RELATIVE MOLECULAR MASS, MOLAR MASS AND PERCENTAGE COMPOSITION

RELATIVE MOLECULAR MASS: -

The relative atomic mass or the relative molecular mass of an element is the number of times one mole of the element is as heavy as one-twelfth (1/12) the mass of one atom of carbon -12. It has no unit.

The relative molecular mass of a compound is the sum of the masses of all the atoms present in one molecule of the compound. e.g.

For NaCl, the relative molecular mass= (23 +35.5) = 58.5

For ethanol = C₂H₅OH (carbon=12, H=1, O =16)

The relative molecular mass of ethanol

= C₂H₅OH

  (12×2) + (1×5) + (16) + (1)

  24 + 5 + 16 +1 = 46

activities 

1. Calculate the relative molecular mass of the following compounds 

a. NaOH     b. CaCO₃   c. Al₂(SO₄)₂ 

[ Na = 23, O =16, S = 32, Ca = 40, Al = 27, C = 12, H = 1]

 Solution 

a.  NaOH 

(Na x 1) + (O x 1) + (H x 1)

(23 x 1) + (16 x 1) + (1 x 1)

 23 + 16 + 1 =

  40

b.  CaCO₃

(Ca x 1) + (C x 1) + (O x 3)

(40 x 1) + (12 x 1) + (16 x3)

 40 + 12 + 48 = 

    100

c. Al₂(SO₄)₃

   (Al x 2) + (S x 3) + (O x 12)

    (27 x 2) + (32 x 3) + (16 x 12)

      54 + 96 + 192 =

             342

THE MOLAR MASS

This is the relative molecular mass expressed in grams. That is, expressing the molecular mass of a compound in grammes.

E.g. the molar mass of ethanol is 46gmol^-1

In 12g of carbon-12, there are 6 × 10²³ atoms of carbon. This is one mole of carbon -12.

THE MOLE CONCEPT

The mole is the amount of a substance that contains as many elementary particles as there are atoms in 12 g of C-12.

Carbon -12 isotope has been taken as an arbitrary standard, and it contains 6×10²³ particles of C-12 atoms and it weighs 12grammes and so the amount of any substance that contains 6× 10²³ is equal to 1 mole of the substance, and the mass will be equivalent to the molar mass of the substance. 

A mole of any substance is the amount of that substance which contains 6× 10²³ particles of that substance e.g. One mole of ethanol has a mass of 46g and contains 6× 10²³ ethanol molecules.

The symbol of any element or the formular of any compound represents one mole of the element or compound. For example, 

     the symbol Na represents 1mole of sodium atom  

                        Na⁺ represents 1mole of sodium ions 

                       O2 represents 1mole od oxygen molecule 

                       H2O represents 1mole of water molecule

and all the examples above contain Avogadro's number of particles that is, 6× 10²³.

NOTE: The relative molecular mass has no unit, but the molar mass of any substance is expressed in grammes per mole (g/mol).

This number 6× 10²³ is known as Avogadro’s number or Avogadro’s constant

Calculations on mole concept 

1. How many moles are there in 20g of sodium atoms (Na)

Solution

1mole of Na =23 (since the symbol of an element represent 1mole)

     Number of moles = mass given 
                                      molar mass (relative atomic/molecular mass)

     i.e mass give = 20g, relative atomic mass of sodium = 23

number of moles = 20 = 0.87mols
                               23

the unit of the mole is mol

2. how many moles are there in 25g of CaCO3

   Solution 

1 mole of CaCO3= 40 + 12 + (16 x 3) = 100

 i.e, 1mole of CaCO3 =100g

        25g of CaCO3 will contain 

            number of moles = mass given                  25 = 0.25mols
                                          molecular mas             100

PERCENTAGE COMPOSITION OF A COMPOUND

The percentage composition of an element in a compound is calculated by dividing the total mass of the element in the compound multiply by 100.

Example

1.To calculate the percentage composition of the elements in ethanol whose molecular formula is C₂H₅OH, given the relative atomic masses of carbon, hydrogen and oxygen. [C= 12, H= 1, and O=16] 

First calculate the molar mass of C₂H₅OH  

Then determine the masses of C   H and O present.

Mass of carbon= 12×2=24g

Mass of hydrogen= 6×1 = 6g

Mass of oxygen = 16× 1 = 16g

(12×2) +(1×6) + (16) = 46gmol^-1

Molar mass of C₂H₅OH= 46g

Therefore, 

         % of C = 24 × 100 = 52.17%
                             46
       

           % of H = 6 × 100 = 13.04%
                               46
         

           % of O = 16 × 100 =    34.78%
                                46

 percentage composition of the elements in sodium hydroxide (Na2CO3)      [Na= 23, O=16, C=12]  

first you calculate the molecular mass of Na2CO3

(Na x 2) + (C x 1) + (O x 3)

(23 x 2) + (12 + 1) + (16 x 3)

  46 + 12 + 48 = 106

      %   of Na in Na2CO3 = 46 x100 = 43.40%
                                               106

     % of C in Na2CO3 = 12 x100 = 11.32%
                                           106

    % of O in Na2CO3 = 48 x100 = 45.28%
                                         106

       

OBJECTIVE QUESTIONS 

1. what is the percentage composition of Sulphur in Al2(SO4)3

a. 

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 simplest 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. 

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 respects 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

                   CH₃                                                                   CH₃       CH₃

(v)        CH₃CHC≡CCHCH₃                                (vi)       CH₃C-C≡C-C-CH₃

                                CH₃                                           CH₃      CH₃

            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

3. 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.

The following test can be performed to distinguished clearly the different classes of hydrocarbons, that is, the alkanes, alkenes and alkynes.

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.