or flocs are allowed to settle in the settling tank to form sediments at the bottom of the tank.
3. Filtration: The water above the sediment still contains some suspended particles. The water is passed through a filter bed to remove the remaining fine dirt particles.
4. Chlorination (Disinfection): Chemicals like chlorine is then added to the water to kill germs. Iodine and fluorine are also added as food supplements to prevent goiter and tooth decay respectively. The treated water is then stored in a reservoir and distributed to the town.

    

GENERAL EVALUATION/REVISION

1. Mention two compounds that causes permanent hardness in water
2. State two ways of removing permanent hardness in water

1. List two advantages of hard water
   1. State Faraday’s second law of electrolysis
2. Using electron dot-cross representation, show the formation of carbon (IV) oxide and name the type of bond formed

READING ASSIGNMENT

New School Chemistry for Senior Secondary School by O.Y.Ababio (6^th edition) pages 296-302

WEEKEND ASSIGNMENT

SECTION A:Write the correct option ONLY

1. Treated town water undergoes the following steps except A. co-agulationB. precipitation C. sedimentation D. chlorination
2. Water is temporarily hard because it contains A. CaSO₄B. MgSO₄C. chlorine D.Ca(HCO₃)₂
3. Temporary hardness of water is removed by the use of one of the following A. boiling B. use of use of Ca(OH)₂C. use of Na₂CO₃D. use of alum
4. A substance that turns white anhydrous CuSO₄ blue is A. water B.liquid ammonia C. hydrochloric acid D. molten sulphur
5. Distilled water is different from deionized water because A. distilled water is a product of condensed steam while deionized water is filtered laboratory water B. distilled water is always pure and sold in packs while deionized is not packaged for consumption C. distilled water is condensed steam but deionized water is produced using ion-exchange resins which absorbs undesired ions. D. distilled water is man-made while deionized water is both natural and artificial

SECTION B

1.  State the steps involved in the treatment of river water for town supply.

2.  Write two equations to show the removal of permanent hardness of water.

WEEK TWO

TOPIC: SOLUBILITY AND SOLUTIONS

CONTENT

• Definition of Terms.
• Calculations based on Solubility.
• Solubility Curves.
• Uses of Solubility.

   

  SOLUTIONS
  

  A solution is a uniform or homogenous mixture of two or more substances.

   Solution = Solvent + Solute

  A solute is a dissolved substance which may be a solid, liquid or gas.

  A solvent is a substance (usually liquid) which dissolves a solute.

TYPES OF SOLUTIONS

1. Aqueous Solution: This is formed when a solute is dissolved in water.
2. Chemical Solution: This is the apparent solution of a solute in a solvent accompanied by a chemical change. For example, magnesium appears to dissolve in dilute hydrochloric acid, what actually happens is that the magnesium attacks the acid to form magnesium chloride, which dissolves in water present.

    

   TRUE SOLUTION AND COLLOIDAL SOLUTION
   

A true solution is formed when solute particles dissolve such that they are able to get in between the solvent particles. Example of true solution is aqueous solution of sodium chloride and copper (II) tetraoxosulphate (VI).

A False or Colloidal solution is one in which the individual particles are larger than the particles of a true solution, but not large enough to be seen by the naked eye. Examples of colloids are starch and albumen.

TYPES OF COLLOIDS

1. Sols and Gels: These are colloids where solid particles are dispersed in liquid medium. Example: starch, glue, jelly, etc
2. Aerosols: In aerosols, liquid particles are dispersed in a gas. Fog, smoke, spray of insecticide is examples of aerosol.
3. Emulsion: For emulsions, a liquid is dispersedin another liquid. Examples of emulsions are milk, hair cream; cleaning action of detergents is due to their ability to form emulsion.

    

   EVALUATION

4. Define the term ‘Solution’.
5. State THREE differences between True solution and False solution.

    

   SOLUBILITY
   

The solubility of a solute (substance) in a solvent at a particular temperature is the maximum amount of solute in moles or grams that will dissolve in 1 dm³ of the solvent at that temperature.

The concentration in moldm^-3 of a saturated solution is termed the solubility of the substance i.e. Solubility (moldm³) = Concentration in gdm³

 Molar mass

Solubility in mol/dm³ can also be expressed as = mass x  1000

Molar mass volume

Solubility in g/dm³ = mass x  1000

volume  1

Solubility of a solid solute in a solvent increases with rise in temperature while solubility of gases decreases with rise in temperature.  

DEFINITION OF TERMS

1. Saturated Solution: A saturated solution at a particular temperature is one which contains as much solute as it can dissolves at that temperature in the presence of undissolved solute particles.
2. Unsaturated solution: This is a solution which contains less of the solute than it can dissolve at a particular temperature.
3. Super saturated solution: This is a solution which contains more of the solute than it can dissolve at a particular temperature.

    

EVALUATION

1. Define Solubility
2. Differentiate between Saturated solution and Unsaturated solution

DETERMINATION OF SOLUBILITY

Solute: KCl, Solvent: water

Method

1. A saturated solution of KCl is prepared by dissolving excess of the solid in water in a beaker
2. Allow the solution in the beaker to settle down to obtain a clear saturated solution
3. Decant a portion of clear solution into another beaker and measures its temperature
4. Transfer the solution into a weighed evaporation dish and record the mass of the solution
5. Evaporate the solution to a complete dryness in a water bath
6. Allow the resulting solid to cool and reweigh the basin with content
7. Obtain mass of the dissolved salt and calculate the mass of the salt that would dissolve in 1dm³ of water at that temperature.

    

   CALCULATION
   

   Mass of basin =  xg

   Mass of basin + solution  =  yg

   Mass of basin + salt =  zg

   Mass of solution =  (y-x)g

   Mass of salt =  (z-x)g

   Mass of water used =  (y-z)g

   :. (y – z)g H₂O dissolves (z – x)g salt

   :. 100g H₂O dissolves (z – x)/(y – z) x 100g salt

   [Density of water = 1gcm³]

   :. No of moles of salt =  100(z – x)

   (y-z) x M.M

    

   :. Moles of salt dissolves in 1 dm³ water = 100(z-x)

   (y-z) x M.M

    

   FACTORS THAT AFFECT SOLUBILITY
   

1.  Nature of solvent and solute

2.  Temperature

3.  Pressure (often neglected)

SOLUBILITY CURVES

These are the graphs of solubility against temperature. The graph provides useful source of information.

USES OF SOLUBILITY CURVES

1.  It provides useful information about suitable solvent and temperature for solvent extraction from natural sources

2.  It provides useful information about temperature for fractional crystallization of a mixture of soluble salts.

3.  The curves enable pharmacists to determine the amount of solid drugs that must be dissolved in a given quantity of solvent to give a prescribed drug mixture.

EVALUATION

1.  Define super-saturated solution

2.  State two applications of solubility curves

CALCULATION ON SOLUBILITY

1.  If 12.2g of Pb(NO₃)₂ were dissolved in 21cm³ of distilled water at 20^oC. Calculate the solubility of the solute in moldm^-3

Solution:

Molar mass of Pb(NO₃)₂ = 331g

 No of moles of Pb(NO₃)₂ = 12.2/331 = 0.037moles

 If 21cm³ of water at 20⁰C dissolved 0.037mole salt

 :. 1000cm³ of water at 20⁰C dissolves 0.037 x 1000/21

 = 176moles Pb(NO₃) per dm³ H₂O

2.  1.0dm³ of an aqueous solution at 90^oC contains 404g of KNO₃ and 245g of KClO₃.

a. Determine which of the two salts will separate out when the solution is cooled to 60^oC

b. mass of salt that will separate out at 60^oC

(Solubility of KNO₃ in H₂O at 60^oC = 5.14moldm^-3, solubility of KClO₃ in H₂O at 60^oC = 1.61moldm^-3)

Solution:

No of moles of KNO₃ = 404/101 = 4.0moles dm^-3

No of moles of KClO₃ = 245/122.5 = 2.0 moldm^-3

The solubility of KClO₃ at 60^oC (5.14 moldm^-3) is higher than the amount in solution (4.0 moldm^-3), then KNO₃ will remain in solution while KClO₃ will crystallize out at 60^oC since the solubility at 60^oC is lower than the amount in solution.

b. Mass of salt that will separate out at 60^oC = 2.0 – 1.61 = 0.39mole

Mass of salt = Number of moles x Molar mass

= 0.39 x 122.5 = 47.78g

3.  The solubility of KNO₃ is exactly 1800g per 1000g water at 83^oC and 700g per 1000g water at 40^oC. Calculate the mass of KNO₃ that will crystallize out of solution if 155g of the saturated solution at 83^oC is cooled to 40^oC.

