EMPIRICAL AND MOLECULAR FORMULAE

 

 EMPIRICAL AND MOLECULAR FORMULAE

Empirical formula: - This is the simplest formula of a compound; it shows the elements present and the ratio to which they are combined together.

The empirical formula of a compound can be calculated from the percentage compositions and the relative atomic mass of each element of the compound.

CALCULATION OF THE EMPIRICAL  FORMULA FROM PERCENTAGE COMPOSITION BY MASS.h

Te simplest empirical formula of a compound can be calculated from the percentage compositions of the various elements that make up the compound. For example, the formula for anhydrous disodium trioxocarbonate (iv)  can be calculated if the percentage composition by mass of each element present is known.

That is,  the percentage composition of the compound was found to be Na=43.40%, C= 11.32% and O = 45.28%. This would mean that in every 100g of the compound, the masses of Na, C and O were 43.40g, 11.32g and 45.28g respectively.

The amount in moles of Na, C and O would be.

      Na                C              O

  43.40g           11.32        45.28
   23                    12             16

 1.88                  0.94        2.83

1.88,  0.94 and 2.83 are the number of moles of each element respectively.

Now we divide by the smallest to get the mole ratio

 1.89:             0.94:         2.83. 

 0.94.             0.94.        0.94

  2:                    1    :        3 

            = Na₂C O₃      

The empirical formula is therefore, Na₂CO3

Example 2. What is the empirical formula of an organic compound whose percentage composition is carbon = 52.2%, hydrogen= 13.1% and oxygen = 34.7% (C = 12, H = 1, O = 16). 

It is very important to note that the addition of all the percentage compositions must be equally to 100

SOLUTION: 

  Carbon    Hydrogen      Oxygen      

  52.2           13.1              34.7

divide by the atomic mass

 52.2          13.1        34.7 
  12             1              16 

4.35           13.1            2.17

Divide by the smallest

    4.35          13.1           2.17  
   2.17          2.17            2.17.

 = 2           = 6               = 1

The empirical formular therefore = C₂H₆O.

Example 2. An organic compound has the following composition 55% of carbon, 9% hydrogen and 36% oxygen. Calculate the empirical formular for the  compound. (C= 12, H= 1, O= 16)

SOLUTION: 

  Carbon    Hydrogen Oxygen            

 55%            9%           36

Divide by atomic mass:

55             9                   36
12              1                   16

 4.58         9.00             2.25

Divide by the smallest: 

 4.58            9              2.25
 2.25           2.25           2.25

  2      :        4        :         1

The empirical formular = C₂H₄O.

Molecular formula of a compound is the actual formula of a compound, it shows the exact number of atoms present in one molecule of the compound.

Most molecular formulas are actually multiples of their empirical formulas, for example, a substance whose empirical formula is CH₂ have a molecular formula C₂H₄, C₄H₈ and so on. So the molecular formulas of a compound is calculated from the it's empirical formulary and it's molecular mass. 

I.e molecula formula = ( empirical formula)n

EXAMPLE 3: An organic compound contains carbon = 62.1%, hydrogen =10.3% and oxygen= 27.6% by mass.

(i) Find the empirical formula of the compound

(ii) If the molar mass of the compound is 58.0g, find its molecular formula. (C = 12, H = 1, O = 16).

SOLUTION: 

 Carbon    Hydrogen    Oxygen 

 62.1        10.3              27.6 
 12             1                  16

=5.1        = 10.3          = 1.8 

Divide by the smallest:

 4.35         13.1             1.8 
  1.8          1.8               1.8

  = 3          = 6                 = 1

Therefore, the empirical formular = C₃H₆O.

(ii) To calculate the molecular formula, relate the empirical formula to the molar mass   

(Empirical formula) n = Molar mass

     (C₃H₆O)n  = 58

(3x12 + 1x6 + 16x1)n   = 58

        58n   =   58
        58          58 

                 n= 1

The molecular formula = (C₃H₆O)n = C₃H₆O.

EXAMPLE 4: A hydrocarbon contains 20.80% of hydrogen and has a relative molar mass of 30, what is the

(i) empirical formula

(ii) molecular formula (C = 12, H = 1).

SOLUTION:

Hydrocarbons contain only two elements carbon and hydrogen. And the percentage composition of all elements in a compound must be equal to 100. Therefore, the percentage composition of carbon which is the second element contained by a hydrocarbon equals 79.20%. i.e 100 – percentage composition of hydrogen (20.80%). 

 Carbon              Hydrogen  

 79.20                  20.80 
  12                          1 

 =6.60                   = 20.80

  4.60                     20.80 
  6.60                      6.60 

    = 1                       = 3                      Therefore the empirical formular = CH₃.

(ii) To calculate the molecular formula, we relate the empirical formula to the molar mass.

