chapter 3 the mole and stoichiometry. chapter 3 table of contents copyright © cengage learning. all...

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


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


Return to TOC


• 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


Return to TOC


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


• Elements occur in nature as mixtures of isotopes.• Carbon = 98.89% 12C

1.11% 13C

< 0.01% 14C

Section 3.2


Return to TOC


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


Return to TOC


• 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


Return to TOC


Schematic Diagram of a Mass Spectrometer

Section 3.2


Return to TOC


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 Mole Counting by Weighing

Return to TOC


• The number equal to the number of carbon atoms in exactly 12 grams of pure 12C.

• 1 mole of anything = 6.022 x 1023 units of that thing (Avogadro’s number).

• 1 mole C = 6.022 x 1023 C atoms = 12.01 g C

• Other units? Dozen? Fathom?? Stone?? Hand?

Section 3.3


Return to TOC


The Mole !!!!!

3

Section 3.3


Return to TOC


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


Return to TOC


How many is a mole?• 6.022 x 1023 Casio EXP below 3, TI30 XA EE above 7,

XIIS 2ndEE above 7, XZ multiview x10n 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 106) to count to a

mole?• If you counted 1 per second, it would take you

2 x 1016 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


Return to TOC


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


Return to TOC


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


Return to TOC


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


Return to TOC


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


Return to TOC


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


Return to TOC


1. To learn to convert among moles, mass, and # of atoms.

2. Unit Analysis! Multi Step!

Objectives 3.3 part 2

Section 3.3


Return to TOC


• 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 1023

• Grams moles # atoms

• X g or 1 mol• 1 mol X g

• 6.02 X 1023 atoms or 1 mol . • 1 mol 6.02 X 1023 atoms

Section 3.3


Return to TOC


• 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 1023 atoms Se =• 78.96 g Se 1 mol Se

• How many atoms in 2.35 mol of #%^$?• 2.35 mol #%^$ 6.02 X 1023 atoms #%^$ =• 1 mol #%^$• Remember that a mole measures quantity!!

Section 3.3


Return to TOC


• 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 1024 C atoms?• What is the mass in g of the above moles of C?• What is the mass in g of 9.4 X 1025 atoms of Mg?• How many atoms are in 365 g of V?

Section 3.3


Return to TOC


Concept Check

Calculate the number of copper atoms in a 63.55 g sample of copper.

6.022×1023 Cu atoms

Section 3.3


Return to TOC


Concept Check

Calculate the number of grams contained in a sample of 1.21 X 1023 Al atoms.

5.42 g Al

Section 3.3


Return to TOC


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


Return to TOC


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


Return to TOC


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


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


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


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


• Conversions Help Page!• 1 mole = 6.02 X 1023

• Grams moles # atoms or molecules

• For molecules…

• 6.02 X 1023 molecules or 1 mol . • 1 mol 6.02 X 1023 molecules

Section 3.4

Molar Mass

Return to TOC


2. Calculations Using Molar Mass (~Atomic Mass)

• What is the Molar Mass of methane gas CH4?

• Use Unit Analysis to convert:• How many molecules are contained within 1 mole of

methane gas CH4?

• How many C atoms are contained within 1 mole of methane gas CH4?

• How many H atoms are contained within 1 mole of methane gas CH4?

• 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



• Find the Molar Mass of SO2

• Find the Molar Mass of NaCl

• Find the Molar Mass of CaCO3

• What would the mass of 4.86 mol CaCO3 be?

• Find the Molar Mass of Na2SO4

• How many moles in 300 g of Na2SO4? (know-go?)

• SO2: 64 g/mol

• NaCl: 58.5 g/mol

• CaCO3: 100 g/mol, 486 g

• Na2SO4: 142 g/mol, 2.11 mol

Section 3.4

Molar Mass

Return to TOC



• How many moles in 1.56 g of Juglone C10H6O3?

