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Chemistry Scheme of Work for SS1 Second Term
Week 1: Revision/Introduction to the mole concept: Molar volume of gases, Avogadro’s number, Percentage of element in a compound.
Writing and Balancing Chemical Equations.
Week 2 – 3: Stoichiometry of Reactions: Calculation of Masses of Reactants and Products, Calculation of Volume of Reacting Gases. Empirical and Molecular formulae.
Week 4 – 5: Chemical Laws and their Verification: Law of Conservation of Mass, Law of Constant Composition, Law of Multiple Proportion.
Week 6 – 7: Chemical Combinations: Electrovalent Bond: Properties of Electrovalent
Compounds, Covalent Bond: Properties of Covalent Compounds. Other Types of Bonding
Week 8 – 9: TheKinetic theoryof Matter and the Gas Laws: Boyle’s Law, Charles’ Law, Ideal Gas Equation, Dalton’s Law of Partial Pressure.
Week 10: Avogadro’s law, Gay-Lussac’s Law of Combining Volumes, Graham’s Law of Diffusion.
- New Chemistry for Senior Secondary School by Osei Yaw Ababio; U.T.M.E Past Questions and Answers.
- Practical Chemistry for Senior Secondary Schools by Godwin Ojokuku
- Outline Chemistry for Schools & Colleges by Ojiodu C.C.
- Chemistry Pass Questions for S.S.C.E and UTME.
Lesson Note on Chemistry for SS1 Second Term
WEEK 1 TOPIC: Revision/Introduction to the mole concept: Molar volume of gases, Avogadro’s number, Percentage of element in a compound.
Lesson Note On Introduction to the mole concept
- Relative atomic mass
- Relative molecular mass
- Molar volume of gases
- Percentage of an element in a compound
A mole is a number of particles of a substance which may be atoms, ions, molecules or electrons. This number of particles is approximately 6.02 x 1023 in magnitude and is known as Avogadro’s number of particles.
The mole is defined as the amount of a substance which contains as many elementary units as there are atoms in 12g of Carbon-12.
RELATIVE ATOMIC MASS
The relative atomic mass of an element is the number of time 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.
The atomic mass of an element contains the same number of atoms which is 6.02 x 1023atoms; 1 mole of hydrogen having atomic mass of 2.0g contains 6.02 x 1023 atoms.
- Define relative atomic mass of an element
- State the relative atomic mass of the following elements: potassium, chlorine, silver, lead, phosphorus and nitrogen
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. It is also called the formula mass. The formula mass refers not only to the relative mass of a molecule but also that of an ion or radical.
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]
- MgCl 2 = 24 + 35.5×2 = 24 + 71 = 95gmol-1
- NaOH = 23 + 16 + 1 = 40gmol-1
- CaCO3 = 40 + 12 +16×3 = 100gmol-1
- What is relative molecular mass of a compound?
- Calculate the relative molecular mass of (a) NaNO3 (b) CuSO4.5H2O
MOLAR VOLUME OF GASES
The volume occupied by 1 mole of a gas at standard conditions of temperature and pressure (s.t.p) is 22.4 dm3. Thus 1 mole of oxygen gas of molar mass 32.0gmol-1 occupies a volume of 22.4dm3 at s.t.p and 1 mole of helium gas of molar mass 4.0gmol-1 occupies a volume of 22.4 dm3 at s.t.p.
Note: When the conditions of temperature and pressure are altered, the molar volume will also change. Also, standard temperature = 273K and standard pressure = 760mmHg.
RELATIONSHIP BETWEEN QUANTITIES
Molar mass = mass(g) i.e. M = m gmol-1
Amount (moles) n
Note: Amount = Number of moles
Molar volume of gas = volume ( cm3 or dm3) i.e. Vm = v dm3mol-1
Amount (mole) n
Amount = Reacting mass (g)
Molar mass (gmol-1)
Also, Amount of substance = Number of particles
But, Avogadro’s constant = 6.02 x 1023
Combining the two expressions:
Reacting mass = Number of particles
Molar mass 6.02 x 1023
- What is the mass of 2.7 mole of aluminium (Al=27)?
Amount = Reacting mass
Reacting mass = Amount x Molar mass
= 2.7mole x 27 gmol-1 = 72.9g.
- What is the number of oxygen atoms in 32g of the gas (O=16, NA = 6.02 x 1023)?
