Lesson Notes By Weeks and Term v3 - Senior Secondary 1
Symbols, Formulae and Equations
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Symbols, Formulae and Equations
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Subject: Chemistry
Class: Senior Secondary 1
Term: 3rd Term
Week: 1
Theme: The Chemical World
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state the symbolsof the first 20elements and othercommon elements; distinguishbetween elements,compounds and mixtures write chemicalformula and chemicalequations calculate betweenthe empirical and molecular for mulae of compound illustrate thatmatter is neithercreated nor destroyed state and illustratethe laws of constantcomposition and multiple proportions
The Chemical World Assume 100g of the compound (so percentages become masses in grams).
2. Convert the mass of each element to moles using its atomic mass (Moles = Mass / Molar Mass).
3. Divide each number of moles by the smallest number of moles obtained.
4. If the ratios are not whole numbers, multiply all ratios by the smallest integer that converts them to whole numbers.
5. Write the empirical formula using these whole number ratios as subscripts. Steps to Determine Molecular Formula from Empirical Formula and Molar Mass:
1. Calculate the molar mass of the empirical formula (EFM).
2. Determine the value of 'n' using the formula: n = (Molar Mass of Compound) / (EFM).
3. Multiply the subscripts in the empirical formula by 'n' to get the molecular formula.
Worked Example 1: Determining Empirical Formula A compound is found to contain 40.0% Carbon, 6.7% Hydrogen, and 53.3% Oxygen. (Relative Atomic Masses: C=12, H=1, O=16).
Step 1: Assume 100g sample. Mass of C = 40.0 g Mass of H = 6.7 g Mass of O = 53.3 g Step 2: Convert to moles. Moles of C = 40.0 g / 12 g/mol = 3.33 mol Moles of H = 6.7 g / 1 g/mol = 6.7 mol Moles of O = 53.3 g / 16 g/mol = 3.33 mol Step 3: Divide by the smallest number of moles. (Smallest is 3.33) C: 3.33 / 3.33 = 1 H: 6.7 / 3.33 ≈ 2 O: 3.33 / 3.33 = 1 Step 4: Check for whole numbers.
Ratios are 1:2:1, which are whole numbers.
Step 5: Write the Empirical Formula.
CH2O Worked Example 2: Determining Molecular Formula If the compound in Example 1 (Empirical Formula CH2O) has a molar mass of 180 g/mol, determine its molecular formula.
Step 1: Calculate EF
M. EFM of CH2O = (1 × 12) + (2 × 1) + (1 × 16) = 12 + 2 + 16 = 30 g/mol Step 2: Determine 'n'. n = (Molar Mass of Compound) / EFM = 180 g/mol / 30 g/mol = 6 Step 3: Write the Molecular Formula. MF = (CH2O)6 = C6H12O6 (This is the formula for glucose, a common sugar found in Nigerian food crops like yam and cassava). 2.
6. Laws of Chemical Combination These laws describe how elements combine to form compounds. 2.6.
1. Law of Conservation of Matter (or Mass)
Statement: Matter can neither be created nor destroyed in any ordinary chemical reaction. The total mass of the reactants before a chemical change is equal to the total mass of the products after the chemical change.
Illustration: Chemical Equations: Balancing chemical equations is a direct application of this law. The number of atoms of each element is the same on both sides, hence the total mass is conserved.
Example: C(s) + O2(g) → CO2(g) Mass of C (12g) + Mass of O2 (32g) = Mass of CO2 (44g) 12 + (2 x 16) = 12 + (2 x 16) => 44g = 44g.
Simple Experiment: Reacting a solution of lead(II) nitrate with a solution of potassium iodide in a closed system (e.g., a sealed conical flask). A yellow precipitate of lead(II) iodide forms. The total mass of the flask and its contents before the reaction will be exactly equal to the total mass after the reaction. This can be demonstrated using a balance.