(Empirical formula) n = Molar mass

       (CH₃)n     =     30

        (12 + 1×3)n = 30

          (15)n    =    30

               n    =    30 
                           15

                 n    =     2

The molecular formula = (CH₃)n = C₂H₆.

EXAMPLE 5: A carbohydrate contains 40% and hydrogen 6.72%, Calculate its empirical formula and the molecular formula, if the molar mass is 180 (C = 12, H = 1, O = 16).

SOLUTION:

Carbohydrate contains the elements carbon, hydrogen and oxygen, but from the question above oxygen is missing, hence the percentage composition of oxygen equals 100 – (percentage composition of carbon and hydrogen)

 = 100 – (40 + 6.72)

= 100 – 46.72 = 53.30% 

  Carbon     Hydrogen     Oxygen

         40          6.72             53.3 
         12            1                 16

   =3.33       = 6.72           = 3.33

Divide by the smallest:               

 3.33              6.72          3.33    
 3.33             3.33           3.33

  = 1             = 2             = 1

Therefore, the empirical formular = CH₂O.

 (ii) To calculate the molecular formula, we relate the empirical formula to the molar mass

.(Empirical formula)n = Molar mass

      (CH₂O)n    = 180

    (12 + 1x2 + 16)n   =  180

          (30)n  =    180

            n = ¹⁸⁰/₃₀

             n =  6

 The molecular formula = (CH₂O)n =(CH2O)6 = C₆H₁₂O₆.

The molecular formula = C₆H₁₂O_6.

OBJECTIVE QUESTIONS 

1. The empirical formula of a compound is: 

A. The actual number of atoms in a molecule
B. The simplest whole-number ratio of atoms present
C. The molecular mass of the compound
D. The structural arrangement of atoms

2. Which of the following represents an empirical formula?

 A. C₂H₄
B. C₆H₁₂O₆
C. CH₂
D. C₄H₈

3. The molecular formula of benzene is C₆H₆. Its empirical formula is: 

A. CH
B. CH₂
C. C₂H₂
D. C₃H₃

4. A compound contains carbon and hydrogen in the ratio 1:2. Its empirical formula is: 

A. CH
B. CH₂
C. C₂H₄
D. C₃H₆

5. The molecular formula of a compound is:

 A. The simplest ratio of atoms present
B. The arrangement of atoms in space
C. The actual number of atoms of each element in a molecule
D. The mass of one mole of the compound

6. The empirical formula of hydrogen peroxide (H₂O₂) is:

 A. HO
B. H₂O
C. H₂O₂
D. HO₂

7. If the empirical formula of a compound is CH₂ and its molecular mass is 42, the molecular formula is:

 A. CH₂
B. C₂H₄
C. C₃H₆
D. C₄H₈

8. What is the empirical formula of glucose (C₆H₁₂O₆)?      

 A. CH₂O

B. C₂H₄O₂

C. C₃H₆O₃
D. CHO

9. A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen. Its empirical formula is:

 A. CH₂O
B. CHO
C. C₂H₄O
D. C₂H₂O₂

10. The formula mass of the empirical formula CH₂O is:

 A. 12
B. 18
C. 30
D. 32

11. If the molecular mass of a compound is twice the empirical formula mass, then: 

A. Molecular formula = empirical formula
B. Molecular formula = 2 × empirical formula
C. Empirical formula = 2 × molecular formula
D. Molecular mass = empirical mass

12. Which of the following pairs have the same empirical formula?

 A. CO and CO₂
B. C₂H₄ and C₃H₆
C. H₂O and H₂O₂
D. CH₄ and C₂H₆

13. The empirical formula of ethyne (C₂H₂) is: 

A. CH
B. CH₂
C. C₂H₂
D. C₂H₄

14. A compound has the molecular formula C₄H₁₀. Its empirical formula is: 

A. CH₂

B. C₂H₅
C. C₄H₁₀
D. CH₃

15. The ratio of carbon to oxygen in carbon monoxide is:

 A. 1:1
B. 1:2
C. 2:1
D. 2:2

16. Which statement is correct?

 A. Every compound has the same empirical and molecular formula.
B. Molecular formula is always the simplest ratio.
C. Empirical formula may be different from molecular formula.
D. Empirical formula gives the actual number of atoms present.

17. If a compound has an empirical formula NH₂ and a molecular mass of 32, its molecular formula is: A. NH₂
B. N₂H₄
C. NH₃
D. N₂H₂

18. The molecular formula of a compound is C₂H₆O. Which of the following is its empirical formula? A. CH₃O
B. CH₂O
C. C₂H₆O
D. CHO

19. Which of the following compounds has identical empirical and molecular formulae? A. H₂O₂
B. C₆H₆
C. CO₂
D. C₂H₄

20. The empirical formula mass of CH is: A. 12
B. 13
C. 14
D. 15

THEORY QUESTIONS 

1. A hydrocarbon contains 92.40% of carbon. If the vapour density of the hydrocarbon is 39. Find

(i) empirical formula

(ii) molecular formula (C = 12, H = 1).