• (Dye and herbicide made from black walnuts)

• 8.96 X 10 -3 mol C10H6O3

Section 3.4

Molar Mass

Return to TOC



• Bees release 1 X 10-6 g of isopentyl acetate (C7H14O2)

during a sting. (Also a banana scent!)• How many moles of isopentyl acetate are released?

• 8 X 10-9 mol C7H14O2

• How many molecules of isopentyl acetate are released?

• 5 X 1015 molecules C7H14O2

Section 3.4

Molar Mass

Return to TOC


2. Calculations Using Molar Mass (~Atomic Mass)• How many molecules are contained in 135 g of

Teflon (C2F4)?

• 8.127 X 1023 molecules of C2F4

Section 3.4

Molar Mass

Return to TOC


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


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


Return to TOC


• Mass percent of an element:

• For iron in iron(III) oxide, (Fe2O3):

mass of element in compoundmass % = × 100%

mass of compound

2( 55.85 g) 111.70 gmass % Fe = = × 100% = 69.94%

2( 55.85 g)+3( 16.00 g) 159.70 g

Section 3.5


Return to TOC


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 of a given element in 1 mol of compound100%

mass of 1 mol of compound Mass percent =

Section 3.5


Return to TOC



• Determine the percent composition by mass of each element in methane gas CH4

• Molar Mass• Each Element • X 100

• 75% C, 25% H

Section 3.5


Return to TOC



• Determine the percent composition by mass of each element in SO2

• Molar Mass

• Each Element

• X 100

• 50/50

Section 3.5


Return to TOC



• Determine the percent composition by mass of each element in NaCl


• 39% Na, 61% Cl

Section 3.5


Return to TOC



• Determine the percent composition by mass of each element in CaCO3


• 40% Ca, 12% C, 48% O

Section 3.6

Determining the Formula of a Compound

Return to TOC


• Device used to determine the mass percent of each element in a compound.

Analyzing for Carbon and Hydrogen

Section 3.6


Return to TOC


• Empirical formula = CH Simplest whole-number ratio

• Molecular formula = (empirical formula)n

[n = integer]

• Molecular formula = C6H6 = (CH)6

Actual formula of the compound

Formulas

Section 3.6


Return to TOC


2. Calculation of Empirical Formulas

Section 3.6


Return to TOC



• 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


Return to TOC



• 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


Return to TOC



• Now, let’s find the empirical formula:

• 0.00671 mol C

• 0.01353 mol H

• 0.00671 mol O

• CH2O(simplest) C2H4O2 ; C4H8O4 ;C5H10O5

Section 3.6


Return to TOC



• 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


Return to TOC


2. Calculation of Empirical Formulas • AlxOy

• 4.151 g Al

• 3.692 g O

• Al2O3

Section 3.6


Return to TOC



• It is observed that 0.664 g of lead combined with 0.2356 g of chlorine. Determine the empirical formula.

• PbCl2

Section 3.6


Return to TOC


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)• PtN2H6Cl2

Section 3.6


Return to TOC



• 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.

• C6NH11O

Section 3.6


Return to TOC


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 n2. (empirical formula) n = molecular formula **You have to be given the molar mass of the

molecular formula!

Section 3.6


Return to TOC


3. Calculation of Molecular Formulas

• Calculate the molecular formula for a compound if the empirical formula is P2O5 and the molecular molar mass is 283.88 g/mol.

• P4O10 Check the mm=(4(31)+10(16))=284

Section 3.6


Return to TOC



• 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


Return to TOC



• 71.65% Cl

• 24.27% C

• 4.07% H

• ClCH2 empirical

Section 3.6


Return to TOC



• ClCH2 (empirical)

• The molar mass is known to be 98.96 g/mol• n = molar mass molecular formula

molar mass empirical formula

• Cl2C2H4

Section 3.6


Return to TOC



• 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


Return to TOC



• 49.48% C

• 5.15% H

• 28.87% N

• 16.49% O

Section 3.6


Return to TOC



• Caffeine (MM = 194.2 g/mol) • n = molar mass molecular formula

molar mass empirical formula

C8H10N4O2

Section 3.6


Return to TOC


Hydrate Lab

Calculate the molar mass

MgSO4 * 7H2O

Mg + S + 4(O) + 7(H2O)

