Chapter Six Chemical Reactions: - PowerPoint Presentation

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Chapter Six

Chemical Reactions:Mole and Mass Relationships

Fundamentals of General, Organic, and Biological Chemistry

7th Edition

Chapter 6 Lecture

© 2013 Pearson Education, Inc.

Julie Klare

Gwinnett Technical College

McMurry, Ballantine, Hoeger, PetersonSlide2

6.1 The Mole and Avogadro’s Number

6.2 Gram–Mole Conversions6.3 Mole Relationships and

Chemical Equations6.4 Mass Relationships and Chemical

Equations6.5 Limiting Reagent and Percent YieldSlide3

Goals

1. What is the mole, and why is it useful in chemistry?  Be able to explain the meaning and uses of the mole and

Avogadro’s number.

2. How are molar quantities and mass quantities related? Be able to convert between molar and mass quantities of an

element or compound. 3. What are the limiting reagent, theoretical yield, and percent yield of a reaction?

Be able to take the

amount of product actually formed in a reaction, calculate the amount that could form theoretically, and express the

results as a percent yield.Slide4

6.1 The Mole and Avogadro’s NumberSlide5

How to balance reactions by counting moleculesSlide6

Counting molecules: Conceptually

T

he reactants in a chemical reaction must be balanced One molecule of hydrogen (H2) reacts with one molecule of iodine (I

2), creating two hydrogen iodide molecules (HI) Slide7

Of course, we can’t visually count molecules to achieve this one-to-one ratio In fact, the only tool available is an analytical mass balance (using grams!) Slide8

Molecular Weight

But how do we relate a sample’s mass in grams to the number of molecules it contains? We begin with the concept of ‘molecular weight’ (molecular mass)

Let’s start by utilizing the Ch 2 concept of mass by ‘amu’ Note: We will be focusing on molecules for a whileSlide9

Recall, atomic weight is the average mass of an element

’s isotope atoms (review next slide)

By analogy, molecular weight (MW)

is the mass of a compound’s molecules A molecule’s molecular weight is the sum of the atomic weights for all the atoms in the molecule A salt’s

formula weight is the sum of the atomic weights for all the atoms in the formula unit Slide10

fyi: Atomic Weight

CalculationC-12 is 98.89% of all natural carbonC-13 is 1.11% of all natural carbon

The mass of C-12 is: 12 amu (by definition)The mass of C-13 is: 13.003355 amu

Atomic weight =  [(% isotope abundance) × (isotope mass)]

[98.89% x 12 amu] + [

1.11% x 13.0034 amu] = (0.9889)(12 amu) + (0.0111)(13.0034 amu) = 12.0111 12.01 amu

11.8668

0.1443Slide11

fyi: Disambiguation

For molecules, these terms are identicalmolar weight molar massmolecular weight molecular massFor atoms, these terms are identicalmolar weight molar mass

atomic weightatomic massSlide12

Molar Mass of N = 14.01 g/mol

Molar Mass of N2 = 28.02 g/mol Molar Mass of H2O = 18.02 g/mol

(2 × 1.008 g) + 16.00 g Molar Mass of Ba(NO

3)2 = 261.35 g/mol 137.33 g + (2 × 14.01 g) + (6 × 16.00 g)Copyright © Cengage Learning. All rights reserved12Molecular Weight (Mass): Mass in grams of one mole of the substance:Slide13

Molecular weight

So if the mass (atomic weight) of one atom of hydrogen (H) is 1.008 amu then the mass (molecular weight) of one hydrogen molecule (H2) is 2.016

amu If the atomic weight of one iodine atom (I) is 126.90 amuthen the molecular weight of one iodine molecule (I2) is 253.80 amu

Finally, the molecular weight of each HI molecule must be: 1.008 amu + 126.90 amu = 127.91 amuSlide14

Counting molecules: by massSlide15

Here is where we left our reaction making two HI molecules from H2 and I2 Slide16

We learned in chapter 2 that the iodine atom (I) is approximately 126 times heavier than the hydrogen atom (H)H

= 1.008 amu I = 126.90 amuSo mathematically, it is necessary that the iodine molecule (I2

) is approximately 126 times heavier than the hydrogen molecule (H2) H2 = 2.016 amu