2. Calculate the empirical formular of an organic compound containing 81.8% carbon and 18.2% hydrogen (C = 12, H = 1).

3. What is the empirical formular of an oxide of phosporius that contains 43.6% phosphorous and 56.4% oxygen (P = 31, O = 16)

4. Determine the empirical formula of an oxide of nitrogen containing 70% oxygen, if the relative molecular mass of the oxide is 92, deduce its molecular formular 

ALKYNES

 

UNSATURATED HYDROCARBON (ALKYNES)

Alkynes are a homologous series of unsaturated hydrocarbons containing triple bond. It has a functional group of (≡) and general molecular formular of C_nH_2n-2 where n= 1,2,3, ... n for successive members of the group. 

The first member of the alkyne family is ethyne (acetylene).

 Alkynes are named by replacing ending –ane  of the corresponding alkane with –yne.

 

NOTE: Since alkynes contain triple bonds between C≡C therefore n=1 is not visible.

When n=	General Molecular Formulae  CnH2n-2	Name
2.	C2H2x2-2 = C2H2	Ethyne
3.	C3H2x3-2 = C3H4	Propyne
4.	C4H2x4-2 = C4H6	Butyne
5.	C5H2x5-2 = C5H8	Pentyne
6.	C6H2x6-2 = C6H10	Hexyne
7.	C7H2x7-2 = C7H12	Heptyne
8.	C8H2x8-2 = C8H14	Octyne
9.	C9H2x9-2 = C9H16	Nonyne
10.	C10H2x10-2 = C10H18	Decyne
11.	C11H2x11-2 = C11H20	Undacyne
12.	C12H2x12-2 = C12H22	Dodecyne
13.	C13H2x13-2 = C13H24	Tridecyne
14.	C14H2x14-2 = C14H26	Tetradecyne
15.	C15H2x15-2 = C15H28	Pentadecyne
16.	C16H2x16-2 = C16H30	Hexadecyne
17.	C17H2x17-2 = C17H32	Heptadecyne
18.	C18H2x18-2 = C18H34	Octadecyne
19.	C19H2x19-2 = C19H36	Nonadecyne
20.	C20H2x20-2 = C20H38	Icosyne/Eiocosyne
21.	C21H2x21-2 = C21H40	Heneicosyne
22.	C22H2x22-2 = C22H42	Docosyne
23.	C23H2x23-2 = C23H44	Tricosyne
24.	C24H2x24-2 = C24H46	Tetracosyne
25.	C25H2x25-2 = C25H48	Pentacosyne
26.	C26H2x26-2 = C26H50	Hexacosyne
27.	C27H2x27-2 = C27H52	Heptacosyne
28.	C28H2x28-2 = C28H54	Octacosyne
29.	C29H2x29-2 = C29H56	Nonacosyne
30.	C30H2x30-2 = C30H58	Triacontyne

 