Solution:

Saturated solution of KNO₃ at 83^oC = 1000 + 1800 = 2800g

Saturated solution of KNO₃ at 40^oC = 1000 + 700 = 1700g

Mass of solute deposited = 2800 – 1700 = 1100g

From 83^oC to 40^oC, 2800 of saturated solution deposited 1100g of solute

155g of saturated solution will deposit 1100 x 155/2800 = 60.80g of salt.

EVALUATION

1. Define the following terms: Solubility, Saturated solution, Unsaturated solution.
2. 1.33 dm³ of water at 70^oC are saturated by 2.25 moles of lead (II) trioxonitrate (V) and 1.33 dm³ of water at 18oc are saturated by 0.53 mole of the same salt. If 4.50dm³ of the saturated solution are cooled from 70^oC to 18^oC, calculate the mount of solute that will be deposited in (a) moles (b) grams.

GENERAL EVALUATION/REVISION

1. Calculate the solubility of KCl in g/dm³ if 5g of the salt was dissolved in 50cm³ of water at 40^oC
2. If 50cm³ of a saturated solution of potassium chloride at 30^oC yielded 18.62g of dry salt, calculate the solubility of the salt in mol/dm³ at 30^oC
3. Define solubility
4. A certain mass of a gas occupies 300cm³ at 35^oC. At what temperature will it have its volume reduced by half, assuming its pressure remains constant?
5. A certain mass of hydrogen gas collected over water at 10oc and 760mm Hg pressure has a volume of 37cm³. Calculate the volume when it is dry at s.t.p. (Saturated vapour pressure of water at 10^oC = 9.2mmHg)

    

    

   READING ASSIGNMENT
   

New School Chemistry for Senior Secondary School by O.Y.Ababio (6^thedition) pages 303-310

WEEKEND ASSIGNMENT

SECTION A: Write the correct option ONLY

1. A saturated solution is a solution a. in which the solute is in equilibrium with the solvent b. in which the solute saturates the solution c. the solvent can still accept more solute except when the temperature is lowered d. whose solvent has low solubility at a given temperature
   1. A graph of solubility against temperature is called a. sigmoid curve

   b. supernant curve c. solubility curve d. dispersion curve

2. On heating 25g of a saturated solution to dryness at 60^oC, 4g of anhydrous salt was recovered. Calculate its solubility in g/dm³. a. 160 b. 180 c. 200 d. 220
3. The solubility of alcohols in water is due to a. their covalent nature b. hydrogen bonding c. their low boiling point d. their ionic character
4. A common solvent of sulphur is a. water b. carbon(IV)sulphide c. alcohol d. ethanoic acid

SECTION B

1.  Define the following:

 (a) Solubility (b) Saturated solution (c) Unsaturated solution

2.  If the solubility of KNO₃ at 0^oC is 1.33mol/dm³, determine whether a solution containing 30.3g/dm3 at 0^oC is saturated or unsaturated.

WEEK THREE

TOPIC: MASS/VOLUME RELATIONSHIP

CONTENT

• Mole and Molar Quantities
• Relative Atomic Mass and Relative Molecular Mass.
• Calculations involving Mass and Volume.

   

  MOLE AND MOLAR QUANTITIES
  

THE MOLE

A mole is a number of particles of a substance which may be atoms, ions, molecules or electrons. This number of particles is approximately 6.02 x 10²³ in magnitude and is known as Avogadro’s number of particles.

The mole is defined as the amount of a substance which contains as many elementary units as there are atoms in 12g of Carbon-12.

RELATIVE ATOMIC MASS

The relative atomic mass of an element is the number of time the average mass of one atom of that element is heavier than one twelfth the mass of one atom of Carbon-12. It indicates the mass of an atom of an element. For e.g, the relative atomic mass of hydrogen, oxygen, carbon, sodium and calcium are 1, 16, 12, 23, and 40 respectively.

The atomic mass of an element contains the same number of atoms which is 6.02 x 10²³atoms; 1 mole of hydrogen having atomic mass of 2.0g contains 6.02 x 10²³ atoms.

RELATIVE MOLECULAR MASS

The relative molecular mass of an element or compound is the number of times the average mass of one molecule of it is heavier than one-twelfth the mass of one atom of Carbon-12

It is the sum of the relative atomic masses of all atoms in one molecule of that substance. It is also called the formula mass. The formula mass refers not only to the relative mass of a molecule but also that of an ion or radical.

CALCULATION

Calculate the relative molecular mass of:

1. Magnesium chloride
2. Sodium hydroxide
3. Calcium trioxocarbonate

 [Mg=24, Cl=35.5, Na=23, O=16, H=1, Ca=40,C=12]

Solution:

1. MgCl ₂ = 24 + 35.5×2 = 24 + 71 = 95gmol^-1
2. NaOH = 23 + 16 + 1 = 40gmol^-1
3. CaCO₃ = 40 + 12 +16×3 = 100gmol^-1

    

EVALUATION

1. What is relative molecular mass of a compound?

1. Calculate the relative molecular mass of (a) NaNO₃ (b) CuSO₄.5H₂O

    

MOLAR VOLUME OF GASES

The volume occupied by 1 mole of a gas at standard conditions of temperature and pressure (s.t.p) is 22.4 dm³. Thus 1 mole of oxygen gas of molar mass 32.0gmol^-1 occupies a volume of 22.4dm³ at s.t.p and 1 mole of helium gas of molar mass 4.0gmol^-1 occupies a volume of 22.4 dm³ at s.t.p.

Note: When the conditions of temperature and pressure are altered, the molar volume will also change. Also, standard temperature = 273K and standard pressure = 760mmHg.

RELATIONSHIP BETWEEN QUANTITIES

Molar mass = mass (g) i.e. M = m gmol^-1

Amount (moles)  n

Note: Amount = Number of moles

Molar volume of gas = volume ( cm³ or dm³) i.e. Vm = v dm³mol^-1

Amount (mole) n

Amount = Reacting mass (g)

Molar mass (gmol^-1)

Also, Amount of substance = Number of particles

Avogadro’s constant

But, Avogadro’s constant = 6.02 x 10²³

Combining the two expressions:

Reacting mass = Number of particles

Molar mass 6.02 x 10²³

CALCULATIONS

1. What is the mass of 2.7 mole of aluminium (Al=27)?

    Solution:

   Amount = Reacting mass

   Molar mass

   Reacting mass = Amount x Molar mass

   = 2.7mole x 27 gmol^-1 = 72.9g.

    

2. What is the number of oxygen atoms in 32g of the gas? (O=16, N_A = 6.02 x 10²³)

   Solution:

   Reacting mass = Number of atoms

   Molar mass  6.02 x 10²³
   

   Number of atoms = Reacting mass x 6.02 x 10²³
   

    Molar mass

   Molar mass of O₂ = 16×2 =32gmol^-1
   

   Number of atoms = 32g x 6.02 x 10²³

   32gmol^-1

    = 6.02 x 10²³

   The number of oxygen atoms is 6.02 x 10²³

    

   EVALUATION

3. Define the molar volume of a gas
4. How many molecules are contained in 1.12dm³ of hydrogen gas at s.t.p?

    

   STOICHIOMETRY OF REACTION
   

The calculation of the amounts (generally measured in moles or grams) of reactants and products involved in a chemical reaction is known as stoichiometry of reaction. In other words, the mole ratio in which reactants combine and products are formed gives the stoichiometry of the reactions.

From the stoichiometry of a given balanced chemical equation, the mass or volume of the reactant needed for the reaction or products formed can be calculated.

CALCULATION OF MASSES OF REACTANTS AND PRODUCTS

1. Calculate the mass of solid product obtained when 16.8g of NaHCO₃was heated strongly until there was no further change.

