Chapter 3 The Mole and Stoichiometry. Chapter 3 Table of Contents Copyright © Cengage Learning. All rights reserved 2 3.1 Counting by Weighing 3.2 Atomic.

Chapter 3 The Mole and Stoichiometry

Chapter 3 Table of Contents Copyright © Cengage Learning. All rights reserved 2 3.1 Counting by Weighing 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass 3.5 Percent Composition by Mass of Compounds 3.6Determining the Formula of a Compound 3.9 Stoichiometric Calculations: Amounts of Reactants and Products 3.10Calculations Involving a Limiting Reactant

Section 3.1 Counting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 3 1.To understand the concept of average mass 2.To learn how counting can be done by massing 3.To understand atomic mass and learn how it is determined 4.To understand the mole concept and Avogadro’s number 5.To learn to convert among moles and mass. Objectives 3.1 – 3.3

Section 3.1 Counting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 4 Need average mass of the object. Objects behave as though they were all identical. Find the average mass of one bean. Find the mass of 1000 beans.

Section 3.1 Counting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 5 A pile of marbles weigh 394.80 g. 10 marbles weigh 37.60 g. How many marbles are in the pile? Average mass of 1 marble = 37.60 g = 3.76g/marble 10 marbles 394.80 g 1 marble = 105 marbles 3.76 g Exercise:

Section 3.2 Atomic MassesCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 6 Elements occur in nature as mixtures of isotopes. Carbon = 98.89% 12 C 1.11% 13 C < 0.01% 14 C

Section 3.2 Atomic MassesCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 7 98.89% of 12 amu + 1.11% of 13.0034 amu = Average Atomic Mass for Carbon (0.9889)(12 amu) + (0.0111)(13.0034 amu) = 12.01 amu exact number

Section 3.2 Atomic MassesCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 8 Even though natural carbon does not contain a single atom with mass 12.01, for stoichiometric purposes, we can consider carbon to be composed of only one type of atom with a mass of 12.01. This average enables us to count atoms of natural carbon by weighing a sample of carbon. Average Atomic Mass for Carbon

Section 3.2 Atomic MassesCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 10 Exercise An element consists of 62.60% of an isotope with mass 186.956 amu and 37.40% of an isotope with mass 184.953 amu. Calculate the average atomic mass and identify the element. 186.2 amu Rhenium (Re)

Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 11 The number equal to the number of carbon atoms in exactly 12 grams of pure 12 C. 1 mole of anything = 6.022 x 10 23 units of that thing (Avogadro’s number). 1 mole C = 6.022 x 10 23 C atoms = 12.01 g C Other units? Dozen? Fathom?? Stone?? Hand?

Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 13 The Mole !!!!!! Let’s use the Periodic Table to determine the Atomic Mass of the following: (Rounding?) H Co Rn Fr W Ne How many atoms do each of the above atomic mass values contain? What are the units?

Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 14 How many is a mole? 6.022 x 10 23 Casio EXP below 3, TI30 XA EE above 7, XIIS 2ndEE above 7, XZ multiview x10 n 3 above #8 Al Foil 1 mole of marbles would cover the earth To a depth of 50 miles Would you accept $1 million ($1 x 10 6 ) to count to a mole? If you counted 1 per second, it would take you 2 x 10 16 years to finish Your hourly wage would be $5 x 10 -15 per hour It would take hundreds of millions of years to earn $0.01

Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 15 How to convert using unit analysis. First, let’s learn to convert from moles to grams. What is the atomic mass (molar mass) of Fe? What are the units of atomic mass (molar mass)? How many grams are contained in 1 mol of Fe? How many grams are contained in 2 mol of Fe? How many grams are contained in 0.5 mol of Fe? How many grams are contained in 2.4 mol of Fe? 2.4 mol Fe 55.85 g Fe = 1 mol Fe Know to Go approach, Unit Analysis!!

Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 16 How to convert using unit analysis. Use Unit Analysis to convert from moles to grams: 0.250 mol Al 1.28 mol Ca 0.371 mol P 10.24 mol Au To convert from moles to grams, multiply by the molar mass.

Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 17 How to convert using unit analysis. Second, let’s learn to convert from grams to moles What is the atomic mass (molar mass) of Al? What are the units of atomic mass (molar mass)? How many moles of atoms do 27 g of Al contain? How many moles of atoms do 54 g of Al contain? How many moles of atoms do 108 g of Al contain? How many moles of atoms do 56 g of Al contain? 56 g Al 1 mole Al = 27 g Al Know to Go approach, Unit Analysis!!

Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 18 How to convert using unit analysis. Use Unit Analysis to convert from grams to moles: 220 g Al 568 g Si 3.61 mg He 26.7 g Li To convert from grams to moles, divide by the molar mass.

Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 19 1.To understand the concept of average mass 2.To learn how counting can be done by massing 3.To understand atomic mass and learn how it is determined 4.To understand the mole concept and Avogadro’s number 5.To learn to convert among moles and mass. Objectives Review 3.1 – 3.3

Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 20 1.To learn to convert among moles, mass, and # of atoms. 2.Unit Analysis! Multi Step! Objectives 3.3 part 2

Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 21 Conversions Help Page! To convert from moles to grams, multiply by the molar mass. To convert from grams to moles, divide by the molar mass. 1 mole = 6.02 X 10 23 Grams  moles  # atoms X g or 1 mol 1 mol X g 6.02 X 10 23 atoms or 1 mol. 1 mol 6.02 X 10 23 atoms

Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 22 Know to Go approach, Unit Analysis!! Let it Guide you! Grams  moles  # atoms How many atoms in 79 g of Se? (Know to go?) 79 g Se 1 mol Se 6.02 X 10 23 atoms Se = 78.96 g Se 1 mol Se How many atoms in 2.35 mol of #%^$? 2.35 mol #%^$ 6.02 X 10 23 atoms #%^$ = 1 mol #%^$ Remember that a mole measures quantity!!

Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 23 Know to Go approach, Unit Analysis!! Let it Guide you! Grams  moles  # atoms How many atoms are contained in 4.17 mol of Al? How many atoms are contained in 4.17 mol of Si? How many atoms are contained in 1.67 mol of Al? How many atoms are contained in 2 X 10 -4 mol of Al? How many moles are present in 2.13 X 10 24 C atoms? What is the mass in g of the above moles of C? What is the mass in g of 9.4 X 10 25 atoms of Mg? How many atoms are in 365 g of V?

Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 24 Concept Check Calculate the number of copper atoms in a 63.55 g sample of copper. 6.022×10 23 Cu atoms

Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 25 Concept Check Calculate the number of grams contained in a sample of 1.21 X 10 23 Al atoms. 5.42 g Al

Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 26 Exercise Rank the following according to number of moles (greatest to least): a) 107.9 g of silver b) 70.0 g of zinc c) 21.0 g of magnesium b) a) c)

Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 27 Exercise Rank the following according to number of atoms (greatest to least): a) 107.9 g of silver b) 70.0 g of zinc c) 21.0 g of magnesium b) a) c)

Section 3.3 The MoleCounting by Weighing Return to TOC Copyright © Cengage Learning. All rights reserved 28 1.To learn to convert among moles, mass, and # of atoms. 2.Unit Analysis! Multi Step! 3.Work Session: Page 117 # 21,27(amu = g),34,35 Objectives Review 3.3 part 2

Section 3.4 Molar Mass Return to TOC Copyright © Cengage Learning. All rights reserved 29 1.To understand the definition of molar mass for atoms and compounds 2.To learn to convert between moles and mass for compounds Objectives

Section 3.4 Molar Mass Return to TOC Copyright © Cengage Learning. All rights reserved 30 1. Molar Mass A compound is a collection of atoms bound together. The molar mass of a compound is obtained by summing the masses of the component atoms.

Section 3.4 Molar Mass Return to TOC Copyright © Cengage Learning. All rights reserved 31 1. Molar Mass For compounds containing ions the molar mass is obtained by summing the masses of the component ions.

Section 3.4 Molar Mass Return to TOC Copyright © Cengage Learning. All rights reserved 32 Conversions Help Page! 1 mole = 6.02 X 10 23 Grams  moles  # atoms or molecules For molecules… 6.02 X 10 23 molecules or 1 mol. 1 mol 6.02 X 10 23 molecules

Section 3.4 Molar Mass Return to TOC Copyright © Cengage Learning. All rights reserved 33 2. Calculations Using Molar Mass (~Atomic Mass) What is the Molar Mass of methane gas CH 4 ? Use Unit Analysis to convert: How many molecules are contained within 1 mole of methane gas CH 4 ? How many C atoms are contained within 1 mole of methane gas CH 4 ? How many H atoms are contained within 1 mole of methane gas CH 4 ? How many g of C are contained within 1 mole of methane gas? How many g of H are contained within 1 mole of methane gas?

Section 3.4 Molar Mass Return to TOC Copyright © Cengage Learning. All rights reserved 34 2. Calculations Using Molar Mass (~Atomic Mass) Find the Molar Mass of SO 2 Find the Molar Mass of NaCl Find the Molar Mass of CaCO 3 What would the mass of 4.86 mol CaCO 3 be? Find the Molar Mass of Na 2 SO 4 How many moles in 300 g of Na 2 SO 4 ? (know-go?) SO 2 : 64 g/mol NaCl: 58.5 g/mol CaCO 3 : 100 g/mol, 486 g Na 2 SO 4 : 142 g/mol, 2.11 mol

Section 3.4 Molar Mass Return to TOC Copyright © Cengage Learning. All rights reserved 35 2. Calculations Using Molar Mass (~Atomic Mass) How many moles in 1.56 g of Juglone C 10 H 6 O 3 ? (Dye and herbicide made from black walnuts) 8.96 X 10 -3 mol C 10 H 6 O 3