 MOLECULAR STRUCTURES OF ALKYNES

N	ALKYNES	STRUCTURAL FORMULAR	MOLECULAR FORMULAR
2.	C2H2 Ethyne	H-C≡C-H	HC≡CH
3.	C3H4 Propyne	H  H-C-C≡C-H       H	CH3C≡CH
4.	C4H6 Butyne	H H  H-C-C-C≡C-H      H H	CH3CH2C≡CH
5.	C5H8 Pentyne	H H H  H-C-C-C-C≡C-H      H H H	CH3(CH2)2C≡CH
6.	C6H10 Hexyne	H H H H  H-C-C-C-C-C≡C-H      H H H H	CH3(CH2)3C≡CH
7.	C7H12 Heptyne	H H H H H  H-C-C-C-C-C-C≡C-H      H H H H H	CH3(CH2)4C≡CH
8.	C8H14 Octyne	H H H H H H  H-C-C-C-C-C-C-C≡C-H       H H H H H H	CH3(CH2)5C≡CH
9.	C9H16 Nonyne	H H H H H H H  H-C-C-C-C-C-C-C-C≡C-H      H H H H H H H	CH3(CH2)6C≡CH
10.	C10H18 Decyne	H H H H H H H H  H-C-C-C-C-C-C-C-C-C≡C-H       H H H H H H H H	CH3(CH2)7C≡CH
11.	C11H20 Undecyne	H H H H H H H  H H  H-C-C-C-C-C-C-C-C-C-C≡C-H       H H H H H H H H  H	CH3(CH2)8C≡CH
12.	C12H22 Dodecyne	H H H H H H H H HH  H-C-C-C-C-C-C-C-C-C-C-C≡C-H       H H H H H H H H HH	CH3(CH2)9C≡CH
13.	C13H24 Tridecyne	H H H H H H H H H H H  H-C-C-C-C-C-C-C-C-C-C-C-C≡C-H       H H H H H H H H H H H	CH3(CH2)10C≡CH
14.	C14H26 Tetradecyne	H H H H H H H H H H H H  H-C-C-C-C-C-C-C-C-C-C-C-C-C≡C-H       H H H H H H H H H H H H	CH3(CH2)11C≡CH
15.	C15H28 Pentadecyne	H H H H H H H H H H H H H  H-C-C-C-C-C-C-C-C-C-C-C-C-C-C≡C-H       H H HH H H H H H H H H H	CH3(CH2)12C≡CH
16.	C16H30 Hexadecyne	H H H H H H H H H H H H H H  H-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C≡C-H       H H H H H H H H H H H H H H	CH3(CH2)13C≡CH
17.	C17H32 Heptadecyne	H H H H H H H H H H H H H H H  H-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C≡C-H       H H H H H H H H H H H H H H H	CH3(CH2)14C≡CH
18.	C18H34 Octadecyne	H H H H H H H H H H H H H H H H  H-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C≡C-H      H H H H H H H H H H H H H H H H	CH3(CH2)15C≡CH
19.	C19H36 Nonadecyne	H H H H H H H H H H H H H H H H H  H-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C≡C-H       H H H H H H H H H H H H H H H H H	CH3(CH2)16C≡CH
20.	C20H28 Eiocosyne	H H H H H H H H H H H H H H H H H H  H-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C≡C-H       H H H H H H H H H H H H H H H H H H	CH3(CH2)17C≡CH

NOMENCLATURE OF ALKYNES

The nomenclature of alkynes is similar to that of alkenes in many ways as shown in the structures below. The only difference lies on the type of bonds, in alkenes (double bond) and alkynes (triple bond).

 

(i)         CH₃-CH₂-C≡CCH₃                     (ii)        CH₃CH₂CH₂C≡CCH₃    

            Pent-2-yne                                           hex-2-yne

            

             CH₃                                                        CH₃
              |                                                              | 
(iii)       CHC≡CCH₃                               (iv)       CH₂-C≡C-CH₂
              |                                                                                 |
             CH₃                                                                           CH₃
             4-methylpent-2-yne                                       hex-3-yne

                   

1.                         CH₃                                                                     CH₃       CH₃
2. (v)                    |                                                                              |            | 
3.                  CH₃CHC≡CCHCH₃                                (vi)       CH₃C-C≡C-C-CH₃
4.                                     |                                                                   |            |
5.                                    CH₃                                                             CH₃      CH₃
6.             2,5-dimethylhex-3-yne                                    2,2,5,5-tetramethylhex-3-yne

                                 CH₃

(vii)      CH₃CH-C ≡CC-CH₂CH₃                                 (viii)     CH₃C≡CCH₂

                   CH₃         CH₃                                                           CH₃

             2,5,5-trimethylhept-3-yne                              pent-2-yne

                      CH₃                                                                              CH₃ 

(ix)       CH≡CC-C=C-------CH—CH₂CH₂C≡CH               (x)        CH₃C-CHC≡CC≡CC≡CH 

                      CH₃              CH₂CH₃                                                    CH₂CH₃  

            6-ethyl,3,3-dimethyldec-1,6-diyne                 8-ethyl, 8-methynon-1,3,5-triyne

                                                                                              Cl

(xi)       CH₃C≡CCHCH₃                                    (xii)      CH₃-C-C≡CH 

                          Cl                                                                  Cl

            4-chloropent-2-yne                                         3,3-dichlorobut-1-yne

(xiii)     CH₃CHC≡CC≡CCHCH₃                          (xiv)     CH₃CHC≡CCHC≡CH  

                   Cl                Br                                                      Cl         Cl

            2-bromo, 7-chlorooct-3,5-diyne                     3,6-dichlorohept-1,3-diyne

                        H    

                    H-C-H      

             H H H           

(xv)  H-C-C-C-CC≡CH

             H H H        

                   H-C-H

                       H

            3,3-dimethylhex-1-yne

LABORATORY PREPARATION OF ETHYNES (ALKYNES)

Ethyne is prepared in the laboratory by adding cold water into calcium dicarbide (CaC₂). Much heat is evolved and sand is placed beneath the flask to protect the flask from breakage. Ethyne is collected over water. The chief impurity, phosphine, PH₃ is absorbed by the acidified CuSO₄ solution.

EQUATION FOR THE REACTION

CaC₂  +  2H₂O → Ca(OH)₂  +  C₂H₂.

