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Chapter 3 : Chemical Formulae and Equations 3.1 Relative Atomic Mass & Relative Molecular Mass 3.2 Relationship between the Number of Moles and the Number of Particles 3.3 Relationship between the Number of Moles of a Substance and its Mass 3.4 Relationship between the Number of Moles of a Gas and its Volume 3.5 Chemical Formulae 3.6 Chemical Equations 3.7 Scientific Attitudes and Values in Investigating Matter

Chapter 3 part 1 ( 3.1 3.4 )

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Page 1: Chapter 3 part 1 ( 3.1   3.4 )

Chapter 3 : Chemical Formulae and Equations

3.1 Relative Atomic Mass & Relative Molecular Mass3.2 Relationship between the Number of Moles and the Number of Particles3.3 Relationship between the Number of Moles of a Substance and its Mass3.4 Relationship between the Number of Moles of a Gas and its Volume3.5 Chemical Formulae3.6 Chemical Equations3.7 Scientific Attitudes and Values in Investigating Matter

Page 2: Chapter 3 part 1 ( 3.1   3.4 )

3.1 Relative Atomic Mass and Relative Molecular Mass

Relative Atomic Mass (Ar) – the number of times one atom of the element is heavier than one twelfth of the mass of a carbon-12 atom

Page 3: Chapter 3 part 1 ( 3.1   3.4 )

Example:

Sodium atom , Na is 23 time heavier than one-twelfth of the mass of one carbon-12 atom. Thus the relative atomic mass of Na is 23.

23atom 12 -carbon one of mass

121

(23) atom Na one of Mass

Page 4: Chapter 3 part 1 ( 3.1   3.4 )

Relative Molecular Mass (Mr) – the number of one molecule of the compound is heavier than one-twelfth of the mass of a carbon-12 atom

Page 5: Chapter 3 part 1 ( 3.1   3.4 )

Example:

A methane molecule , CH4 is 16 time heavier than one-twelfth of the mass of one carbon-12 atom. Thus the relative molecular mass of CH4 is 16.

16atom 12 -carbon one of mass

121

(16) molecule CH one of Mass 4

Page 6: Chapter 3 part 1 ( 3.1   3.4 )

3.2 Relationship between the Number of Moles and the Number of Particles

Mole – the amount of substance which contains the same number of particles (atoms/ions/molecules) as there are in 12 grams of carbon-12.

The number of atoms in 12grams of carbon-12 is 6.02 x 1023 (Avogadro’s Number or Avogadro’s Constant (NA)

Page 7: Chapter 3 part 1 ( 3.1   3.4 )

Example:- 1 mol of gold contains 6.02 x 1023 of gold atoms- 1 mol of magnesium ions contains 6.02 x 1023 Mg2+ ions

- 1 mol of magnesium chloride (MgCl2) contains 6.02 x 1023 Mg2+ ions and 2 x 6.02 x 1023 Cl- ions.- 1 mol of carbon dioxide contains 6.02 x 1023 CO2 molecules

CO2 is a covalent compounds; chemical bond that involves the sharing of electron pairs between atoms.

Page 8: Chapter 3 part 1 ( 3.1   3.4 )

Conversion of the number of moles to the numbers of particles and vice versa:-

Number of Particles = Number of mole x NA

Number of Moles = Number of particles ÷ NA

Example:-Calculate the number of particles in 0.75 mol of aluminium atoms,Al.Solution:0.75 mol x 6.02 x 1023 Al atoms = 4.52 x 1023

Al atoms.

Page 9: Chapter 3 part 1 ( 3.1   3.4 )

3.3 Relationship between the Number of Moles of a Substance and Its Mass Molar mass – mass of a substance that

contains one mole of the substance The molar mass of any substance contains

6.02 x 1023 particles The mole atom = relative atomic mass of an

atom but expressed in gram.Eg: 1 mole atom of Al = 27g

The mole molecule = relative molecular mass of a compund expressed in gram.

Eg: 1 mole molecule of water ,H2O = 18g

Page 10: Chapter 3 part 1 ( 3.1   3.4 )

Conversion of the number of moles of a substance to its mass and vice versa:-

Number of mole-atom = mass in gram ÷ relative atomic mass

Number of mole-molecule = mass in gram ÷ relative molecular mass

Mass in gram = Number of mole x relative atomic mass or relative molecular mass

Page 11: Chapter 3 part 1 ( 3.1   3.4 )

Example:

Calculate the number of moles of 23.5g of copper (II) nitrate , Cu(NO3)2.

[ RAM: Cu = 64, N=14, O=16]

Solution:

1 mol of Cu(NO3)2 = 64 + 2[14+3(16)] g

= 188 g RMM

23.5 g of Cu(NO3)2 =

= 0.125 mol

mol 1188

5.23

Page 12: Chapter 3 part 1 ( 3.1   3.4 )

Example:

Determine the mass for 0.08 mol of ascorbic acid , C6H8O6.