Section 3.4 Molar Mass Return to TOC Copyright © Cengage Learning. All rights reserved 36 2. Calculations Using Molar Mass (~Atomic Mass) Bees release 1 X 10 -6 g of isopentyl acetate (C 7 H 14 O 2 ) during a sting. (Also a banana scent!) How many moles of isopentyl acetate are released? 8 X 10 -9 mol C 7 H 14 O 2 How many molecules of isopentyl acetate are released? 5 X 10 15 molecules C 7 H 14 O 2

Section 3.4 Molar Mass Return to TOC Copyright © Cengage Learning. All rights reserved 37 2. Calculations Using Molar Mass (~Atomic Mass) How many molecules are contained in 135 g of Teflon (C 2 F 4 )? 8.127 X 10 23 molecules of C 2 F 4

Section 3.4 Molar Mass Return to TOC Copyright © Cengage Learning. All rights reserved 38 1.To understand the definition of molar mass for atoms and compounds 2.To learn to convert between moles and mass for compounds 3.Work Session: 118 # 37,39,41,43,45,47,49 Objectives Review

Section 3.5 Percent Composition of Compounds Return to TOC Copyright © Cengage Learning. All rights reserved 39 1.To learn to calculate the mass percent of an element in a compound 2.To learn to calculate empirical formulas 3.To learn to calculate the molecular formula of a compound Objectives 3.5 and 3.6

Section 3.5 Percent Composition of Compounds Return to TOC Copyright © Cengage Learning. All rights reserved 41 1. Percent Composition of Compounds Rules to determine the percent composition by mass of each element: 1. Determine the molar mass of the compound 2. Divide the mass of each element type by the total mass of the compound (#1) 3. Multiply by 100 for percent Mass percent =

Section 3.5 Percent Composition of Compounds Return to TOC Copyright © Cengage Learning. All rights reserved 42 1. Percent Composition of Compounds Determine the percent composition by mass of each element in methane gas CH 4 Molar Mass Each Element X 100 75% C, 25% H

Section 3.5 Percent Composition of Compounds Return to TOC Copyright © Cengage Learning. All rights reserved 43 1. Percent Composition of Compounds Determine the percent composition by mass of each element in SO 2 Molar Mass Each Element X 100 50/50

Section 3.5 Percent Composition of Compounds Return to TOC Copyright © Cengage Learning. All rights reserved 44 1. Percent Composition of Compounds Determine the percent composition by mass of each element in NaCl Molar Mass Each Element X 100 39% Na, 61% Cl

Section 3.5 Percent Composition of Compounds Return to TOC Copyright © Cengage Learning. All rights reserved 45 1. Percent Composition of Compounds Determine the percent composition by mass of each element in CaCO 3 Molar Mass Each Element X 100 40% Ca, 12% C, 48% O

Section 3.6 Determining the Formula of a Compound Return to TOC Copyright © Cengage Learning. All rights reserved 46 Device used to determine the mass percent of each element in a compound. Analyzing for Carbon and Hydrogen

Section 3.6 Determining the Formula of a Compound Return to TOC Copyright © Cengage Learning. All rights reserved 47 Empirical formula = CH  Simplest whole-number ratio Molecular formula = (empirical formula) n [n = integer] Molecular formula = C 6 H 6 = (CH) 6  Actual formula of the compound Formulas

Section 3.6 Determining the Formula of a Compound Return to TOC Copyright © Cengage Learning. All rights reserved 49 2. Calculation of Empirical Formulas Patch’s Interpretation: Calculate moles if you don’t already have them. Divide all by the smallest # of moles Get to Whole Number Multiples.

Section 3.6 Determining the Formula of a Compound Return to TOC Copyright © Cengage Learning. All rights reserved 50 2. Calculation of Empirical Formulas Determine the empirical formula if an analysis of a C, H, and O compound resulted in the following : 0.0806 g C 0.01353 g H 0.1074 g O 0.00671 mol C, 0.01353 mol H, 0.00671 mol O

Section 3.6 Determining the Formula of a Compound Return to TOC Copyright © Cengage Learning. All rights reserved 51 2. Calculation of Empirical Formulas Now, let’s find the empirical formula: 0.00671 mol C 0.01353 mol H 0.00671 mol O CH 2 O(simplest) C 2 H 4 O 2 ; C 4 H 8 O 4 ;C 5 H 10 O 5

Section 3.6 Determining the Formula of a Compound Return to TOC Copyright © Cengage Learning. All rights reserved 52 2. Calculation of Empirical Formulas Determine the empirical formula if an analysis of a Ni and O compound resulted in the following : 0.2636 g Ni 0.0718 g O NiO

Section 3.6 Determining the Formula of a Compound Return to TOC Copyright © Cengage Learning. All rights reserved 54 2. Calculation of Empirical Formulas It is observed that 0.664 g of lead combined with 0.2356 g of chlorine. Determine the empirical formula. PbCl 2

Section 3.6 Determining the Formula of a Compound Return to TOC Copyright © Cengage Learning. All rights reserved 55 2. Calculation of Empirical Formulas Determine the empirical formula of cisplatin, a cancer tumor treatment drug that has the following % composition by mass: 65.02% Pt 9.34% N 2.02% H 23.63% Cl Assume some mass (100g) PtN 2 H 6 Cl 2

