1.2 MOLE CONCEPT 11/11/2018 MATTER

Learning Outcome At the end of this topic, students should be able : (a) Define mole in terms of mass of carbon-12 and Avogadro constant, NA. (b) Interconvert between moles, mass, number of particles, molar volume of gas at s.t.p. and room temperature. 11/11/2018 MATTER

(c) Determine empirical and molecular formulae from mass composition or combustion data. 11/11/2018 MATTER

(d) Define and perform calculation for each of the following concentration measurements : i) molarity (M) ii) molality (m) iii) mole fraction, X iv) percentage by mass, % w/w v) percentage by volume, %V/V 11/11/2018 MATTER

(e) Determine the oxidation number of an element in a chemical formula. (f) Write and balance : i) chemical equation by inspection method ii) redox equation by ion-electron 11/11/2018 MATTER

(g) Define limiting reactant and percentage yield. (h) Perform stoichiometric calculations using mole concept including reactant and percentage yield. 11/11/2018 MATTER

1.2 Mole Concept A mole is defined as the amount of substance which contains equal number of particles (atoms / molecules / ions) as there are atoms in exactly 12.000g of carbon-12. Pair = 2 11/11/2018 MATTER

One mole of carbon-12 atom has a mass of exactly 12 One mole of carbon-12 atom has a mass of exactly 12.000 grams and contains 6.02 x 1023 atoms. The value 6.02 x 1023 is known as Avogadro Constant. NA = 6.02 x 1023 mol-1 Dozen = 12 11/11/2018 MATTER

Example 1.0 mole of chlorine atom = 6.02 x 1023 chlorine atoms = 35.5 g Cl 1.0 mole of chlorine molecules = 6.02 x 1023 chlorine molecules = 71.1 g Cl2 = 6.022 x 1023 x 2 chlorine atoms 1.0 mole of NH3 = 6.02x 1023 molecules = 6.02 x 1023 x 4 atoms = 6.02 x 1023 N atom = 6.02 x 1023 X 3 H atoms 11/11/2018 MATTER

Molar Mass The mass of one mole of an element or one mole of compound is referred as molar mass. Unit : g mol-1 Example: - molar mass of Mg = 24 g mol-1 - molar mass of CH4 = (12 + 4) gmol-1 = 16 g mol-1 11/11/2018 MATTER

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1.2.1 Mole Concept of Gases Molar volume of any gas at STP = 22.4 dm3 mol-1 s.t.p. = Standard Temperature and Pressure Where, T = 273.15 K P = 1 atm 11/11/2018 MATTER

1 mole of gas has a volume of 22.4 dm3 at s.t.p At s.t.p, volume of gas (dm3) = number of mole X 22.4 dm3 mol-1 1 mole of gas has a volume of 24.0 dm3 at room temperature At room temperature, volume of gas (dm3) = number of mole X 24.0 dm3 mol-1 11/11/2018 MATTER

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Exercise A sample of CO2 has a volume of 56 cm3 at STP. Calculate: The number of moles of gas molecules 0.0025 mol The number of molecular 1.506 x 1021 molecules The number of oxygen atoms in the sample 3.011x1021atoms Note: 1 dm3 = 1000 cm3 1 dm3 = 1 L 11/11/2018 MATTER

Empirical And Molecular Formulae Empirical formula is a chemical formula that shows the simplest ratio of all elements in a molecule. Molecular formula is a formula that show the actual number of atoms of each element in a molecule. 11/11/2018 MATTER

The relationship between empirical formula and molecular formula is : Molecular formula = n ( empirical formula ) Where ; 11/11/2018 MATTER

Example A sample of hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its molar mass is 56. Determine the empirical formula and molecular formula of the compound. 11/11/2018 MATTER

Solution : Empirical formula = CH2 C H mass 85.7 14.3 Number of mol 12 7.1417 1 Simplest ratio 2 11/11/2018 MATTER

n = 56 14 = 4 molecular formula = C4H8 11/11/2018 MATTER

1.2.2 Concentration of Solution When an amount of solute dissolved completely in a solvent and it will form a homogeneous mixture. 11/11/2018 MATTER

Exercise A combustion of 0.202 g of an organic sample that contains carbon, hydrogen and oxygen produce 0.361g carbon dioxide and 0.147 g water. If the relative molecular mass of the sample is 148, what is the molecular formula. Ans : C6H12O4 11/11/2018 MATTER

Units of concentration of a solution: A. Molarity B. Molality C. Mole Fraction D. Percentage by Mass E. Percentage by Volume 11/11/2018 MATTER