Solution:

The equation for the reaction is:

2NaHCO_3(s) → Na₂CO_3(s) + H₂O_(g) CO_2(g)

Molar mass of NaHCO₃ = 23 + 12 + 16×3 = 84gmol^-1

Molar mass of Na₂CO₃ = 23×2 +12+16×3 = 106gmol^-1

From the equation:

2 moles NaHCO₃ produces 1 mole Na₂CO₃

2x84g NaHCO₃ produces 106g Na₂CO₃

16.8g NaHCO₃ will produce Xg Na₂CO₃

Xg Na₂CO₃ = 106g x 16.8g  =10.6g

2x84g

Mass of solid product obtained = 10.6g

1. Calculate the number of moles of CaCl₂ that can be obtained from 25g of limestone [CaCO₃] in the presence of excess acid.

Solution:

The equation for the reaction is:

CaCO_3(s) + 2HCl → CaCl_2(s) + H₂0_(l) + CO_2(g)

Number of moles = Reacting mass

Molar mass

Molar mass of CaCO₃ = 40 + 12 + 16×3 = 100gmol^-1

Number of moles of CaCO₃ = 25g = 0.25 mole

100gmol^-1

From the equation of reaction,

1 mole CaCO₃ yields 1 mole CaCl₂

Therefore, 0.25 mole CaCO₃ yielded 0.25 mole CaCl_2.

EVALUATION

1. What does the term ‘Stoichiometry of reaction’ mean?

   Ethane [C₂H₆] burns completely in oxygen. What amount in moles of CO₂will be produced when 6.0g of ethane are completely burnt in oxygen?

    

   CALCULATION OF VOLUME OF REACTING GASES
   

2. In an experiment, 10cm³ of ethene [C₂H₄] was burnt in 50cm³ of oxygen.
   1. Which gas was supplied in excess? Calculate the volume of the excess gas remaining at the end of the reaction.
   2. Calculate the volume of CO₂ gas produced

Solution:

The equation for the reaction is:

C₂H_4(g) + 3O_2(g) → 2CO_2(g) + 2H₂O_(g)

1. From the equation,

   1 mole of ethene reacts with 3mole of oxygen

   1 volume of ethene reacts with 3 volumes of oxygen

   10cm³ of ethene will react with 30cm³ of oxygen

   Since 50cm³ of oxygen was supplied, oxygen was in excess

   Hence volume of the excess gas = initial volume – volume used up = 50-30 = 20cm³

2. 1 volume of ethene produces 2 volumes of CO₂

   10 cm3 of ethene will produce 20cm3 of CO₂

   Therefore, 20cm³ of CO₂ was produced

    

1. 20cm³ of CO was mixed and sparked with 200cm³ of air containing 21% of O₂. If all the volumes are measured at s.t.p, calculate the total volume of the resulting gases.

    

Solution:

In 200cm³ of air,

Volume of O₂ = 21 x 200cm³ = 42cm³

100

Volume of N₂ and rare gases = 200-42 = 158cm³

The equation for the reaction is:

2CO_(g) + O_2(g) → 2CO_2(g)

Volume ratio 2 : 1 : 2

Before sparking 20cm³ 42cm³

Reacting volume 20cm³ 10cm³

After sparking 32cm³ 20cm³

Volume of resulting gases = 32 + 20 + 158 = 210cm³

GENERAL EVALUATION/REVISION

1. Find the volume of oxygen produced by 1 mole of KClO₃ at s.t.p in the following reaction: 2KClO_3(s) → 2KCl_(s) + 30_2(g)
   
2. Define the term ‘Relative atomic mass’
   
3. Balance the following redox equations I^– + MnO₄^– IO₃^– + MnO₂ in basic medium
4. Write the symbols of the following elements: mercury, silver, gold, lead, tin, antimony.
5. Define valency.

READING ASSIGNMENT

New School Chemistry for Senior Secondary School byO. Y. Ababio, Pg 156-164

WEEKEND ASSIGNMENT

SECTION A: Write the correct option ONLY

1. Amount of a substance is expressed in a. mole b. grams c. kilograms d. mass
2. Determine the mass of CO₂ produced by burning 104g of ethyne [C₂H₂]a. 256g b.352g c. 416g d. 512g
3. The mole ratio in which reactants combine and products are formed is known as a. rate of reaction b. stoichiometry of reaction C. equation of reaction d. chemical reaction
4. The unit for relative molecular mass is a. mole b. gmol^-1 c. grams d. mass
5. What mass of Pb(NO₃)₂ would be required to 9g of PbCl₂ on the addition of excess NaCl solution? [Pb=207, Na=23, O=16, N=14] a. 10.7g b. 1.2g c. 6.4g d. 5.2g

SECTION B

1. Calculate the number of molecules of CO₂ produced when 10g of CaCO₃ is treated with 100cm³ of 0.20moldm^-3HCl
2. Calculate the volume of nitrogen that will be produced at s.t.p from the decomposition of 9.60g ammonium dioxonitrate (III), NH₄NO₂.

WEEK FOUR

TOPIC: ACID/BASE REACTIONS

CONTENT

• Preparation of Standard Solutions.
• Indicators
• Calculations based on Acid-Base Titration.

   

TITRATION

There are two types of quantitative analysis namely: volumetric and gravimetric analysis. Volumetric analysis is based on volume measurement while gravimetric analysis involves direct mass measurement.

Volumetric analysis is carried art using Titration. In titration, a standard solution (one of known concentration must be using be used to react with a solution of unknown of concentration)

PREPARATION OF A STANDARD SOLUTION

A standard solution is a solution of which the concentration is known .A standard solution is prepared by weighing a pure solute, for instance, and dissolving it in a suitable solvent, usually water, and making up the solution to a definite volume in a volumetric flask.

For instance, a solution known to contain exactly 10.6g of anhydrous sodium trioxocarbonate (IV), Na₂CO₃, in 1 dm3 of solution is a standard solution.

Preparation of 0.1mol/dm³NaOH

40g NaOH dissolved in 1 dm³ of the water gives 1.0mol/dm³ solution

 XgNaOH will be dissolved in 1 dm3 of water to give 0.1mol/dm³

 Xg = 40g x 0.1mol/dm³

1.0mol/dm³

= 4g

Therefore, 4g of sodium hydroxide pellet is measured, dissolved in water and made up to 1dm³ mark to obtain 0.1mol/dm³NaOH

Preparation of 0.1mol/dm³HCl

To prepare 0.1mol/dm³HCl, the dilution formula is used to determine the volume of the stock acid that will be measured and dissolved in water to obtain the desired concentration.

The dilution formula is C₁V₁ = C₂V₂

Where C₁ = concentration of stock acid = 11.6mol/dm³ (for HCl)

 V₁ = volume of stock acid  

 C₂ = desired concentration of acid = 0.1mol/dm³

 V₂ = volume of water = 1000cm³ (1dm³)

 V₁ = C₂V₂  = 0.1 x 1000 = 8.6cm³

  C₁   11.6

Thus, 8.6cm³ of the stock acid is measured using a measuring cylinder and added to water, then made up to 1dm³ to obtained 0.1 mol/dm³HCl.

EVALUATION

1. Describe how to prepare 0.05mol/dm³ H₂SO₄
2. List the apparatuses used during volumetric analysis(titration)

INDICATORS FOR ACID/BASE TITRATION

Acid-base indicators are dyes that change colour when according to the pH of the medium. The table below shows some titration and their suitable indicator:

Acid / base Indicator

Strong acid and strong base  methyl orange or phenolphthalein

Strong acid and weak base  methyl orange

Weak acid and strong base  phenolphthalein

Weak acid and weak base  No suitable indicator

CONCENTRATION

The concentration of a solution is the amount of solute in a given volume of the solution. It can be expressed as mol/dm³ or g/dm³.

