Chapter 3. Mass Relations in Chemistry; Stoichiometry
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1 Chapter 3 Mass Relations in Chemistry; Stoichiometry Copyright 2001 by Harcourt, Inc. All rights reserved. Requests for permission to make copies of any part of the work should be mailed to the following address: Permissions Department, Harcourt, Inc Sea Harbor Drive, Orlando, Copyright Florida by Harcourt, Inc. All rights reserved. 3.1
2 Atomic and Formula Masses Meaning of Atomic Masses Atomic Masses from Isotopic Composition Masses of Individual Atoms Copyright 2001 by Harcourt, Inc. All rights reserved. 3.2
3 Meaning of Atomic Masses Give relative masses of atoms Based on C-12 scale Most common isotope of carbon is assigned an atomic mass of 12 amu. Copyright 2001 by Harcourt, Inc. All rights reserved. 3.3
4 The mass spectrometer Copyright 2001 by Harcourt, Inc. All rights reserved. 3.4
5 Meaning of Atomic Masses A nickel atom is / = times as heavy as a calcium ion It is / = times as heavy as a boron ion element B Ca Ni atomic mass amu amu amu Copyright 2001 by Harcourt, Inc. All rights reserved. 3.5
6 Mass spectrum of chlorine Copyright 2001 by Harcourt, Inc. All rights reserved. 3.6
7 Atomic Masses from Isotopic Composition A.M. = (A.M. isotope 1)( ) % 100 % (A.M. isotope 2)( ) + Isotope Atomic Mass Percent Ne amu Ne amu 0.26 Ne amu 8.82 Copyright 2001 by Harcourt, Inc. All rights reserved. 3.7
8 Atomic Masses from Isotopic Composition A.M. Ne = (0.9092) (0.0026) (0.0882) = amu Copyright 2001 by Harcourt, Inc. All rights reserved. 3.8
9 Masses of Individual Atoms The atomic masses of H, Cl, and Ni are H = amu Cl = amu Ni = amu Therefore g H, g Cl, and g Ni all have the same number of atoms: N A N A = Avogadro s number = Copyright 2001 by Harcourt, Inc. All rights reserved. 3.9
10 Masses of Individual Atoms (cont.) Mass of H atom: g H 1 H atom = g atoms Number of atoms in one gram of nickel: atoms Ni 1.00 g Ni = atoms g Ni Copyright 2001 by Harcourt, Inc. All rights reserved. 3.10
11 The Mole Meaning Molar Mass Mole-Mass Conversions Copyright 2001 by Harcourt, Inc. All rights reserved. 3.11
12 Meaning 1 mol = items 1 mol H = atoms; mass = g 1 mol Cl = atoms; mass = g 1 mol Cl 2 = molecules; mass = g 1 mol HCl = molecules; mass = g Copyright 2001 by Harcourt, Inc. All rights reserved. 3.12
13 Molar Mass Generalizing from the previous examples, the molar mass, M, is numerically equal to the sum of the atomic masses sum of atomic molar mass ( M) masses CaCl amu g/mol C 6 H 12 O amu g/mol Copyright 2001 by Harcourt, Inc. All rights reserved. 3.13
14 Mole-Mass Conversions Calculate mass in grams of 13.2 mol CaCl 2 mass = 13.2 mol CaCl g CaCl 2 1 mol CaCl 2 = g Calculate number of moles in 16.4 g C 6 H 12 O 6 moles = 16.4 g C 6 H 12 O 6 1 mol C 6 H 12 O g C 6 H 12 O 6 = mol Copyright 2001 by Harcourt, Inc. All rights reserved. 3.14
15 Formulas Mass % from Formula Simplest Formula from % Composition Simplest Formula from Analytical Data Molecular Formula from Simplest Formula Copyright 2001 by Harcourt, Inc. All rights reserved. 3.15
16 Mass % from Formula Percent composition of K 2 CrO 4? molar mass = ( ) g / mol = g / mol %K = 100% = 40.27% %Cr = 100% = 26.78% %O = 100% = 32.96% Note that percents must add to 100 Copyright 2001 by Harcourt, Inc. All rights reserved. 3.16
17 Simplest Formula from % Composition Find mass of each element in sample of compound. Find numbers of moles of each element. Find mole ratio. Copyright 2001 by Harcourt, Inc. All rights reserved. 3.17
18 Simplest Formula from % Composition Simplest formula of compound containing 26.6% K, 35.4% Cr, 38.0% O Work with 100 g sample: 26.6 g K, 35.4 g Cr, 38.0 g O. 1 mol moles K = 26.6 g = mol K g moles Cr = 35.4 g 1 mol g = mol Cr Copyright 2001 by Harcourt, Inc. All rights reserved. 3.18
19 Simplest Formula from % Composition (cont.) moles O = 38.0 g 1 mol g = 2.38 mol O Note that 2.38 / = 3.50 = 7 / 2 Simplest formula: K 2 Cr 2 O 7 Copyright 2001 by Harcourt, Inc. All rights reserved. 3.19