Section 3.6 Determining the Formula of a Compound Return to TOC Copyright © Cengage Learning. All rights reserved 56 2. Calculation of Empirical Formulas The most common form of nylon (Nylon-6) is 63.68% C, 12.38% N, 9.80% H, and 14.4% O by mass. Calculate the empirical formula for Nylon-6. C 6 NH 11 O

Section 3.6 Determining the Formula of a Compound Return to TOC Copyright © Cengage Learning. All rights reserved 57 3. Calculation of Molecular Formulas The molecular formula is the exact formula of the molecules present in a substance. The molecular formula is a whole number multiple (n) of the empirical formula. n = molar mass molecular formula molar mass empirical formula 1.Find n 2.(empirical formula) n = molecular formula **You have to be given the molar mass of the molecular formula!

Section 3.6 Determining the Formula of a Compound Return to TOC Copyright © Cengage Learning. All rights reserved 58 3. Calculation of Molecular Formulas Calculate the molecular formula for a compound if the empirical formula is P 2 O 5 and the molecular molar mass is 283.88 g/mol. P 4 O 10 Check the mm=(4(31)+10(16))=284

Section 3.6 Determining the Formula of a Compound Return to TOC Copyright © Cengage Learning. All rights reserved 59 3. Calculation of Molecular Formulas Calculate the empirical and molecular formulas for a compound that gives the following mass percentages upon analysis: 71.65% Cl 24.27% C 4.07% H The molar mass is known to be 98.96 g/mol

Section 3.6 Determining the Formula of a Compound Return to TOC Copyright © Cengage Learning. All rights reserved 61 3. Calculation of Molecular Formulas ClCH 2 (empirical) The molar mass is known to be 98.96 g/mol n = molar mass molecular formula molar mass empirical formula Cl 2 C 2 H 4

Section 3.6 Determining the Formula of a Compound Return to TOC Copyright © Cengage Learning. All rights reserved 62 3. Calculation of Molecular Formulas Caffeine (MM = 194.2 g/mol) contains by mass: 49.48% C 5.15% H 28.87% N 16.49% O Determine the molecular formula.

Section 3.6 Determining the Formula of a Compound Return to TOC Copyright © Cengage Learning. All rights reserved 64 3. Calculation of Molecular Formulas Caffeine (MM = 194.2 g/mol) n = molar mass molecular formula molar mass empirical formula C 8 H 10 N 4 O 2

Section 3.6 Determining the Formula of a Compound Return to TOC Copyright © Cengage Learning. All rights reserved 65 Hydrate Lab Calculate the molar mass MgSO 4 * 7H 2 O Mg + S + 4(O) + 7(H 2 O) Calculate the % mass of waters… 7(H 2 O). X 100 Mg + S + 4(O) + 7(H 2 O) MM = 246g/mol : % H 2 O = 51.22% Silica Gel Packaging…

Section 3.6 Determining the Formula of a Compound Return to TOC Copyright © Cengage Learning. All rights reserved 66 1.To understand the meaning of empirical formula 2.To learn to calculate empirical formulas 3.To learn to calculate the molecular formula of a compound 4.Work Session: pg 119 # 60, 67, 69, 68, 74 Objectives Review 3.5 and 3.6

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 69 1.To understand the information given in a balanced equation 2.To use a balanced equation to determine relationships between moles of reactant and products (mole ratio) (Stoichiometry) Objectives 3.9 Part 1

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 70 1.Let’s create a sandwich recipe…PHeTPHeT 2.Slice, dozen, 100, 6.02 X 10 23, one mole Sandwich Shoppe!

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 72 1. Information Given by Chemical Equations The coefficients of a balanced equation give the relative numbers of molecules. A balanced chemical equation gives relative numbers (or moles) of reactant and product molecules that participate in a chemical reaction.

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 73 2. Mole-mole Relationships A balanced equation can predict the moles of product that a given number of moles of reactants will yield. How many moles 0 2 from __ moles H 2 O?

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 74 The mole ratio allows us to convert from moles of one substance in a balanced equation to moles of a second substance in the equation. 2. Mole-mole Relationships

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 75 2. Mole-mole Relationships The mole ratio is determined by: Mole ratio = new mole substance A balanced mole substance A How many mole ___ from ___ moles ___? 2H 2 O  2H 2 + O 2 mole mole ratio mole

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 76 1.Balance the equation for the reaction. 2. 3.Use the balanced equation to set up the appropriate mole ratios. 4.Use the appropriate mole ratios to calculate the number of moles of desired reactant or product. 5. Calculating Masses of Reactants and Products in Reactions

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 77 2. Mole-mole Relationships Must first have a BALANCED equation __C 3 H 8(g) + __O 2  __CO 2(g) + __H 2 O (g) + heat 4.30 C 3 H 8 +__O 2  __ CO 2 + __ H 2 O + heat Calculate the number of moles of CO 2 formed when 4.30 mol C 3 H 8 combust. mole mole ratio mole