Reacting mass= Number of atoms
Molar mass 6.02 x 1023
Number of atoms = Reacting mass x 6.02 x 1023
Molar mass of O2 = 16×2 =32gmol-1
Number of atoms = 32g x 6.02 x 1023
= 6.02 x 1023
The number of oxygen atoms is 6.02 x 1023
- Define the molar volume of a gas
- How many molecules are contained in 1.12dm3 of hydrogen gas at s.t.p?
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
- What is the percentage by mass of nitrogen in NH4NO3 ( H=1, N=14, 0=16)?
Molar mass of NH4NO3 = 14×2 + 1×4 + 16×3 = 80gmol-1
Percentage by mass of N2 = Mass of N2 x 100
Molar mass of NH4NO3 1
= 28 x 100= 35%
- Calculate the percentage by mass of water of crystallization in MgSO4.7H2O
(Mg=24, S=32, 0=16, H=1)
Molar mass of MgSO4.7H20 = 24 + 32 + 16×4 + 9(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
- What is the number of molecules in 6.4g of SO2 (NA = 6.02 X 1023)?
- What is the volume in cm3 of 2.2g of CO2 at s.t.p ( C=12, O=16)?
- Determine the percentage by mass of oxygen in Al2(SO4).2H2( Al=27, S=32, O=16, H=1)
New School Chemistry for Senior Secondary Schools by O. Y Ababio, Pg 28-31
- What is the relative atomic mass of potassium A. 40 B. 39 C. 32 D. 24
- An element with relative atomic mass 108 is A. Ca B. Cl C. Ag D. Al
- Modern standard element with which chemist define relative atomic mass is A.12C B.C133H D.16O
- Calculate the relative molecular mass of CH3A. 60gmol-1B. 70gmol-1C. 80gmol-1D. 90gmol-1
- How many moles are there in 12g of CO2 (C=12, 0=16)?A. 0.27 B. 0.47 C. 0.16 D. 0.32
- Calculate the actual number of atoms contained in 2.8dm3 of chlorine (Molar volume of gas = 22.4dm3, NA = 6.02 X 1023)
- How many moles are there in 10g of iron (II) tetraoxosulphate (VI)?
Related Lesson Notes:
Lesson Note on Chemistry for SS1 Second Term
WEEK 2 TOPIC: Stoichiometry of Reactions: Calculation of Masses of Reactants and Products, Calculation of Volume of Reacting Gases. Empirical and Molecular formulae.
Lesson Note On Stoichiometry of Reactions: Calculation of Masses of Reactants and Products, Calculation of Volume of Reacting Gases. Empirical and Molecular formulae.
Chemical equations are representation of chemical reactions in terms of the symbols and formulae of the elements and compounds involved. In a chemical equation, the reactants are always written on the left hand side while the products are written on the right hand side. For instance, if A and B combines together to give C and D, the equation of the reaction is written as:
A + B → C + D
BALANCING CHEMICAL EQUATIONS
All equations must be balanced in order to comply with the law of conservation of matter. Equations are balanced through the use of coefficients in front of the formula and not by changing the subscript numbers within the formulae of the products.
Example 1: Write a balanced equation for the combustion of ammonia gas in air.
Step I: Write the reactants and predict the products
NH3(g) + O2(g) → NO(g) + H2O(g)
Step II: The equation is not balanced. Therefore the equation can be balanced by placing the right coefficient in front of each molecule to balance the number of atoms. Thus, the balanced equation is:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
Example 2: Write a balanced equation for the combustion of ethane in oxygen.
The general formula for the combustion of Alkanes is
CxHy + (x + y/4) O2 → XCO2 + y/2 H2O
The molecular formula for ethane is C2H6, so, x=2 and y=6
Substituting x and y into the formula above gives
C2H6 + (2 + 6/4) O2 →2CO2 + 6/2 H2O
C2H6 + 7/2 O2 →2CO2 + 3H2O
The equation is balanced. However, equations are written with whole number coefficients. By multiplying the entire equation by 2, we get
2C2H6 + 7O2 →4CO2 + 6H2O
IMPORTANCE OF CHEMICAL EQUATIONS
- It gives us information on the product that can be formed from the combination of two or more reactants in a particular reaction.
- It tells us the physical states of the reactants and products.
- It indicates the direction of the reaction and whether the reaction is reversible.
- It tells us the stoichiometry of the reaction (i.e. the relationship between the amount of reactants and products) in terms of mole ratio of the reactants and products involved.
Consider the table below: …………………..
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