Molar concentration

The molar concentration of a compound is one which contains one mole or the molar mass of the compound in 1dm³ of the solution. Unit of molar concentration is mol/dm³

Mass concentration

The mass concentration of a compound is the mass of the compound contained in 1 dm³ of solution. The unit is g/dm³

Relationship between Molar concentration and Mass concentration

Concentration = number of moles = n/V………………(i)

volume

Number of moles, n = C x V ……………………..(ii)

But, number of moles, n= m/M

Where M = molar mass and m = mass

Substituting n=m/M into …….(ii)

We have m/M = C x V

That is, m/V = C x M

But m/V = mass concentration

Therefore, mass concentration = molar concentration x molar mass = C x M

TITRATION REPORT

Average volume of acid used = 23.20 + 23.20 + 23.10

3

 = 23.17cm³

GENERAL EVALUATION/REVISION

1. Which substance is added to the base during titration experiment?
   
2. What is the colour of methyl orange in a base?
3. What is the point at which the titration experiment appears to complete called?
4. State three differences between electrolytic cell and electrochemical cell
5. Calculate the standard e.m.f of a cell given that the e.m.f of Zn²⁺/Zn and Cu²⁺/Cu are -0.76V and +0.34V respectively

READING ASSIGNMENT: New School Chemistry SSS by O. Y. Ababio (6^th edition) pages 165-168

WEEKEND ASSIGNMENT

SECTION A: Write the correct option ONLY

1.  The indicator used when titrating a weak acid against a strong base isA. methyl orange B. phenolphthalein C. methyl redD. any indicator

2.  The colour of phenolphthalein in acids is A. blue B. red C.colourlessD. yellow

3.  Which of the following formulae is direct for amount n? A. n=C/p B. n=M/m C. n=C x V D. n=C x m

4.  The mass concentration of a substance can be expressed as A. mass/densityB. molar concentration/molar mass C. mass/volume D. number of moles x volume

5.  At the end point there is A. a colour change B. no change of colourC. decrease in mass D. an increase in mass

SECTION B

1. Define the following terms (a) molar concentration (b) Equivalent point
2. 160cm³ of distilled water is added to 40cm³ of 0.500mol/dm³ H₂SO₄ solution. Determine the concentration of the diluted solution.

WEEK FIVE

TOPIC: INTRODUCTION TO ORGANIC CHEMISTRY

CONTENT

• Unique Nature of Carbon.
• Characteristics Features of Organic Compounds
• Classification of Hydrocarbons.
• Definition of Terms used in Organic Chemistry
  

Organic chemistry is the chemistry of carbon compounds with the exception of compounds such as carbon (II) oxide, carbon (IV) oxide the trioxocarbonate (IV).Carbon has unique ability to form numerous organic compounds because it has ability to catenate. Catenation is the ability of atoms of an element to form bonds between its own atoms and produce long chain structure.

All organic compounds contain carbon as the main element together with one or more other elements such as hydrogen, oxygen, chlorine, nitrogen and sulphur.

UNIQUE NATURE OF CARBON

1. The valency of carbon: the electronic configuration of carbon is as follows:

   C = 6: 1s² 2s¹ 2p_x¹ 2p_y¹ 2p_z¹

    Carbon form covalent bonds and after bond formation it has neither vacant orbital nor lone pair of electrons. This makes many carbon compounds chemically stable.

2. The bond between carbon and hydrogen is non-polar. Thus, hydrogen atoms attached to carbon do not weaken carbon-carbon bonds.
3. The types of orbital hybridization available to carbon: carbon can form three types of hybridizations: SP3, SP2 and SP hybridization. Hence, it has ability to form single, double or triple covalent bonds between its atoms.
4. Large amount of energy is required to break carbon single bond.

CHARACTERISTICS FEATURE OF ORGANIC COMPOUNDS

1. Organic compounds are covalent in nature.
2. They are non-polar substances and are insoluble in polar solvents.
3. They have low melting and boiling points.
4. They are highly flammable.
5. Their reactions are relatively slow compared to inorganic chemical reactions.

CLASSIFICATION OF ORGANIC COMPOUNDS

Organic compounds are classified into Aliphatic and Aromatic compounds.

Aliphatic Compounds: These are compounds whose molecules are composed of chains of carbon atoms. They can be

1. Straight chain compounds e.g pentane
2. Branched chain compounds e.g 2-methylbutane

Straight and branched chain aliphatic compounds exist as open chain and are called ACYCLIC compounds. Aliphatic compounds which exist as closed chain are called the CYCLIC compounds e.gcyclo propane.

Aromatic Compounds: Benzene, C₆H₆, is a typical aromatic compound. Other aromatics compounds are derivatives of benzene e.g C₆H₅OH.

EVALUATION

1. List four reasons why carbon forms numerous organic compounds

1. State five characteristic features of organic compounds.

DEFINITION OF TERMS USED IN ORGANIC CHEMISTRY

HOMOLOGOUS SERIES

Homologous series is a family of organic compounds which follows a regular structural pattern and in which each successive member differs in its molecular formula by –CH₂– group.

The simplest series of compounds in organic chemistry is the Alkane series. The general molecular formula of the alkane series is C_nH_2n+2. It is the parent series from which every other series is obtained. Other homologous series include the Alkenes, Alkynes, Alkanols, Alkanoic acids, etc.

CHARACTERISTICS OF HOMOLOGOUS SERIES

1. All members conform to general molecular formula , e.g for alkanes, C_nH_2n+2
   
2. Each successive member differs in molecular formula by – CH₂– group.
3. All members undergo similar chemical reactions.
4. The physical properties of members change gradually along the series.
5. All members are prepared by the same method.

    

ALKYL AND FUNCTIONAL GROUPS

Alkyl groups: Alkyl groups are all groups derived from the alkanes by the loss of a hydrogen atom. Alkyl groups have a general formula of CnH₂n+1.They are named after the parent alkane by replacing the ending –ane by –yl. The alkyl group derived from the first two members of the parent alkane series are:

Parent alkane Alkyl group Formula

Methane, CH₄ Methyl -CH₃

Ethane, C₂H₆ Ethyl -C₂H₅

Functional groups: The substituent of hydrogen atom in the alkane series to form the alkyl group determines the chemical properties of the compound formed thereafter. This substituent is called FUNCTIONAL GROUP.

A functional group is an atom, a radical or a bond common to a homologous series and which determines the chemical properties of the series.

Examples of functional groups include: Hydroxyl group -OH, amino group NH₂, carboxyl group -COO, double and triple bonds.

Alkyl group in a compound determines the physical properties of the compound; while functional group determine the chemical properties of the compound.

EVALUATION

1.  Define a homologous series.

2.  Define a functional group and give two examples.

SATURATED AND UNSATURATED COMPOUNDS

Saturated compounds are compounds containing atoms joined together by single covalent bond. Alkanes are saturated compounds, e.g ethane, C₂H₆

H H

 H C C H

H H

Unsaturated compounds are compounds containing atoms joined together by double or triple bonds. Alkenes and alkynes are unsaturated compounds,

e.gEthene, C₂H₄ H H

C C

 H H

Ethyne: H C C H

FORMULAE OF ORGANIC COMPOUNDS

Organic compounds are characterized by the following formulae

1.  Empirical formula

2.  Molecular formula

3.  Structural formula

-Empirical Formula is the simplest formula which indicates the component elements and ratio of combination of atoms in a compound.

– Molecular Formula is a chemical formula of a compound which indicates the actual number of atoms of each element in a compound.

-Structural Formula is a formula which indicates how atoms are arranged within the molecule of a substance.

Structural formula can be

1. Open structural formula
2. Condensed structural formula

Open structure Condensed structure

  H H

H C C H CH₃CH₃

H H

GENERAL EVALUATION/REVISION

1. Differentiate between saturated and unsaturated compounds.
2. Write the open and condensed structural formula of pentane[C₅H₁₂]_.
3. Define the following terms: Homologous series and functional group.

1. Determine the oxidation number of Cl in each of the following compounds and give the IUPAC name of the compound (a) NaOCl (b) KClO₃
2. Split the following redox equations into oxidation and reduction half equation (a) Cu_(s) + 2Ag⁺_(aq) → Cu²⁺_(aq) + 2Ag_(s)

    (b) Cl_2(g) + 2I^–_(aq) → 2Cl^– + I_2(s)

READING ASSIGNMENT

New School Chemistry for Senior Secondary School by O .Y. Ababio (6^th edition), pages 514-520

WEEKEND ASIGNMENT

SECTION A: Write the correct option ONLY.