A. Molarity (M) The number of moles of solute per cubic decimetre (dm3) or litre (L) of solution. Note: 1 dm3 = 1000 cm3 1 L = 1000 mL 11/11/2018 MATTER

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B. Molality (m) Molality is the number of moles of solute dissolved in 1 kg of solvent Note: Mass of solution = mass of solute + mass of solvent Volume of solution ≠ volume of solvent 11/11/2018 MATTER

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C. Mole Fraction (X) Mole fraction is the ratio of the number of moles of one component to the total number of moles of all component present. 11/11/2018 MATTER

It is always smaller than 1 The total mol fraction in a mixture (solution) is equal to one. XA + XB + XC = 1 11/11/2018 MATTER

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D. Percentage by Mass (%w/w) Percentage by mass is defined as the percentage of the mass of solute per mass of solution. 11/11/2018 MATTER

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E. Percentage By Volume (%V / V) Percentage by volume is defined as the percentage of volume of solute in milliliter per volume of solution in milliliter. 11/11/2018 MATTER

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1.2.3 Balancing Chemical Equation A chemical equation shows a chemical reaction using symbols for the reactants and products. The formulae of the reactants are written on the left side of the equation while the products are on the right. 11/11/2018 MATTER

Example: x A + y B z C + w D Reactants Products 11/11/2018 MATTER

The methods to balance an equation: Inspection Method The total number of atoms of each element is the same on both sides in a balanced equation. The number x, y, z and w, showing the relative number of molecules reacting, are called the stoichiometric coefficients. The methods to balance an equation: Inspection Method 11/11/2018 MATTER

Inspection Method Write down the unbalanced equation. Write the correct formulae for the reactants and products. Balance the metallic element, followed by non-metallic atoms. Balance the hydrogen and oxygen atoms. Check to ensure that the total number of atoms of each element is the same on both sides of equation. 11/11/2018 MATTER

Example NH3 + CuO → Cu + N2 + H2O Balance the chemical equation by applying the inspection method. NH3 + CuO → Cu + N2 + H2O 11/11/2018 MATTER

Exercise Balance the chemical equation below by applying inspection method. a. Fe(OH)3 + H2SO4 → Fe2(SO4)3 + H2O b. C6H6 + O2 → CO2 + H2O c. N2H4 + H2O2 → HNO3 + H2O d. ClO2 + H2O → HClO3 + HCl 11/11/2018 MATTER

1.2.4 Redox Reaction Redox reaction is a reaction that involves both reduction and oxidation. 11/11/2018 MATTER

The substance loses one or more elactrons. Oxidation The substance loses one or more elactrons. Increase in oxidation number Act as an reducing agent (reductant) 11/11/2018 MATTER

The substance gains one or more elactrons. Reduction The substance gains one or more elactrons. decrease in oxidation number Act as an oxidising agent (oxidant) 11/11/2018 MATTER

Oxidation numbers of any atoms can be determined by applying the following rules: In a free element , as an atom or a molecule the oxidation number is zero. Example: Na = 0 Cl2 = 0 Br2 = 0 O2 = 0 Mg = 0 11/11/2018 MATTER

For monoatomic ion, the oxidation number is equal to the charge on the ion. Example: Na+ = +1 Mg2+ = +2 Al3+ = +3 S2- = -2 11/11/2018 MATTER

Oxidation number of F in NaF = -1 Oxidation number of Cl in HCl = -1 Fluorine and other halogens always have oxidation number of -1 in its compound. Only have a positive number when combine with oxygen. Example: Oxidation number of F in NaF = -1 Oxidation number of Cl in HCl = -1 Oxidation number of Cl in Cl2O7 = +7 11/11/2018 MATTER

Oxidation number of H in HCl = +1 Oxidation number of H in NaH = -1 Hydrogen has an oxidation number of +1 in its compound except in metal hydrides which hydrogen has an oxidation number of -1 Example: Oxidation number of H in HCl = +1 Oxidation number of H in NaH = -1 Oxidation number of H in MgH2 = -1 11/11/2018 MATTER

Oxygen has an oxidation number of -2 in most of its compound. Example: Oxidation number of O in MgO = -2 Oxidation number of O in H2O = -2 11/11/2018 MATTER

However there are two exceptional cases: - in peroxides, its oxidation number is -1 Example: Oxidation number of O in H2O2 = -1 - When combine with fluorine, posses a positive oxidation number Oxidation number of O in OF2 = +2 11/11/2018 MATTER

Oxidation number of H2O = 0 Oxidation number of HCl = 0 In neutral molecule, the sum of the oxidation number of all atoms that made up the molecule is equal to zero. Example: Oxidation number of H2O = 0 Oxidation number of HCl = 0 Oxidation number of KMnO4 = 0 11/11/2018 MATTER

