Chapter 9 Introduction to General, Organic, and Biochemistry 10e John Wiley & Sons, Inc Morris Hein, Scott Pattison, and Susan Arena Calculations from

Chapter 9

Introduction to General, Organic, and Biochemistry 10eJohn Wiley & Sons, Inc

Morris Hein, Scott Pattison, and Susan Arena

Calculations from Chemical Equations

Accurate measurement and calculation of the correct dosage are important in dispensing the correct medicine to patients throughout the world.

Chapter Outline

Copyright 2012 John Wiley & Sons, Inc 9-2

9.1 A Short Review

9.2 Introduction to Stoichiometry

9.3 Mole-Mole Calculations

9.4 Mole-Mass Calculations

9.5 Mass-Mass Calculations

9.6 Limiting Reactant and Yield Calculations

Objectives for Today Review the “mole” concept Determine relationships

between moles using stoichiometry

Develop mole and mass relationships

9-3

REVIEWING MOLES

9-4

Molar Mass – sum of atomic masses of all atoms in 1 mole of an element or compound; the units are g/mol.

6.022x1023 molecules6.022x1023 formula units6.022x1023 atoms6.022x1023 ions

9-5

1 mole =

What is the molar mass of Al(ClO3)3?

Al 1(26.98 g)3Cl 3(35.45 g)9O 9(16.00 g)

Al(ClO3)3 277.33 g/mol

7-6

atomic mass

Al 26.98

Cl 35.45

O 16.00

3 3

3 3

277.33 g Al(ClO )

1 mol Al(ClO )

Calculate the mass of 2.5 moles of aluminum chlorate.

9-7

3 32.5 mol Al(ClO ) 3 3 = 690 g Al(ClO )

Plan 2.5 mol Al(ClO3)3 g Al(ClO3)3

Calculate

1 mol Al(ClO3)3 = 277.33 g Al(ClO3)3

Calculate the moles of 3.52g of aluminum chlorate.

9-8

3 3

3 3

1 mol Al(ClO )

277.33 g Al(ClO )

3 33.52 g Al(ClO ) 23 3 = 1.27 10 mol Al(ClO )

Plan 3.52 g Al(ClO3)3 mol Al(ClO3)3

Calculate

1 mol Al(ClO3)3 = 277.33 g Al(ClO3)3

23

3 3

6.022 10 formula units

277.33 g Al(ClO )

Calculate the number of formula units contained in 12.4 g aluminum

chlorate.

9-9

3 312.4 g Al(ClO ) 22= 2.69 10 formula units

Plan 12.4 g Al(ClO3)3 formula units Al(ClO3)3

1 mol Al(ClO3)3 = 277.33 g = 6.022x1023 formula units

Calculate

Your Turn!

What is the mass of 3.61 moles of CaCl2?

a. 3.61 gb.272 gc. 2.17 × 1024 gd.401 g

9-10

atomic mass

Ca 40.08

Cl 35.45

Your Turn!

How many moles of HCl are contained in 18.2 g HCl?

a. 1.00 molb.0.500 molc. 0.250 mold.0.125 mol

9-11

atomic mass

H 1.01

Cl 35.45

Your Turn!

What is the mass of 1.60×1023 molecules of HCl?a. 9.69 gb.137 gc. 0.729 gd.36.5 g

9-12

atomic mass

H 1.01

Cl 35.45

USING STOICHIOMETRY

9-13

Stoichiometry deals with the quantitative relationships between the reactants and products in a balanced chemical equation.

9-14

9-15

1N2(g) + 3I2(s) 2NI3(s)

1 mol N2 + 3 mol I2 2 mol NI3

Mole ratios come from the coefficients in the balanced equation:

The 3 other possibilities are the inverse of these ratios.

2

2

3 mol I

1 mol N2

3

3 mol I

2 mol NI2

3

1 mol N

2 mol NI

Your Turn!

Which of these statements is not true about the reaction?

1N2(g) + 3I2(s) 2NI3(s)

a. 1 mole of nitrogen is needed for every 3 moles of iodine

b.1 gram of nitrogen is needed for every 3 grams of iodine

c. Both statements are true

9-16

Calculate the number of moles of NI3 that can be made from 5.50

mol N2 in the reaction: 1N2(g) + 3I2(s) 2NI3(s)

9-17

25.50 mol N 3

2

2 mol NI

1 mol N

Plan

Calculate

5.50 mol N2 mol NI3

3 = 11.0 mol NI

Set-Upmoles of desired substance in equation

mole ratio = moles of starting substance in equation

3

2

2 mol NImol ratio =

1 mol N

Calculate the number of moles of I2 needed to react with 5.50 mol N2 in

the reaction: 1N2(g) + 3I2(s) 2NI3(s)


25.50 mol N 2

2

3 mol I

1 mol N

Plan

Calculate

5.50 mol N2 mol I2

2 = 16.5 mol I

Set-Up 2

2

3 mol Imole ratio =

1 mol N

How many moles of HF will be produced by the complete reaction of

1.42 moles of H2 in the following equation?

H2 + F2 2HF

a. 0.710b.1.42c. 2.00d.2.84

9-19

Problem Solving Strategy for stoichiometry problems:

1. Convert starting substance to moles.2. Convert the moles of starting substance to

moles of desired substance.3. Convert the moles of desired substance to

the units specified in the problem.

9-20


USING STOICHIOMETRY IN MOLE-MOLE CALCULATIONS

9-22

How many moles of Al are needed to make 0.0935 mol of H2?

2Al(s) + 6HCl (aq) 2AlCl3(aq) + 3H2(g)

9-23

2

2 mol Al

3 mol H

Plan

Calculate

0.0935 mol H2 mol Al

2 0.0935 mol H =.0623 mol Al

Set-Up2

2 mol Almole ratio =

3 mol H

How many moles of HCl are needed to make 0.0935 mol of H2?


9-24

2

6 mol HCl

3 mol H

Plan

Calculate

0.0935 mol H2 mol HCl

2 0.0935 mol H = 0.187 mol HCl

6 mol HClmole ratio =

3 mol H2Set-Up

Your Turn!

How many moles of H2 are made by the reaction of 1.5 mol HCl with excess aluminum?


a. 0.75 mol b.3.0 molc. 6.0 mold.4.5 mol


Your Turn!

How many moles of carbon dioxide are produced when 3.00 moles of oxygen react completely in the following equation?

C3H8 + 5O2 3CO2 + 9H2O

a. 5.00 molb.3.00 molc. 1.80 mold.1.50 mol


Your Turn!

How many moles of C3H8 are consumed when 1.81x1023 molecules of CO2 are produced in the following equation?

C3H8 + 5O2 3CO2 + 4H2O

a. 0.100b.0.897c. 6.03 × 1022

d.5.43 × 1023 Copyright 2012 John Wiley & Sons, Inc 9-27

Objectives for Today Review the “mole” concept Determine relationships

between moles using stoichiometry

Develop mole and mass relationships

9-28

MOLE-MASS CALCULATIONS

9-29

What mass of H2 (2.02 g/mol) is made by the reaction of 3.0 mol HCl with

excess aluminum?


9-30

Plan

Calculate

3.0 mol HCl mol H2 g H2

3.0 mol HCl 23 mol H

6 mol HCl

2

2

2.02 g H

1 mol H

21.5 mol H

21.5 mol H 2 3.0 g H

How many moles of HCl are needed to completely consume 2.00 g Al

(26.98g/mol)?


9-31

Plan

Calculate

2.00 g Al mol Al mol HCl

2.00 g Al 1 mol Al

26.98 g Al

6 mol HCl

2 mol Al

= 0.0741 mol Al

0.0741 mol Al 0.0222 mol HCl

What mass of Al(NO3)3 (213g/mol) is needed to react with .093 mol Na2CO3?

3Na2CO3(aq) + 2Al(NO3)3(aq) Al2(CO3)3(s) + 6NaNO3(aq)

9-32

Plan

Calculate

0.093 mol Na2CO3 mol Al(NO3)3 g Al(NO3)3

2 3.093 mol Na CO 3 3

2 3

2 mol Al(NO )

3 mol Na CO

3 3

3 3

213.00g Al(NO )

1 mol Al(NO )

3 3.062 mol Al(NO )

3 30.062 mol Al(NO ) 3 313 g Al(NO )

How many moles of Al2(CO3)3 are made by the reaction of 3.45g Na2CO3

(105.99 g/mol) with excess Al(NO3)3?

3Na2CO3(aq) + 2Al(NO3)3(aq) Al2(CO3)3(s) + 6NaNO3(aq)

9-33

Plan

Calculate

3.45g Na2CO3 mol Na2CO3 g Al2(CO3)3

2 33.45g Na CO 2 3

2 3

1 mol Na CO

105.99g Na CO

2 3 3

2 3

1 mol Al (CO )

3 mol Na CO

2 30.0326 mol Na CO

2 30.0326 mol Na CO 2 30.0109 mol Na CO

Your Turn!

How many moles of oxygen are consumed when 38.0g of aluminum oxide are produced in the following equation?

4Al(s) + 3O2(g) 2Al2O3(s)

a. 0.248b.0.559c. 1.50d.3.00


atomic mass

Al 26.98

O 16.00

Your Turn!

What mass of HCl is produced when 1.81x1024 molecules of H2 react completely in the following equation?

H2(g) + Cl2(g) 2HCl(g)

a. 54.7gb.72.9gc. 109gd.219g


atomic mass

H 1.01

Cl 35.45

Objectives for Today Further analyze moles and

mass using stoichiometry Use stoichiometry to examine

limiting reagents and yield

9-36

MASS-MASS CALCULATIONS

9-37

Mass-Mass CalculationsNow we will put it all together.

9-38

What mass of Br2 (159.80 g/mol) is needed to completely consume 7.00 g

Al (26.98 g/mol)?2Al(s) + 3Br2(l) 2AlBr3(s)

9-39

Plan

Calculate

7.00 g Al mol Al mol Br2 g Br2

7.00 g Al 1 mol Al

26.98g Al

23 mol Br

2 mol Al

0.259 mol Al

0.259 mol Al 20.389 mol Br

2

2

159.80g Br

1 mol Br

20.389 mol Br 262.2 g Br

What mass of Fe2S3 (207.91g/mol) can be made from the reaction of 9.34 g

FeCl3 (162.20 g/mol) with excess Na2S?2FeCl3(aq) + 3Na2S(aq) Fe2S3(s) + 6NaCl(aq)

9-40

Plan

Calculate

9.34 g FeCl3 mol FeCl3 mol Fe2S3 g Fe2S3

39.34 g FeCl 3

3

1 mol FeCl

162.20g FeCl

2 3

3

1 mol Fe S

2 mol FeCl

30.0576 mol FeCl

30.0576 mol FeCl 2 30.0288 mol Fe S

2 3

2 3

207.91g Fe S

1 mol Fe S

2 30.0288 mol Fe S 2 35.99 g Fe S

Your Turn!

What mass of oxygen is consumed when 54.0g of water is produced in the following equation?

2H2 + O2 2H2O

a. 0.167 gb.0.667 gc. 1.50 gd.47.9 g


atomic mass

H 1.01

O 16.00

Your Turn!

What mass of H2O is produced when 12.0g of HCl react completely in the following equation?

6HCl + Fe2O3 2FeCl3 + 3H2O

a. 2.97 gb.39.4 gc. 27.4 gd.110. g


atomic mass

H 1.01

O 16.00

Cl 35.45

LIMITING REACTANT

9-43

Determine the number of that can be made given these quantities of reactants and the reaction equation:

9-44

+

+

The limiting reactant is the reactant that limits the amount of product that can be made.

The reaction stops when the limiting reactant is used up.

What was the limiting reactant in the reaction:

The small blue balls.


+

The excess reactant is the reactant that remains when the reaction stops.

There is always left over excess reactant.

What was the excess reactant in the reaction:

The excess reactant was the larger blue ball.


+

Technique for solving limiting reactant problems:

1.Convert reactant 1 to moles or mass of product

2.Convert reactant 2 to moles or mass of product

3.Compare answers. The smaller answer is the maximum theoretical yield.


2H2(g) + O2(g) 2H2O(g)

1. Calculate the theoretical yield of H2O assuming H2 is the limiting reactant and that O2 is the excess reactant.

2. Calculate the theoretical yield of H2O assuming that O2 is the limiting reactant and that H2 is the excess reactant.

9-48

Calculate the number of moles of water that can be made by the reaction of 1.51 mol H2 with 0.932 mol O2

Assuming that H2 is limiting and O2 is excess:

9-49

2 2 2

2

2 mol H O1.51 mol H =1.51 mol H O

2 mol H

2 2 2

2

2 mol H O0.932 mol O =1.86 mol H O

1 mol 0

So what is the maximum yield of H2O?

Assuming that O2 is limiting and H2 is excess:

How much H2 and O2 remain when the reaction stops?

H2: Limiting Reactant – None remains. It was used up in the reaction.

O2: Excess Reactant – Calculate the amount of O2 used in the reaction with H2. Then subtract that from the original amount.

9-50

2 2 2

2

1 mol O1.51 mol H x =0.755 mol O

2 mol H

2 2 20.932 mol O to start - 0.755 mol O = 0.177 mol of excess O

Calculate the mass of copper that can be made from the combination of 15.0

g aluminum with 25.0 g copper(II) sulfate.

2Al(s) + 3CuSO4(aq) Al2(SO4)3(aq) + 3Cu(s)


Plan 15 g Al mol Al mol Cu g Cu

25 g CuSO4 mol CuSO4 mol Cu g Cu

Compare answers. The smaller number is the right answer.

2Al(s) + 3CuSO4(aq) Al2(SO4)3(aq) + 3Cu(s)

9-52

1 mol Al

15.0 g Al x 26.98 g Al

3 mol Cu

2 mol Al

63.55 g Cu=

1 mol Cu

1. Assume Al is limiting and CuSO4 is in excess.

2. Assume CuSO4 is limiting and Al is in excess.4

44

1 mol CuSO25.0 g CuSO

159.58 g CuSO

4

3 mol Cu

3 mol CuSO

63.55 g Cu=

1 mol Cu

53.0 g Cu

9.96 g Cu

3. Compare answers.

CuSO4 is the limiting reagent. The theoretical yield of Cu is 9.96 g.

Your Turn!

True/False:You can compare the quantities of reactant

when you work a limiting reactant problem. The reactant you have the least of is the limiting reactant.

a. Trueb.False


Your Turn!

Which is the limiting reactant when 3.00 moles of copper are reacted with 3.00 moles of silver nitrate in the following equation?

Cu + 2AgNO3 Cu(NO3)2 + 2Ag

a. Cub.AgNO3

c. Cu(NO3)2

d.AgCopyright 2012 John Wiley & Sons, Inc 9-54

Your Turn!

What is the mass of silver (107.87 g/mol) produced by the reaction of 3.00 moles of copper with 3.00 moles of silver nitrate?

Cu + 2AgNO3 Cu(NO3)2 + 2Ag

a. 162gb.216gc. 324gd.647g


PERCENT YIELD

9-56

Actual Yield% Yield = 100

Theoretical Yield

The theoretical yield is the result calculated using stoichiometry.

The actual yield of a chemical reaction is the experimental result, which is often less than the theoretical yield due to experimental losses and errors along the way.

9-57

Calculate the % yield of PCl3 that results from reacting 5.00 g P with excess Cl2 if only 17.2 g of PCl3 were recovered. 2P + 3Cl2 2PCl3

Compute the expected yield of PCl3 from 5.00 g P with excess Cl2.

1 mol P

5.00 g P x 30.97 g P

32 mol PCl

2 mol P

3

3

137.33 g PCl=

1 mol PCl

22.2g PCl3

Compute the % Yield.

Actual Yield 17.2 g% Yield = 100%= 100%=77.5%

Theoretical Yield 22.2 g


Your Turn!

In a reaction to produce ammonia, the theoretical yield is 420. g. What is the percent yield if the actual yield is 350. g?

A. 83.3% B. 20.0% C. 16.7% D. 120.%


Objectives for Today Further analyze moles and

mass using stoichiometry Use stoichiometry to examine

limiting reagents and yield

9-60

Documents

Chapter 9 Introduction to General, Organic, and Biochemistry 10e John Wiley & Sons, Inc Morris Hein, Scott Pattison, and Susan Arena Calculations from