Calculate the % mass of waters…

7(H2O) . X 100

Mg + S + 4(O) + 7(H2O)

MM = 246g/mol : % H2O = 51.22%

Silica Gel Packaging…

Section 3.6


Return to TOC


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


Section 3.9


Return to TOC


Section 3.9


Return to TOC


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


Return to TOC


1. Let’s create a sandwich recipe…PHeT

2. Slice, dozen, 100, 6.02 X 1023, one mole

Sandwich Shoppe!rutherford-scattering_en.jar

Section 3.9


Return to TOC


Quinoa Side Dish www.allrecipes.com

Section 3.9


Return to TOC


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


Return to TOC


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 02 from __ moles H2O?

Section 3.9


Return to TOC


• 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.


Section 3.9


Return to TOC



• The mole ratio is determined by:

• Mole ratio = new mole substance A

• balanced mole substance A

• How many mole ___ from ___ moles ___?

• 2H2O 2H2 + O2

• mole mole ratio mole

Section 3.9


Return to TOC


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


Return to TOC



• Must first have a BALANCED equation

• __C3H8(g) + __O2 __CO2(g) + __H2O(g) + heat

• 4.30 C3H8 +__O2 __ CO2 + __ H2O + heat

• Calculate the number of moles of CO2 formed when 4.30 mol C3H8 combust.


Section 3.9


Return to TOC



• From this new mole value, we can calculate the new recipe!

• C3H8(g) + 5O2 3CO2(g) + 4H2O(g) + heat

• 4.30 C3H8 +__O212.9 CO2 + __ H2O + heat

• Calculate the number of moles of H2O formed when 4.30 mol C3H8 combust.


Section 3.9


Return to TOC



• For all reactants AND products!

• C3H8(g) + 5O2 3CO2(g) + 4H2O(g) + heat

• 4.30 C3H8 +__O212.9 CO2 + 17.2 H2O + heat

• Calculate the number of moles of O2 consumed when 4.30 mol C3H8 combust.


Section 3.9


Return to TOC



• C3H8(g) + 5O2 3CO2(g) + 4H2O(g) + heat

• Calculate the number of moles of all reactants and products when 2.70 mol C3H8 combust.

• 2.7 C3H8 +__O2 __ CO2 + __ H2O + heat

• 2.7 C3H8 +13.5 O2 8.1 CO2 + 10.8 H2O + heat

Section 3.9


Return to TOC



• Balance the following equation and determine how much NH3 can be made from 1.3 mol H2.

• __N2 + __H2 __NH3

• N2 + 3H2 2NH3

• How much N2 will be needed to completely react the 1.3 mol H2?

Section 3.9


Return to TOC



• Using the balanced equation, determine the new mole recipe given the new mole value.

• 2 K + 2 H2O H2 + 2 KOH

• ___K + 1.7 H2O ___ H2 + ___KOH

• 1.7 K + 1.7 H2O 0.85 H2 + 1.7 KOH

• 4 NH3 + 5 O2 4 NO + 6 H2O

• ___ NH3 + ___ O2 0.6 NO + ___ H2O

• 0.6 NH3 + 0.75 O2 0.6 NO + 0.9 H2O

Section 3.9


Return to TOC



2.



5.


Section 3.9


Return to TOC




Objectives Review 3.9 Part 1

Section 3.9


Return to TOC


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


Return to TOC


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


Return to TOC



• N2 + 3H2 2NH3

• How much NH3 will be produced when 2.6 g H2 completely reacts with N2?

• **Can’t do a mass ratio!!**• Must first convert from grams to moles.• Don’t forget the mole!

• grams grams


Section 3.9


Return to TOC



• N2 + 3H2 2NH3

• How much NH3 will be produced when 2.6 g H2 completely reacts with N2?