1. Exceptional large number of carbon compounds is essentially due to the ability of

   (a). carbon to catenate liberally (b).various groups to catenate

   (c). nitrogen, hydrogen, phosphorus and the halogens to catenate with themselves

   (d). hydrocarbons to dominate other groups.

2. Which of the following is not a characteristics feature of organic compounds?

   (a). Covalent in nature (b). They dissolve in all polar solvents (c).Low melting and boiling points (d).Highly flammable

3. Functional groups in organic compounds (a).determine the chemical properties of the homologous series (b).does not modify the other when they are more than one in a molecule (c).have a general formula which may include the functional group (d). are responsible for the physical properties.
4. Homologues have the same (a).empirical formula (b).structural formula (c).general formula (d). molecular formula
5. The four main classes of hydrocarbons are (a).methane, ethene, ethyne and benzene(b).ethane, ethene, ethyne and toluene (c).cycloalkane, cycloalkene, alkynes and arenes(d). alkanes, alkenes, alkynes and aromatics

SECTION B

1. Define the following terms: a. Functional group b. Homologous series
2. Write the open chain structure of the following
   1. CH₃C(CH₃)₂CH₂CH(CH₃)CH₂CH₃
   2. (CH₃)₂CHCH₂CH(CH₂Cl)CH₃
   3. CH₃C(Br)₂CH₂CH₃

WEEK SIX

TOPIC: ALKANES

CONTENT

• IUPAC Nomenclature
• Preparation, Properties and Uses.

   

The alkanes are aliphatic hydrocarbons. Their general molecular formula is C_nH_2n+2. Hence

For n=1 CH₄ Methane

n=2 C₂H₆  Ethane

n=3 C₃H₈  Propane

n=4 C₄H₁₀  Butane

n=5 C₅H₁₂  Pentane and so on.

There is no functional group in the alkane series.

THE IUPAC NOMENCLATURE FOR ALIPHATIC COMPOUNDS

In IUPAC nomenclature, every name of organic compound consists of a ROOT, SUFFIX and PREFIX names.

Root name: Name of the parent aliphatic hydrocarbon of the longest carbon chain in a molecule.

Suffix name: Name of the principal functional group on the longest carbon chain in a molecule.

Prefix name: Name of the other substituents on the longest carbon chain which are not functional groups. For example, 1- chloroethane- 1- ol; has 1-chloro as prefix, ethane as root and -1-ol as suffix.

RULES FOR THE IUPAC NOMENCLATURE

1. Select the longest continuous carbon chain as parent hydrocarbon.

1. Number the longest carbon chain starting from the end that gives lowest possible number to the suffix (functional group).
2. Indicate the other substituents by prefixes preceded by numbers to show their position on the carbon chain.
3. If the same alkyl or other substituents group occurs more than once as side chain, show how many are present by using prefix di, tri, tetra etc and indicate by various numbers the position of each group on the longest carbon chain.
4. If there are several different alkyl groups attached to the parent chain, name them in alphabetical order.

Examples: H HH

H C¹ C² C³ H

H CH₃ H 2- methyl propane

H HH CH₃ H

H C⁵C⁴ C³ C² C¹ H

H HHHH 2- methylpentane

H HHH CH₃ C₂H₅ H H

H C⁸ C⁷ C⁶ C⁵ C⁴C³C² C¹ H

H HH C₃H₇H C₂H ₅H H

3,3- diethyl-4-methyl-5-propyloctane

6. If there are halogens together with alkyl groups attached to the parent chain, name the halogens first in alphabetical order and the alkyl group as explained earlier.

Example

  H Cl

H C C OH

Br CH₃

1- bromo-2- chloropropan-2-ol

EVALUATION

Name the following compounds

1. H HH 2. CH₃CHClCH₂OH

H C CC H

H HH

METHANE (CH₄)

LABORATORY PREPARATION

Methane is prepared in the laboratory by heating ethanoate salt with corresponding alkalis e.g sodium ethanoate and soda-lime. Soda-lime is quick lime slaked with a concentrated solution of sodium hydroxide. It is used in preference to caustic soda because it is not deliquescent and does not attack glass so readily.

PHYSICAL PROPERTIES

1.  Methane is a colourless and odourless gas.

2.  It is slightly soluble in water.

3. It is less dense than air

4. It has no action on litmus paper

CHEMICAL PROPERTIES

1. Combustion: Methane burns in air or oxygen to produce steam, carbon(IV) oxide and a lot of heat

   CH_4(g) + 2O_2(g) 2H₂O_(l) + CO_2(g).

The general equation for combustion of alkanes is represented as

C_xH_y(g) + ( x + ^y/₄)O₂  ^y/ ₂H₂O_(g) + xCO_2(g).

1. Substitution reaction: Substitution reaction involves the direct displacement of atom or group by another atom or group. Methane reacts with chlorine in the presence of ultra-violet light to yield a mixture of products.

   CH_4(g) + Cl_2(g) CH₃Cl_(g) + HCl_(g)  

   Chloromethane

USES OF METHANE

1.  Methane is used as fuel by itself or mixed with other gases.

2.  It is used for making hydrogen gas.

3.  It is used in making carbon black.

4.  Trichloromethane is used in surgical operations as an anaesthetic.

Evaluation

1.  Describe laboratory preparation of methane.

2.  Write two equations to show the chemical properties of methane.

Isomerism

This is the existence of two or more organic compounds with the same molecular formula but different molecular structures.

TYPES OF ISOMERISM

1.  Structural Isomerism

2.  Stereoisomerism.

The structural isomerism occurs in organic compound with the same molecular formula but different structural arrangement of the carbon atom.

TYPES OF STRUCTURAL ISOMERISM

1. Chain isomerism:This is the kind of isomerism which occurs due the differences in the way by which the carbon atoms are arranged in the chain.Example
   

 H  HHH H  H  H  

H- C –   C – C – C – H H-  C – C –  C – H

H HHHH H

 Butane H – C – H  

H

  2- methyl propane

1. Functional isomerism: This is the kind of isomerism which is due to the differences in functional group.Example

H H H  H

H- C – C – OH   H- C – O- C- H

H HHH

Ethanol Methoxymethane.

1. Positional isomerism: This is the kind of isomerism which occurs as a result of the difference in the way the functional group is positioned. Example:  

H HH  H  H

H- C =C- C- C- H  H – C – C = C – C – H

  H HHHHHHH

But-1-ene But-2- ene

STEREO ISOMERISM: This arises as a result of differences in spatial orientation of atoms or groups of atoms about a carbon-carbon double bond or ring structure or a carbon atom surrounded by four different groups.

TYPES OF STEREO ISOMERISM

1. Geometric isomerism : This is the existence of compound with the same
   

molecular formula but different spatial structural formula.Example

CH₃CH₃  CH₃ H

C = C  C = C

H  HH CH₃

Cis but -2-ene   Trans but-2-ene

1. Optical isomerism: They have different configuration and they rotate plane polarized light.Example

H H

CH₃ – C – COOH HOOC – C – CH₃

OH OH

GENERAL EVALUATION/REVISION

Give the structure of the following organic compounds:

1. 1-chloro-2-methyl pentane
2. 2,2,4-trimethyl hexane
3. 1-bromo-2-methyl butane
4. a. What is meant by the term isomerism?

   b. Draw the structure of the two isomers of the compound with the molecular formula C₂H₆O. Give the name of each of the isomers.

   c. State the major difference between the isomers.