Lavoissier's experiment: Burning a candle in a closed container. Even though the candle appears to "disappear", if all products (CO2, H2O) are collected, the total mass remains constant. 2.6.
2. Law of Constant Composition (or Definite Proportions)
Statement: A pure chemical compound always contains the same elements combined in the same fixed ratio by mass, regardless of its source or method of preparation.
Illustration: * Water (H2O): Always consists of Hydrogen and Oxygen in a mass ratio of 1:8 (2g H to 16g O) regardless of whether it's from a river in Nigeria, distilled water from a laboratory, container. Even though the candle appears to "disappear", if all products (CO2, H2O) are collected, the total mass remains constant. 2.6.
2. Law of Constant Composition (or Definite Proportions)
Statement: A pure chemical compound always contains the same elements combined in the same fixed ratio by mass, regardless of its source or method of preparation.
Illustration: Water (H2O): Always consists of Hydrogen and Oxygen in a mass ratio of 1:8 (2g H to 16g O) regardless of whether it's from a river in Nigeria, distilled water from a laboratory, or formed by burning hydrogen.
Carbon Dioxide (CO2): Always consists of Carbon and Oxygen in a mass ratio of 12:32 or 3:8, whether produced from burning charcoal, respiration, or decomposition of limestone.
Sodium Chloride (NaCl): Always contains Sodium and Chlorine in a fixed mass ratio (23:35.5) regardless of its origin (e.g., sea salt from the Nigerian coast or rock salt). 2.6.
3. Law of Multiple Proportions Statement: When two elements combine to form more than one compound, if the mass of one element is fixed, then the masses of the other element in the different compounds will bear a simple whole number ratio to one another.
Illustration: Carbon and Oxygen: Form two common compounds: Carbon Monoxide (CO) and Carbon Dioxide (CO2).
In CO: 12g of C combines with 16g of
O. In CO2: 12g of C combines with 32g of O. If we fix the mass of Carbon (12g), the masses of Oxygen that combine with it are 16g (in CO) and 32g (in CO2).
The ratio of these masses of Oxygen (16:32) is 1:2, a simple whole number ratio.
Sulphur and Oxygen: Form Sulphur Dioxide (SO2) and Sulphur Trioxide (SO3).
In SO2: 32g of S combines with 32g of
O. In SO3: 32g of S combines with 48g of
O. Fixing the mass of Sulphur (32g), the masses of Oxygen (32g and 48g) are in the ratio 32:48, which simplifies to 2:3, a simple whole number ratio.
3. Teaching and Learning Activities 3.
1. Introduction (10 minutes)
Teacher Activity: Engage students by asking them how scientists communicate chemical information. Introduce the idea of a universal "chemical language" comprising symbols, formulae, and equations. Briefly connect the topic to understanding everyday chemical products (e.g., ingredients on food packaging, common household chemicals, industrial materials used in Nigeria).
Student Activity: Respond to questions, participate in a brief discussion on communication and chemistry. Listen attentively and take notes. 3.
2. Elements and Symbols (20 minutes)
Teacher Activity: Define elements and explain the rules for writing symbols with examples (one-letter, two-letter, Latin/Greek origins). Present a chart or flashcards of the first 20 elements and other common elements. Lead a call-and-response session for memorizing symbols. Provide examples of common elements found in Nigeria (e.g., Fe in Itakpe, Sn in Jos Plateau).
Student Activity: Copy symbols and names of elements. Practice writing symbols for given elements. Participate in call-and-response drills. 3.
3. Distinguishing Elements, Compounds, and Mixtures (25 minutes)
Teacher Activity: Provide clear definitions and examples for elements, compounds, and mixtures. Use a table to highlight the key differences (composition, properties, separation methods, energy changes).
Illustrate with local examples: Elements: Iron (from scrap metal), Carbon (charcoal).
Compounds: Water, Sugar (sucrose), Common salt.
Mixtures: Sand and gravel (heterogeneous), palm oil and water (heterogeneous), air (homogeneous), 'kunu' (a local beverage - heterogeneous).