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 78 2. Mole-mole Relationships From this new mole value, we can calculate the new recipe! C 3 H 8(g) + 5O 2  3CO 2(g) + 4H 2 O (g) + heat 4.30 C 3 H 8 +__O 2  12.9 CO 2 + __ H 2 O + heat Calculate the number of moles of H 2 O formed when 4.30 mol C 3 H 8 combust. mole mole ratio mole

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 79 2. Mole-mole Relationships For all reactants AND products! C 3 H 8(g) + 5O 2  3CO 2(g) + 4H 2 O (g) + heat 4.30 C 3 H 8 +__O 2  12.9 CO 2 + 17.2 H 2 O + heat Calculate the number of moles of O 2 consumed when 4.30 mol C 3 H 8 combust. mole mole ratio mole

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 80 2. Mole-mole Relationships C 3 H 8(g) + 5O 2  3CO 2(g) + 4H 2 O (g) + heat Calculate the number of moles of all reactants and products when 2.70 mol C 3 H 8 combust. 2.7 C 3 H 8 +__O 2  __ CO 2 + __ H 2 O + heat 2.7 C 3 H 8 +13.5 O 2  8.1 CO 2 + 10.8 H 2 O + heat

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 81 2. Mole-mole Relationships Balance the following equation and determine how much NH 3 can be made from 1.3 mol H 2. __N 2 + __H 2  __NH 3 N 2 + 3H 2  2NH 3 How much N 2 will be needed to completely react the 1.3 mol H 2 ?

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 82 2. Mole-mole Relationships Using the balanced equation, determine the new mole recipe given the new mole value. 2 K + 2 H 2 O  H 2 + 2 KOH ___K + 1.7 H 2 O  ___ H 2 + ___KOH 1.7 K + 1.7 H 2 O  0.85 H 2 + 1.7 KOH 4 NH 3 + 5 O 2  4 NO + 6 H 2 O ___ NH 3 + ___ O 2  0.6 NO + ___ H 2 O 0.6 NH 3 + 0.75 O 2  0.6 NO + 0.9 H 2 O

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 83 1.Balance the equation for the reaction. 2. 3.Use the balanced equation to set up the appropriate mole ratios. 4.Use the appropriate mole ratios to calculate the number of moles of desired reactant or product. 5. Calculating Masses of Reactants and Products in Reactions

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 84 1.To understand the information given in a balanced equation 2.To use a balanced equation to determine relationships between moles of reactant and products (mole ratio) (Stoichiometry) Objectives Review 3.9 Part 1

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 85 1.To learn to use Stoichiometry to relate masses of reactants and products in a chemical reaction 2.To perform mass calculations that involve scientific notation 3.Calculate the theoretical predications for a reaction that will be performed in the lab Objectives 3.9 Part 2

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 86 1. Mass Calculations Stoichiometry is the process of using a balanced chemical equation to determine the relative masses (quantities) of reactants and products involved in a reaction. –Scientific notation can be used for the masses of any substance in a chemical equation. Mole ratio = new mole substance A balanced mole substance A

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 87 1. Mass Calculations N 2 + 3H 2  2NH 3 How much NH 3 will be produced when 2.6 g H 2 completely reacts with N 2 ? **Can’t do a mass ratio!!** Must first convert from grams to moles. Don’t forget the mole! grams mole mole ratio mole

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 88 1. Mass Calculations N 2 + 3H 2  2NH 3 How much NH 3 will be produced when 2.6 g H 2 completely reacts with N 2 ? 2.6 grams H 2 grams mole mole ratio mole 0.867 mol NH 3 : 14.71 g NH 3

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 89 1. Mass Calculations N 2 + 3H 2  2NH 3 __ N 2 + 1.3 H 2  0.867 NH 3 (new mol recipe) __ g N 2 2.6 g H 2 14.71 g NH 3 (new g recipe) How much N 2 will be consumed when 2.6 g H 2 completely reacts? Use the same mole ratio— 0.433 mol N 2 : 12.12 g N 2

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 90 1. Mass Calculations N 2 + 3H 2  2NH 3 0.433 N 2 + 1.3 H 2  0.867 NH 3 (new mol recipe) 12.12 g N 2 2.6 g H 2 14.71 g NH 3 (new g recipe) How does this relate to the Law of Conservation of Mass? Does the mass reactants = mass products? 12.12 g + 2.6 g = 14.72 g Compared to 14.71 g Why the 0.01 g difference?