                                                 Ethyne

 

 

PHYSICAL PROPERTIES OF ETHYNE

1. It is colourless gas

2. It has sweet smell when pure

3. Almost insoluble in water

4. It is neutral to litmus

5. It is strongly exothermic

CHEMICAL PROPERTIES OF ETHYNE

 Alkynes such as ethyne also undergoes addition reaction – a reaction in which one molecule of a compound is simply added on to the alkynes at the position of the carbon – carbon triple bond (C≡C) and this is converted to carbon – carbon single bond (C-C) that is, the alkanes. Examples of addition reaction are:

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

                                       Ni
            CH≡CH + 2H₂   →   CH₃CH₃   
            ethyne                           ethane

2. Reaction of ethyne with bromine to produce 1,1,2,2-tetrabromoethane. The reddish brown colour of bromine is destroyed.

            CH≡CH + 2Br₂ → CHBr₂-CHBr₂

Ķ. Reaction of ethyne with chlorine to produce hydrogen chloride

            CH≡CH + Cl₂ → 2C+ 2HCl

4. Reaction of ethyne with oxygen or combustion reaction of ethyne (alkynes) to produce carbon(iv)oxide and water

            2CH≡CH + 5O₂  →4CO₂ + 2H₂O

5. Polymerization reaction of ethyne to produce benzene.

            3C₂H_{2 →} C₆H₆ 

6. Reaction of ethyne with water in the presence of dilute H₂SO₄ and mercury as a catalyst to produce ethanal

            CH≡CH + H₂O →CH₃CHO

7. Reaction of ethyne with KMnO₄ to produce 1,2-ethan-diol (glycol)

            CH≡CH  +KMnO₄ →CH₂-CH₂
                                                 |        | 
                                                 OH   OH
                                               1,2-ethan-diol

USE OF ETHYNE

1. In oxy-acetylene flame for welding and cutting of metals

2. In oxy-acetylene torch

3. In preparation of acetic acid

4. as a starting material for making polyvinylchloride (PVC) which is used in electrical insulation and water proofing.

TESTS TO DISTINGUISHED BETWEEN ALKANES, ALKENES AND ALKYNES.

To distinguishe between the three aliphatic hydrocarbons, the following test can be performed to clearly the differentiate the classes of hydrocarbons, that is, the alkanes, alkenes and alkynes.

NOTE: All alkanes are saturated compounds while both alkenes and alkynes are unsaturated.

TEST 1: 

To the suspected hydrocarbons, add an acidified solution of KMnO₄ or K₂Cr­₂O₇ solution. Alkanes have no effect in any of these solutions while both alkenes and alkynes decolorized. Acidified KMnO₄ solution changes from purple to colourless, while K₂Cr₂O₇ changes from orange to green.

TEST 2: 

To the suspected hydrocarbons, add the solution of Ammonical copper (I) chloride. Alkanes and alkenes have no effect, but alkynes form a yellowish or reddish –brown precipitate.

2NH₄OH_(aq)+ 2CuCl  + C₂H₂ → CuC₂  + 2NH₄Cl  + 2H₂O.

TEST: To the suspected hydrocarbon, add solution of Ammonical silver tronitrate (V). Alkanes and alkenes have no effect, but alkynes form a yellowish precipitate.  

2NH₄OH + 2AgNO₃ + C₂H₂ → 2AgC + 2NH₄NO₃ + 2H₂O.

EQUILLIBRIUM note for students

                       EQUILLIBRIUM

EQUILLIBRIUM is said to be established in a reversible reaction when the rate of forward reaction is equally to the rate of backward reaction.

 We can describe equilibrium here as a state where there is no observable change.

In this Chapter we will be looking at Equilibrium in Chemistry and try to explain it as best as we can

Types of Equilibrium

Equilibrium can be Static and Dynamic

I. Static equilibrium: - In static equilibrium there is basically no movement within the system and the substances involved are at the same level. An example of a static equilibrium is when two children on a seesaw are suspended in space.

You will observe that both children suspended in space have the same potential energy. But any slight disturbance and the equilibrium will be lost.

Ii. Dynamic equilibrium: - In dynamic equilibrium, there is actually movements, but the changes are not observable. For example, a child going up an escalator that is actually descending at the same speed as the child, the child would seem to remain at a particular spot even though he is moving up.

Dynamic equilibrium can be grouped into two

I. Physical equilibrium and 

Ii. Chemical equilibrium

I. Dynamic physical equilibrium: -In dynamic equilibrium no new substance is formed (just like a physical change). For example, when sodium chloride is dissolved in water to give a saturated solution, the dissolved crystals will crystallize out of the solution just as undissolved crystals are dissolving into the solution, with these forward and backward reactions occurring at the same rate, we say the saturated solution is in equilibrium. 

