CHAPTER 3

a.Relative Atomic Mass

Relative Mass

The relative mass of an object is the comparison of the mass of the object to the mass of a standard object.

Relative Atomic Mass

The relative atomic mass (Ar) of an element is the average mass of one atom of the element when compared with 1/12 of the mass of an atom of carbon-12, which taken as 12 units.

1. The mass of an atom when compared to another is known as the relative atomic mass (Ar).
2. The relative atomic mass (Ar) of an element is the average mass of one atom of the element when compared with  1/12 of the mass of an atom of carbon-12, which taken as 12 units.
3. 1/12 of the mass of an atom of carbon-12 is named as 1 atomic mass unit (amu).
4. The mass of one carbon atom is 12 amu.
5. 1/12 of the mass of an atom of carbon-12 is named as 1 atomic mass unit (amu).
6. The mass of one carbon atom is 12 amu.

Example 1

The mass of a sodium atom is 23 times greater than 1/12 of the mass of carbon-12 atom. What is the relative atomic mass of sodium?

Answer:

23

Example 2:
The mass of element A is twice of the mass of carbon, therefore its relative atomic mass is __________. (Relative atomic mass of carbon = 12)

Answer:

Relative Atomic Mass of Element A = 2 x 12 = 24

Example 3:
An atom of element X is 13 times heavier than one atom of helium. Calculate the relative atomic mass of X.( Ar: He = 4 )

Answer:

Relative Atomic Mass of X = 13 x 4 = 52

Example 4:
How many times that the mass of 2 bromine atoms are greater than 4 neon atoms? (Ar: Ne = 20; Br = 80 )

Answer:

2(80)4(20)=2$$\frac{2(80)}{4(20)}=2$$

The mass of 2 bromine atoms is 2 time greater than the mass of 4 neon atoms.

Example 5
4 atoms of element L have same mass as 1 tellurium atom. Find the relative atomic mass of L. (Ar: Te = 128 )

Answer:

4L=1(128)L=1284=32$$\begin{array}{l}4L=1(128)\\ L=\frac{128}{4}=32\end{array}$$

The relative atomic mass of L = 32

b.Relative Molecular Mass

The relative molecular mass (Mr) of an element is the average mass of one molecule of the element/compound when compared with   of the mass of an atom of carbon-12, which taken as 12 units.

The relative molecular mass of a molecule is equal to the sum of the relative atomic mass of all the atoms in the molecule.

Example
Find the relative molecular mass of carbon dioxide.
[ JAR: C = 12; O = 16 ]

Answer:

The formula of carbon dioxide = CO2

Relative molecular mass of CO2 = 1 x 12 + 2 x 16 = 44

Example
What is the relative molecular mass of  aluminium sulphate [ Al2(SO4)3]?
( Ar: O = 16, S = 32; Al = 27 )
Answer:

Relative molecular mass of Al2(SO4)3 = 2 x 27 + 3( 32 + 4 x 16) = 342

C.Concept Of Mole

1. A mole is defined as the amount of substance which contains the Avogadro Number of particles.
2. The Avogadro Number (or Avogadro Constant) is defined as the number of atoms in 12 g of the carbon-12 isotope, which is equal to 6.02 x 10²³.

1 mole = 6.02 x 10²³

Example:
1 mol of atoms = 6.02 x 10²³ atoms
½ mol of atoms = 3.01 x 10²³ atoms
¼ mol of atoms = 1.505 x 10²³ atoms.

Note:
The particles in a material can be atoms, molecules or ions.
Therefore
1 mol potassium atom = 6.02 x 10²³ potassium atom
1 mol carbon dioxide molecules = 6.02 x 10²³ of carbon dioxide molecules.
1 mol sulphate ions = 6.02 x 10²³ sulphate ions.

Example:
Find the number of atoms in:

1. 2 mol ferum 
2. 3.6 mol zink 
3. 2.8 mol zink 
4. ¼ mol ferum 

Answer:

1. 2 mol iron = 2 x 6.02 x 10²³ = 1.204 x 10²⁴  iron atoms
2. 3.6 mol zink =   3.6 x 6.02 x 10²³ = 2.167 x 10²⁴  zink atoms
3. 2.8 mol zink = 2.8 x 6.02 x 10²³ = 1.686 x 10²⁴ zink atoms
4. ¼ mol iron = ¼ x 6.02 x 10²³ = 1.505 x 10²³  iron atoms.

