Chapter 3 Stoichiometry
Chapter 3 Table of Contents (3.1) (3.2) (3.3) (3.4) (3.5) (3.6) Counting by weighing Atomic masses The mole Molar mass Learning to solve problems (can read on own, not responsible for on exam) Percent composition of compounds
Chapter 3 Table of Contents (3.7) (3.8) (3.9) (3.10) (3.11) Determining the formula of a compound Chemical equations Balancing chemical equations Stoichiometric calculations: Amounts of reactants and products The concept of limiting reactant
Section 3.1 Counting by Weighing Chemical Stoichiometry Study of the amounts of materials consumed and produced in chemical reactions Copyright Cengage Learning. All rights reserved 4
Section 3.2 Atomic Masses Average Atomic Mass Since elements occur in nature as mixtures of isotopes, atomic masses are average of isotopic masses Copyright Cengage Learning. All rights reserved 5
Section 3.2 Atomic Masses Average Atomic Mass of Carbon Natural carbon is a mixture of 12 C, 13 C, and 14 C Atomic mass of carbon is an average value of these three isotopes Composition of natural carbon: 12 C atoms (mass = 12 u {most texts use amu}) - 98.89% 13 C atoms (mass = 13.003355 u) - 1.11% Do NOT memorize for example & info only. Copyright Cengage Learning. All rights reserved 6
Section 3.2 Atomic Masses Average Atomic Mass of Carbon (continued) Calculation of average atomic mass for natural carbon 98.89% of 12 u + 1.11% of 13.0034 u = 0.9889 12 u + 0.0111 13.0034 u ( )( ) ( )( ) = 12.01 u For stoichiometric purposes, assume that carbon is composed of only one type of atom with a mass of 12.01 Copyright Cengage Learning. All rights reserved 7
Section 3.3 The Mole Mole (mol) Unit of measure established for use in counting atoms Number equal to the number of carbon atoms in exactly 12 grams of pure 12 C Techniques such as mass spectrometry determine this number to be 6.02214 10 23 Avogadro s number: One mole of anything consists of 6.022 10 23 units of anything (eggs, people, cars, etc) Copyright Cengage Learning. All rights reserved 8
Section 3.3 The Mole Figure 3.4 - One-Mole Samples of Several Elements
Section 3.3 The Mole Table 3.1 - Comparison of 1 Mole Samples of Various Elements atomic mass of elements
Section 3.3 The Mole Interactive Example 3.5 - Calculating the Number of Moles and Mass Cobalt (Co) is a metal that is added to steel to improve its resistance to corrosion Calculate both the number of moles (?) in a sample of cobalt containing 5.00 10 20 atoms and the mass of the sample (?) End class 9/24/18 Monday
Section 3.3 The Mole Here is the definition of the mole for Co: 1 mole Co = 58.93 g of Co = 6.022 x 10 23 atoms Co Use the above definition as conversion factors. My version of the answer on doc camera.
Section 3.3 The Mole Interactive Example 3.5 - Solution (continued 1) How do we get there? (textbook answer) Note that the sample of 5.00 10 20 atoms of cobalt is less than 1 mole (6.022 10 23 atoms) of cobalt What fraction of a mole it represents can be determined as follows: using dimensional analysis 1 mol Co 6.022 10 atoms Co 20 5.00 10 atoms Co 23 4 = 8.30 10 mol Co
Section 3.3 The Mole Interactive Example 3.5 - Solution (continued 2) Since the mass of 1 mole of cobalt atoms is 58.93 g, the mass of 5.00 10 20 atoms can be determined as follows: using dimensional analysis 4 8.30 10 mol Co 58.93 g Co 2 = 4.89 10 g Co 1 mol Co
Section 3.4 Molar Mass Molar Mass of a Compound (formula unit or molecule) Mass in grams of one mole of the compound Traditionally known as molecular weight (molecular mass) Obtained by summing the masses of the component atoms of a compound Molar Mass of an Element is just the atomic mass directly from periodic table Copyright Cengage Learning. All rights reserved 15