  NaCl(s) ⇌ NaCl(aq)

II. Dynamic Chemical Equilibrium: - In this equilibrium system a new substance is formed (just like a chemical change).

An example is the reaction between Nitrogen and Hydrogen to yield Ammonia

N2(g) + 3H2(g) ⇌ 2NH3(g)

In dynamic equilibrium a new substance is formed.

Properties Of a System in Equilibrium

If a system is in equilibrium, then it will possess the following properties

I. The reaction must be a reversible reaction 

Ii. The rate of forward reaction must be equal to the rate of backward reaction 

III. ∆G must be equal to zero. (delta G is the Gibbs free energy)

IV. The equilibrium can be approached from either side of the reaction.

V. The reaction must occur in a closed system.

Factors affecting a system in equilibrium position of a reversible reaction

I. Temperature 

II. Concentration 

III. Pressure

IV. Catalyst

 

Le Chatelier's Principle: States that if a system is in equilibrium and an external factor ( such as a change in temperature, pressure or concentration) is impose on the equilibrium, the equilibrium will adjust itself so as to cancel the effect of the change

The effect in equilibrium position brought about by a change in any of the factors mentioned above was predicted and may be explained by Le Chatelier's principle 

i. How Temperature Affects equilibrium position of a reversible reaction: - Given a reaction where the forward reaction is exothermic, then the backward reaction will be endothermic.  For such a reaction, an increase in temperature will shift the position of Equilibrium to the left, that is, it will favor the backward reaction on the other hand, a decrease in temperature will favor the forward reaction since it does not require much heat.

How Concentration Affects the equilibrium position of a reversible reaction: - When any one of the reactants in a reversible reaction is increased this will cause the equilibrium position to shift to the right, favouring the forward reaction.   This is because an equilibrium has already been established and adding more reactants will alter that equilibrium, and so according to Le Chatelier's principle the reactants will be quickly used up and be converted to product. 

If the products are removed from the system as they produced, this can also cause the equilibrium to shift to the right.

For example, let us consider the equilibrium reaction of the Haber process, that the reaction between H2 and N2 the produce NH3

 

 N2(g) + 3H2(g) ⇌ 2NH3(g)

An increase in the concentration of the N2 or H2 will cause the equilibrium position to shift to the right, favouring the forward reaction.

How Pressure Affects the equilibrium position of a reversible reaction: - For Pressure to affect the equilibrium position of a reversible reaction, at least one of the reactants or products must be a gas and the total number of mols (vol.) of the reactants must be different from the total number of mols (vol.) of the product. 

For example, considering the two reactions 

  1. Fe(s) + 4H2O(g) ⇌ Fe2O3 + 4H2(g)        4vol.          £    4vol  2.  N2O4(g) ⇌ 2NO2(g)

1vol.        2vol.

Lq

Pressure cannot affect theqqq position of equilibrium of equation 1 because the volumes of both the gaseous reactant and gaseous product are the same (i.e. 4vol.)

However, in equation 2. the total volume of reactants is different from the volume of product, so an increase in pressure for such a reaction will lead to a decrease in volume (Boyle's law) hence causing the equilibrium to shift the left (the position that has a small volume) favouring the backward reaction. Similarly, a decrease in pressure will result to an increase volume and this will cause the equilibrium to shift to the right (area with larger volume) favouring the forward reaction.

Equilibrium Constant 

The equilibrium constant for a reaction in equilibrium, is the product of the molar concentrations (or pressures) of the products divided by the product of the reactants raised to the power of the coefficient of each reactant.

For a reaction 

            aA +bB ⇌ cC + dD

The equilibrium constant for the reaction will be 

                Kc = [C]^c[D]^d
                        [A]^a[B]^b

         

🧪 OBJECTIVE QUESTION

1. Chemical equilibrium is the state in which
A. reactants are completely used up
B. products stop forming
C. the rates of forward and backward reactions are equal
D. reactions no longer occur

2. A reaction at equilibrium is said to be
A. static
B. one-directional
C. dynamic
D. finished

3. Which of the following changes will shift equilibrium to the right?
A. Removing products
B. Adding products
C. Decreasing temperature
D. Adding a catalyst

4. Le Chatelier’s principle states that when a system at equilibrium is disturbed, it will
A. stop reacting
B. explode
C. oppose the disturbance
D. increase pressure

5. Increasing the concentration of reactants will
A. have no effect
B. shift equilibrium to the left
C. shift equilibrium to the right
D. stop the reaction

6. Which of the following does NOT affect chemical equilibrium?
A. Temperature
B. Pressure
C. Concentration
D. Catalyst

7. Increasing pressure favours the side of the reaction with
A. more gas molecules
B. fewer gas molecules
C. no molecules
D. equal molecules

8. At equilibrium, the concentration of reactants and products
A. are always equal
B. are always zero
C. remain constant
D. keep changing