Oxidation number of KMnO4- = -1 For polyatomic ions, the total oxidation number of all atoms that made up the polyatomic ion must be equal to the nett charge of the ion. Example: Oxidation number of KMnO4- = -1 Oxidation number of Cr2O72- = -2 Oxidation number of NO3- = -1 11/11/2018 MATTER

Assign the oxidation number of Cr in Cr2O72-. Solution : Example : Assign the oxidation number of Cr in Cr2O72-. Solution : Cr2O7 = -2 2 Cr + 7 (-2) = -2 2 Cr = + 12 Cr = + 6 11/11/2018 MATTER

Exercise Assign the oxidation number of Mn in the following chemical compounds. i. MnO2 ii. MnO4- Assign the oxidation number of Cl in the following chemical compounds. i. KClO3 ii. Cl2O72- Assign the oxidation number of following: i. Cr in K2Cr2O7 ii. U in UO22+ iii. C in C2O42- 11/11/2018 MATTER

1.2.4.1 Balancing Redox Reaction Redox reaction may occur in acidic and basic solutions. Follow the steps systematically so that equations become easier to balance. 11/11/2018 MATTER

Balancing Redox Reaction In Acidic Solution Fe2+ + MnO4- → Fe3+ + Mn2+ Divide the equation into two half reactions, one involving oxidation and the other reduction i. Fe2+ → Fe3+ ii. MnO4- → Mn2+ 11/11/2018 MATTER

Balance each half-reaction a. first, balance the element other than oxygen and hydrogen i. Fe2+ → Fe3+ ii. MnO4- → Mn2+ 11/11/2018 MATTER

and hydrogen by adding H+ i. Fe2+ → Fe3+ ii. MnO4- + 8H+ → Mn2+ + 4H2O b. second, balance the oxygen atom by adding H2O and hydrogen by adding H+ i. Fe2+ → Fe3+ ii. MnO4- + 8H+ → Mn2+ + 4H2O c. then, balance the charge by adding electrons to the side with the greater overall positive charge. i. Fe2+ → Fe3+ + 1e ii. MnO4- + 8H+ + 5e → Mn2+ + 4H2O 11/11/2018 MATTER

____________________________________________ 3. Multiply each half-reaction by an interger, so that number of electron lost in one half-reaction equals the number gained in the other. i. 5 x (Fe2+ → Fe3+ + 1e) 5Fe2+ → 5Fe3+ + 5e ii. MnO4- + 8H+ + 5e → Mn2+ + 4H2O 4. Add the two half-reactions and simplify where possible by canceling species appearing on both sides of the equation. i. 5Fe2+ → 5Fe3+ + 5e ____________________________________________ 5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O 11/11/2018 MATTER

5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O Check the equation to make sure that there are the same number of atoms of each kind and the same total charge on both sides. 5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O Total charge reactant = 5(+2) + (-1) + 8(+1) = + 10 - 1 + 8 = +17 Total charge product = 5(+3) + (+2) + 4(0) = + 15 + (+2) 11/11/2018 MATTER

Example: In Acidic Solution C2O42- + MnO4- + H+ → CO2 + Mn2+ + H2O Solution: i. Oxidation : C2O42- → CO2 ii. Reduction : MnO4- → Mn2+ 2. i. C2O42- → 2CO2 ii. MnO4- + 8H+ → Mn2+ + 4H2O 3. i. C2O42- → 2CO2 + 2e ii. MnO4- + 8H+ + 5e→ Mn2+ + 4H2O 11/11/2018 MATTER

ii. 2 x (MnO4- + 8H+ + 5e→ Mn2+ + 4H2O) 4. i. 5 x (C2O42- → 2CO2 + 2e) → 5C2O42- → 10CO2 + 10e ii. 2 x (MnO4- + 8H+ + 5e→ Mn2+ + 4H2O) → 2MnO4- + 16H+ + 10e→ 2Mn2+ + 8H2O 5. i. 5C2O42- → 10CO2 + 10e ii. 2MnO4- + 16H+ + 10e→ 2Mn2+ + 8H2O _________________________________________________ 5C2O42- + 2MnO4- + 16H+ → 10CO2 + 2Mn2+ + 8H2O 11/11/2018 MATTER

Balancing Redox Reaction In Basic Solution Firstly balance the equation as in acidic solution . Then, add OH- to both sides of the equation so that it can be combined with H+ to form H2O. The number of hydroxide ions (OH-) added is equal to the number of hydrogen ions (H+) in the equation. 11/11/2018 MATTER

