Chemistry -Mole concept

THE MOLE CONCEPTS 

PERFORMANCE OBJECTIVES

By the end of the lesson, the learners should be able to;

● Know how to measure entities in chemistry (Remembering)

● Explain the terms “relative atomic mass” and “relative molecular mass” of a substance (remembering & Analyze)

● Determine the relative molecular mass of a substance (apply)

● Define the term “mole as unit of measurement and its mathematical expression” (Remembering)

● Solve for the amount of a substance and the percentage mass of element in compounds (apply)

● Weigh the mass of substance (NaOH) given, determine its amount (mole) and the number of atoms present in the   

              laboratory (evaluate)

HOW TO MEASURE ENTITIES IN CHEMISTRY 

Just as the solid object and liquid are measure in grams and litres respectively, similarly, mole is used to measure entities on the microscopic level e.g. atoms, ions, electrons and molecules.

RELATIVE ATOMIC MASS OF AN ELEMENT

The relative atomic mass of an element is the number of times the average mass of one atom of that element is heavier than one-twelfth (1/12) the mass of one atom of Carbon-12 i.e. 

It is simply the mass of an atom of that element e.g the relative atomic mass of Hydrogen (H), Oxygen (O), Carbon (C) and Sodium (Na) are approximately 1, 16, 12 and 23 respectively.

The atomic mass of an element contains the same number of atoms which is 6.02 x1023 atoms

e.g 1 mole of hydrogen having atomic mass of 2.0g contains 6.02 x1023 atoms

RELATIVE MOLECULAR MASS OF A MOLECULE/COMPOUND

The relative molecular mass of a molecule or compound is the number of times the average mass of one molecule of it is heavier than one-twelfth (1/12) the mass of one atom of Carbon-12. It is the sum of the relative atomic mass of all atoms in one molecule of that substance.

Example 1

1a. Calculate the relative molecular mass of limestone, CaCO3 (Ca =40, C=12, O=16)

Solution:

RMM of CaCO3 = (1 x Ar of Ca) + (1 x Ar of C) + (3 x Ar of O)

                           = (1 x 40) + (1 x 12) + (3 x 16) = 40 + 12 + 48

                           = 100g or 100g/mol

1b. Calculate the relative molecular mass of Sodium trioxocarbonate (IV) pentahydrate, Na2CO3.5H2O

      (Na= 23, C=12, O=16, H=1)

Solution:

RMM of Na2CO3.5H2O  = (2 x Ar of Na) + (1 x Ar of C) + (3 x Ar of O) + 5((2 x Ar of H) + (1 x Ar of O))

                                        = (2 x 23) + (1 x 12) + (3 x 16) + (2 x 1) + (1 x 16) = 46 + 12 + 48 + 5(2 + 16)

                                        = 46 + 12 + 48 + 5(18) = 46 + 12 + 48 + 90

                                      = 196g or 196g/mol

Class Activity 1: Calculate the relative molecular mass of lead (II) nitrate (Pb=108, N=14, O=16)

    MOLE AND ITS MATHEMATICAL EXPRESSION

● A mole is a number of particles of a substance which may be atoms, ions, molecules or electrons. This number is approximately 6.02 x1023 in magnitude and is known as Avogadro’s number of particles i.e.

● It can be defined as the amount of a substance which contains as many elementary units as there are atoms in 12g of Carbon -12

● Mole can be also expressed as the ratio of  the mass of a substance to its molar mass

          Note: Molar mass is also known as relative molecular mass of the substance

DETERMINATION OF MOLES

Example 2: What is the mass of 3 moles of oxygen, gas O2 (O = 16)?

Solution:

Mass of 1 mole of O2 = (2 x 16.0g) = 32.0g

Mass of 3 moles of O2= (3 x 32.0g) = 96.0g

Example 3: How many atoms are there in 6g of carbon, C (1 mole = 6.02 x1023, C=12)

Solution:

Method 1

             12g of C = 1 mole 

             6g of C = ½ mole 

Since 1 mole = 6.02 x 1023 atoms

          ½ mole = ½ x 6.02 x 1023 atoms

                     = 3.01 x 1023 atom

Example 4: How many moles are there in 20g of CaCO3?  (CaCO3 = 100g/mol)

Solution:

Class Activity 2

i.          Calculate the number of sodium atoms in 0.121mol Na?  (1 mole = 6.02 x1023  atoms)  

 Answer:

ii.         What is the mass in grams of 2.01 x1022 atoms of Sulphur?

Answer:

 PERCENTAGE COMPOSITION OF ELEMENT IN COMPOUNDS

Percentage by mass of a component element of a compound from its formula using mole concept

Class activity 3

Calculate the percentage by mass of all the component elements in sodium trioxonitrate (V), NaNO3

(Na= 23, N=14, O=16)

LABORATORY WORK - WEIGHING OF SAMPLE C (Na2CO3)

Apparatus & reagent: weighing/analytical balance, weighing dish (filter paper/watch glass/beaker),

       spatula, sample.

Procedure:     1. With nothing on the balance, set to zero.

                             2.  Place your weighing dish on the balance

                         3. Set the balance with beaker to zero again

                             4.  Use clean spatula to transfer the given sample to determine its mass

Result

Mass of the sample C                                            =   _______________

Mole of the sample C                                                =   _______________

Number of atoms in the sample C                       =   _______________

Evaluation

Click the mole concept for multiple choice questions

Thanks for listening!