I2 = 253.80 amuSlide17

Copyright © Cengage Learning. All rights reserved17

Meaning that this reaction

is balanced – in amu units

H2(g) + I2(g) → 2HI(g) 2.016 amu 253.80 amu 1 : 126Slide18

Copyright © Cengage Learning. All rights reserved18

So by

the law of identical ratios, this reaction must also be balanced – in gram units

H2(g) + I2(g) → 2HI(g) 2.016 grams 253.80 grams 1 : 126Now notice

we have gonefrom amu to gramsSlide19

Copyright © Cengage Learning. All rights reserved19

And again by The Law of Identical Ratios, this must also be balancedH

2(g) + I2(g

) → 2HI(g) 1.00 g 126 gSlide20

Mass / Number relationship for I2 & H2

So in order control the number ratio of I2 to H2 molecules at 1:1 we can measure out any of the following

so long we keep the mass ratio at 1:126 H2(g) + I

2(g) → 2HI(g) 2.016 amu 253.80 amu 2.016 g 253.80 g 1.00 g 126 g 0.0079 g 1.0000 g Slide21

But how many molecules is that?W

e know that: 2.016 amu of H2 contains one molecule, and253.80 amu of I

2 contains one molecule Our final question is, how many molecules are contained in: 2.016 g

of H2 253.80 g of I2 Slide22

That amount we call The MoleSlide23

A mole is the amount of substance whose mass in grams is numerically equal to its molecular weight

2.016 g of H2 contains one mole of H2 molecules 253.80 g of I2 contains one

mole of I2 molecules The mole (NA) has now been measured One

mole of any molecule contains 6.022 × 1023 molecules One mole of any salt contains 6.022 × 1023 formula units Slide24

Getting from amu to grams

The number ‘126.904,’ typically placed as shown, functions BOTH: as the mass in amu per atomand the mass in grams per mole of atoms So I has a mass of

126.904 grams per mole of I atomsI = 126.904 g/mol Slide25

The technical definition of a mole Slide26

How many silicon atoms are in 0.532 moles of silicon?

8.83

× 10

−25 atoms3.81 × 1023 atoms2.6 × 1023 atoms

3.8 × 1023 atomsSlide27

6.2 Gram–Mole ConversionsSlide28

fyi: Synonyms The terms Molecular Weight

(section 6.1) and Molar Mass (this section – 6.2) are identical Also, the term Molecular Mass is allowed Also, the term Molar Weight is allowed When we defined Molecular Weight in Section 6.1, we had not yet define the Mole so you couldn’t have understood the more common term “Molar” MassSlide29

The molar mass of water is 18.02 g/molSo how many moles of water are there in

27 grams?Most importantly, molar

mass serves as a conversion factor between numbers of moles and mass in grams for use in dimensional analysisSlide30

Molar Mass: Mole to Gram

Conversion Ibuprofen is a pain reliever used in Advil. Its molecular weight is 206.3

g/mol. If a bottle of Advil contain 0.082 mole of ibuprofen, how many grams of ibuprofen does it contain?

Worked Example 6.3 Slide31

WORKED EXAMPLE

6.3 Molar Mass: Mole to Gram Conversion (Continued)Slide32

We now have the tools to convert:

grams → moles →

number of atoms grams → moles

→ number of molecules grams → moles → number of formula unitsAnd we can go backwards Slide33

33Slide34

What is the mass in grams of 3.2

× 1022 molecules of water?

0.0029 g

339 g0.90 g0.96 gSlide35

6.3 Mole Relationships and Chemical Equations

Molar coefficients:

using Mole ratios as conversion factorsSlide36

Coefficients in a

balanced chemical equation tell us the necessary ratio of moles of

reactants and how many moles of each product are

formed Slide37

Especially useful, coefficients can be put in the form of

mole ratios which act as

conversion factors when setting up dimensional analysis calculations Slide38

This reaction represents the reaction of A

2

(red) with B

2 (blue)

a) Write a balanced equation for the reaction

b) How many moles of product can be made from 1.0 mole of A

2? From 1.0 mole of B2? Mole to mole Slide39

0.67 mol

2.00 mol 3.33 mol 7.50 mol

How many moles of NH

3 can be produced from 5.00 moles of H2 according to the following equation? N2 + 3 H2  2 NH3Slide40

6.4 Mass Relationships and Chemical Equations

Putting it all together Slide41

The actual amounts of substances used in the laboratory

must be weighed out in grams Furthermore, customers generally want the quantity of product reported in terms of gramsSlide42