Example: In Basic Solution Cr(OH)3 + IO3- + OH- → CrO32- + I- + H2O Solution: i. Oxidation : Cr(OH)3 → CrO32- ii. Reduction : IO3- → I- 2. i. Cr(OH)3 → CrO32- + 3H+ ii. IO3- + 6H+ → I- + 3H2O 3. i. Cr(OH)3 → CrO32- + 3H+ + 1e ii. IO3- + 6H+ + 6e → I- + 3H2O 11/11/2018 MATTER

i. 6 x (Cr(OH)3 → CrO32- + 3H+ + 1e) → 6Cr(OH)3 → 6CrO32- + 18H+ + 6e ii. IO3- + 6H+ + 6e → I- + 3H2O 5. i. 6Cr(OH)3 → 6CrO32- + 18H+ + 6e ________________________________________________ 6Cr(OH)3 + IO3- → 6CrO32- + I- + 12H+ + 3H2O 6. 6Cr(OH)3 + IO3- + 12OH- → 6CrO32- + I- + 12H+ + 3H2O + 12OH- 7. 6Cr(OH)3 + IO3- + 12OH- → 6CrO32- + I- + 15H2O 11/11/2018 MATTER

Exercise Balance the following redox equations: a. In Acidic Solution i. Cu + NO3 + H+→ Cu2+ + NO2 + H2O ii. MnO4- + H2SO3 → Mn2+ + SO42- + H2O + H+ iii. Zn + SO42- + H+ → Zn2+ + SO2 + H2O b. In Basic Solution i. ClO- + S2O32- → Cl- + SO42- ii. Cl2 → ClO3- + Cl- iii. NO2 → NO3 + NO 11/11/2018 MATTER

1.2.5 Stoichiometry Stoichiometry is the quantitative study of reactants and products in a chemical reaction. 11/11/2018 MATTER

CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l) Example: CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l) 1 mole of CaCO3 reacts with 2 moles of HCl to yield 1 mole of CaCl2, 1 mole of CO2 and 1 mole of H2O. Stoichiometry can be used for calculating the species we are interested in during a reaction. 11/11/2018 MATTER

Example 1 How many moles of hydrochloric acid, HCl do we need to react with 0.5 moles of zinc? 11/11/2018 MATTER

Example 2 How many moles of H2O will be formed when 0.25 moles of C2H5OH burns in oxygen? 11/11/2018 MATTER

Exercise 1 A 16.50 mL 0.1327 M KMnO4 solution is needed to oxidise 20.00mL of a FeSO4 solution in an acidic medium. What is the concentration of the FeSO4 solution? The net ionic equation is: 5Fe 2+ + MnO4- +8H+ Mn 2+ +5Fe 3+ +4H2O Answer : 0.5474 M 11/11/2018 MATTER

Exercise 2 How many mililitres of 0.112 M HCl will react exactly with the sodium carbonate in 21.2 mL of 0.150 M Na2CO3 according to the following equation? 2HCl(aq)+Na2CO3(aq) 2NaCl(aq)+CO2(g)+H2O(l) Answer : 56.8 mL 11/11/2018 MATTER

1.2.5.1 Limiting Reactant A limiting reactant is the reactant that is completely consumed in a reaction and limits the amount of products formed. An excess reactant is the reactant that is not completely consumed in a reaction and remains at the end of the reaction. 11/11/2018 MATTER

Example 1 S + 3F2 → SF6 If 4 mol of S reacts with 10 mol of F2 , which of the two reactants is the limiting reagent? 11/11/2018 MATTER

Example 2 C is prepared by reacting A and B : A + 5B → C In one process, 2 mol of A react with 9 mol of B. Which is the limiting reactant? Calculate the number of mole(s) of C? How much of the excess reactant (in mol) is left at the end of the reaction? 11/11/2018 MATTER

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Percentage yield The percentage yield is the ratio of the actual yield (obtained from experiment) to the theoretical yield (obtained from stoichiometry calculation) multiply by 100% 11/11/2018 MATTER

Percentage yield = actual yield x 100% theoretical yield 11/11/2018 MATTER

Exercise In a certain experiment, 14.6g of SbF3 was allowed to react with CCl4 in excess. After the reaction was finished, 8.62g of CCl2F2 was obtained. 3 CCl4 + 2 SbF3 3 CCl2F2 + 2 SbCl3 [ Ar Sb = 122, F = 19, C= 12, Cl = 35.5 ] What was the theoretical yield of CCl2F2 in grams ? b) What was the percentage yield of CCl2F2 ? Ans : a) 11.6 g b) 74.31 % 11/11/2018 MATTER