D.Mole and mol

1. Mole is the unit of amount of substance.
2. “mol” is the symbol of mole.

E.Number Of Mole And Number Of Particles

1. We have just learn that, mole is a quantity, and it is equal to 6.02 x 10²³. The number 6.02 x 10²³ is called the Avogadro constant.  
2. Therefore, if we are given the number of mole of substance, and asked to find the number of particles (atoms, molecules or ions) in it, we multiply the number of mole by the Avogadro constant. 
3. Likewise, if we are given the number of particles, and asked to find the number of mole of the particles, we divide the number of particles by the Avogadro constant.

Example
Which contains more atoms, 1 mol of helium or 1 mol of uranium? Which has a greater mass? [ RAM: He=4; U=238 ]

Answer:

1 mol of helium and 1 mol of uranium has equal number of atoms.

The mass of one Uranium atom ss greater than the mass of one helium atom.

Example
Find the number of atoms in 2.5 mol of gold.

Answer:

Number of atoms

= Number of mole x Avogadro constant

= 2.5 x 6.02 x 10²³ = 1.505 x 10²⁴ 

Example
How many moles of magnesium that contain 2.76 x 10²³ of magnesium atom?

Answer:

Number of mole

= Number of atoms ÷ Avogadro constant

= 2.76 x 10²³ ÷ 6.02 x 10²³

= 0.46 mol

F.Number Of Mole Of Atoms And Number Of Mole Of Molecules

1. All molecules contain more than one atom.
2. For example, in a carbon dioxide molecule (CO2), there are 3 atoms - 1 carbon atom and 2 oxygen atoms.
3. Therefore, in 5 carbon dioxide molecules, there will be 15 atoms  - 5 carbon atom and 10 oxygen atoms.
4. Similarly,  in 1 mole carbon dioxide molecules, there will be 3 mole atoms  - 1 mole carbon atoms and 2 mole oxygen atoms, and in 3 mole carbon dioxide molecules, there will be 9 mole atoms  - 3 mole carbon atoms and 6 mole oxygen atoms, so on and so forth.

Example:

1. How many hydrogen molecules can be made with 20 mol of hydrogen atoms?
2. How many hydrogen atoms are there in 20 mol of hydrogen molecules?

Answer:

a. 

Each hydrogen molecules (H2) consist of 2 hydrogen atoms.

Therefore, 20 mol of hydrogen atoms can make 10 mole of hydrogen molecules.

b. Number of mole of hydrogen atoms in 20 mole of hydrogen molecules

= 2 x 20 mole = 40 mole.

Example:

1. Find the number of carbon atom in 0.75 mol of carbon dioxide.
2. Find the number of oxygen atom in 0.75 mol of carbon dioxide.

Answer:

a. 

Each carbon dioxide molecules (CO2) consist of 1 carbon atom and 2 oxygen molecules.

Therefore, 0.75 mole of carbon dioxide contain 0.75 mole carbon atoms.

b.

Number of mole of oxygen atoms in 0.75 mole of carbon dioxide

= 2 x 0.75 mole = 1.50 moles.

G.Number Of Mole And Mass Of A Substance

Molar Mass

1. The mass of one mole of atoms is its relative atomic mass in grams, and is called a molar mass.
2. A Molar mass is the mass of a substance containing the Avogadro Constant of particles.

Element	Relative atomic mass	Mass of 1 mol of particle	Molar Mass
Hydrogen	1	1 g	1 g mol-1
Carbon	12	12 g	12 g mol-1
Oxygen	16	16 g	16 g mol-1
Copper	64	27 g	27 g mol-1
Iron	56	56 g	56 g mol-1

Example

1. The relative atomic mass of copper is 64, therefore the mass of 1 mole copper is 64g
2. The RAM of sodium is 23, therefore the mass of 2 mole sodium is 46g
3. The RAM of nitrogen is 14, therefore the mass of 2.5 mole l nitrogen is 35g
4. The RAM of calcium is 20, therefore the mass of 0.3 mole calcium is 6g

Conclusion 

Note:
The relationship between the mass of a substance and the number of mole of the particles in the substance can be summarised by using the following equation:

n=mMolar Mass$$n=\frac{m}{\text{Molar Mass}}$$

If you are given the mass of substance and asked to find the number of mole of the substance (or vice versa), the problem can be solved by using this equation.

