CHEMISTRY SS1 SECOND TERM LESSON NOTE
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CHEMISTRY SS1 SECOND TERM LESSON NOTE
E-NOTE FOR CHEMISTRY SS1
SCHEME OF WORK FOR SECOND TERM SS1 CHEMISTRY.
WEEKS TOPICS
1. REVISION OF FIRST TERM /INTRODUCTION TO MOLE CONCEPT.
2. CALCULATIONS IN TERMS OF AVOGADRO’S CONSTANT,EMPIRICAL FORMULA AND MOLECULAR FORMULA
3. WRITING AND BALANCING OF CHEMICAL EQUATIONS.
4&5 STATE, ILLUSTRATIONS AND VERIFICATION OF CHEMICAL LAWS.
6. CHEMICAL COMBINATIONS OR BONDING.
7. THE KINETIC THEORY OF MATTER.
8. GAS LAWS AND CALCULATIONS.
9. GAY-LUSSAC’S LAW
10. AIR & FLAME.
CHEMISTRY SS1 SECOND TERM LESSON NOTE
WEEK 1
INTRODUCTION TO MOLE CONCEPT
Matter is defined as anything that has mass and occupies space it is made up of discrete of tiny particles such as atoms, molecules and ions.
ATOM: This is the smallest particle of an element which can participate/take part in a chemical reaction. E.g., O, N, F, S, Cl, Na, He etc.
NOTE: Atoms cannot exist separate.
MOLECULE: This is the smallest particle of a substance that is capable of independent existence and still retains the chemical properties of that substance. Examples H2, 02, F2, S8, Cl2, H2O, NaCl etc.
NOTE: A molecule is formed when two or more atoms are chemically joined together. The combination of atoms of the same type produces molecules of an element while the combination of different types of atoms produce molecule of a compound. The molecule does not lose its identity.
ATOMICITY: This is the number of atoms in a molecule of an element, most gaseous elements are diatomic. Examples comprise of noble gases and metals respectively. Example He, ,Ar, Ne, Ca, Mgetc.P4 tetra atomic element 03 triatomic element S8 polyatomic element.
CHEMISTRY SS1 SECOND TERM LESSON NOTE
IONS: These are atoms or group of atoms which possesses an electric charge. There are two types of ions namely:
i. Cations which results from the loss of electrons by the atom of a metal to become positively charge. Example, K+, Na+, Ca2+, Al3+, Cr3+, Ag+, Pb2+, NH4+
ii. Anions which results from the gain of electrons by the atom of non metal to become negatively charge. Example 0-2, Cl-, F-, S-2, N-3, 0H-, S042-, Cr2072-, MnO4- etc.
RELATIVE FORMULAE MASS
The relative formulae mass of a substance can be defined as the number of times the mass of one
Formulae unit of the substance is heavier than one twelfth of the mass of one atom of carbon 12
i.e.
RELATIVE FORMULAE MASS
Mass of 1 formulae unit of a substance 1/12 x mass of atom of Carbon -12.
RELATIVE ATOMIC MASS (RAM)
The relative atomic mass (RAM) of an element is the mass of one atom of the element. R.A.M
not a whole number because of the existence of the phenomenon of isotopy.
NOTE: R.A.M has no unit.
CHEMISTRY SS1 SECOND TERM LESSON NOTE
DEFINITION:
The relative atomic mass of an element can be defined as the number of times the average mass
Of atom of the element is heavier than one twelfth of the mass of one atom of carbon-12.
R.A.M.=mass of I atom of the element
1/12 x mass of I atom of carbon
NOTE:
a. Carbon is used as a standard
b. The instrument used for measuring relative atomic mass is called mass spectrometer
RELATIVE MOLECULAR MASS (Mr/Rmm) of a substance is the mass of one molecule of the substance. The r.m.m. is the sum of the R.A.M. of all the atoms present in the molecule.
NOTE: The r.m.m has no unit.
DEFINITION
The relative molecule mass of an element or a compound is the number of time the mass of one
Molecule of the element or compound is heavier than one twelfth of the mass of one atom of
Carbon-12.
R.m.m. = Mass of 1molecule of substance
1/12 x mass of atom of carbon -12
= mass of 1 molecule of substance x 12
Mass o atom of carbon -12
Example
1. Calculate the relative molecular masses of each of the following
a) H2S04 (b) Al2 (S04) 3
c) Fe S04 7H20
(H=1, S=32, 0=16, Al=27, Fe = 56)
SOLUTION
a. = 2 H + 1S + 40
= 2 x 1 + 32 + 4 x 16
2 + 32 + 64 = 98
b. Al2 (S04)3
= 2Al + 3S + 12 0
= (2x27) + (3x32) + (12x16)
CHEMISTRY SS1 SECOND TERM LESSON NOTE
54 + 96 + 192
= 342
c. FeS04. 7H20
= Fe+ S+40 + 7 (2H+0)
= 56 + 32 + (4x16) + (7x18)
= 56 + 32 + 64 + 126
= 278
Exercises
Calculate the R.m.m. of the following compounds
i. Calcium hydroxide Ca (OH) 2
ii. Lead (ii) trioxonitrate Pb (N03)2
iii. Ammoniumtrioxocarbonate (IV) (NH4)2C03
iv. Iron (IV) tetraoxo sulphate (VI) Fe2 (S04)3
v. FeS04.Al2 (S04)3.12H20
(R.a.m = Ca=40, 0=16, H=1,Pb = 207, N=14, C=12, Al=27, S=32, Fe =56)
MOLE
The mole can be defined as the amount of substance which contains Avogadro’s number of particles. The particles may be of different kinds, which may be atoms, molecules, ions, electrons, protons neutrons etc. Hence it is very necessary to state the type of particle involved.