[RAM: C=12,H=1,O=16]

Solution:

1 mol of C6H8O6 = 6(12) + 8(1)+ 6(16)

= 176g

0.08 mol of C6H8O6 = 0.08 x 176g

=14.08g

Page 13: Chapter 3 part 1 ( 3.1   3.4 )

3.4 Relationship between the Number of Moles of a Gas and Its Volume

One mole of any gas at room temperature and 1 atm presure occupies a volume of 24dm3 ( 24000 cm3)

At standard temperature and pressure,s.t.p ( 0oC and 1 atm), one mole of gas occupies a volume of 22.4dm3 ( 22400 cm3).

Molar volume – volume occupied by one mole of any gas.

Page 14: Chapter 3 part 1 ( 3.1   3.4 )

Conversion of the number of moles of a gas to its volume and vice versa:-

Number of mole of a gas = volume of gas ÷ molar volume

Volume of gas = Number of mole of a gas x molar volume

Page 15: Chapter 3 part 1 ( 3.1   3.4 )

Example:

Calculate the number of moles of 4.8dm3 of chlorine gas at room temperature.

[1 mol of gas occupies a volume of 24dm3 at room temperature]

Solution:

Number of moles=

= 0.2 mol

mol 1dm24

dm8.43

3

Page 16: Chapter 3 part 1 ( 3.1   3.4 )

Example:

Calculate the volume of 0.75 mol of nitrogen gas at s.t.p.

[ 1 mol of gas occupies a volume of 22.4dm3 at s.t.p]

Solution:

Volume of nitrogen gas = 0.75 mol x

22.4dm3

= 16.8dm3

Page 17: Chapter 3 part 1 ( 3.1   3.4 )
Page 18: Chapter 3 part 1 ( 3.1   3.4 )

3.5 Chemical Formulae

Used to represent a chemical compound It shows:-

- the elements (denoted by symbols)

- the relative numbers (indicated by subscript after the symbol)

Example:-

H2O

Page 19: Chapter 3 part 1 ( 3.1   3.4 )

Chemical formulae of some covalent compounds

Name of compound

Chemical formula

Number of each element in the compound

Oxygen O2 2 oxygen atoms

Water H2O 2 hydrogen atoms

1 oxygen atom

Page 20: Chapter 3 part 1 ( 3.1   3.4 )

Chemical formulae of some ions (cations)

Charge Cation Symbol

+1 Sodium ion Na+

+2 Magnesium ion Mg 2+

+3 Iron(III) ion Fe3+

Page 21: Chapter 3 part 1 ( 3.1   3.4 )

Chemical formulae of some ions (anions)

Charge Anion Symbol

-1 Fluoride ion F-

-2 Oxide ion O2-

-3 Nitride ion N3-

Page 22: Chapter 3 part 1 ( 3.1   3.4 )

To write the chemical formula of an ionic compounds:-

- write the formula of the ions involved in forming the compound

- balance the positive and negative charge (use subscript)

- Write the chemical formula of the ionic compound without the charges.

Page 23: Chapter 3 part 1 ( 3.1   3.4 )

Example:-

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Formulae of some ionic compounds

Cation Anion Chemical formula

Na+ Cl- NaCl

Ca2+ Cl- CaCl2

Al3+ N3- AlN

Page 25: Chapter 3 part 1 ( 3.1   3.4 )

Empirical formulae of a compound:-

- shows the simplest ratio of the atoms of the elements that combine to form a compound

- steps to determine the empirical formula of a compound:-

1) write the mass / percentage of each element in the compound

2) calculate the number of moles for each element

3) Divide each number by the smallest number to obtain simplest ratio

4) Write the empirical formula of the compound

Page 26: Chapter 3 part 1 ( 3.1   3.4 )

Example:-

Page 27: Chapter 3 part 1 ( 3.1   3.4 )

Molecular formulae of the compound:-

- shows the actual numbers of the atoms of the elements that combine to form the compound

Compound Molecular formula

Simplest ratio of the elements

Empirical formula

Water H2O H:O = 2:1 H2O

Ethene C2H4 C:H=1:2 CH2

Glucose C6H12O6 C:H:O=1:2:1 CH2O

Page 28: Chapter 3 part 1 ( 3.1   3.4 )

3.6 Chemical Equations Chemical reaction can be represented by a

chemical equation Reactants – chemicals that are reacting.

Written on LHS. Products – chemicals formed in the reaction.

Written on the RHS

Page 29: Chapter 3 part 1 ( 3.1   3.4 )

Writing a chemical equation:-

1) write the correct formulae of all reactants on the LHS of the equation

2) write the correct formulae of all products on the RHS of the equation

3) the equation is then balanced.

4) make sure the number of atoms before and after reaction are the same

5) Write the physical state of each reactants and products

Page 30: Chapter 3 part 1 ( 3.1   3.4 )

Example:-