• 2.6 grams H2 grams

• mole mole ratio mole• 0.867 mol NH3: 14.71 g NH3

Section 3.9


Return to TOC


1. Mass Calculations• N2 + 3H2 2NH3

• __ N2 + 1.3 H2 0.867 NH3 (new mol recipe)

• __ g N2 2.6 g H2 14.71 g NH3 (new g recipe)

• How much N2 will be consumed when 2.6 g H2 completely reacts?

• Use the same mole ratio—

• 0.433 mol N2: 12.12 g N2

Section 3.9


Return to TOC



• N2 + 3H2 2NH3

• 0.433 N2 + 1.3 H2 0.867 NH3 (new mol recipe)

• 12.12 g N2 2.6 g H2 14.71 g NH3 (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


Return to TOC



2. Convert the known mass of the reactant or product to moles of that substance.



5. Convert from moles back to grams if required by the problem.


Section 3.9


Return to TOC



• __Al(S) + __I2(g) __AlI3(s)

• Balance the equation and calculate the mass of I2 needed to completely react with 35.0 g Al. Also calculate the mass of AlI3 that will be formed.

• grams grams

• mole mole ratio mole (next)

Section 3.9


Return to TOC


• 2Al(S) + 3I2(g) 2AlI3(s)

• __Al(S) + __I2(g) __AlI3(s) (new mole)

• 35.0 g • Balance the equation and calculate the mass of I2

needed to completely react with 35.0 g Al. Also calculate the mass of AlI3 that will be formed.

• 35.0 g Al (all new moles)

Section 3.9


Return to TOC


• 2 Al(S) + 3 I2(g) 2 AlI3(s)

• 1.3 Al(S) + 1.95 I2(g) 1.3 AlI3(s) (new mole)

• 35.0 g • Balance the equation and calculate the mass of I2


• Now convert mole grams for each

• 1.95 mol I2 =

• 1.3 mol AlI3 =

Section 3.9


Return to TOC


• 2 Al(S) + 3 I2(g) 2 AlI3(s)

• 1.3 Al(S) + 1.95 I2(g) 1.3 AlI3(s) (new mole)

• 35.0 g 495.3 g 530.4 g• Balance the equation and calculate the mass of I2


• Check the Law of Conservation of Mass!

Section 3.9


Return to TOC


2. Scientific Notation (CO2 scrubber on space vehicles)

• __LiOH(s) + __CO2(g) __Li2CO3(s) + __H2O(l)

• What mass of CO2 can 1 X 103 g LiOH absorb?

• What mass of H2O will be available for use?

• What mass of Li2CO3 will be generated?

• First Balance! Then, below!!

• grams grams


Section 3.9


Return to TOC



• 2 LiOH(s) + CO2(g) Li2CO3(s) + H2O(l)

• __LiOH(s) + __CO2(g) __Li2CO3(s) + __H2O(l)

• What mass of CO2 can 1 X 103 g LiOH absorb?

• Do new mole recipe next…

• grams grams


Section 3.9


Return to TOC




• 41.67LiOH + 20.8CO2 20.8Li2CO3 + 20.8H2O

• Now convert all moles to grams.

• 20.8 mol CO2 =

• 20.8 mol Li2CO3 =

• 20.8 mol H2O =

Section 3.9


Return to TOC




• 41.67LiOH + 20.8CO2 20.8Li2CO3 + 20.8H2O• 1 X 103 g 915.2 g 1539.2 g 374.4 g• What mass of CO2 can 1 X 103 g LiOH absorb?• What mass of H2O will be available for use?• What are some possible uses for the H2O?• What mass of Li2CO3 will be generated?• What will be done with the Li2CO3?• Conserved?

Section 3.9


Return to TOC


2. Mass Calculations Using Scientific Notation

• Hydrofluoric acid, an aqueous solution containing dissolved hydrogen fluoride, is used to etch glass by reacting with silica, SiO2, in the glass to produce gaseous silicon tetrafluoride and liquid water. The unbalanced reaction is:

• __HF(aq) + __SiO2(s) __SiF4(g) + __H2O(l)

• Calculate the mass of HF that is needed and the products that are produced to react completely with 5.68 g SiO2.