   5.  Give the two reasons why soda-lime is used instead of caustic soda in the laboratory preparation of methane.

READING ASSIGNMENT

New School Chemistry for Senior Secondary School by O.Y. Ababio(6^th edition) Pages 520-525, 530-532

Weekend Assignment

SECTION A: Write the correct option ONLY

1. The name of C(CH₃)₄ is a. butane b. 2-methyl propane c. methyl propane d. 2,2-dimethyl propane 2
2. The structure of the organic compound 1,1-dichloro-2-methyl pentane is

A. H HH CH₃ClB. H H CH₃ H

H C CCCC H H C CCC H

H HHHHHHHH

C. H HHHH  D. Cl H HHH

H C CCCC H H C CCCC H

ClCl H CH₃ H Cl CH₃H HH

1. The carbon atoms in alkanes are  a. not hybridized b. sp³ hybridized c. sp² hybridized d. sp hybridized
2. The general formula for the alkanes is a. RCHO b. C_nH_2n+1 c. C_nH_2n+2 d. RCOOR
3. C₄H₁₀ belongs to the same homologous series as a. C₄H₈ b. C₂H₂ c. C₃H₈ d. C₅H₁₀

    

   SECTION B
   

4. a. Define the term isomerism

   b. Name the alkanol that is isomeric with Methoxy methane (CH₃OCH₃)

2  a. Name the following compound:(i) CH₃(CH₂)₃CH₃ (ii) C₂H₄Cl₂

 b. Write the structure of the following compounds

(i) 1-methyl cyclopropane (ii) 2-bromo-4-methyl pentane

WEEK SEVEN

TOPIC: ALKENES

CONTENT

• Nomenclature
• Preparation, Properties and Uses.

   

UNSATURATED HYDROCARBONS

These are hydrocarbons in which carbon atoms join with each other by multiple bonds. The multiple bonds can be double bonds e.gAlkenes or triple bonds e.gAlkynes.

NOMENCLATURE

The process of naming in alkenes is obtained by substituting “ane” in alkane with ‘ene’ e.g Ethane changes to Ethene, propane to propene

LABORATORY PREPARATION

Ethene is prepared by heating ethanol with excess concentrated tetraoxosulphate(VI) acid at 170^o C. The acid acts as a dehydrating agent by removing water from the ethanol.  Thus the process is called dehydration.

The reaction occurs in two stages

 C₂H₅OH_(aq) + H₂SO_4(aq) C₂H₅HSO_4(aq) + H₂O_(l)

 C₂H₅HSO_4(aq) C₂H_4(g) + H₂SO₄.

The overall reaction is represented by the equation

C₂H₅OH(aq)^H₂^SO₄C₂H_4(g) + H₂SO_4(aq)

 -H₂O

PHYSICAL PROPERTIES

1.  Ethene is a colourless gas with faint sweetish smell.

2. It is sparingly soluble in water.

3. It is slightly less dense than air.

4. It has no action on litmus paper.

Evaluation

1. How would you prepare a jar of ethene gas in the laboratory?
2. Mention four physical properties of Ethene.

CHEMCIAL PROPERTIES

1.  Combustion: Ethene undergoes combustion in air or in the presence of oxygen and produce carbon (IV) oxide and steam.

 C₂H_4(g) + 3O_2(g) 2CO_2(g) + 2H₂O_(l)

2.  Addition reaction: This is a reaction in which two molecules combine to form one molecule.

 a. Reaction with hydrogen (Hydrogenation):

H H H H

H-C= C – H + H₂ H – C – C – H

Ethene

 H H Ethane

b. Reaction with halogens (Halogenation):

H  H  HH

H – C = C- H + Cl₂ H – C – C – H

Ethene ClCl 1,2- dichloroethane

c.  Reaction with hydrogen halides(Halohydrogenation):

H  H

H- C = C – H + HBr H – C  C – H

H HH Br

Ethene 1-bromoethane.

d.  Reaction with acidified /Alkaline KMnO₄ (Hydroxylation): It decolourises acidified KMnO₄, but turns alkaline KMnO₄ to green and ethane -1,2- diol is formed.

 OH OH

H – C = C – H + KMnO₄ H – C – C – H

H H   H H

Ethane-1,2-diol (glycol)

e.  Reaction with Hydrogen peroxide in the presence of osmium trioxde to form ethan -1,2- diol.

OH OH

 H – C = C – H + H₂O₂ H – C – C – H

H HHH

Ethane -1,2- diol.

f.  Reaction with concentrated H₂SO₄produces a fuming liquid (ethyl hydrogen sulphate)

C₂H₄ + H₂SO₄ C₂H₅HSO₄

When ethyl hydrogen sulphate is hydrolyzed, tetraoxosulphate (VI) acid and ethanol are produced.

C₂H₅SO₄  C₂H₅OH + H₂SO₄

g. Ethene gas decolourizes bromine water to produce bromoethanol.

H H  

  H – C = C – H+HBrO   H – C –C – H

  H H( brown)  Br OH

Bromoethanol

h. Polymerization of ethane to produce polythene.

H HHHH  H  

C = C- C – C – C – C-

n H HHHHH  n

 Polyethene

i. Ethene can also undergo addition reaction with oxygen in the presence of silver catalyst at about250^oC to form epoxy ethane.

H H  HH

H – C = C – H + ½ O2 H – C – C – H

O

USES OF ETHENE: Ethene is used

• In the manufacture of plastics.
  • In making synthetic rubber.
  • To hasten the ripening of fruits.
• In the production of other organic compounds e.g halo-alkane, ethane and ethanol.

GENERAL Evaluation/REVISION

1. Write balanced equations to show the reaction of ethene with the following:
   1. Bromine water
   2. Chlorine water
   3. Acidified KMnO₄
2. State four uses of ethene.
3. Why is an empty flask inserted between the flat bottom flask and the conical flask holding the drying agent in the laboratory preparation of ethene?
4. State THREE factors that determine the spontaneity of a chemical reaction.
5. 0.92g of ethanol raised the temperature of 100g of water from 298K to 312.3K when

   burned completely. What is the heat of combustion of ethanol?

READING ASSIGNMENT

New School Chemistry for Senior Secondary School by O.Y. Ababio (6^thedition) Pages 532-535

Weekend Assignment

SECTION A: Write the correct option ONLY

1.  The name of the organic compound with the structure below is

CH₃   H

C = C

H CH₃

A. Cis- but-2-eneB.Trans –cis-but-2-eneC.Trans-1-2- but-2-eneD.1,2- dimethyl ethane.

2. In the reaction given below:

 C₂H₅OH ^{Conc H}₂^SO₄ C₂ H₄Conc H₂SO₄is acting as

-H₂O

A.  oxidizing agent B. reducing agent  C. dehydrating agent D. drying agent.

3. One of the following is not a chemical property of ethene.

A. Polymerization B. Substitution reaction C. Hydration D. Addition reaction

4. Function of the empty bottle during the preparation of ethane gas is

A. to remove oxygen B. to remove CO₂C. to prevent sucking back of the gas D. None of the above

5. Addition reaction of hydrogen and ethene is known as

A. polymerization B. halogenation C. combustion D. hydrogenation

SECTION B

1. Write and name the geometric isomers of compound with the molecular formula C₅H₁₀

2. Write balanced chemical equation to show how ethene reacts with the following:

a. concentrated H₂SO₄b. bromine water c. acidified KMnO₄

WEEK EIGHT

TOPIC:ALKYNES

CONTENT

• Nomenclature
• Preparation, Properties and Uses
• Aromatic Hydrocarbons: Benzene Structure
  • Preparation, Properties and Uses

NOMENCLATURE

Alkynes are the homologous series of unsaturated hydrocarbon with a general molecular formula C_nH_2n-2.

Alkynes show a high degree of unsaturation than alkenes, hence, they are chemically more reactive than the corresponding alkenes or alkanes.

They are named by replacing the ‘ane’ of alkanes with ‘yne’.

Examples

H

 H – C C – H H – C – C C- H

H

 Ethyne Prop-1-yne

ETHYNE

Ethyne is the first member of the alkynes series. It has a molecular formula,

C₂H₂, and a structural formula of HC = CH.

LABORATORY PREPARATION

Ethyne is usually prepared in the laboratory by the action of cold water on calcium carbide. The reaction is carried out on a heap of sand to prevent the flask from cracking as a result of the large quantity of heat evolved.

Evaluation

1.  Write and name all possible structure of C₆H₈

2. How can you prepare a few jars of ethyne in the laboratory?

PHYSICAL PROPERTIES

1.  Ethyne is a colourless gas with a characteristic sweet smell when pure.

2.  It is only sparingly soluble in water

3.  It is slightly less dense than air.

4.  It is unstable and may explode on compression to liquid.

CHEMICAL PROPERTIES

1. Combustion: It undergoes combustion reaction in air to form water and carbon(IV) oxide

 2C₂H_2(g) + 5O_2(g) 2H₂O_(l) + 4CO_2(g)

NB: In limited air, it burns with very smoky and luminous flame because of its high carbon content. But in plenty of air and appropriate proportion, it burns with non-luminous very hot flame of about 3000^oC.