20 Chemical Equations Balancing Mass Relations in Reactions Copyright 2001 by Harcourt, Inc. All rights reserved. 3.20
21 Balancing Must have same number of atoms of each type on both sides. Achieve this by adjusting coefficients in front of formulas. Copyright 2001 by Harcourt, Inc. All rights reserved. 3.21
22 Balancing Example: Combustion of propane in air to give carbon dioxide and water: Balance C: Balance H: Balance O: C 3 H 8 (g) + O 2 (g) CO 2 (g) + H 2 O(l) C 3 H 8 (g) + O 2 (g) 3CO 2 (g) + H 2 O(l) C 3 H 8 (g) + O 2 (g) 3CO 2 (g) + 4H 2 O(l) C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l) Meaning: 1 mol C 3 H 8 reacts with 5 mol O 2 to form 3 mol CO 2 and 4 mol H 2 O. Copyright 2001 by Harcourt, Inc. All rights reserved. 3.22
23 Mass Relations in Reactions Example: How many moles of CO 2 are produced when 1.65 mol C 3 H 8 burns? Use coefficients of balanced equation to obtain conversion factor: 1.65 mol C 3 H 8 = 4.95 mol CO 2 3 mol CO 2 1 mol C 3 H 8 Copyright 2001 by Harcourt, Inc. All rights reserved. 3.23
24 Mass relations in Reactions (cont.) Example: What mass of O 2 is required to react with 12.0 g of C 3 H 8? 1 mol C 12.0 g C 3 H 8 3 H g O g C 3 H 8 1 mol O 2 5 mol O 2 1 mol C 3 H 8 = 43.6 g O 2 Copyright 2001 by Harcourt, Inc. All rights reserved. 3.24
25 Yield of Product in a Reaction Limiting Reactant, Theoretical Yield Actual Yield, Percent Yield Copyright 2001 by Harcourt, Inc. All rights reserved. 3.25
26 Limiting Reactant, Theoretical Yield Ordinarily, reactants are not present in the exact ratio required for reaction. Instead, one reactant is in excess; some of it is left when the reaction is over. The other, limiting reactant, is completely consumed to give the theoretical yield product. Copyright 2001 by Harcourt, Inc. All rights reserved. 3.26
27 Limiting Reactant, Theoretical Yield (cont.) To calculate the theoretical yield and identify the limiting reactant: Calculate the yield expected if the first reactant is limiting Repeat this calculation for the second reactant The theoretical yield is the smaller of these two quantities The reactant that gives the smaller theoretical yield is the limiting reactant Copyright 2001 by Harcourt, Inc. All rights reserved. 3.27
28 Limiting Reactant, Theoretical Yield (cont.) 2Ag(s) + I 2 (s) 2ΑgI(s) Calculate the theoretical yield of AgI and determine the limiting reactant starting with 1.00 g Ag and 1.00 g I 2. Copyright 2001 by Harcourt, Inc. All rights reserved. 3.28
29 Limiting Reactant, Theoretical Yield (cont.) Theoretical yield if Ag is limiting: g AgI 1.00 g Ag = 2.18 g AgI g Ag Theoretical yield if I 2 is limiting: g AgI 1.00 g I 2 = 1.85 g AgI g I 2 Theoretical yield = 1.85 g AgI; I 2 is limiting reactant Copyright 2001 by Harcourt, Inc. All rights reserved. 3.29
30 Actual Yield, Percent Yield actual yield % yield = 100 theoretical yield Suppose actual yield of AgI were 1.50 g: 1.50 % yield = 100 = Copyright 2001 by Harcourt, Inc. All rights reserved. 3.30
31 Simplest Formula from Analytical Data A sample of acetic acid (C, H, O atoms) weighing g burns to give g CO 2 and g H 2 O. Simplest formula? Solution: Find mass of C in sample (from CO 2 ) Find mass of H in sample (from H 2 O) Obtain mass of O by difference Copyright 2001 by Harcourt, Inc. All rights reserved. 3.31
32 Simplest Formula from Analytical Data mass C = g CO g C g CO 2 = g C mass H = g H 2 O 2.02 g H g H 2 O = g H mass O = 1.00 g g g = g Copyright 2001 by Harcourt, Inc. All rights reserved. 3.32
33 Simplest Formula from Analytical Data moles C = g C moles H = g H 1 mol C g C 1 mol H g H = mol C = mol H moles O = g O Simplest formula is CH 2 O 1 mol O g O = mol O Copyright 2001 by Harcourt, Inc. All rights reserved. 3.33
34 Molecular Formula from Simplest Formula Must know molar mass For acetic acid: M = 60 g/mol 60 / 30 = 2 Molecular formula = C 2 H 4 O 2 Copyright 2001 by Harcourt, Inc. All rights reserved. 3.34
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