Section 3.9


Return to TOC


2. Mass Calculations Using Scientific Notation

• 4 HF(aq) + SiO2(s) SiF4(g) + 2 H2O(l)

• __ HF(aq) + __ SiO2(s) __ SiF4(g) + __ H2O(l)

• 5.68 g SiO2 =

• Mole ratio

• grams grams • • mole mole ratio mole

Section 3.9


Return to TOC


2. Mass Calculations Using Scientific Notation • 4 HF(aq) + SiO2(s) SiF4(g) + 2 H2O(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 grams


Section 3.9


Return to TOC


2. Mass Calculations Comparisons

• Which is the better antacid? Baking Soda,NaHCO3,or MOM, Mg(OH)2?

• How many moles of HCl will react with 1.00 g of each antacid?

• NaHCO3(s) + HCl(aq) NaCl(aq) + H2O(l) + CO2(g) Demo

• Mg(OH)2(s) + 2HCl(aq) 2H2O(l) + MgCl2(aq)

• grams grams


Section 3.9


Return to TOC


2. Mass Calculations Comparisons • Which is the better antacid? Baking Soda,NaHCO3,or

MOM, Mg(OH)2?• How many moles of HCl will react with 1.00 g of each

antacid? • NaHCO3(s) + HCl(aq) NaCl(aq) + H2O(l) + CO2(g)

• 1.0 g/1.19 X 10-2 mol NaHCO3

• 1.19 X 10-2 mol HCl• Mg(OH)2(s) + 2HCl(aq) 2H2O(l) + MgCl2(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


Return to TOC


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 SrCl2.

• ___SrCl2 + ___Na2CO3

•

Section 3.9


Return to TOC


2. Mass Calculations for lab

• SrCl2 + Na2CO3 SrCO3 + 2NaCl

• 0.01 SrCl2 + ___Na2CO3___SrCO3+___NaCl

• 0.01SrCl2 + 0.01Na2CO3 0.01SrCO3 + 0.02NaCl

• 1.59 g + 1.06 g 1.48 g + 1.17 g

• Conservation of mass?

• Theoretical vs Real…

Section 3.9


Return to TOC


2. Mass Calculations for lab

• Calculate the amount of grams contained in 0.01 mol the following hydrated crystals:

• SrCl2 x 6H2O

• Na2CO3 X H2O

• 2.67 g SrCl2 x 6H2O 1.24 g Na2CO3 X H2O

Section 3.9


Return to TOC


2. Mass Calculations for lab• 0.01SrCl2 x 6H2O + 0.01Na2CO3 X H2O 0.01SrCO3 + 0.02NaCl

• 2.67 g + 1.24 g 1.48 g + 1.17 g

Section 3.9


Return to TOC



2. Convert the known mass of the reactant or product to moles of that substance.



5. Convert from moles back to grams if required by the problem.


Section 3.9


Return to TOC




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


Return to TOC


Work Session Challenge Question

Methane (CH4) reacts with the oxygen in the

air to produce carbon dioxide and water.

Ammonia (NH3) reacts with the oxygen in the

air to produce nitrogen monoxide and water.

Write balanced equations for each of these reactions.

Section 3.9


Return to TOC


Work Session Challenge Question

Methane (CH4) reacts with the oxygen in the

air to produce carbon dioxide and water.

Ammonia (NH3) 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


Return to TOC


• 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


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


Return to TOC


• 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


Return to TOC


1. The Concept of Limiting Reactants

• Stoichiometric mixture (balanced)

– N2(g) + 3H2(g) 2NH3(g)

Section 3.10


Return to TOC


• Limiting reactant mixture (runs out first)

1. The Concept of Limiting Reactants

– N2(g) + 3H2(g) 2NH3(g)

Section 3.10


Return to TOC


Mixture of CH4 and H2O Molecules ReactingCH4 + H2O 3H2 + CO

Section 3.10


Return to TOC


CH4 + H2O 3H2 + CO

Section 3.10


Return to TOC


• 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


Return to TOC


Limiting Reactants

Section 3.10


Return to TOC


2. Calculations Involving a Limiting Reactant

• What mass of water is needed to react with 249 g methane (CH4)?