2.  Addition Reaction:Ethyne undergoes addition reaction to produce unsaturated product with double bonds and then a saturated compound with single bond.

a.  Reaction with hydrogen in the presence of nickel as catalyst.

H HHH  

 H- C C – H + H₂ H – C = C – H H_2(g) H – C – C – H

EtheneH H Ethane

b.  Reaction with halogens:

ClClClCl

 H- CC – H + Cl₂   H – C = C – H Cl₂ H – C – C – H

ClCl

1,2- dichloroethene1,1,2,2-tetrachloroethane

c. Reaction with hydrogen halide: Hydrogen halide reacts with ethyne toproduce halo-alkene and further halogenation produce halo-alkane.

ClCl

H – C = C – H +HCl H – C = C – H HCl H – C – C – H

H ClH H

Chloroethene   1,2- dichloro ethane.

d.  Reaction with water: When ethyne is passed through dilute tetraoxosulphate (vi) acid in the presence of mercury (II) tetraoxosulphate (VI) as catalyst, addition of water takes place to form ethanal.

  H OH H  

 H-C=C-H + H₂O H- C = C – H H- C – C – H

Ethenol  H O Ethanal

e.  Reaction with acidified KMnO₄: If ethyne is added to acidified KMnO4, it decolourises it. But with alkaline KMnO4, the solution turns to green.

O O  

H – C = C – H + 4[O] HO – C – C- OH

KMnO₄Ethanedioic acids

3.  Polymerization: In the presence of complex organic –nickel as catalyst ethyne polymerizes to produce benzene.

3 C₂H₂ C₆H₆

4.  Substitution Reaction

a.  Ethyne reacts with ammoniacal solution of copper (1) chloride to form reddish brown solution of copper (I) dicarbide

 C₂H₂ + 2CuCl Cu₂C₂ + 2HCl  

b.  With ammoniacal silver trioxonitrate (v), ethyne forms white silver dicarbide

 C₂H₂ + 2AgNO₃Ag₂C₂ + 2HNO₃

 These reactions to form dicarbide are used to distinguish ethyne from ethene.

USES OF ETHYNE: Ethyne is

1. Mixed with oxygen to produce oxy ethyne flame for cutting and welding of metals.
   1. Used in the manufacture of PVC plastics.
   2. Used in miner’s lamp as fuel.
   3. Used in making synthetic fiber.

       

TEST FOR UNSATURATION

Unsaturated compound decolorizes bromine water.

Evaluation

1. Give a chemical test to distinguish between ethyne and ethene.
2. Write two balanced equations to show addition reaction of ethyne.

AROMATIC HYDROCARBONS

These are hydrocarbons that have the same structure as benzene.

Benzene is a typical aromatic compound with molecular formula of C₆H₆.

STRUCTURE OF BENZENE

Over the years, there has been a controversy on the structure of benzene. But in 1865, August Kekule suggested a structure for benzene. Kekule proposed that benzene has a ring structure with alternate single and double carbon-carbon bonds as shown below:

which can be conveniently represented as

These two forms of benzene structure are known as resonance forms. Resonance occurs when two forms of the same compound have the same arrangement of atoms but differ in the arrangement of electrons that form the bonds.

The Kekule structure of benzene accounted for the stability of the six carbon atoms but it was unable to explain why a highly unsaturated compound failed to undergo many of the addition reactions like decolourising bromine water, reaction with hydrogen halides etc; characteristic of alkenes.

Benzene undergoes mostly substitution reactions. Thus, the structural formula with threedouble bonds describing the benzene molecule does not agree with the chemical behaviour of benzene. Therefore, the bonding in benzene cannot be described as three double bonds and three single bonds as proposed by Kekule but rather the bonding must be considered as a delocalized electron cloud spread out over the whole benzene ring. Hence, the modern structure of benzene is considered to be a plain hexagon with an inscribed ring which represents the electron cloud spread out over the whole benzene ring as shown below:

PREPARATION: Benzene can be prepared from:

1.  Coal tar: The destructive distillation of coal produced coal tar which contains benzene.

2.  Petroleum: The dehydrogenated of alkane using vanadium (V) oxide (V₂O₅) as catalyst at 500^oC and 20 atm gives benzene.

 C₆H₁₄ V₂O₂C₆H₆ + 4H₂

The process is known as catalytic reforming.

3.   Polymerization of ethyne

 3 ( H – C = C – H ) C₆H₆

Evaluation

1. Describe three methods of preparing benzene.

1. Draw the resonance structures of benzene structure of benzene.

PHYSICAL PROPERTIES

1.  It has a pleasant smell.

2.  It has boiling point of 80^oC.

3.  It is insoluble in water.

4.  It burns with sooty flame.

CHEMICAL PROPERTIES

Benzene can undergo both addition reaction and substitution reaction.

1. Addition Reaction:

   i. Hydrogenation: Benzene reduces to cyclo-hexane if hydrogen gas is passed through finely divided nickel at 150^oC.

 ii. Halogenation: In the presence of ultra-violet light, benzene reacts with halogen to produce cyclic compound.

2.  Substitution Reaction: Benzene undergoes substitution reaction due to presence of single bonds in its structure.

i.  Halogenation:

ii.   Nitration: This occurs in the mixture of HNO₃ and H₂SO₄ together with benzene

iii. Sulphonation: Benzene react with concentrated H₂SO₄ to form benzene sulphonic acid.

iv. Alkylation:- It involves reactions of benzene with halo-alkanes in the presence of AlCl₃.

USES

1.  It is used as a solvent to dissolve organic.

2.  It is used as fuel in petrol.

3.  It is used in the manufacture of aromatic compound e.g. benzoic acid.

GENERAL Evaluation/REVISION

1. How would you obtain ethanal from ethyne? Give the equation for the reaction.
   1. Describe how to prepare ethyne in the laboratory.
   2. What is resonance? Give the resonance structure of benzene.
2. Explain why hydrogen fluoride exists as a liquid whereas hydrogen chloride is a gas at room temperature.
3. Explain why HCl in water conducts electricity but HCl in methyl benzene does not conduct electricity.

    

READING ASSIGNMENT

New School Chemistry for Senior Secondary School by .O.Y. Ababio (6^th edition), pages 535-539.

WEEKEND ASSIGNMENT

SECTION A: Write the correct option ONLY

1.  Which of the following hydrocarbons is alkyne?

a. C₂H₄  b. C₂H₆ c. C₂H₂ d. C₃H₈

2.  The final product of complete reaction between ethyne and hydrogen gas is

a. ethane b. methyl ethane c. ethane  d. hydroethyne

3.  Ethynepolymerizes in the presence of organo nickel complex as catalyst to form a. polythene b. benzene c. polythene d. methyl benzene

4.  Which of these compounds exhibits resonance? a. Ethanol b. Ethane

 c. Benzene d. Butyne

5.  Which of these is an aromatic hydrocarbon? a. Benzene b. Cyclohexane

 c. Ethene d. Methylamine

SECTION B

1  a.With the aid of a labeled diagram, describe the laboratory preparation of ethyne.

 b. Give a chemical test to distinguish between ethane and ethyne.

2  a.What is resonance? Draw two resonance structure of benzene.

 b. Write balanced equation to the following reactions of benzene:

1. Reaction with ethene(ii) Reaction with chlorine.

WEEK NINE

TOPIC: ALKANOLS

CONTENT

• Types and Classes
• Industrial Production by Fermentation
• Properties and Uses

Alkanols is a homologous series with general molecular formula of C_nH_2n+1OH or ROH.

The functional group in alkanols is the hydroxyl (-OH) group.

NOMENCLATURE:

The names of alkanols are obtained by substituting “e” in alkanes with “ol”.

Example:

Methanol – CH₃OH, Ethanol – CH₃CH₂OH

CLASSIFICATION

The alkanols are classified based on the number of alkyl groups directly linked to the carbon atom carrying the hydroxyl group.

