1 CHAPTER 7 Chemical Calculations 7.1-7.7. 2 The Mole A number of atoms, ions, or molecules that is large enough to see and handle. A mole = number of.

1

CHAPTER 7Chemical Calculations 7.1-7.7

2

The Mole

A number of atoms, ions, or molecules that is large enough to see and handle.

A mole = number of things Just like a dozen = 12 things One mole = 6.022 x 1023 things

Avogadro’s number = 6.022 x 1023 Symbol for Avogadro’s number is NA.

3

The Mole

4

The Mole

How do we know when we have a mole? count it out weigh it out

Molar mass - mass in grams numerically equal to the atomic weight of the element in grams.

H has an atomic weight of 1.00794 g 1.00794 g of H atoms = 6.022 x 1023 H atoms

Mg has an atomic weight of 24.3050 g 24.3050 g of Mg atoms = 6.022 x 1023 Mg atoms

5

The Mole

6

The Mole

Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures.

Mgg ?

7

The Mole


atom Mg1 Mgg ?

8

The Mole


atoms Mg 106.022

atoms Mg mol 1atom Mg 1Mg g ?

23

9

The Mole

Example 2-1: Calculate the mass of a single Mg atom, in grams, to 3 significant figures.

Mg g 104.04atoms Mg mol 1

Mgg24.30

atoms Mg 106.022

atoms Mg mol 1atom Mg 1Mg g ?

23

23

10

The Mole

Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures.

atoms Mg?

11

The Mole


Mgg 24.30

Mg mol 1 Mgg 101.00atoms Mg? 6

12

The Mole


atoms Mgmol 1

atoms Mg106.022

Mgg 24.30

Mg mol 1 Mgg 101.00atoms Mg?

23

6

13

The Mole


atoms Mg102.48atoms Mgmol 1

atoms Mg106.022

Mgg 24.30

Mg mol 1 Mgg 101.00atoms Mg?

1623

6

14

The Mole

Example 2-3. How many atoms are contained in 1.67 moles of Mg?

atoms Mg ?

15

The Mole


Mg mol 1.67atoms Mg ?

16

The Mole


Mg mol 1

atoms Mg 106.022 Mg mol 1.67atoms Mg ?

23

17

The Mole


atoms Mg 101.00

Mg mol 1

atoms Mg 106.022 Mg mol 1.67atoms Mg ?

24

23

18

The Mole


atoms Mg101.00

Mgmol 1

atoms Mg106.022Mg mol 1.67atoms Mg?

24

23

19

The Mole

Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg?

You do it!You do it!

20

The Mole


Mg g 4.73Mg mol ?

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The Mole


Mg g 24.30

atoms Mg mol 1 Mg g 4.73Mg mol ?

22

The Mole


Mg mol 02.3

Mg g 24.30

atoms Mg mol 1 Mg g 4.73Mg mol ?

IT IS IMPERATIVE THAT YOU KNOWHOW TO DO THESE PROBLEMS

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Formula Weights, Molecular Weights, and Moles

How do we calculate the molar mass of a compound? add atomic weights of each atom

The molar mass of propane, C3H8, is:

amu 44.11 mass Molar

amu 8.08 amu 1.01 8H 8

amu 36.03amu 12.01 3C 3

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The molar mass of calcium nitrate, Ca(NO3)2 , is:


25


amu 164.10 massMolar

amu 96.00 amu 16.006 O6

amu 28.02 amu 14.012 N2

amu 40.08 amu 40.081 Ca1

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One Mole of Contains Cl2 or 70.90g 6.022 x 1023 Cl2 molecules

2(6.022 x 1023 ) Cl atoms C3H8 You do it!You do it!

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One Mole of Contains Cl2 or 70.90g 6.022 x 1023 Cl2 molecules 2(6.022 x 1023 ) Cl atoms C3H8 or 44.11 g 6.022 x 1023 C3H8 molecules 3 (6.022 x 1023 ) C atoms 8 (6.022 x 1023 ) H atoms

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Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane.

8383 HC g 74.6molecules HC ?

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83

83

8383

HC g 44.11

HC mole 1

HC g 74.6molecules HC ?

30



83

8323

83

83

8383

HC g 44.11

molecules HC 106.022

HC g 44.11

HC mole 1


31



molecules 10 02.1

HC g 44.11

molecules HC 106.022

HC g 44.11

HC mole 1


24

83

8323

83

83

8383

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Example 2-8. Calculate the number of O atoms in 26.5 g of Li2CO3.


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Example 2-8. Calculate the number of O atoms in 26.5 g of Li2CO3.

atoms O 106.49

COLi unit formula 1

atoms O 3

COLi mol 1

COLi unitsform.106.022

COLi g 73.8

COLi mol 1COLi g 26.5atoms O ?

23

3232

3223

32

3232

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Percent Composition % composition = mass of an individual

element in a compound divided by the total mass of the compound x 100%

Determine the percent composition of C in C3H8.

81.68%

100%g 44.11

g 12.013

100%HC mass

C mass C %

83

35

Percent Composition and Formulas of Compounds

What is the percent composition of H in C3H8?


36

Percent Composition

What is the percent composition of H in C3H8?

81.68%100%18.32%

or

%18.32100%g 44.11

g 1.018

100%HC

H8

100%HC mass

H massH %

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83

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Percent Composition

Example 2-10: Calculate the percent composition of Fe2(SO4)3 to 3 significant figures.

You do it!

38

Percent Composition Example 2-10: Calculate the percent

composition of Fe2(SO4)3 to 3 sig. fig.

100% Total

O 48.0% 100%g 399.9

g 16.012 100%

)(SOFe

O12 O %

S 24.1% 100%g 399.9

g 32.13 100%

)(SOFe

S3 S %

Fe 27.9% 100%g 399.9

g 55.82 100%

)(SOFe

Fe2Fe %

342

342

342

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Percent Composition

40

Empirical and Molecular Formulas

Empirical Formula - smallest whole-number ratio of atoms present in a compound

Molecular Formula - actual numbers of atoms of each element present in a molecule of the compound

We determine the empirical and molecular formulas of a compound from the percent composition of the compound.

41

Empirical And Molecular Formulas

42

Empirical Formulas

Example 2-11: A compound contains 24.74% K, 34.76% Mn, and 40.50% O by mass. What is its empirical formula?

Make the simplifying assumption that we have 100.0 g of compound.

In 100.0 g of compound there are: 24.74 g of K 34.76 g of Mn 40.50 g of O

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Empirical Formulas

K mol 0.6327K g 39.10

K mol 1K g 24.74 K mol ?

44

Empirical Formulas

Mn mol 0.6327Mn g 54.94

Mn mol 1Mn g 34.76 Mn mol ?

K mol 0.6327K g 39.10


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Empirical Formulas

rationumber wholesmallest obtain

O mol 2.531O g 16.00

O mol1O g 40.50 O mol ?

Mn mol 0.6327Mn g 54.94


K mol 0.6327K g 39.10


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Empirical Formulas

Mn 10.6327

0.6327Mnfor K 1

0.6327

0.6327Kfor


O mol 2.531O g 16.00


Mn mol 0.6327Mn g 54.94


K mol 0.6327K g 39.10


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Empirical Formulas

4KMnO is formula chemical thethus

O 40.6327

2.531Ofor

Mn 10.6327

0.6327Mnfor K 1

0.6327

0.6327Kfor


O mol 2.531O g 16.00


Mn mol 0.6327Mn g 54.94


K mol 0.6327K g 39.10


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Empirical Formulas

Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound?


49

Empirical Formulas Example 2-12: A sample of a compound

contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound?

ratio number wholesmallest find

O mol 0.1480O g 16.00

O mol1O g 2.368O mol ?

Co mol 0.1110Cog 58.93

Co mol 1Co g 6.541Co mol ?

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Empirical Formulas Example 2-12: A sample of a compound

contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound?

43OCo

:is formula scompound' theThus

O 43O 1.333 Co 33Co 1

number wholetofraction turn to 3 by both multipy

O1.3330.1110

0.1480O for Co 1

0.1110

0.1110Co for

51

Molecular Formulas Example 2-13: A compound is found to

contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula? short cut method

H of g 8.100.1437g 56.1

C of g 48.00.8563g 56.1

H is 14.37% and C is 85.63%

g 56.1 contains mol 1

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Determination of Molecular Formulas

84HC

:is formula theThus

H mol 8H g 1.01

H mol 1H of g 8.10

C mol 4C g 12.0

C mol 1C of g 48.0

moles tomassesconvert

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Calculations Based on Chemical Equations

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Example 3-1: How many CO molecules are required to react with 25 formula units of Fe2O3?

CO of molecules 75

unit formula OFe 1

molecules CO 3OFe units formula 25 = molecules CO ?

3232

Fe O + 3 CO 2 Fe + 3 CO2 3 2

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Example 3-2: How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?

325 OFe units formula 102.50=atoms Fe ?

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OFe units formula 1

atoms Fe 2

OFe units formula 102.50=atoms Fe ?

32

325

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atoms Fe 105.00 OFe units formula 1

atoms Fe 2

OFe units formula 102.50=atoms Fe ?

5

32

325

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Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide?

32

3232 OFe g 7.159

OFe mol 1OFe g 146 = CO g ?

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3232

3232 OFe mol 1

CO mol 3

OFe g 7.159


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CO g 8.76CO mol 1

CO g 28.0

OFe mol 1

CO mol 3

OFe g 7.159


3232

3232

61


Example 3-4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with carbon monoxide?

OFe mol 1

CO mol 3OFe mol 540.0CO g ?

32

2322

62



2

2

32

2322 CO mol 1

CO g 0.44

OFe mol 1

CO mol 3OFe mol 540.0CO g ?

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? g CO mol Fe O3 mol CO

1 mol Fe O

g CO

mol CO

= 71.3 g CO

2 2 32

2 3

2

2

2

0 54044 0

1.

.

64


Example 3-5: What mass of iron (III) oxide reacted with carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams?


65


3232

32

2

32

2

2232

O Feg 5.10O Femol 1

O Feg 7.159

CO mol 3

O Femol1

CO g 44.0

molCO 1CO g 8.65O Feg ?

Example 3-5: What mass of iron (III) oxide reacted with carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams?

66


Example 3-6: How many pounds of carbon monoxide would react with 125 pounds of iron (III) oxide?


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CO lb 7.65CO g 454

CO lb 1

CO mol 1

CO g 28

OFe mol 1

CO mol 3

OFe g 7.159

OFe mol 1

OFe lb 1

OFe g 454OFe lb 125 = CO lb ?

3232

32

32

3232

YOU MUST BE PROFICIENT WITH THESE

TYPES OF PROBLEMS!!!

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Limiting Reactant Concept

Kitchen example of limiting reactant concept.1 packet of muffin mix + 2 eggs + 1 cup of milk

12 muffins How many muffins can we make with the

following amounts of mix, eggs, and milk?

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Mix Packets Eggs Milk1 1 dozen 1 gallon

limiting reactant is the muffin mix2 1 dozen 1 gallon3 1 dozen 1 gallon4 1 dozen 1 gallon5 1 dozen 1 gallon6 1 dozen 1 gallon7 1 dozen 1 gallon

limiting reactant is the dozen eggs

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Example 3-7: Suppose a box contains 87 bolts, 110 washers, and 99 nuts. How many sets, each consisting of one bolt, two washers, and one nut, can you construct from the contents of one box?

numbersmallest by the determined

sets 55 is make can number we maximum the

sets 99nut 1set 1nuts 99

sets 55 washers2set 1 washers110

sets 87bolt 1set 1bolts 87

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Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?

2222 SO 2 CO O 3 CS

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mol 2 mol 1 mol 3 mol 1

SO 2 CO O 3 CS 2222

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CS O CO 2 SO

1 mol 3 mol 1 mol 2 mol

76.2 g 3(32.0 g) 44.0 g 2(64.1 g)

2 2 2 2 3

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Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen? Determine which mass makes the most product

g 76.2

CS mol 1CS g 6.95 SO mol ?

SO 2 CO O 3 CS

222

2222

75


22

2

2

2

222

2222

SO g 161SO mol 1

SO g 1.64

CS mol 1

SO mol 2

g 76.2


SO 2 CO O 3 CS

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22

2

2

2

2

222

22

2

2

2222

2222

SO g 147SO mol 1

SO g 1.64

O mol 3

SO mol 2

O g 32.0

O mol 1O g 110SO mol ?

SO g 161SO mol 1

SO g 1.64

CS mol 1

SO mol 2

g 76.2


SO 2 CO O 3 CS

Which is limiting reactant? Limiting reactant is O2. What is maximum mass of sulfur dioxide? Maximum mass is 147 g.

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Percent Yields from Reactions Theoretical yield is calculated by assuming that

the reaction goes to completion. Determined from the limiting reactant calculation.

Actual yield is the amount of a specified pure product made in a given reaction. In the laboratory, this is the amount of product that is

formed in your beaker, after it is purified and dried. Percent yield indicates how much of the product is

obtained from a reaction.

% yield = actual yield

theoretical yield100%

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Percent Yields from Reactions

Example 3-9: A 10.0 g sample of ethanol, C2H5OH, was boiled with excess acetic acid, CH3COOH, to produce 14.8 g of ethyl acetate, CH3COOC2H5. What is the percent yield?

yieldal theoretic theCalculate 1.

O HHCOOCCH OH HC +COOH CH 2523523

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523

52

52352523

2523523

HCOOCCH g 1.19

OHHC g 0.46

HCOOCCH g 88.0 OHHC g 10.0= HCOOCCH g ?

yield al theoretic theCalculate 1.

OH HCOOCCH OHHC + COOHCH

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yield.percent theCalculate .2

HCOOCCH g 1.19

OHHC g 0.46




523

52

52352523

2523523

81


%5.77%100HCOOCCH g 19.1

HCOOCCH g 14.8= yield %

yield.percent theCalculate .2

HCOOCCH g 1.19

OHHC g 0.46




523

523

523

52

52352523

2523523

1 CHAPTER 7 Chemical Calculations 7.1-7.7. 2 The Mole A number of atoms, ions, or molecules that is large enough to see and handle. A mole = number of.

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