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 91 1.Balance the equation for the reaction. 2.Convert the known mass of the reactant or product to moles of that substance. 3.Use the balanced equation to set up the appropriate mole ratios. 4.Use the appropriate mole ratios to calculate the number of moles of desired reactant or product. 5.Convert from moles back to grams if required by the problem. Calculating Masses of Reactants and Products in Reactions

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 92 1. Mass Calculations __ Al (S) + __ I 2(g)  __ AlI 3(s) Balance the equation and calculate the mass of I 2 needed to completely react with 35.0 g Al. Also calculate the mass of AlI 3 that will be formed. grams mole mole ratio mole (next)

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 93 2 Al (S) + 3 I 2(g)  2 AlI 3(s) __ Al (S) + __ I 2(g)  __ AlI 3(s) (new mole) 35.0 g Balance the equation and calculate the mass of I 2 needed to completely react with 35.0 g Al. Also calculate the mass of AlI 3 that will be formed. 35.0 g Al (all new moles)

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 94 2 Al (S) + 3 I 2(g)  2 AlI 3(s) 1.3 Al (S) + 1.95 I 2(g)  1.3 AlI 3(s) (new mole) 35.0 g Balance the equation and calculate the mass of I 2 needed to completely react with 35.0 g Al. Also calculate the mass of AlI 3 that will be formed. Now convert mole  grams for each 1.95 mol I 2 = 1.3 mol AlI 3 =

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 95 2 Al (S) + 3 I 2(g)  2 AlI 3(s) 1.3 Al (S) + 1.95 I 2(g)  1.3 AlI 3(s) (new mole) 35.0 g 495.3 g 530.4 g Balance the equation and calculate the mass of I 2 needed to completely react with 35.0 g Al. Also calculate the mass of AlI 3 that will be formed. Check the Law of Conservation of Mass!

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 96 2. Scientific Notation (CO 2 scrubber on space vehicles) __LiOH (s) + __CO 2(g)  __Li 2 CO 3(s) + __H 2 O (l) What mass of CO 2 can 1 X 10 3 g LiOH absorb? What mass of H 2 O will be available for use? What mass of Li 2 CO 3 will be generated? First Balance! Then, below!! grams mole mole ratio mole

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 97 2. Scientific Notation (CO 2 scrubber on space vehicles) 2 LiOH (s) + CO 2(g)  Li 2 CO 3(s) + H 2 O (l) __LiOH (s) + __CO 2(g)  __Li 2 CO 3(s) + __H 2 O (l) What mass of CO 2 can 1 X 10 3 g LiOH absorb? Do new mole recipe next… grams mole mole ratio mole

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 98 2. Scientific Notation (CO 2 scrubber on space vehicles) 2 LiOH (s) + CO 2(g)  Li 2 CO 3(s) + H 2 O (l) 41.67LiOH + 20.8CO 2  20.8Li 2 CO 3 + 20.8H 2 O Now convert all moles to grams. 20.8 mol CO 2 = 20.8 mol Li 2 CO 3 = 20.8 mol H 2 O =

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 99 2. Scientific Notation (CO 2 scrubber on space vehicles) 2 LiOH (s) + CO 2(g)  Li 2 CO 3(s) + H 2 O (l) 41.67LiOH + 20.8CO 2  20.8Li 2 CO 3 + 20.8H 2 O 1 X 10 3 g 915.2 g 1539.2 g 374.4 g What mass of CO 2 can 1 X 10 3 g LiOH absorb? What mass of H 2 O will be available for use? What are some possible uses for the H 2 O? What mass of Li 2 CO 3 will be generated? What will be done with the Li 2 CO 3 ? Conserved?

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 100 2. Mass Calculations Using Scientific Notation Hydrofluoric acid, an aqueous solution containing dissolved hydrogen fluoride, is used to etch glass by reacting with silica, SiO 2, in the glass to produce gaseous silicon tetrafluoride and liquid water. The unbalanced reaction is: __HF (aq) + __SiO 2(s)  __SiF 4(g) + __H 2 O (l) Calculate the mass of HF that is needed and the products that are produced to react completely with 5.68 g SiO 2.

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 101 2. Mass Calculations Using Scientific Notation 4 HF (aq) + SiO 2(s)  SiF 4(g) + 2 H 2 O (l) __ HF (aq) + __ SiO 2(s)  __ SiF 4(g) + __ H 2 O (l) 5.68 g SiO 2 = Mole ratio grams mole mole ratio mole

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 102 2. Mass Calculations Using Scientific Notation 4 HF (aq) + SiO 2(s)  SiF 4(g) + 2 H 2 O (l) 0.378 9.45 X10 -2  9.45 X10 -2 + 0.189 (moles) 7.56 g 5.68 g  9.828 g 3.41 g (grams) Was mass Conserved? grams mole mole ratio mole

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 103 2. Mass Calculations Comparisons Which is the better antacid? Baking Soda,NaHCO 3,or MOM, Mg(OH) 2 ? How many moles of HCl will react with 1.00 g of each antacid? NaHCO 3(s) + HCl (aq)  NaCl (aq) + H 2 O (l) + CO 2(g) Demo Mg(OH) 2(s) + 2HCl (aq)  2H 2 O (l) + MgCl 2(aq) grams mole mole ratio mole

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 104 2. Mass Calculations Comparisons Which is the better antacid? Baking Soda,NaHCO 3,or MOM, Mg(OH) 2 ? How many moles of HCl will react with 1.00 g of each antacid? NaHCO 3(s) + HCl (aq)  NaCl (aq) + H 2 O (l) + CO 2(g) 1.0 g/1.19 X 10 -2 mol NaHCO 3 1.19 X 10 -2 mol HCl Mg(OH) 2(s) + 2HCl (aq)  2H 2 O (l) + MgCl 2(aq) 1.0 g/1.71 X 10 -2 mol Mg(OH) 2 3.42 X 10 -2 mol HCl MOM is better!