NOTE:
. Mole is a unit of measurement
. The Avogadro’s number is constant and the value is 6.02 x1023 atoms or molecules or ions or electrons or protons or neutrons etc.
. For a reaction to occur, the particles of reactants must come together to form certain number of particles of product.
. Very large numbers of particles are to be worked with; hence it is difficult to measure individual particles in the reaction. To this effect, a unit for measuring the amount of particles in a given mass of a substance is designed and this is called a MOLE.
. From experimental work, it was found that 1 mole of substance = 6.02 x 1023 particles. This number is called AVOGADRO’S NUMBER.
. Therefore, a mole of any substance is the amount of that substance that contains the Avogadro’s number. Example just as 1 dozen of egg = 12 eggs so is 1 mole of oxygen
atoms contain 6.02 x 1023 oxygen atoms.
. 302 moles of oxygen molecules or 6 moles of oxygen atoms.
. 3MgS04 = 3 moles of MgS04 molecules or 3 moles of magnesium atom, 3 moles of sulphur atoms and 12 mole of oxygen atoms.
. 2H2 + 02 2H20 2moles of hydrogen molecules reacts with 1 mole of oxygen molecule to produce 2 moles of water.
MOLAR MASS
The mass of one mole of any substance expressed in grams unit = g/mol
Examples:
Chlorine gas (Cl2)
= 35.5 x 2, = 71g /mole
Carbon dioxide gas (C02)
= 12 + (16 x 2) = 44g/mol
THE MOLE CONCEPT
The mole can be expressed in the following ways:
i. The mole in terms of formula
ii. The mole in term of relative molecular man (R.A.M)
iii. The mole in terms of Avogadro’s number
The mole in term of the molar volume
NOTE: The expression of the mole in different ways mentioned above is known as the MOLE CONCEPT.
A. MOLE IN TERMS OF THE FORMULA
The mole can be expressed as:
Element: Mole = mass of the element
• Relative Atomic Mass of Element
OR
Compound: = Mole mass in grams
Or molecule R.M.M
Example:
Calculate the number of moles of atoms, present in 40g of calcium carbonate or calciumtrioxo
Carbonate (IV) (CaC03)
Solution:
Mass of CaC03 = 40g
R.M.M. CaC03 = (40 + 12 +16)
= 100glmol
Mole (n) = Mass is g
R.M.M CaC03
N = 40g
100g/mole
= 0.4 mole
QUESTION
1.A molecule is the smallest particle of
(A) a matter that can exist in Free State
(B) an element that can exist in Free State
(C) a radical that can exist in Free State
(D) a lattice that can exist in Free State
2. 3NH3 is
(A) three moles of ammonium
(B) three moles of ammonia
(C) six moles of ammonia
(D) six moles of ammonium
1. How many moles of substance are present is
a. 35g of oxygen gas (02)
b. 140g of sodium chloride (NaCl)
c. 150g of carbon dioxide gas (C02)
(0 – 16, Na = 23, C1 = 35.5, C =12)
2. Calculate the grams molecular mass for the following compounds
(a) Na2C03 (b) H3 P04
(c) Fe2 (S04)3 (d) Al2 03
(Ca = 40, C=16, H = 1,
P=31, Fe = 56, S =32, Al = 27)
3.Determine the number of grammes of substance contained in 0.5 moles of hydrogen chloride gas (HCl) (Cl=35.5, H=1).
WEEK 2
MOLE IN TERMS OF THE RELATIVE ATOMIC MASS OR RELATIVE MOLECULAR MASS OF ASUBSTANCE.
The mole can be expressed in terms of the R.A.M of an element or the R.M.M. of a substance/molecule/compound this:
1 mole of any substance = the R.A.M. of the substance or the R.M.M of the substance.
NOTE:
1 mole of Na (g) = 23g mole
1 mole of 02 (aq) = 16g mole
1 mole of 02 (g) = 16 x 2 = 32glmol
1 mole of C02 (g) = 12 + 16 x 2 = 44g1mol
1 mole of H2S04 (aq) = (2 x 1) + 32 +64 = 98glmol
Example:
1. Calculate the number of moles present in 11g of carbon dioxide or carbon (iv) oxide (C02) gas
Solution:
1 mole C02 (g) = Rmm of C02 (g)
1 mole of C02 = 12 + (16x2) = 44g g/mol
:. 44g of C02 = 1 mole of C02 (g)
1 g of C02 = 1/44 mole of C02 (g)
:. 11g of C02 = 1/44 x 11 mole of C02 (g)
= 0.25 mole of C02 (g)
• Determine the number of grammes of substance present in 0.05 of sodium carbonate (sodium trioxocarboante IV) (Na2 C03).
Solution:
1 mole Na2 C03 = Rmm of Na2 C03
1 mole Na2C03 = (23 x 2) + 12 + (16 x 3)
= 106glmol
:. 0.05 mole Na2 C03
Of Na2 C03
= 0.055moles
CHEMISTRY SS1 SECOND TERM LESSON NOTE
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