We already have the necessary tools

Mass to

mole conversions

Mole to mole conversions Mole to mass conversions Slide43

Mole to mole conversions

(section 6.3) are carried out using mole ratios

as conversion factors

If you have 9.0 moles of H2, how many moles of NH3 can you make?Slide44

2) Mole-to-mass and mass-to-mole conversions

(from section 6.2) are carried out using molar mass

as a conversion factor Slide45

3) Mass to mass conversions

cannot be carried out

directly

If you know the mass of A and need to find the mass of B first convert the mass of A into moles of A then carry out a mole to mole conversion to find moles of B then convert moles of B into the mass of B Slide46

So after collecting the fundamental data, always follow this three step process for mass to mass

1) convert grams of A to moles via molar mass2) convert moles of A to B via mole ratio 3) convert mole of B to grams via molar massSlide47

47Let’s go back to our hydrogen / iodine reaction

giving hydrogen iodide Slide48

48

I2 = 253.80 g/

mol H2 = 2.016 g/molHI = 127.90 g/mol

10.0 grams of HI requires how many grams of H2Slide49

How many grams of oxygen are needed to react with 25.0 g of K according to the following equation?

4 K(s)

+ O2(g)

 2 K2O(s) 10.2 g O22.56 g O28.66 g O25.12 g O2Slide50

6.5 Limiting Reagent and Percent Yield

Vocabulary:Reactant = ReagentSlide51

limiting reactant (reagent)

Reactants are not always perfectly balanced When running a chemical reaction,

we generally ‘overcharge

’ one of the reactants As our mechanic has ‘overcharged’ tires Slide52

limiting reactant (reagent)

The limiting reactant is

the reactant that runs

out first The reactant that never runs out is called the excess reactant So which is ‘limiting here?’ Tires or car bodies.Slide53

limiting reactant (reagent)

In order to make this

‘reaction’ balanced stoichiometrically

, how many tires should our mechanic actually have on hand? 32 tires would perfectly ‘balance’ this ‘reaction’Slide54

Back to chemistrySlide55

A stoichiometric mixture of reactants

contains

the relative amounts (in moles) of reactants that matches the coefficients in

the balanced equation N2(g) + 3H2(g)  2NH3(g)Copyright © Cengage Learning. All rights reserved55Slide56

A stoichiometric mixture of reactants

Contains the relative amounts (in moles) of reactants that matches the

coefficients in the balanced equation N

2(g) + 3H2(g)  2NH3(g)Copyright © Cengage Learning. All rights reserved56Slide57

A limiting reactant mixture

Copyright © Cengage Learning. All rights reserved57

on the other hand, contains an excess of one of the reactants The other reactant is thus limiting

N2(g) + 3H2(g)  2NH3(g)Slide58

A limiting reactant mixture

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Which reactant is in excess in this reaction mixture?Which reactant is limiting?

on the other hand, contains an excess of one of the reactants The other reactant is thus limiting N2(g) + 3H2(g)  2NH3(g)Slide59

Definition: Limiting Reactant

N2(g) + 3H2(

g)  2NH3(g

)The limiting reactant (reagent) is the reactant that runs out first and thus constrains (limits) the amounts of product(s) that can form H2 in the previous example is limiting Copyright © Cengage Learning. All rights reserved59Slide60

But the limiting reactant must be identified before the correct amount of product can be calculated

Copyright © Cengage Learning. All rights reserved60

2H

2 + O2  H2OSlide61

Example

10.0 g of Chemical A are reacted with 10.0 g of B What information do you need to calculate the mass of

the product (C) that will be produced?

The mole ratios between A, B, and C ie, the balanced reaction equation The molar masses of A, B, and C Copyright © Cengage Learning. All rights reserved61Slide62

Example

10.0 g of Chemical A are reacted with 10.0 g of B What information do you need to calculate the mass of

the product (C) that will be produced?A + 3B  2C

molar masses: A is 10.0 g/molB is 20.0 g/molC is 25.0 g/mol? 62Slide63

Example

10.0 g of Chemical A are reacted with 10.0 g of B A + 3B  2C1)

Convert known masses of reactants to moles 63Slide64

Example

10.0 g of Chemical A are reacted with 10.0 g of B A + 3B  2C2) Convert to the number of moles of product

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So B is limiting reactantSlide65

Example

10.0 g of Chemical A are reacted with 10.0 g of B A + 3B  2CChoose the least number of moles of product formed as limiting reactant:

B is limiting 65Slide66

Example

10.0 g of Chemical A are reacted with 10.0 g of B A + 3B  2C3) Convert moles of C to grams of C using the molar mass

66Slide67

shortcut to limiting reactantHow do we determine quickly which reactant is limiting and by deduction, which is in excess?