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 105 2. Mass Calculations for Lab- don’t round too much! Predict the charged balanced products and their solubility. Then balance the equation, and calculate the number of grams of all reactants and products needed to completely react with 0.01 mol SrCl 2. ___SrCl 2 + ___Na 2 CO 3 

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 106 2. Mass Calculations for lab SrCl 2 + Na 2 CO 3  SrCO 3 + 2NaCl 0.01 SrCl 2 + ___Na 2 CO 3  ___SrCO 3 +___NaCl 0.01SrCl 2 + 0.01Na 2 CO 3  0.01SrCO 3 + 0.02NaCl 1.59 g + 1.06 g  1.48 g + 1.17 g Conservation of mass? Theoretical vs Real…

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 107 2. Mass Calculations for lab Calculate the amount of grams contained in 0.01 mol the following hydrated crystals: SrCl 2 x 6H 2 O Na 2 CO 3 X H 2 O 2.67 g SrCl 2 x 6H 2 O 1.24 g Na 2 CO 3 X H 2 O

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 108 2. Mass Calculations for lab 0.01SrCl 2 x 6H 2 O + 0.01Na 2 CO 3 X H 2 O  0.01SrCO 3 + 0.02NaCl 2.67 g + 1.24 g  1.48 g + 1.17 g

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 109 1.Balance the equation for the reaction. 2.Convert the known mass of the reactant or product to moles of that substance. 3.Use the balanced equation to set up the appropriate mole ratios. 4.Use the appropriate mole ratios to calculate the number of moles of desired reactant or product. 5.Convert from moles back to grams if required by the problem. Calculating Masses of Reactants and Products in Reactions

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 110 1.To understand the information given in a balanced equation 2.To use a balanced equation to determine relationships between moles of reactant and products (mole ratio) (Stoichiometry) 3.Calculate the theoretical predications for a reaction that will be performed in the lab 4.Work Session: pg 121 # 89, 91, 94, 95 and Challenge Question next 3 slides. Objectives Review 3.9 Part 2

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 111 Work Session Challenge Question Methane (CH 4 ) reacts with the oxygen in the air to produce carbon dioxide and water. Ammonia (NH 3 ) reacts with the oxygen in the air to produce nitrogen monoxide and water.  Write balanced equations for each of these reactions.

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 112 Work Session Challenge Question Methane (CH 4 ) reacts with the oxygen in the air to produce carbon dioxide and water. Ammonia (NH 3 ) reacts with the oxygen in the air to produce nitrogen monoxide and water.  What mass of ammonia would produce the same amount of water as 1.00 g of methane reacting with excess oxygen?

Section 3.9 Stoichiometric Calculations: Amounts of Reactants and Products Return to TOC Copyright © Cengage Learning. All rights reserved 113 Where are we going?  To find the mass of ammonia that would produce the same amount of water as 1.00 g of methane reacting with excess oxygen. How do we get there?  We need to know: How much water is produced from 1.00 g of methane and excess oxygen. How much ammonia is needed to produce the amount of water calculated above. Let’s Think About It

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 114 1.To understand the concept of limiting reactants 2.To learn to recognize the limiting reactant in a reaction 3.To learn to use the limiting reactant to do stoichiometric calculations 4.To learn to calculate percent yield Objectives

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 115 Limiting reactant – the reactant that is consumed first and therefore limits the amounts of products that can be formed. Determine which reactant is limiting to calculate correctly the amounts of products that will be formed. Limiting Reactants

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 116 1. The Concept of Limiting Reactants Stoichiometric mixture (balanced) –N 2 (g) + 3H 2 (g)  2NH 3 (g)

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 117 Limiting reactant mixture (runs out first) 1. The Concept of Limiting Reactants –N 2 (g) + 3H 2 (g)  2NH 3 (g)

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 120 The amount of products that can form is limited by the methane. Methane is the limiting reactant. Water is in excess. Limiting and Excess Reactants depend on the actual amounts of reactants present and must be calculated for each different reaction. (Water won’t always be in excess!) Limiting Reactants

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 122 2. Calculations Involving a Limiting Reactant What mass of water is needed to react with 249 g methane (CH 4 )? CH 4 + H 2 O  3H 2 + CO 249 g CH 4 How many g of H 2 and CO are produced? 279 g H 2 O, 93 g H 2, 434 g CO

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 123 2. Calculations Involving a Limiting Reactant How many g of H 2 and CO are produced when 249 g methane (CH 4 ) reacts with 300 g H 2 O? CH 4 + H 2 O  3H 2 + CO 249 g + 279 g  93 g + 434 g (last slide) 249 g + 300 g  93 g + 434 g The 249 g CH 4 will react with only 279 g H 2 O, thereby leaving 21 g H 2 O in excess. CH 4 is the reactant that will run out first and is the LIMITING REACTANT.

