Week 1, 2nd Term
SECOND TERM E-LEARNING NOTE
SUBJECT: CHEMISTRY SECOND TERM E-LEARNING NOTE
SUBJECT: CHEMISTRY CLASS: SS 1
SCHEME OF WORK
WEEK TOPIC
RESUMPTION TEST / MOLE CONCEPT 1 : Relative Molecular mass, Percentage of element in a compound.
2 MOLE CONCEPT 2 : Relative molecular mass, molar volume of a gas, No of particles, Empirical and molecular formular.
3. Writing and balancing Chemical Equation.
4. CHEMICAL LAWS and their Verification: calculation based on chemical laws.
5. CHEMICAL COMBINATIONS / BONDING : Electrovalent Bond: Properties of Electrovalent
Compounds, Covalent Bond: Properties of Covalent Compounds. Other Types of Bonding
6. . THE KINETIC THEORY OF MATTER : Meaning, state of matter, use of kinetic theory to explain Bownian movement, diffusion, osmosis.
7. OPEN DAY/ MID TERM
8. KINETIC THEORY OF GASES ; Postulations of kinetic theory, Gas laws « Boyle’s Law, Charles’ Law, Ideal Gas Equation, graphs and calculation.
9. GAS LAW CONTINUED ; Dalton’s Law of Partial Pressure, Avogadro’s law, Gay-Lussac’s Law of Combining Volumes, Graham’s Law of Diffusion.
10. GAS LAW : Experiment / Project
11. WATER ( introduction)
12. Revision
13. Examination
WEEK ONE
TOPIC: MOLE CONCEPT 1
THE MOLE
The mole is defined as the amount of a substance which contains as many elementary units as there are atom in 12g of Carbon-12.
Molar Mass
Concentration
1mole = 22.4 dm3
1 Avogadros ( 6.02 X 1023 )
1 Faraday ( 96500C)
Mole (n) = Mass (m)
Molar Mass
For gas at STP ( Standard Temperature and Pressure )
1 mole = 22.4 dm3
For solution
I mole (n) = Concentration (mol/dm3) X Volume (dm3)
For ideal gas mole (n) = PV
RT
P= pressure V= Volume, T = Temperature, R = general gas constant
For elementary particles 1 mole (n) = 6.02 x 1023 in magnitude and is known as Avogadro’s number of particles.
RELATIVE ATOMIC MASS
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 the mass of one atom of Carbon-12. It indicates the mass of an atom of an element. For e.g, the relative atomic mass of hydrogen, oxygen, carbon, sodium and calcium are 1, 16, 12, 23, and 40 respectively.
RELATIVE MOLECULAR MASS
The relative molecular mass of an element or compound is the number of times the average mass of one molecule of it is heavier than one-twelfth the mass of one atom of Carbon-12
It is the sum of the relative atomic masses of all atoms in one molecule of that substance.
CALCULATION
Calculate the relative molecular mass of:
Magnesium chloride
Sodium hydroxide
Calcium trioxocarbonate
[Mg=24, Cl=35.5, Na=23, O=16, H=1, Ca=40, C=12]
Solution:
MgCl 2 = 24 + 35.5x2 = 24 + 71 = 95gmol-1
NaOH = 23 + 16 + 1 = 40gmol-1
CaCO3 = 40 + 12 +16x3 = 100gmol-1
CLASS WORK
Calculate the relative molecular mass of:
H2SO4
Al2(SO4)3
NaNO3
[ H =1, S = 32, O = 16, Al =27, Na = 23, N=14]
PERCENTAGE OF AN ELEMENT IN A COMPOUND
The percentage composition of an atom in a compound is the amount of the atom expressed in percentage.
Percentage of an element in a compound = Mass of element in the compound x 100 Molar mass of compound 1
CALCULATIONS
1. What is the percentage by mass of nitrogen in NH4NO3 ( H=1, N=14, 0=16)?
Solution:
Molar mass of NH4NO3 = 14x2 + 1x4 + 16x3 = 80gmol-1
Percentage by mass of N2 =
Mass of N2 x 100
Molar mass of NH4NO3 1
= 28 x 100
80 1
= 35%
2. Calculate the percentage by mass of water of crystallization in MgSO4.7H2O
(Mg=24, S=32, 0=16, H=1)
Solution:
Molar mass of MgSO4.7H20 = 24 + 32 + 16x4 + 7(2+16) = 246gmol-1
7 moles of water of crystallization = 126g
Percentage by mass of water =
Mass of H2O x 100
Molar mass of MgSO4.7H2O 1
= 126g x 100
246gmol-1 1
= 51.2%
CLASS WORK;
Determine the percentage by mass of oxygen in Al2(SO4).2H2O.
( Al=27, S=32, O=16, H=1)
ASSIGNMENT: 1. Calculate the relative molecular mass of CH3COOH.
Determine the percentage by mass of Sulphur in Al2(SO4).2H2O.( Al=27, S=32, O=16, H=1)
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