Conduct a quick practical demonstration: Show a sample of iron filings (element). Show a sample of water (compound). Mix sand and iron filings (mixture) and demonstrate separation using a magnet. Discuss how a compound (e.g., rust, iron oxide) cannot be separated by physical means.
Student Activity: Engage in discussion, ask clarifying questions. Categorize given substances as elements, compounds, or mixtures. Observe the demonstration and discuss their observations. 3.
4. Writing Chemical Formulae (30 minutes)
Teacher Activity: * Briefly explain the concept of valency as combining power. Provide a list of common valencies for SS1 (including polyatomic filings (element). Show a sample of water (compound). Mix sand and iron filings (mixture) and demonstrate separation using a magnet. Discuss how a compound (e.g., rust, iron oxide) cannot be separated by physical means.
Student Activity: Engage in discussion, ask clarifying questions. Categorize given substances as elements, compounds, or mixtures. Observe the demonstration and discuss their observations. 3.
4. Writing Chemical Formulae (30 minutes)
Teacher Activity: Briefly explain the concept of valency as combining power. Provide a list of common valencies for SS1 (including polyatomic ions). Explain the criss-cross method for writing formulae step-by-step using several examples (e.g., binary compounds, compounds with polyatomic ions). Emphasize the use of parentheses for polyatomic ions. Provide practice exercises on the board.
Student Activity: Copy valencies and examples. Work through examples on their own, guided by the teacher. Practice writing formulae for compounds given their constituent elements and valencies. 3.
5. Chemical Equations and Balancing (40 minutes)
Teacher Activity: Define chemical equations and identify components (reactants, products, arrow, state symbols, coefficients). Introduce the Law of Conservation of Matter as the basis for balancing equations. Demonstrate the step-by-step method for balancing equations using several examples, starting from simple ones and progressing to slightly more complex ones (e.g., combustion of methane, reaction of metals with acids). Emphasize that only coefficients can be changed, not subscripts. Explain the significance of state symbols.
Student Activity: Copy definitions and steps for balancing. Attempt to balance equations independently or in pairs, guided by the teacher. Actively participate in identifying the number of atoms on each side. 3.
6. Empirical and Molecular Formulae Calculations (45 minutes)
Teacher Activity: Define empirical and molecular formulae and explain their relationship. Present the step-by-step procedure for calculating empirical formulae from percentage composition. Work through a detailed example on the board, explaining each step clearly (e.g., a compound found in a local agricultural product or industrial waste). Explain how to calculate the molecular formula using the empirical formula and molar mass. Work through a detailed example. Provide atomic masses for elements required in calculations.
Student Activity: Take detailed notes on definitions and calculation steps. Follow the worked examples, performing calculations alongside the teacher. Attempt a new problem independently or in small groups. 3.
7. Laws of Chemical Combination (40 minutes)
Teacher Activity: State and explain the Law of Conservation of Matter, linking it to balancing equations. Discuss a simple experiment (e.g., reacting vinegar and baking soda in a sealed bottle on a balance) or use a diagram/video. State and explain the Law of Constant Composition. Use water and carbon dioxide as examples, emphasizing fixed mass ratios regardless of source. Relate this to quality control in Nigerian industries (e.g., ensuring purity of water or cement). State and explain the Law of Multiple Proportions. Use Carbon Monoxide and Carbon Dioxide as key examples, carefully showing the fixed mass of one element and the simple whole number ratio of the other. Engage students in a Q&A session to check understanding of each law.
Student Activity: Listen, take notes, and ask questions for clarification. Suggest real-life examples or applications of the laws. * Discuss observations from diagrams or video illustrations.
4. Guided Practice (With Solutions)
1. Question: Write the chemical symbols for the following elements: a. Iron b. Sulphur c. Potassium d. Chlorine e. Lead f. Silver g. Zinc h.