9. A catalyst affects equilibrium by
A. changing the equilibrium position
B. stopping the reaction
C. speeding up both forward and backward reactions
D. increasing product yield

10. Which of the following best describes a reversible reaction?
A. It goes only forward
B. It can go forward and backward
C. It stops after some time
D. It is always complete

11 what will happen if more heat is applied to the following system in equilibrium

   

     X_2(g) + 3Y_2(g) ⇌ 2XY_3(g); ∆H= -xkjmol^-1

   

A. The yield of XY₃ will increase.

B. More of will XY₃ decompose 

C. more of X₂ will react 

D. The forward reaction will go to completion

12. What is the expression for the equilibrium constant (Kc) for the following reaction?  N_2(g) + O_2(g) ⇌ 2NO_(g)

A.      [NO]²
      [N₂] + [O₂]  

 

b.       [2NO]²
         [N₂]]O₂]

 

c.     [N2] + [O]
          2[NO]²

 

d.       [NO]²

        [N₂][O₂]

13. A reaction represented by the equation below  

     A_2(g) + B_(2(g) ⇌ 2AB_(g); ∆H = +X kjmol^-1             

which of the following statement about the system is correct

a. The forward reaction is exothermic 

b. The reaction goes to completion at equilibrium 

c. Pressure has no effect on the equilibrium mixture

d. At equilibrium increase in temperature favours the reverse reaction.

 

14.  The position of equilibrium in a reversible reaction is affected by 

       a). Particle size of the reactants 

       b). Change in concentration of the reactants 

       c). Change in the size of the reaction vessel 

       d). Vigorous stirring of the reaction mixture.

THEORY QUESTIONS

1(a). When few drops of aqueous KSCN are added to a solution of iron (III) salt the following equilibrium is _{The equilibrium mixture has a pale colour}

                                          

                                       ^{Yellow      colourless         deep red} 

(i). Explain what will happen if more KSCN were added to the equilibrium mixture 

(ii). Which of the ions in the equilibrium mixture forms an insoluble hydroxide with NAOH (aq)?   Write an equation for the reaction

(iii).  State two changes observed on adding NaOH to the equilibrium mixture.

(b) i. Given the equation:

 N2(g) + O2(g) ⇋ 2NO(g)  △H=+xkjol/mol

I. state one factor that will bring greater yield of the product.

II. Give a reason for your answer

 

2. Consider the reaction represented by the following equation 

    Q(s) ⇌ Q(l)    △H = -xkJmol-1.  

 what will be the effect of decrease in temperature on the system at equilibrium

ATOMIC NUMBER and ATOMIC MASS at a glance

CONSTITUENTS OF AN ATOM

The atom as explained earlier is made up of three sub-atomic particles. they are the 

i. Protons, 

ii. Electrons and 

iii. Neutrons. 

the Protons are positively charged. 

the Electrons are negatively charged    while 

the Neutrons have no charge (neutral)

     sub-atomic particles         charge                mass

1.             Protons                           + (positive)            1  (unit)

2.             Electrons                         - (negative)          0.00005

3.             Neutrons                          0 (neutral)            1 (unit)                                   

Atomic number: - Of an element is the number of protons in the nucleus of the atom of the element.

atomic numbers are whole numbers and there are NO two elements in the world that  have the same atomic number. 

Mass number or atomic mass: - of an element is the sum of the protons and neutrons in the nucleus of one atom of the element.

that is,

Atomic Mass = Number of protons + Number of neutrons

In chemistry an element X can be represented as    AZX      

where             A = Atomic mass or mass number

                      Z = Atomic number

Example          4020Ca     mass no = 40 (20 protons + 20 neutrons)

                                          atomic no = 20 (number of protons)

     Elements     Num. of  P        Num. of  E         Num. of  N

        40
1.    20 Ca                  20                    20                    20

   

     14
2.   7 N                       7                       7                    14

     

       39
3.    19 K                   19                   19                    20

THEORY QUESTIONS 

1.a(i)          State the constituents of an atom.

     (ii)           What is the number of protons, neutrons and electrons in the following elements

        (a).   115B   (b) 126C   (c.) 2311Na  (d) 3216S

1a(i) How many neutrons are present in the isotope 37Cl17 ?

(ii) State the number of electrons, protons and neutrons present in the following atoms/ions

a)    Ca         b) S2-         c) Al3+         d) P

b(i) Determine the number of electrons, protons and neutrons in each of the following: 39K19, 63.5Cu29

QUANTUM NUMBERS at a glance

 

Rules guiding the arrangement of electrons in an atom

When electrons are arranged or filled into the atoms of elements, certain RULES are considered  0 and obeyed. These rules are the Aufbau's Principle, Pauli's exclusion principle and Hund’s rule of maximum multiplicity.

PAULI EXCLUSION PRINCIPLE 

states that NO two electrons in the same atom have the sets of the four quantum numbers {n, l, m and s in an atom}.

