Lecture 1 Chemical Quantities - Pharm D, E-JUST PDF

Summary

These lecture notes cover Chemical Quantities and Aqueous Reactions, part of the Pharmaceutical Analytical Chemistry-1 course for Pharm D students at the Egypt-Japan University of Science and Technology (E-JUST). The document includes course details, a course coordinator's contact information, and course grading details.

Full Transcript

Program: Pharm D Course: Pharmaceutical Analytical Chemistry-1 Code: PMC 111 Lecture 1 Chemical Quantities and Aqueous Reactions Dr. Ahmed Sayed Saad 1 Course Coordinator Dr. Ahmed Sayed Saad Associate Professor...

Program: Pharm D Course: Pharmaceutical Analytical Chemistry-1 Code: PMC 111 Lecture 1 Chemical Quantities and Aqueous Reactions Dr. Ahmed Sayed Saad 1 Course Coordinator Dr. Ahmed Sayed Saad Associate Professor of Analytical Chemistry Email: [email protected] Dr. Ahmed Sayed Saad 2 Course credits and Grading scheme Credit Hours Lecture Practical Total 2 1 3 2 hours lecture 2 hours laboratory Topic Marks remarks Midterm exam 15 Practical 25 Practical exam (20) + Lab work (5) Written exam 50 Objective & Subjective Questions Oral Exam 10 Total 100 Dr. Ahmed Sayed Saad 3 Course description Dr. Ahmed Sayed Saad 4 Textbook Textbook: Chemistry: A molecular Approach Author: Nivaldo J. Tro Publisher: Pearson ISBN-10: 0-13-487437-4 / ISBN-13: 978-0-13- 487437-1 Dr. Ahmed Sayed Saad 5 Atomic structure Dr. Ahmed Sayed Saad 6 A Boy And His Atom: The World's Smallest Movie IBM researchers used a scanning tunneling microscope (STM) to move thousands of carbon monoxide molecules. They used scanning Tunneling microscope to magnify the atoms 100 million times. Atomic composition of our bodies Dr. Ahmed Sayed Saad 8 Molecules and Compounds Molecules and Compounds Dr. Ahmed Sayed Saad 9 10 Dr. Ahmed Sayed Saad 11 Molecules and Compounds When two or more elements combine to form a compound, an entirely new substance results. What will happen to water if we add an extra Oxygen Atom → H2O2 Dr. Ahmed Sayed Saad 12 Molecules and Compounds Highly reactive Toxic gas When two or more elements combine to form a compound, an entirely new substance results. 13 Dr. Ahmed Sayed Saad Mixtures vs. Compounds Dr. Ahmed Sayed Saad 14 Chemical Bonds Compounds are composed of atoms held together by chemical bonds. Ionic bonds Covalent bonds Occur between metals and nonmetals Occur between two or more nonmetals Involve the transfer of electrons from Involve the sharing of electrons between one atom to another. two atoms Dr. Ahmed Sayed Saad 15 Chemical Bonds Dr. Ahmed Sayed Saad 16 Refresh Dr. Ahmed Sayed Saad 17 Chemical Formulas Chemical formula = Symbol for each element + Subscript indicating the relative number of atoms of the element. Subscript of 1 is omitted. Chemical formulas list the more metallic (or more positively charged) elements first, followed by the less metallic (or more negatively charged) elements. Formula Compound Represent H2O Water molecule Hydrogen and oxygen in ratio 2:1 NaCl Sodium chloride Sodium & Chloride ions in ratio 1:1 CCl4 Carbon tetrachloride Carbon and chlorine in ratio 1:4 CO Carbon monoxide ? CO2 Carbon dioxide ? Dr. Ahmed Sayed Saad 18 Chemical Formulas Molecular Formula Empirical Formula Structural Formula Gives the actual number Gives the relative number Use lines to represent of atoms of each element of atoms of each element covalent bonds and shows in a molecule of a in a compound how atoms in a molecule compound connect or bond to each other. H2O2 HO H-O-O-H CO2 CO2 O=C=O Dr. Ahmed Sayed Saad 19 Refresh Dr. Ahmed Sayed Saad 20 Elements vs. Compounds Type of elements Count Bond Dr. Ahmed Sayed Saad 21 Chemical Formulas Dr. Ahmed Sayed Saad 22 Refresh Dr. Ahmed Sayed Saad 23 Formulas for ionic compounds Summarizing Ionic Compound Formulas: Ionic compounds always contain positive and negative ions. Sum of the charges of the positive ions (cations) = Sum of the charges of the negative ions (anions). The formula of an ionic compound reflects the smallest whole-number ratio of ions. (The simplest charge neutral collection of ions) Dr. Ahmed Sayed Saad 24 Problem Dr. Ahmed Sayed Saad 25 Home Activity Dr. Ahmed Sayed Saad 26 Chemical Quantitates Dr. Ahmed Sayed Saad 27 Amount, Volume, and Concentration Amount Volume Concentration Gram (g) Liter (L) g/L (g L-1) Mole (mol) mol/L (mol L-1 or M) Dr. Ahmed Sayed Saad 28 Mole A mole (abbreviated mol) is the amount of material containing 6.022×1023 particles. 6.022×1023 ➔ Avogadro’s number One mole of anything is 6.022×1023 units of that thing. Dr. Ahmed Sayed Saad 29 Converting between number of Moles and number of Atoms Example Calculate the number of copper atoms in 2.45 mol of copper. 1 mol of Cu = 6.022×1023 atoms Dr. Ahmed Sayed Saad 30 Converting between number of Moles and mass The mass of one mole of atoms of Mass (g) an element is its molar mass. Molar n mass Element’s molar mass in grams per mole = The element’s atomic mass in atomic mass units. For example: copper has an atomic mass of 63.55 amu and a molar mass of 63.55 g/mol. The value of the mole is equal to the number of atoms in exactly 12 g of pure carbon-12 (12 g C=1 mol C atoms =6.022×1023) Since the masses of all other elements are defined relative to carbon-12, the same relationship holds for all elements. Dr. Ahmed Sayed Saad 31 Converting between number of Moles and mass Dr. Ahmed Sayed Saad 32 Refresh Q1. Which mole contains more atoms? Q2. Which mole is heavier? Dr. Ahmed Sayed Saad 33 Converting between number of Moles and mass Dr. Ahmed Sayed Saad 34 Converting between number of Moles and mass Example Calculate the amount of carbon (in moles) contained in a 0.0265 g pencil “lead.” molar mass of carbon is 12.01 g/mol (Assume that the pencil lead is made of pure graphite, a form of carbon.) Note. Amount of matter ≡ 𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 Mass (g) Molar n mass n Amount (mol) Number of units Mass (g) 1 mol of C 6.022×1023 C atoms 12.01 g of C 0.0265 0.0265 g of C ?= 12.01 = 2.21 × 10−3 𝑚𝑜𝑙 𝐶 Dr. Ahmed Sayed Saad 35 Converting between mass and number of atoms Example How many copper atoms are in a copper penny with a mass of 3.10 g? molar mass of Cu is 63.55 g/mol (Assume that the penny is composed of pure copper.) Mass (g) Molar n mass n Amount (mol) Number of units Mass (g) 1 mol of Cu 6.022×1023 Cu atoms 63.55 g of Cu 3.10 ×6.022×1023 3.10 g of Cu ?