1.  Primary alkanols (1⁰): It has only one alkyl group attached to the carbon atom that carries the hydroxyl group. Example

H  H  H

  H  C  C OH H C OH

H  H  H

  Ethanol   Methanol

1. Secondary alkanols (2⁰): They have two alkyl groups directly linked with the carbon atom carrying the hydroxyl group. Example

    H  H  H  

   

    HC  C  C  H  

   

   H  OH  H

    Propan – 2- ol

2. Tertiary alkanols (3⁰ ): The alkanols here have three alkyl groups attached to the carbon atom holding the hydroxyl group

CH₃  

CH₃  C CH₃  

OH

2 – methylpropan -2 –ol

TYPES OF ALKANOLS

The type of alkanols is determined by the number of the hydroxyl group –OH, present in the molecule.

1. Monohydric alkanols: This type has only one hydroxyl(–OH), present in its molecule.

   Example: C₂H₅OH, C₃H₇OH.

2. Dihydric alkanols: This type has two of hydroxyl group per molecule.

H  H

 H  C  OH H  C  OH

 H  C  OH H  C  H

H H  C  OH  

H

Ethan -1, 2- diol Propan- 1,3-diol

1. Polyhydric alkanols: This type has three or more hydroxyl groups per molecule.

Example

H  

 H  C  OH  

 H  C  OH

 H  C  OH

H

EVALUATION

1.  Name the functional group in the alkanol.

2.  Give an example each by writing the structure and names of the classes of alkanols.

ETHANOL

LABORATORY PREPARATION

1. Hydrolyzing ethyl esters with hot alkali
2. Reducing ethanol with nascent hydrogen

COMMERCIAL PREPARATION

1. From ethene: Ethene is obtained by the cracking of petroleum. It is then absorbed in 95% H₂SO₄ at 80⁰C and 30 atm to form ethyl hydrogen tetraoxosulphate (VI)

   C₂H₄ + H₂SO₄ C₂H₅HSO₄

The ethyl hydrogen tetraoxosulphate (VI) is hydrolyse by boiling in water to produce ethanol.

C₂H₅HSO₄+ H₂O  C₂H₅OH + H₂SO₄.

The ethanol is distilled off leaving the acid behind which can be used again.

1. Preparation by fermentation: Ethanol is prepared industrially from raw materials containing starch or sugar by the process of fermentation. Fermentation is an enzymatic process which involves the decomposition of large organicmolecules to simple molecule by micro-organism.The common micro-organism used is YEAST

    

   PRODUCTION OF ETHANOL FROM STARCHY FOOD

   Ethanol can be prepared from starchy food like rice, potatoes, maize etc.

   The following steps are involved;

• Crush and pressure cook the starchy materials.
  • Extract the starch granules by mixing with water.
• Allow the starch granules to settle and decant
• Treat the starch granules with malt (partially germinated barley which contains the enzyme, DIASTASE) at 50⁰C for one hour.
• The starch is then converted to MALTOSE.

  2(C₆H₁₀O₅)n + nH₂O nC₁₂H₂₂O₁₁

• Then yeast is added at room temperature for some time (at least one day). Yeast contains two enzymes, namely MALTASE and ZYMASE. Maltase converts maltose to two glucose units, while Zymase converts the glucose to ethanol and carbon (IV) oxide.

  C₁₂H₂₂O₁₁ + H₂O ^maltase2C₆H₁₂O₆

  C₆ H₁₂ O₆  ^Zymase  2C₂H₃OH + CO₂

 Ethanol

EVALUATION

1.  Describe fully, the production of ethanol from a named starchy material/food.

2.  What type of chemical reaction is involved in fermentation of sugar?

PHYSICAL PROPERTIES

1.  Ethanol is a colourless volatile liquid.

2.  It is soluble in water.

3.  It has boiling point of 78⁰C.

4.  It has no action on litmus paper.

CHEMICAL PROPERTIES

1. Combustion: The lower members of alkanols burn with clean flames in plenty air

2CH₃OH + 3O₂  2CO₂+H₂O

2.  Oxidation: The products of oxidation depend on the structure of the alkanol.

–  Primaryalkanols are oxidized to alkanal first and then to alkanoic acid in the presence of oxidizing agent e.g KMnO₄

 CH₃CH₂OH  ^O  CH₃CHO  ^O  CH₃COOH

–  Secondary alkanols oxidize to alkanone. Example

CH₃  CH₃

CH₃  C  OH   [O ]   CH₃  C = O + H₂O

propanone

 H

Propon-2-ol

–  Tertiary alkanols are not oxidized because there is no carbon-hydrogen bond to be broken for the oxidation to take place.

Note: The colour change of oxidizing agent if acidified is purple KMnO₄ change to colourless and range K₂Cr₂O₇ turns green.

3.  Esterification:

This is the reversible reaction between alkanol and alkanoic and to produce sweet smelling compound known as ester. The reaction is catalysed by concentrated H₂SO₄. Example

CH₃CH₂OH +CH3COOH   H⁺   CH₃COOCH₂CH₃ + H₂O

4.  Dehydration:

 Alkanols are dehydrated to alkenes in the presence of concentrated H₂SO₄.

CH₃CH₂OH + H₂SO₄   CH₃CH₂HSO₄ + H₂O

CH₃CH₂HSO₄₁₇₀^o_CC₂H₄ + H₂SO₄

5.  Reaction with sodium and potassium: Sodium and potassium react vigorously with alkanols to liberate hydrogen gas and form corresponding organic salt of sodium and potassium.

2C₂H₃OH + Na   2C₂H₃ONa + H₂

6.  Reactions with the chlorides of phosphorus: Ethanol reacts vigorously with PCl₅ in the cold to produce fumes of HCl and chloroethanevapour.

C₂H₅OH + PCl₅C₂H₅Cl + POCl₃ + HCl

 PCl₃ gives a similar reaction, but less vigorous.

C₂H₅OH + PCl₃  3C₂H₅Cl + H₃PO₃

USES OF ETHANOL

1.  It is used as organic solvent.

2.  It is the main constituent of methylated spirit used to clean wounds and to dissolve paint.

3.  It is used as petrol addictive for use as fuel in vehicles.

4.  It is used to manufacture other chemicals such as ethanol and ethanoic acid.

5.  It is used as ingredient in making alcoholic drinks e.g. beers, wines and spirits.

6.  It is used as anti-freeze in automobile radiator because of its low freezing point (-117⁰C).

GENERAL EVALUATION/REVISION

1.  Describe how ethanol can be prepared from cane sugar.

2.  Using balanced equations, state five chemical properties of ethanol.

3.  Describe a test to identify an unknown solution to be ethanol.

4.  What is the number of oxygen atoms in 32g of the gas? [N_A = 6.02 x 10²³]

5.  5.6dm3 of oxygen gas was evolved at the anode during the electrolysis of dilute copper (II) tetraoxosulphate (VI) using platinum electrodes. What mass of copper is deposited at the cathode during the process? [Cu = 64, Molar volume of a gas at s.t.p = 22.4dm³, 1F = 96500C]

READING ASSIGNMENT

New School Chemistry for Senior Secondary School by O.Y.Ababio (6^th edition) pages 539-544.

WEEKEND ASSIGNMENT

SECTION A:Write the correct option ONLY.

1.  The functional groups of the alkanolis(a).double bond (b). carboxyl group (c). hydroxyl group (d). triple bond

2.  Primaryalkanols are oxidized to carboxylic acid; secondary alkanols are oxidized to alkanones while tertiary alkanols are (a). oxidized to alkanols(b). oxidized to alkanones(c). not oxidized (d).oxidized to alkenes

3.  The solubility of alkanols in water is due to (a). the covalent nature(b).hydrogenbonding

(c).their low melting point (d).their low melting point

4.  When acidified KMnO₄ is used as oxidizing agent for alkanol, the colour change observed is(a). yellow to red (b). purple to colourless(c). orange to green (d).white to black

5. Which of the following enzymes converts glucose to ethanol?(a). maltose (b).zymase

(c).diastase (d).amylase

SECTION B

1  (a).Write the structural formula of two named primaryalkanols.

 (b). Explain the structural different between secondary and tertiary alkanols giving one example each.

2  (a).What is fermentation?

 (b). Describe the preparation of ethanol from table sugar.