Nitrogen Solution: a. Iron - Fe b. Sulphur - S c. Potassium - K d. Chlorine - Cl e. Lead - Pb f. Silver - Ag g. Zinc - Zn h. Nitrogen - N
Commentary: This assesses basic recall of common element symbols, including those with Latin origins.
2. Question: Classify the following substances as an Element (E), Compound (C), or Mixture (M). a. Air b. Water (H2O) c. Glucose (C6H12O6) d. Oxygen gas (O2) e. Salt solution f. Gold (Au) g. Crude oil (found in Niger Delta) h. Limestone (Calcium Carbonate, CaCO3)
Solution: a. Air - M (Homogeneous mixture of gases)
Sodium Chloride:
Symbols: Na Cl
Valencies: Na¹ Cl¹
Criss-cross: NaCl₁ (1s cancel)
Formula: NaCl
Magnesium Oxide:
Symbols: Mg O
Valencies: Mg² O²
Criss-cross: Mg₂O₂ (2s cancel to 1:1 ratio)
Formula: MgO
Aluminium Oxide:
Symbols: Al O
Valencies: Al³ O²
Criss-cross: Al₂O₃
Formula: Al₂O₃
Calcium Hydroxide:
Symbols: Ca OH
Valencies: Ca² (OH)¹
Criss-cross: Ca₁(OH)₂
Formula: Ca(OH)₂
Ammonium Sulphate:
Symbols: NH₄ SO₄
Valencies: (NH₄)¹ (SO₄)²
Criss-cross: (NH₄)₂SO₄
Formula: (NH₄)₂SO₄ (This is used as a fertilizer in Nigerian agriculture).
2. 4. Chemical Equations
Definition: A chemical equation is a shorthand representation of a chemical reaction using symbols and chemical formulae. It shows the reactants (starting materials) on the left side, the products (substances formed) on the right side, separated by an arrow (→) which indicates the direction of the reaction.
Components of a Chemical Equation:
Reactants: Substances undergoing change (left side).
Products: Substances formed (right side).
Arrow (→): "Yields" or "produces".
Coefficients: Numbers placed in front of chemical formulae to balance the equation.
Subscripts: Numbers within a formula indicating the number of atoms of each element.
State Symbols: (s) for solid, (l) for liquid, (g) for gas, (aq) for aqueous solution (dissolved in water).
Reaction Conditions (optional): e.g., heat (Δ), catalyst, pressure, temperature, written above or below the arrow.
Balancing Chemical Equations:
Principle: A chemical equation must obey the Law of Conservation of Matter, meaning the number of atoms of each element must be the same on both sides of the equation (reactants and products).
Steps for Balancing (Trial and Error Method):
Write the correct unbalanced chemical equation with state symbols.
Start with elements that appear only once on each side of the equation.
Balance metals first, then non-metals (excluding H and O).
Balance oxygen (O) atoms.
Balance hydrogen (H) atoms.
If polyatomic ions appear unchanged on both sides, balance them as a unit.
Check that the number of atoms of each element is equal on both sides. Use coefficients (whole numbers only) in front of the formulae, never change the subscripts within a formula.
Worked
Examples:
Formation of Water: H₂(g) + O₂(g) → H₂O(l)
Unbalanced: H = 2, O = 2 (LHS); H = 2, O = 1 (RHS)
Balance O: H₂(g) + O₂(g) → 2H₂O(l) (Now O = 2 on both sides)
Balance H: H = 2 (LHS); H = 4 (RHS). Need 4 H on LH
S. Balanced: 2H₂(g) + O₂(g) → 2H₂O(l)
Combustion of Methane (Cooking gas): CH₄(g) + O₂(g) → CO₂(g) + H₂O(g)
Unbalanced: C=1, H=4, O=2 (LHS); C=1, H=2, O=3 (RHS)
Balance C: Already balanced.
Balance H: CH₄(g) + O₂(g) → CO₂(g) + 2H₂O(g) (Now H=4 on both sides)
Balance O: O=2 (LHS); O=2 (from CO₂) + 2 (from 2H₂O) = 4 (RHS).