= = 2.95 × 1022 𝐶𝑢 𝑎𝑡𝑜𝑚𝑠 63.55 Dr. Ahmed Sayed Saad 36 Refresh Dr. Ahmed Sayed Saad 37 Refresh For different elements or compounds Same Mass → have different Number of units Same Number of units → Have different masses Dr. Ahmed Sayed Saad 38 Chemical Equation Dr. Ahmed Sayed Saad 39 Writing and Balancing Chemical Equations 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 ⟶ 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 Example: Combustion reaction: Substance + O2 → Oxygen containing compound(s) Combustion of Natural Gas 𝐶𝐻4 (𝑔) + 𝑂2(𝑔) ⟶ 𝐶𝑂2 𝑔 + 𝐻2 𝑂(𝑔) Dr. Ahmed Sayed Saad 40 Combustion reaction A combustion reaction involves the reaction of a substance with O2 to form one or more oxygen-containing compounds. Combustion reactions also emit heat. C Combustion CO2 H Combustion H2O Dr. Ahmed Sayed Saad 41 Combustion reaction Example of Combustion reactions Dr. Ahmed Sayed Saad 42 Law of conservation of mass - When a chemical reaction occurs, the total mass of the substances involved in the reaction does not change. Dr. Ahmed Sayed Saad 43 Writing and Balancing Chemical Equations Problem: Unbalanced equation → Violate the low of conservation of mass. New atoms do not form during a reaction, nor do atoms vanish—matter must be conserved. Dr. Ahmed Sayed Saad 44 Writing and Balancing Chemical Equations We cannot change the subscripts when balancing a chemical equation because changing the subscripts changes the substance itself, while changing the coefficients changes the number of molecules of the substance. For example, 2 H2O is simply two water molecules, but H2O2 is hydrogen peroxide, a totally different compound. Dr. Ahmed Sayed Saad 45 Writing and Balancing Chemical Equations Check No. of each type of atom on both sides of the equation is now equal → Balanced equation The balanced equation tells us that one CH4 molecule reacts with two O2 molecules to form one CO2 molecule and two H2O molecules. Dr. Ahmed Sayed Saad 46 Refresh Dr. Ahmed Sayed Saad 47 How to Balance a Chemical Equaiton? Example 1. Combustion of gaseous butane (C4H10) (portable stoves and grills) with gaseous oxygen to form gaseous carbon dioxide and gaseous water. 1. Skeletal Equation Dr. Ahmed Sayed Saad 48 How to Balance a Chemical Equaiton? 2. Atoms in compounds Dr. Ahmed Sayed Saad 49 3. Atoms in pure elements 5. Check 4. Remove fractions Dr. Ahmed Sayed Saad 50 How to Balance a Chemical Equation? Example 2. Write a balanced equation for the reaction between solid cobalt(III) oxide and solid carbon to produce solid cobalt and carbon dioxide gas. 1. Skeletal Equation Dr. Ahmed Sayed Saad 51 How to Balance a Chemical Equation? 2. Atoms in compounds Dr. Ahmed Sayed Saad 52 How to Balance a Chemical Equation? 3. Atoms in pure elements Dr. Ahmed Sayed Saad 53 How to Balance a Chemical Equation? 4. Remove fractions ??? 5. Check Dr. Ahmed Sayed Saad 54 How to Balance a Chemical Equation? Example 3. Write a balanced equation for the combustion of gaseous ethane (C2H6) with gaseous oxygen to form carbon dioxide and water. 1. Skeletal Equation 𝐶2 𝐻6 (𝑔) + 𝑂2(𝑔) ⟶ 𝐶𝑂2 𝑔 + 𝐻2 𝑂(𝑔) 2. Atoms in compounds 3. Atoms in pure elements H.W. 4. Remove fractions 5. Check Dr. Ahmed Sayed Saad 55 How to Balance a Chemical Equation? Example 4. Write a balanced equation for the reaction between silicon dioxide and carbon to produce silicon carbide and carbon monoxide gas. ∆ 1. Skeletal Equation 𝑆𝑖𝑂2 (𝑠) + 𝐶(𝑠) 𝑆𝑖𝐶 + 𝐶𝑂(𝑔) 𝑆 2. Atoms in compounds 3. Atoms in pure elements H.W. 4. Remove fractions 5. Check Dr. Ahmed Sayed Saad 56 Refresh Dr. Ahmed Sayed Saad 57 Reaction Stoichiometry The coefficients in a balanced chemical equation specify the relative amounts in moles of each of the substances involved in the reaction. Combustion of Octane 𝑃𝑟𝑜𝑑𝑢𝑐𝑒𝑠 Burning 2 moles of Octane 16 moles of CO2 Mole-to-Mole Conversions Stoichiometric ratio 2 𝑚𝑜𝑙 𝐶8 𝐻18 ∶ 16 𝑚𝑜𝑙 𝐶𝑂2 Problem How many moles of CO2 are formed 22×16 22 𝑚𝑜𝑙 𝐶8 𝐻18 ∶ ? 𝑚𝑜𝑙 𝐶𝑂2 = = 176 𝑚𝑜𝑙 𝐶𝑂2 when burning 22 moles of octane ? 2 58 Refresh Reaction Stoichiometry Mass-to-Mass Conversions Octane Oxygen Carbon dioxide Water Mole ratio 2 mol 25 mol 16 mol 18 mol Molar mass (8×12)+(18×1)= (2×16) = (1×12)+(2×16)= (2×1)+(1×16)= 114 g/mol 32 g/mol 44 g/mol 18 g/mol Mass ratio 2×114 = 25×32= 16×44= 18×18= Moles × Molar mass 228 g 800 g 704 g 324 g 60 Reaction Stoichiometry Mass-to-Mass Conversions Problem Calculate the mass of CO2 emitted upon the combustion of 4.0×1015 g of octane. Octane Carbon dioxide Mole ratio 2 mol 16 mol Formula mass (8×12)+(18×1)= (1×12)+(2×16)= 114 g/mol 44 g/mol Mass ratio 2×114 = 16×44= Moles × Formula mass 228 g 704 g Problem 4.0×1015 g 4×1015 ×704 ? = = 228 1.2 × 1016 g 61 Problem Carbon dioxide Glucose Mole ratio 6 mol 1 mol Formula mass ((1×12)+(2×16)= (6×12)+(12×1)+(6×16)= 44 g/mol 180 g/mol Mass ratio 6×44 = 1×180= Moles × Formula mass 264 g 180 g Problem 37.8 g 37.8×180 ?= = 25.8 g Dr. Ahmed Sayed Saad 264 62 Home Activity Problem Problem H.W. Dr. Ahmed Sayed Saad 63 Refresh Dr. Ahmed Sayed Saad 64 Refresh Dr. Ahmed Sayed Saad 65 Limiting reactant vs. Reactant in Excess Reactant in excess Reactant in excess Limiting Least Amount of Product → Theoretical Yield Reactant Dr. Ahmed Sayed Saad 66 Theoretical vs. Actual Yield & Percent Yield Theoretical Yield Actual Yield Based on the limiting reactant Percent Yield Dr. Ahmed Sayed Saad 67 Theoretical vs. Actual Yield & Percent Yield Dr. Ahmed Sayed Saad 68 Theoretical vs. Actual Yield & Percent Yield - What is the limiting reactant and theoretical yield of CO2 if we start with 5 CH4 molecules and 8 O2 molecules ? Dr. Ahmed Sayed Saad 69 Refresh Dr. Ahmed Sayed Saad 70 Reaction Stoichiometry Calculating Limiting Reactant, Theoretical Yield, and Percent Yield from Initial Reactant Masses Mg O2 MgO Mole ratio 2 mol 1 mol 2 mol Formula mass 24.31 g/mol 32.00 g/mol (1×24.31)+(1×16)= 40.31 g/mol Mass ratio 2×24.31 = 48.62 g 1×32=32 g 2×40.31= 80.62 g Based on Mg 42.5 g 42.5×80.62 ?= = 70.5 g 48.62 Based on O2 33.8 g 33.8 ×80.62 ? = = 85.15 g 32 Dr. Ahmed Sayed Saad 71 Reaction Stoichiometry Dr. Ahmed Sayed Saad 72 Dr. Ahmed Sayed Saad 73

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