Three step program Calculate the number of moles of each reactantdivide the number of moles of each reactant by its coefficient from the balanced equationthe reactant with the smaller result is limiting Slide68

If 56 g of K is reacted with 56 g of oxygen gas according to the equation below, indicate the mass of product that can be made and identify the limiting reactant.

4 K

(s) + O2(g)

 2 K2O(s) 67g K2O; K is the limiting reactant.270 g K2O; K is the limiting reactant.

270 g K2O; O2 is the limiting reactant.67g K2O; O2 is the limiting reactant.

K

2O = 94.2g/molSlide69

Theoretical vs actual yield

Theoretical YieldThe maximum amount of a given product that can be formed once a limiting

reactant has been completely consumed – it is a calculation Actual

yield The amount of product that is actually produced in a reaction – it is a measurementIt is usually less than the maximum expected (theoretical yield) Slide70

Theoretical yield is found by using the amount of limiting

reactant calculated in a mass-to-mass calculation For the chemist in the lab, the

actual yield is found by weighing the amount of product obtained FYISlide71

Theoretical Yield

For our auto mechanic, the theoretical yield is 8 finished carsActual yield If

one car body was damaged beyond repair, the actual yield would have been 7 finished carsSlide72

Percent yield is the percent of the theoretical yield actually obtained from a chemical

reaction

For our car mechanic: Slide73

The combustion of acetylene gas (C

2H2) produces carbon dioxide and water as indicated in the following

reaction 2 C2

H2 (g) + 5 O2 (g) → 4 CO2 (g) + 2 H2O (g) When 26.0 g of acetylene is burned in sufficient oxygen for complete reaction, the theoretical yield of CO2

is 88.0 g Calculate the percent yield for this reaction if the actual yield is only 72.4 g CO2 Analysis—The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100 WORKED EXAMPLE

6.8 Percent YieldSlide74

WORKED EXAMPLE

6.8 Percent Yield (finished)Slide75

The element boron is produced commercially by the reaction of boric oxide with magnesium at high temperature.

B2O

3 (l) + 3 Mg (s

) → 2 B (s) + 3 MgO (s) What is the theoretical yield of boron when 2350 g of boric oxide is reacted with 3580 g of magnesium? The molar masses of boric oxide and magnesium are 69.6 g/mol and 24.3 g/mol, respectively WORKED EXAMPLE 6.9

Mass to Mole Conversions: Limiting Reagent and Theoretical Yield Slide76

WORKED EXAMPLE

6.9 Mass to Mole Conversions: Limiting Reagent and Theoretical Yield (Continued) Slide77

If 28.56 g of K

2O is produced when 25.00 g K is reacted according to the following equation, what is the percent yield of the reaction?

4 K(s) + O

2(g)  2 K2O(s) 87.54%95%94.82%88%Slide78

Chapter SummarySlide79

What is the mole, and why is it useful in chemistry?

A mole refers to Avogadro

’s number 6.022 ×

1023 formula units of a substance. One mole of any substance has a mass (a molar mass) equal to the molecular or formula weight of the substance in grams. Because equal numbers of moles contain equal numbers of formula units, molar masses act as conversion factors between numbers of molecules and masses in grams.Slide80

How are molar quantities and mass quantities related?

The coefficients in a balanced chemical equation represent the numbers of moles of reactants and products in a reaction. The ratios of coefficients act as mole ratios that relate amounts of reactants and/or products.

By using molar masses and mole ratios in factor-label calculations, unknown masses or molar amounts can be found from known masses or molar amounts. Slide81

What are the limiting reagent, theoretical yield, and percent yield of a reaction?

The limiting reagent is the reactant that runs out first.

The theoretical yield is the amount of product that would be formed based on the amount of the limiting reagent.

The actual yield of a reaction is the amount of product obtained. The percent yield is the amount of product obtained divided by the amount theoretically possible and multiplied by 100%. Slide82

The official name for what we are doing

is Stoichiometry Slide83

Copyright © Cengage Learning. All rights reserved83

amu

to grams

Recall: in Ch 2 we calculated the masses in amu needed to balance this reaction H2(g) + I2(g) → 2HI(g) 2.016 amu 253.80 amuSo the

mass ratio of I2 to H2 needed is actually: 125.90