Combustion of 2.30g of an organic sample, X, yields 3.30 g CO2 and 1.80 g of H2O. Determine the empirical formula of X. Mass of H = Mass of H = 0.20 g H Mass of O = 2.30 – 0.90 – 0.20 Mass of C + H + O = 2.30 g Mass of O = 1.20 g Same as previous method 2.0 gmol−1 18.0 gmol−1 × 1.80 g 51 Mass of H = ? Mass of O = ?
CONCENTRATION UNITS Concentration ➢ is the amount of solute present in a given quantity of solvent in a solution Solution ➢ is a homogenous mixture of 2 or more substances that formed when amount of solute dissolves completely in a solvent Solute ➢ is the substance being dissolved and present in the smaller amount Solvent ➢ is the substance doing the solving and present in the larger amount 52
WATER SUGAR SUGAR SOLUTION 53 EXAMPLE 1: Solute Solvent sugar water Solution +
Expression of Concentration 54 concentration Molar concentration (or molarity) Molal concentration (or molality) Percentage by mass (% w/w) Percentage by volume (% v/v) Mole fraction, X
MOLARITY, M ❑ The number of mole of solute dissolved in 1 L of solution. 1 L = 1 dm3 1 ml = 1 cm3 1 dm3 = 1000 cm3 a ❑ Unit: mol L −1 @ mol dm−3 @ molar, M 55 = () () ,
1.46 molar glucose (C6H12O6 ) 1.46 mole of glucose (C6H12O6 ) solute 1 L of the solution 1.46 mol/L glucose (C6H12O6 ) or 56 EXAMPLE 2:
A matriculation student prepared a solution by dissolving 5.528 x 10-3 mol of sodium carbonate, Na2CO3 in 250.0 cm3 of water. Calculate its molarity. ➢ V of solution: 250.0 cm3 = 250 ml = 0.25 L = 0.0221 mol L -1 = sodium carbonate, Na2CO3 57 EXAMPLE 3: Solute Solution Solvent Answer = sodium carbonate solution,Na2CO3 + H2O =water, H2O ➢ = () () = 5.528 × 10−3 0.25
A student prepared a solution of NaCl by dissolving 1.461 g of NaCl in a 250 mL volumetric flask. What is the molarity of this solution. 58 EXAMPLE 4:
= 0.0250 mol NaCl 59 Answer ➢ = () ( −1) = 1.461 58.5 −1 Solute = ? Solvent = ? Solution = ? NaCl H2O NaCl + H2O
= 0.100 mol/L NaCl solution @ 250.0 cm3 ➢ V of NaCl solution : = 250 ml = 0.25 L ➢ Moles of NaCl = 0.0250 mol = 0.100 M NaCl solution 60 , = () () = 0.0250 0.250
How many grams of calcium chloride, CaCl2 should be used to prepare 250.00 mL solution with a concentration of 0.500 M. 61 EXAMPLE 5: Answer Solute = ? Solvent = ? Solution = ? CaCl2 CaCl H2O 2 + H2O
➢ nCaCl2 = MCaCl2 x Vsolution = 0.500 mol/L x 250 x 10-3 L nCaCl2 = 0.125 mol ➢ Mass of CaCl2 = nCaCl2 x molar mass CaCl2 = 0.125 mol x 111.1 g/mol Mass of CaCl2 = 13.9 g 62 Answer
❑ the number of moles of solute per 1 kg of solvent in a solution. ➢ (unit: mol kg−1 @ molal @ m) MOLALITY Volume of solution volume of solvent Mass of solution = mass of solute + mass of solvent 63 = () () , b
Calculate the molality of a solution prepared by dissolving 0.2880 mol of CaCl2 in 271 g of water? = 0.2880 mol 271 × 10−3 ➢ Moles of CaCl2 = ? Mass of water = ? ➢ Moles of CaCl2 = 0.2880 mol Mass of water = 271/1000 kg 64 EXAMPLE 6: ➢ , = () () Answer Solute = ? CaCl2 Solvent = ? H2O = 1.06 mol kg −
Calculate the molal concentration of ethylene glycol (C2H6O2 ) solution containing 8.40 g of ethylene glycol in 200 g of water. The molar mass of ethylene glycol is 62 g/mol. 65 EXAMPLE 7:
➢ , = () () ➢ No. of moles of C2H6O2 = 8.40 g 62 g/mol = 0.1355 mol = 0.1355 0.2 = 0.68 mol kg-1 ➢ Moles of C2H6O2 = ? mol Mass of water = (200/1000) kg 66 Answer Solute = ? C2H6O2 Solvent = ? H2O ➢ , = 262 () ()