AUFBAU PRINCIPLE states that "when electrons go into atoms they fill orbitals of lower energy first before filling orbitals of higher energy and each orbital may hold up to two electrons.

 

HUND’S RULE OF MAXIMUM MULTIPLICITY state that " When electrons fill degenerate orbitals they go in singly first before pairing up occurs.

 Degenerate Orbitals are orbitals that are at the same energy level. example of degenerate orbitals is the P-orbital, the d-orbital or the f-orbital

Examples of degenerate orbitals are the P-orbitals,  the d-orbitals  and the f-orbitals 

QUANTUM NUMBERS

The quantum numbers are a set of numbers that describes the position of an electron in an atom. 

Studies have showed  that the energy of an electron may be characterized by four quantum numbers. These quantum numbers help to locate the position of electrons in an atom

1. The principal quantum number represented by n with integral values of 1,2,3,4 e.t.c.

This quantum number describes the shell ( k, l, m....)

2. The subsidiary or Azimuthal quantum number represented by l with integral values ranging from 0 to (n-1).

This quantum number describes the sub-shells ( s,p,d,f,)

3.  The magnetic quantum number represented by m with integral values ranging from –l ,0, +l.

4. The spin quantum number represented by s with integral values – ¹/₂ and + ¹/₂.

Element   At. Numb.  Elect. Conf.

 H .            1;             1s¹

He              2;         1s²

Li.               3;         1s² 2s¹

Be              4;          1s² 2s²

B =.           5;          1s² 2s² 2p¹

C =            6;          1s² 2s² 2p²

N =.          7 ;         1s²2s²2p3      

O=            8 ;       1s² 2s² 2p⁴

F=.           9;         1s² 2s² 2p⁵

Ne=        10;         1s² 2s² 2p⁶

Na=.         11;        1s² 2s² 2p⁶ 3s¹

Mg=          12;     1s² 2s² 2p⁶ 3s²

Al=.        13;   1s² 2s² 2p⁶ 3s² 3p¹

Si =       14;    1s² 2s² 2p⁶ 3s² 3p²

P=.       15;     1s² 2s² 2p⁶3s²³3p³

S =.       16;   1s² 2² 2p⁶3s² 3p⁴

Cl =.    17;     1s² 2s² 2p⁶ 3s² 3p⁵

Ar =.   18      1s² 2s² 2p⁶ 3s²3p⁶

K=.    19  1s² 2s² 2p⁶ 3s²3p⁶ 4s¹

Ca =.  20  1s² 2s² 2p⁶ 3s²3p⁶ 4s²

OBJECTIV QUESTION

1.  Which of the following orbitals is spherical in shape?

     (a) s

      (b) p

      (c) d 

      (d) f

  

2.     Which of the following shells have a maximum of eight electrons?

             (a)  K

             (b) L 

             (c) M 

             (d) N

3.     1s² 2s² 2p⁶ 3p¹ is the electronic configuration of

              (a)  potassium

               (b) calcium 

               (c) sodium

              (d) aluminum.

4  . “Two electrons in an atom cannot have the same set for all four quantum numbers”. This statement is

              (a)  Aufbau principle 

               (b) Pauli exclusion 

               (c) Hund’s rule

              (d) Rutherford’s model.

5.    Which of the quantum number is represented by L?

              (a)  principal quantum number

              (b) subsidiary quantum number                  

              (c) magnetic quantum number

              (d) spin quantum.

6.    Pauli exclusion principles related 

a). quantity of electrons in the valence shell

b). filling the orbitals with lower energy first 

c). the filling of degenerate orbitals 

d). quantum numbers of electrons.

7. Atomic orbital is 

a). the circular path through which electrons which electrons revolve round the nucleus 

b). a region around the nucleus where electrons are most likely to be found 

c). the path around the nucleus through which electrons move 

d). the path around the nucleus through which protons move.

THEORY QUESTIONS 

  1(a)(i)what are the quantum numbers 

   (ii). The models below represent the filling of orbitals in an atom

   

State which rule(s) is/are violated or obeyed by each model

(iii).     State the following principle (a) Pauli exclusion principle. (b) Aufbau principle

(b). Write the electronic configuration of 

(i) Oxygen   (ii) Calcium (iii) Fluoride ion (Cl^-) (iv) Potassium ion  (K⁺)    (v). Aluminum ion (Al³⁺

2.a(i). List the quantum numbers that are assigned to an electron in an atom.

(ii) what is the maximum number of electrons that can occupy the 3d orbital?

3.(a)State 

I.  Pauli's exclusion principle 

ii. Hund's rule of maximum multiplicity

iii. Write the electronic configuration of of each of the following ions of copper I. Cu+ 

II. Cu2+.       [29Cu]