• CH4 + H2O 3H2 + CO

• 249 g CH4

• How many g of H2 and CO are produced?

• 279 g H2O, 93 g H2, 434 g CO

Section 3.10


Return to TOC



• How many g of H2 and CO are produced when 249 g methane (CH4) reacts with 300 g H2O?

• CH4 + H2O 3H2 + CO• 249 g + 279 g 93 g + 434 g (last slide)• 249 g + 300 g 93 g + 434 g• The 249 g CH4 will react with only 279 g H2O,

thereby leaving 21 g H2O in excess. CH4 is the reactant that will run out first and is the LIMITING REACTANT.

Section 3.10


Return to TOC



• 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


Return to TOC



Section 3.10


Return to TOC


2. Calculations Involving a Limiting Reactant• Suppose 2.50 X 104 g N2 reacts with 5 X 103 g H2.

Determine the limiting reactant.

• ___N2 + ___H2 ___NH3

• 2.50 X 104 g N2

• 5 X 103 g H2

• From N2 1785.72 mol NH3

• From H2 1666.67 mol NH3 ; 28333.3 g NH3

• Not just because original mass of H2 was less.

Section 3.10


Return to TOC



• How much N2 will be produced when 18.1 g NH3 and 90.4 g CuO react? Determine the limiting reactant.

• __NH3 + __CuO __N2 + __Cu + __H2O

• 18.1 g NH3

• 90.4 g CuO

• NH3 0.53 mol N2

• CuO 0.38 mol N2 *limiting reactant* 10.64 g N2

Section 3.10


Return to TOC


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


Return to TOC


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


Return to TOC


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


Return to TOC


3. Percent Yield • From the previous set of calculations, we

determined that 10.64 g of N2 would theoretically be produced from the reaction:

• 2NH3 + 3CuO N2 + 3Cu + 3H2O

• What would the Percent Yield be if 7.25 g of N2 were actually produced?

• 7.25 g N2 X 100 = 68.14 %

• 10.64 g N2

Section 3.10


Return to TOC


3. Percent Yield

• Consider the following reaction:

• __CO(g) + __H2(g) __CH3OH(l)

• If 6.85 X 104 g CO reacts with 8.6 X 103 g H2 to produce an actual yield of 3.57 X 104 g CH3OH, determine the limiting reactant, the theoretical yield, and the percent yield.

Section 3.10


Return to TOC


3. Percent Yield

• CO(g) + 2H2(g) CH3OH(l)

• 2446.4 mol CO 4300 mol H2

• 2446.4 mol CH3OH 2150 mol CH3OH *Limiting

• 2150 mol CH3OH *32g/mol = 68800 g CH3OH

• % yield = 3.57 X 104 g/ 6.88 X 104 g = 51.89 %

• 28 g/mol CO: 2 g/mol H2: 32 g/mol CH3OH

Section 3.10


Return to TOC


3. Percent Yield

• Consider the following reaction:

• __TiCl4 + __O2 __TiO2 + __Cl2• If 6.71 X 103 g TiCl4 reacts with 2.45 X 103 g O2 with

a percent yield of 75%, determine the limiting reactant, the theoretical yield, and the actual yield of TiO2.

• TiCl4 is limiting

• 2119.6 g TiO2 is actually produced in this reaction.

• 189.68 g/mol TiCl4: 32 g/mol O2: 79.88 g/mol TiO2

Section 3.10


Return to TOC


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


Return to TOC


• 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 3.9


Return to TOC



Section 3.10


Return to TOC


• 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


Return to TOC


• An important indicator of the efficiency of a particular laboratory or industrial reaction.

Percent Yield

yieldpercent%100 yieldlTheoretica

yieldActual

chapter 3 the mole and stoichiometry. chapter 3 table of contents copyright © cengage learning. all...

Documents