Need 4 O on LHS: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
Balanced: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
Reaction of Aluminium with Hydrochloric acid: Al(s) + HCl(aq) → AlCl₃(aq) + H₂(g)
Unbalanced: Al=1, H=1, Cl=1 (LHS); Al=1, Cl=3, H=2 (RHS)
Balance Cl: Al(s) + 3HCl(aq) → AlCl₃(aq) + H₂(g)
Balance H: H=3 (LHS); H=2 (RHS). Find LCM of 3 and 2, which is
6. To get 6 H on LHS: Al(s) + 6HCl(aq) → AlCl₃(aq) + H₂(g)
To get 6 H on RHS: Al(s) + 6HCl(aq) → AlCl₃(aq) + 3H₂(g)
Now Cl is 6 on LHS but 3 on RHS. Need to balance AlCl₃.
Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g) (Now Cl=6 on both sides)
Balance Al: Al=1 (LHS); Al=2 (RHS).
Balanced: 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)
2. 5. Empirical and Molecular Formulae
Empirical Formula (EF): The simplest whole number ratio of atoms of each element in a compound. It is determined from experimental data, such as percentage composition by mass.
Molecular Formula (MF): The actual number of atoms of each element in a molecule of the compound. It is a multiple of the empirical formula.
Relationship: Molecular Formula = n × Empirical Formula, where n = (Molar Mass of Molecular Formula) / (Molar Mass of Empirical Formula). 'n' must be a whole number.
Steps to Determine Empirical Formula from Percentage Composition:
Assume 100g of the compound (so percentages become masses in grams).
Convert the mass of each element to moles using its atomic mass (Moles = Mass / Molar Mass).
Divide each number of moles by the smallest number of moles obtained.
If the ratios are not whole numbers, multiply all ratios by the smallest integer that converts them to whole numbers.
Write the empirical formula using these whole number ratios as subscripts.
Steps to Determine Molecular Formula from Empirical Formula and Molar Mass:
Calculate the molar mass of the empirical formula (EFM).
Determine the value of 'n' using the formula: n = (Molar Mass of Compound) / (EFM).
Multiply the subscripts in the empirical formula by 'n' to get the molecular formula.
Worked Example 1: Determining Empirical Formula
A compound is found to contain 40.0% Carbon, 6.7% Hydrogen, and 53.3% Oxygen. (Relative Atomic Masses: C=12, H=1, O=16).
Step 1: Assume 100g sample.
Mass of C = 40.0 g
Mass of H = 6.7 g
Mass of O = 53.3 g
Step 2: Convert to moles.
Moles of C = 40.0 g / 12 g/mol = 3.33 mol
Moles of H = 6.7 g / 1 g/mol = 6.7 mol
Moles of O = 53.3 g / 16 g/mol = 3.33 mol
Step 3: Divide by the smallest number of moles. (Smallest is 3.33)
C: 3.33 / 3.33 = 1
H: 6.7 / 3.33 ≈ 2
O: 3.33 / 3.33 = 1
Step 4: Check for whole numbers.
Ratios are 1:2:1, which are whole numbers.
Step 5: Write the Empirical Formula. CH₂O
Worked Example 2: Determining Molecular Formula
If the compound in Example 1 (Empirical Formula CH₂O) has a molar mass of 180 g/mol, determine its molecular formula.
Step 1: Calculate EFM.
EFM of CH₂O = (1 × 12) + (2 × 1) + (1 × 16) = 12 + 2 + 16 = 30 g/mol
Step 2: Determine 'n'.
n = (Molar Mass of Compound) / EFM = 180 g/mol / 30 g/mol = 6
Step 3: Write the Molecular Formula.
MF = (CH₂O)₆ = C₆H₁₂O₆ (This is the formula for glucose, a common sugar found in Nigerian food crops like yam and cassava).
2. 6. Laws of Chemical Combination