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 124 2. Calculations Involving a Limiting Reactant So, how do we figure this out using Stoich? 1) Balance Reaction 2) Do Stoich to relate EACH reactant to one product 3) Whichever reactant produces the LEAST amount of product is the LIMITING REACTANT

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 126 2. Calculations Involving a Limiting Reactant Suppose 2.50 X 10 4 g N 2 reacts with 5 X 10 3 g H 2. Determine the limiting reactant. ___N 2 + ___H 2  ___NH 3 2.50 X 10 4 g N 2 5 X 10 3 g H 2 From N 2  1785.72 mol NH 3 From H 2  1666.67 mol NH 3 ; 28333.3 g NH 3 Not just because original mass of H 2 was less.

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 127 2. Calculations Involving a Limiting Reactant How much N 2 will be produced when 18.1 g NH 3 and 90.4 g CuO react? Determine the limiting reactant. __NH 3 + __CuO  __N 2 + __Cu + __H 2 O 18.1 g NH 3 90.4 g CuO NH 3  0.53 mol N 2 CuO  0.38 mol N 2 *limiting reactant* 10.64 g N 2

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 128 Concept Check You know that chemical A reacts with chemical B. You react 10.0 g of A with 10.0 g of B.  What information do you need to know in order to determine the mass of product that will be produced?

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 129 Exercise You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol? They react according to the equation: A + 3B  2C

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 130 3. Percent Yield Theoretical Yield –The maximum amount of a given product that can be formed when the limiting reactant is completely consumed. Just what we’ve been calculating! The actual yield (amount produced is usually given) of a reaction is usually less than the max theoretical yield because of side or incomplete reactions. Percent Yield The actual amount of a given product as the percentage of the theoretical yield.

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 131 3. Percent Yield From the previous set of calculations, we determined that 10.64 g of N 2 would theoretically be produced from the reaction: 2NH 3 + 3CuO  N 2 + 3Cu + 3H 2 O What would the Percent Yield be if 7.25 g of N 2 were actually produced? 7.25 g N 2 X 100 = 68.14 % 10.64 g N 2

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 132 3. Percent Yield Consider the following reaction: __CO(g) + __H 2 (g)  __CH 3 OH(l) If 6.85 X 10 4 g CO reacts with 8.6 X 10 3 g H 2 to produce an actual yield of 3.57 X 10 4 g CH 3 OH, determine the limiting reactant, the theoretical yield, and the percent yield.

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 133 3. Percent Yield CO(g) + 2H 2 (g)  CH 3 OH(l) 2446.4 mol CO 4300 mol H 2 2446.4 mol CH 3 OH 2150 mol CH 3 OH *Limiting 2150 mol CH 3 OH *32g/mol = 68800 g CH 3 OH % yield = 3.57 X 10 4 g/ 6.88 X 10 4 g = 51.89 % 28 g/mol CO: 2 g/mol H 2 : 32 g/mol CH 3 OH

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 134 3. Percent Yield Consider the following reaction: __TiCl 4 + __O 2  __TiO 2 + __Cl 2 If 6.71 X 10 3 g TiCl 4 reacts with 2.45 X 10 3 g O 2 with a percent yield of 75%, determine the limiting reactant, the theoretical yield, and the actual yield of TiO 2. TiCl 4 is limiting 2119.6 g TiO 2 is actually produced in this reaction. 189.68 g/mol TiCl 4 : 32 g/mol O 2 : 79.88 g/mol TiO 2

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 135 1.To understand the concept of limiting reactants 2.To learn to recognize the limiting reactant in a reaction 3.To learn to use the limiting reactant to do stoichiometric calculations 4.To learn to calculate percent yield 5.Work Session: pg 121 # 97, 101, 104, 105 Objectives Review

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 136 We cannot simply add the total moles of all the reactants to decide which reactant mixture makes the most product. We must always think about how much product can be formed by using what we are given, and the ratio in the balanced equation. Notice

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 138 Where are we going?  To determine the mass of product that will be produced when you react 10.0 g of A with 10.0 g of B. How do we get there?  We need to know: The mole ratio between A, B, and the product they form. In other words, we need to know the balanced reaction equation. The molar masses of A, B, and the product they form. Let’s Think About It

Section 3.10 The Concept of Limiting Reagent Return to TOC Copyright © Cengage Learning. All rights reserved 139 An important indicator of the efficiency of a particular laboratory or industrial reaction. Percent Yield

Chapter 3 The Mole and Stoichiometry. Chapter 3 Table of Contents Copyright © Cengage Learning. All rights reserved 2 3.1 Counting by Weighing 3.2 Atomic.

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Chapter 3 The Mole and Stoichiometry. Chapter 3 Table of Contents Copyright © Cengage Learning. All rights reserved 2 3.1 Counting by Weighing 3.2 Atomic.

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Presentation on theme: "Chapter 3 The Mole and Stoichiometry. Chapter 3 Table of Contents Copyright © Cengage Learning. All rights reserved 2 3.1 Counting by Weighing 3.2 Atomic."— Presentation transcript:

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