Example:
Find the number of mol of atoms in 4.6g sodium [Relative atomic mass: Na=23]

Answer:
Number of mole,

n=4.623=0.2mol$$n=\frac{4.6}{23}=0.2mol$$

Example
How many moles of each substance are there in 191 g NaOH [Relative atomic mass: Na=23, O=16, H=1]

Answer:
Relative Formula Mass of NaOH = 23 + 16 + 1 = 40
Number of mole of 191g NaOH,

n=19140=4.775mol$$n=\frac{191}{40}=4.775mol$$

H.Mass Of Substance And Number Of Particles

1. Sometime, you may be given the mass of a substance, and asked to find the number of particles (or vice versa). 
2. To solve the problem, we must find the number of mole of the substance then only we can find the number of particles (or mass of the substance).

Example:

How many molecules are there in 16 g of oxygen (O2)? [ Ar: O =.16, Avogadro Number = 6 x 10²³]

Answer:
The relative molecular mass of  O2 = 2(16) = 32

Number of mole of oxygen molecules=Mass of oxygenMolar mass of oxygen=16 g32g mol−1=0.5 mol

I.Number Of Mole And Volume Of Gas

Molar Volume of Gas

1. Equal amount of gas occupies same volume of space.
2. Molar volume of a gas is defined as the volume occupied by one mol the gas at certain condition
3. At room temperature(25ºC) and pressure (1 atm), one mole of any gas occupies a volume of 24 dm³ (24 000 cm³ )
4. At standard temperature(0ºC) and pressure (1 atm) one mole of any gas occupies a volume of 22.4 dm³ (22400cm³).

Number of mole of gasses	Volume at s.t.p. (dm3)
1 mol of O2 gas	1 x 22.4 = 22.4
1 mol of H2 gas	1 x 22.4 = 22.4
2 mol of He gas	2 x 22.4 = 44.8
3.5 mol of N2 gas	3.5 x 22.4 = 78.4

Number of mole of gasses	Volume at r.t.p. (dm3)
1 mol of O2  gas	1 x 24 = 24
1 mol of H2 gas	1 x 24 = 24
2 mol of He gas	2 x 24 = 48
3.5 mol of N2 gas	3.5 x 24 = 84

Example:
What volume (at room temperature and pressure r.t.p.) would 2 moles of oxygen gas occupy? (Molar Gas Volume at r.t.p. = 24 dm³)

Answer:
Volume of gas = 2 x 24 = 48 dm³

Example:
A sample of ozone gas has a volume of 960cm³ at room temperature and pressure. Find the number of mole of the ozone. [Molar volume at r.t.p. = 24.0dm³]

Answer:

Number of mole=960cm324000cm3=0.04 mol

CHEMICAL FORMULAE

1. 
   
2.  A chemical formulae A representation of a chemical substance using letters for atom and subscript numbers to show the numbers of each type of atoms that are present in 
3. the substance.
4. 
   
5. 
6. 
   
7. 
   
8. 
   
9. 
10. 
    
11.  Compound can be represented by two types: 
12. 1. Empirical formula
13.  2. Molecular formula
14. 
    
15.  Empirical Formula 
16. 
    
17. Meaning
18.  Formula that show the simplest whole number ratio of atoms of each element in the compound.
19. 
    
20. Example:
21.  A sample of aluminium oxide contains 1.08 g of aluminium and 0.96 g of oxygen. What is the empirical formula of this compound? [Relative atomic mass: O = 16; Al = 27]
22. 
    
23. 
24. 
    

1. Experiment of Magnesium Oxide Empirical Formula

Procedure of the experiment:

• Magnesium ribbon is cleaned with sandpaper to remove the oxide layer on its surface.
• The lid is removed at intervals to allow oxygen to enter the crucible and react with magnesium.
• The crucible is then quickly covered with its lid to prevent the white fumes of MgO from escaping.
• Heating, cooling and weighing are repeated until a constant mass is obtained to ensure that the magnesium ribbon reacts completely with oxygen gas.

The white powder is magnesium oxide

1.  Molecular Formula
2. 
   
3.   Meaning
4.  Formula that show the actual number of atoms of each element that are present in a molecule of the compound.
5.  Molecular formula = (Empirical formula)n
6.  Example: 
7. (CH3)n = 30 n [12 + 3(1) ] 
8. = 30 15n 
9. = 30 n
10.  = 30/15
11.  = 2 
12. Molecular formula = (CH3)2 = C2H6
13. 
    

Ionic formulae

1.  
2. 
   
3. 

1.  Chemical Equation 

• Chemical equation is a representation of a chemical reaction in word or using chemical formulae.
• It is a precise description of a chemical reaction.
• Chemical reaction occurs when the reactants react to produce products.

A balanced chemical equation has the following characteristic:

• the correct formulae for reactants and products of the reaction
• the same number of atom of each elements on the right and the left