Section 3.4 Molar Mass Ionic Substances Contain simple ions and polyatomic ions Fundamental unit of an ionic compound is called a formula unit Example - NaCl is the formula unit for sodium chloride
Section 3.4 Molar Mass Interactive Example 3.7 - Calculating Molar Mass II a. Calculate the molar mass of calcium carbonate CaCO 3 b. A certain sample of calcium carbonate contains 4.86 moles What is the mass in grams of this sample? What is the mass of the CO 3 2 ions present? Examples on doc camera End 9/26/18 W
Section 3.4 Molar Mass Interactive Example 3.7 - Solution (a) Calcium carbonate is an ionic compound composed of Ca 2+ and CO 3 2 ions In 1 mole of calcium carbonate, there are 1 mole of Ca 2+ ions and 1 mole of CO 3 2 ions Molar mass is calculated by summing the masses of the components
Section 3.4 Molar Mass Interactive Example 3.7 - Solution (a) (continued) 2+ 1 Ca : 1 40.08 g = 40.08 g 1 CO : 2 3 1 C: 1 12.01 g = 12.01 g 3 O: 3 16.00 g = 48.00 g Mass of 1 mol CaCO = 100.09 g Thus, the mass of 1 mole of CaCO 3 (1 mole of Ca 2+ plus 1 mole of CO 3 2 ) is 100.09 g This is the molar mass 3
Section 3.4 Molar Mass Interactive Example 3.7 - Solution (b) Mass of 1 mole of CaCO 3 is 100.09 g (conversion factor) 100.09 g CaCO3 4.86 mol CaCO3 1 mol CaCO 3 = 486 g CaCO 3 To find the mass of carbonate ions (CO 3 2 ) present in this sample, realize that 4.86 moles of CaCO 3 contains 4.86 moles of Ca 2+ ions and 4.86 moles of CO 3 2 ions
Section 3.4 Molar Mass Interactive Example 3.7 - Solution (b) (continued) Mass of 1 mole of CO 3 2 ions is calculated as follows: 1 C: 1 12.01 = 12.01 g 3 O: 3 16.00 = 48.00 g Mass of 1 mol CO = 60.01 g 2 3 Thus, the mass of 4.86 moles of CO 3 2 ions is use as conversion factor 2 60.01 g CO 4.86 mol CO 3 1 mol CO 2 3 2 3 2 = 292 g CO 3
Section 3.4 Molar Mass Interactive Example 3.8 - Molar Mass and Numbers of Molecules Isopentyl acetate (C 7 H 14 O 2 ) is the compound responsible for the scent of bananas Interestingly, bees release about 1 μg (1 10 6 g) of this compound when they sting Resulting scent attracts other bees to join the attack How many molecules of isopentyl acetate are released in a typical bee sting? How many atoms of carbon are present? example on doc camera
Section 3.4 Molar Mass Interactive Example 3.8 - Solution Where are we going? To calculate the number of molecules of C 7 H 14 O 2 and the number of carbon atoms in a bee sting What do we know? Mass of C 7 H 14 O 2 in a typical bee sting is 1 microgram = 1 10 6 g
Section 3.4 Molar Mass Interactive Example 3.8 - Solution (continued 1) How do we get there? We must first compute the molar mass of C 7 H 14 O 2 g 7 mol C 12.01 = 84.07 g C mol 14 mol g H 1.008 mol = 14.11 g H 2 mol g O 16.00 mol = 32.00 g O Molar mass of C H O = 130.18 g 7 14 2
Section 3.4 Molar Mass Interactive Example 3.8 - Solution (continued 2) 1 mole of C 7 H 14 O 2 = 6.022 10 23 molecules = 130.18 g To find the number of molecules released in a sting, we must first determine the number of moles of C 7 H 14 O 2 in 1 10 6 g: 1 mol C H O 9 = 8 10 mol C H O 130.18 g C H O 6 7 14 2 1 10 g C7H14O 2 7 14 2 7 14 2
Section 3.4 Molar Mass Interactive Example 3.8 - Solution (continued 3) Since 1 mole is 6.022 10 23 units, we can determine the number of molecules end 9/28/18 F 9 8 10 mol C7H14O 2 23 6.022 10 molecules 1 mol C H O 7 14 2 To determine the number of C atoms present, multiply the number of molecules by 7 Since each molecule of C 7 H 14 O 2 contains seven C atoms: 15 7 carbon atoms 16 5 10 molecules = 4 10 carbon atoms molecule 15 = 5 10 molecules