A solution containing 8.89 g glycerol(C3H8O3 ) in 75.0 g of ethanol (C2H6O). What is the molality of the solution? 67 EXERCISE 1: Answer: 1.29 molal
❑ Mole fraction is : the ratio of the number of moles of one component to the total of number of moles present. MOLE FRACTION, X ➢ nA = the no. of moles of component A in the mixture Where; ➢ nTotal = the total no. of moles in all component in the mixture = nA + nB + …… 68 , = c
❑ If a solution containing A, B and C: ✓ No unit for mole fraction ✓ Mole fraction always smaller than 1.000 ✓ Total mole fraction in mixture, nT = 1.000 69 ➢ , = ++ =
= 0.30 0.30 + 40.0 What is the mole fraction of CuCl2 in a solution prepared by dissolving 0.30 mole of CuCl2 in 40.0 mole of H2O ? = 0.0074 UNIT 70 EXAMPLE 8: Answer ➢ 2 = 2 (2 +2)
A sample of ethanol, C2H5OH contains 200.0 g of ethanol and 150.0 g of water. Calculate the mole fraction of; a) Ethanol b) water in the solution. 71 EXAMPLE 9:
= 1 - 0.3429 72 Answer ➢ ℎ = () ( −1) = 200.0 46.0 −1 = . ➢ = () ( −1) = 150.0 18.0 −1 = . ➢ = + = 4.3478 (4.3478 + 8.3333) = . ➢ = − = 0.6571
❑ The ratio of the mass of a solute to the mass of the solution, multiplied by 100 d Since, Mass of solution = mass of solute + mass of solvent 73 ➢ , % Τ = () () × 100 ➢ , % Τ = () + () × 100 @
Given that 10% percent by mass of NaOH in the solution. 10 g of NaOH dissolved in 100 g of solution 10 g of NaOH dissolved in 90 g of solvent (water) From equation : 74 Assume mass of solution = 100 g Mass of solute, NaOH = 10 g mass of solvent = Mass of solution - mass of solute = 100 g - 10 g = 90 g Example 10: ➢ , % Τ = () () × 100
A sample of 0.892 g of potassium chloride, KCl is dissolved in 54.3 g of water. What is the percent by mass of KCl in this solution? = 1.616 % Solute = ? ; Solvent = ? ; Solution = ? Solute = KCl ; Solvent = H2O ; = KCl + H2O 75 Example 11: , % Τ = () () × 100 % Τ = 0.892 (0.892 + 54.3) × 100
A solution is made by dissolving 4.2 g of sodium chloride, NaCl in 100.00 mL of water. Calculate the mass percent of sodium chloride in the solution. Exercise 76 EXERCISE 2: Answer : 4.0%
❑ the ratio of the volume of a solute to the volume of the solution, multiplied by 100 e • Most often used for liquids and gas 77 ➢ , % Τ = × 100
A 200 mL of perfume contains 28 mL of alcohol. What is the % concentration of alcohol by volume in this solution? = 14 % 78 Answer Example 12: ➢ , % Τ = × 100 = ℎ × 100 = 28 200 × 100
A 350 cm3 sample of vinegar contains 2.10 cm3 of ethanoic acid. What is the concentration of ethanoic acid by volume in this sample? = 0.6 % 79 Answer Example 13: ➢ , % Τ = × 100 % Τ = 2.10 3 350 3 × 100
Suppose you have 265.5 mL of an aqueous ethanol solution that is 30.0 % ethanol by volume. How much ethanol (in mL) is in the bottle? Answer:79.65 mL 80 EXERCISE 3:
Unit Concentration Formula Assumption (Suggestion) ❑ For question that give concentration and density, can make assumptions based on concentration. 81 % Τ % Τ () () × 100 () () × 100 () () () () = = = =
An aqueous solution of ethylene glycol used as an automobile engine coolant is 40.0% HOCH2CH2OH by weight and has a density of 1.05 g/ml. What are the; a) Molarity b) Molality of HOCH2CH2OH in the solution a) Molarity,M = 6.774 mol/L b) Molality, m = 10.753 mol/kg 82 EXERCISE 4: Answer:
83
1. An 8.00%(w/w) aqueous solution of ammonia has a density of 0.9651 g mL-1 . Calculate the a) molality b) Molarity c) mole fraction of the NH3 solution 1 : a) 5.12 mol kg -1 , b) 4.54 mol L -1 Answer ,c) 0.0843 2. How many grams of solute are present in 783 mL of an aqueous solution of 0.35 M KOH solution? 3. A 15.00 mL sample of 0.450 M K2CrO4 is diluted to 100.00 mL. What is the concentration of the new solution? 3: 0.0675 M K2CrO4 solution 2: 15.37 g KOH solution 84 EXERCISE 5:
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Edited By : NZSH, NAAN ……………………… Revised By : MPJ ……………………… Approved By : ZA ………………………… 86
1.3 STOICHIOMETRY
LEARNING OUTCOMES At the end of this topic, students should be able to: a) Write a balanced chemical equation : a) by inspection method b) by ion-electron method (redox equation) (C3) b) Define: (C1) i. limiting reactant ii. percentage yield. c) Perform stoichiometric calculations using mole concept including limiting reactant and percentage yield. (C3, C4)
Chemical equation denotes a chemical reaction : xA + yB ⎯⎯→ zC + wD x, y, z, and w = stoichiometric coefficients Balancing Chemical Equation (reactant) (product) 89
90 Two methods of balancing chemical equation: a)Inspection method b) Ion-electron method
Write down the unbalanced equation. Write the correct formulae for the reactants and products. 91 Inspection Method Balance element/atom i. Balance the metallic element, followed by non-metallic atoms. ii. 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.
CO2 + H2O C2H6 + O2 CO2 + H2O ➢ start with C 2 C atom 1 C atom C2H6 + O2 Reactants Products 4 C 12 H 14 O 4 C 12 H 14 O 4 3 7 2 2C2H6 + O2 CO2 + 6H2O 2 7 6 H atoms 2 H atoms 7 O atoms 92 EXAMPLE 1:
Al + H2SO4 Al2 (SO4 )3 + H2 1 SO4 2- ions 3 SO4 2- ions ➢ Balance polyatomic ions as a unit ➢ Balance pure elements such as Al. Al + H2SO4 Al2 (SO4 ) 3 3 + H2 1 Al atom 2 Al atoms 2 6 H atoms 3 Reactants Products 2 Al 6 H 12 O 2 Al 6H 12 O 3 S 3 S ➢ Balance H. 3 H atoms 93 EXAMPLE 2:
Balance these equations. a) Al(s) + O2 (g) → Al2O3 (s) b) N2 (g) + H2 (g) → NH3 (g) c) C6H6 (l) + O2 (g) → CO2 (g) + H2O(l) 94 EXERCISE 1:
95 Answer
Balance the following chemical equation : 1. Fe2O3 + HCl 3. C2H5OH + O2 CO2 + H2O FeCl3 + H2O 2. NH3 + CuO Cu + N2 + H2O 4. AgNO3 + Na2CrO4 Ag2CrO4 + NaNO3 5. Al2 (SO4 )3 + KOH Al(OH)3 + K2SO4 96 EXERCISE 2:
97 Answer
Balancing Redox Equation Redox Reaction Oxidation An increase in oxidation number Loss of electrons OIL ❑ Redox reaction is a reaction that involves both reduction and oxidation reactions. - Oxidation Is the Loss of electrons - Reduction Is the Gain of electrons Reduction A decrease in oxidation number Gain of electrons 98 RIG
99 1. For monoatomic ions, oxidation number = the charge on the ion Example: Monoatomic ions oxidation number Na+ +1 Cl- -1 Al3+ +3 S 2- -2 Oxidation Number ❑ Oxidation numbers of any atoms can be determined by applying the following rules:
2. For free elements, oxidation number on each atom = 0 Example: 3. For most cases, oxidation number for; 100 Elements oxidation number Na 0 O2 0 Br2 , P4 , S8 0 Atom oxidation number O -2 H +1 F, Cl, Br, I -1 Example: