6.1 atomic mass and formula mass6...45 molar mass conversion factors • are written from molar...

1

Chapter 6

Chemical Quantities

6.1

Atomic Mass and Formula Mass

Copyright © 2008 by Pearson Education, Inc.

Publishing as Benjamin Cummings

2

Atomic Mass

Atomic mass is the

• Mass of a single atom in

atomic mass units (amu).

• Mass of an atom

compared to a 12C atom.

• Number below the

symbol of an element.

3

Periodic Table and Atomic Mass

Ag has

atomic

mass =

107.9 amu

C has atomic mass

= 12.01 amu

S has

atomic mass

= 32.07 amu

4

Atomic Mass Factors

The atomic mass

• Can be written as an equality.

Example: 1 P atom = 30.97 amu

• Can be written as two conversion factors.

Example: 1 P atom and 30.97 amu

30.97 amu 1 P atom

5

Uses of Atomic Mass Factors

The atomic mass factors are used to convert

• A specific number of atoms to mass (amu).

• An amount in amu to number of atoms.

6

Using Atomic Mass Factors

1. What is the mass in amu of 75 P atoms?

75 P atoms x 30.97 amu = 2323 amu (2.323 x103 amu)

1 P atom

2. How many Cu atoms have a mass of 4.500 x 105

amu?

4.500 x 105 amu x 1 Cu atom = 7081 Cu atoms

63.55 amu

7

Learning Check

What is the mass in amu of

75 silver atoms?

1) 107.9 amu

2) 8093 amu

3) 1.439 amu



8

Solution

What is the mass in amu of 75 silver atoms?

2) 8093 amu

75 Ag atoms x 107.9 amu = 8093 amu

1 Ag atom

9

Learning Check

How many gold atoms have

a mass of 1.85 x 105 amu?

1) 939 Au atoms

2) 3.64 x 107 Au atoms

3) 106 Au atoms



10

Solution

How many gold atoms have a mass of

1.85 x 105 amu?

1) 939 Au atoms

1.85 x 105 amu x 1 Au atom = 939 Au atoms

197.0 amu

11

Formula Mass

The formula mass is

• The mass in amu of a compound.

• The sum of the atomic masses of the elements

in a formula.



12

Calculating Formula Mass

To calculate the formula mass of Na2SO4

• Multiply the atomic mass of each element by its subscript, then total the masses of the atoms.

2 Na x 22.99 amu = 45.98 amu

1 Na

1 S x 32.07 amu = 32.07 amu 142.05 amu

1 S

4 O x 16.00 amu = 64.00 amu

1 O

13

Learning Check

Using the periodic table, calculate the formula

mass of aluminum sulfide Al2S3.

14

Solution

Using the periodic table, calculate the formula

mass of aluminum sulfide Al2S3.

2 Al x 26.98 amu = 53.96 amu

1 Al

3 S x 32.07 amu = 96.21 amu

1 S 150.17 amu

15

Chapter 6

Chemical Quantities

6.2

The Mole



16

Collection Terms

A collection term states a specific number of items.

• 1 dozen donuts = 12 donuts

• 1 ream of paper = 500 sheets

• 1 case = 24 cans



17

A mole (mol) is a collection that contains

• The same number of particles as there are

carbon atoms in 12.01 g of carbon.

• 6.022 x 1023 atoms of an element (Avogadro’s

number).

1 mol C = 6.022 x 1023 C atoms

1 mol Na = 6.022 x 1023 Na atoms

1 mol Au = 6.022 x 1023 Au atoms

A Mole of Atoms

18

A mole

• Of a covalent compound has Avogadro’s number

of molecules.

1 mol CO2 = 6.022 x 1023 CO2 molecules

1 mol H2O = 6.022 x 1023 H2O molecules

• Of an ionic compound contains Avogadro’s

number of formula units.

1 mol NaCl = 6.022 x 1023 NaCl formula units

1 mol K2SO4 = 6.022 x 1023 K2SO4 formula units

A Mole of A Compound

19

Samples of One Mole

Quantities

Table 6.1

Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

20

Avogadro’s number 6.022 x 1023 can be written as

an equality and two conversion factors.

As an equality:

1 mol = 6.022 x 1023 particles

As conversion Factors:

6.022 x 1023 particles and 1 mol

1 mol 6.022 x 1023 particles

Avogadro’s Number

21

Using Avogadro’s Number

Avogadro’s number

• Converts moles of a substance to the number of particles.

How many Cu atoms are in 0.50 mol

Cu?

0.50 mol Cu x 6.022 x 1023 Cu atoms

1 mol Cu

= 3.0 x 1023 Cu atoms Copyright © 2008 by Pearson Education, Inc.


22

Using Avogadro’s Number

Avogadro’s number

• Converts particles of a substance to moles.

How many moles of CO2 are

2.50 x 1024 CO2 molecules?

2.50 x 1024 CO2 x 1 mol CO2

6.022 x 1023 CO2

= 4.15 mol CO2

23

1. The number of atoms in 2.0 mol Al is

A. 2.0 Al atoms

B. 3.0 x 1023 Al atoms

C. 1.2 x 1024 Al atoms

2. The number of moles of S in 1.8 x 1024 atoms S is

A. 1.0 mol S atoms

B. 3.0 mol S atoms

C. 1.1 x 1048 mol S atoms

Learning Check

24

C. 1.2 x 1024 Al atoms

2.0 mol Al x 6.022 x 1023 Al atoms

1 mol Al

B. 3.0 mol S atoms

1.8 x 1024 S atoms x 1 mol S

6.022 x 1023 S atoms

Solution

25

Subscripts and Moles

The subscripts in a formula state

• The relationship of atoms in the formula.

• The moles of each element in 1 mol of

compound.

Glucose

C6H12O6

In 1 molecule: 6 atoms C 12 atoms H 6 atoms O

In 1 mol: 6 mol C 12 mol H 6 mol O

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Subscripts State Atoms and

Moles

1 mol C9H8O4 = 9 mol C 8 mol H 4 mol O



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Factors from Subscripts

Subscripts used for conversion factors

• Relate moles of each element in 1 mol compound.

• For aspirin C9H8O4 can be written as:

9 mol C 8 mol H 4 mol O

1 mol C9H8O4 1 mol C9H8O4 1 mol C9H8O4 and

1 mol C9H8O4 1 mol C9H8O4 1 mol C9H8O4

9 mol C 8 mol H 4 mol O

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Learning Check

A. How many moles O are in 0.150 mol aspirin

C9H8O4?

B. How many O atoms are in 0.150 mol aspirin

C9H8O4?

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Solution

A. How many mol O are in 0.150 mol aspirin C9H8O4?

0.150 mol C9H8O4 x 4 mol O = 0.600 mol O

1 mol C9H8O4

subscript factor

B. How many O atoms are in 0.150 mol aspirin C9H8O4?

0.150 mol C9H8O4 x 4 mol O x 6.022 x 1023 O atoms

1 mol C9H8O4 1 mol O subscript Avogadro’s

factor Number = 3.61 x 1023 O atoms

30

Learning Check

How many O atoms are in 0.150 mol aspirin C9H8O4?

31

Solution

How many O atoms are in 0.150 mol aspirin C9H8O4?

0.150 mol C9H8O4 x 4 mol O x 6.022 x 1023 O atoms

1 mol C9H8O4 1 mol O

subscript Avogadro’s

factor number

= 3.61 x 1023 O atoms

32

Chapter 6

Chemical Quantities

6.3

Molar Mass



33

Molar Mass

The molar mass

• Is the mass of one

mol of an element or

compound.

• Is the atomic mass

expressed in grams.



34

Molar Mass from the Periodic

Table

Molar mass

• Is the

atomic

mass

expressed

in grams.



35

Give the molar mass for:

A. 1 mol K atoms = ________

B. 1 mol Sn atoms = ________

Learning Check

36

Give the molar mass for:

A. 1 mol K atoms = 39.10 g

B. 1 mol Sn atoms = 118.7 g

Solution

37

Molar Mass of a Compound

The molar mass of a compound is the sum of the

molar masses of the elements in the formula.

Example: Calculate the molar mass of CaCl2.

Element Number

of Moles

Atomic Mass Total Mass

Ca 1 40.08 g/mol 40.08 g

Cl 2 35.45 g/mol 70.90 g

CaCl2 110.98 g

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Molar Mass of K3PO4

Calculate the molar mass of K3PO4.

Element Number

of Moles

Atomic Mass Total Mass

in K3PO4

K 3 39.10 g/mol 117.3 g

P 1 30.97 g/mol 30.97 g

O 4 16.00 g/mol 64.00 g

K3PO4 212.3 g

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Some One-Mol Quantities

32.07 g 55.85 g 58.44 g 294.20 g 342.30 g



40

What is the molar mass of each of the following?

A. K2O

B. Al(OH)3

Learning Check

41

A. K2O 94.20 g/mol

2 mol K (39.10 g/mol) + 1 mol O (16.00 g/mol)

78.20 g + 16.00 g

= 94.20 g

B. Al(OH)3 78.00 g/mol

1 mol Al (26.98 g/mol) + 3 mol O (16.00 g/mol)

+ 3 mol H (1.008 g/mol)

26.98 g + 48.00 g + 3.024 g

= 78.00 g

Solution

42

Prozac, C17H18F3NO, is an antidepressant that

inhibits the uptake of serotonin by the brain. What

is the molar mass of Prozac?

1) 40.06 g/mol

2) 262.0 g/mol

3) 309.4 g/mol

Learning Check

43

Prozac, C17H18F3NO, is an antidepressant that

inhibits the uptake of serotonin by the brain. What is

the molar mass of Prozac?

3) 309.3 g/mol

17C (12.01) + 18H (1.008) + 3F (19.00)

+ 1N (14.01) + 1 O (16.00) =

204.2 +18.14 + 57.00 + 14.01 + 16.00

= 309.4 g/mol

Solution

44

Chapter 6

Chemical Quantities

6.4

Calculations Using

Molar Mass



45

Molar mass conversion factors

• Are written from molar mass.

• Relate grams and moles of an element or

compound.

Example: Write molar mass factors for methane

CH4 used in gas cook tops and gas heaters.

Molar mass:

1 mol CH4 = 16.04 g

Conversion factors:

16.04 g CH4 and 1 mol CH4

1 mol CH4 16.04 g CH4

Molar Mass Factors

46

Acetic acid C2H4O2 gives the sour taste to

vinegar. Write two molar mass conversion

factors for acetic acid.

Learning Check



47

Acetic acid C2H4O2 gives the sour taste to vinegar.

Write two molar mass factors for acetic acid.

Calculate molar mass:

24.02 + 4.032 = 32.00 = 60.05 g/mol

1 mol of acetic acid = 60.05 g acetic acid

Molar mass factors

1 mol acetic acid and 60.05 g acetic acid

60.05 g acetic acid 1 mol acetic acid

Solution

48

Molar mass factors are used to convert between the

grams of a substance and the number of moles.

Calculations Using Molar Mass

Grams

Molar mass factor Moles

49

Aluminum is used to build lightweight bicycle

frames. How many grams of Al are 3.00 mol Al?

Molar mass equality: 1 mol Al = 26.98 g Al

Setup with molar mass as a factor:

3.00 mol Al x 26.98 g Al = 80.9 g Al 1 mol Al molar mass factor

for Al

Moles to Grams

50

Learning Check

Allyl sulfide C6H10S is a

compound that has the

odor of garlic. How many

moles of C6H10S are in

225 g C6H10S?



51

Calculate the molar mass of C6H10S.

(6 x 12.01) + (10 x 1.008) + (1 x 32.07)

= 114.21 g/mol

Set up the calculation using a mole factor.

225 g C6H10S x 1 mol C6H10S

114.21 g C6H10S

molar mass factor(inverted)

= 1.97 mol C6H10S

Solution

52

Grams, Moles, and Particles

A molar mass factor and Avogadro’s number

convert

• Grams to particles

molar mass Avogadro’s

number

(g mol particles)

• Particles to grams

Avogadro’s molar mass

number

(particles mol g)

53

Learning Check

How many H2O molecules are in 24.0 g H2O?

1) 4.52 x 1023

2) 1.44 x 1025

3) 8.02 x 1023

54

Solution

How many H2O molecules are in 24.0 g H2O?

3) 8.02 x 1023

24.0 g H2O x 1 mol H2O x 6.022 x 1023 H2O molecules

18.02 g H2O 1 mol H2O

= 8.02 x 1023 H2O molecules

55

Learning Check

If the odor of C6H10S can be detected from 2 x

10-13 g in one liter of air, how many molecules

of C6H10S are present?

56

Solution

If the odor of C6H10S can be detected from

2 x 10-13 g in one liter of air, how many molecules

of C6H10S are present?

2 x 10-13 g x 1 mol x 6.022 x 1023 molecules

114.21 g 1 mol

= 1 x 109 molecules C6H10S

57

Chapter 6 Chemical Quantities

6.5

Percent

Composition

and

Empirical

Formulas



58

Percent Composition

Percent composition

• Is the percent by mass of each element in a

formula.

Example: Calculate the percent composition of CO2.

CO2 = 1 C(12.01g) + 2 O(16.00 g) = 44.01 g/mol)

12.01 g C x 100 = 27.29 % C

44.01 g CO2

32.00 g O x 100 = 72.71 % O

44.01 g CO2 100.00 %

59

What is the percent composition of lactic acid,

C3H6O3, a compound that appears in the blood

after vigorous activity?

Learning Check

60

STEP 1

3C(12.01) + 6H(1.008) + 3O(16.00) = 90.08 g/mol

36.03 g C + 6.048 g H + 48.00 g O

STEP 2

%C = 36.03 g C x 100 = 40.00% C

90.08 g

%H = 6.048 g H x 100 = 6.714% H

90.08 g

%O = 48.00 g O x 100 = 53.29% O

90.08 g

Solution

61

Learning Check

The chemical isoamyl

acetate C7H14O2 gives

the odor of pears. What

is the percent carbon in

isoamyl acetate?

1) 7.102 %C

2) 35.51 %C

3) 64.58 %C



62

3) 64.58 %C

Molar mass C7H14O2 = 7C(12.01) + 14H(1.008)

+ 2O(16.00) = 130.18 g/mol

Total C = 7C(12.01) = g

% C = total g C x 100 total g % C = 84.07 g C x 100 = 64.58 % C 130.18 g

Solution

63

The empirical formula

• Is the simplest whole number ratio of the atoms.

• Is calculated by dividing the subscripts in the

actual (molecular) formula by a whole number

to give the lowest ratio.

C5H10O5 5 = C1H2O1 = CH2O

actual (molecular) empirical formula

formula

Empirical Formulas

64

Some Molecular and Empirical

Formulas • The molecular formula is the same or a multiple of

the empirical.

Table 6.3


65

A. What is the empirical formula for C4H8?

1) C2H4 2) CH2 3) CH

B. What is the empirical formula for C8H14?

1) C4H7 2) C6H12 3) C8H14

C. Which is a possible molecular formula for CH2O?

1) C4H4O4 2) C2H4O2 3) C3H6O3

Learning Check

66

A. What is the empirical formula for C4H8?

2) CH2 C4H8 4

B. What is the empirical formula for C8H14?

1) C4H7 C8H14 2

C. Which is a possible molecular formula for CH2O?

2) C2H4O2 3) C3H6O3

Solution

67

A compound has an empirical formula SN. If

there are 4 atoms of N in one molecule, what is

the molecular formula? Explain.

1) SN

2) SN4

3) S4N4

Learning Check

68

A compound has an empirical formula SN. If

there are 4 atoms of N in one molecule, what

is the molecular formula? Explain.

3) S4N4

In this molecular formula 4 atoms of N and 4

atoms of S and N are related 1:1. Thus, it has

an empirical formula of SN.

Solution

69

A compound contains 7.31 g Ni and 20.0 g Br.

Calculate its empirical (simplest) formula.

Learning Check

70

Convert 7.31 g Ni and 20.0 g Br to moles.

7.31 g Ni x 1 mol Ni = 0.125 mol Ni

58.69 g Ni

20.0 g Br x 1 mol Br = 0.250 mol Br

79.90 g Br

Divide by smallest:

0.125 mol Ni = 1 Ni 0.250 mol Br = 2 Br

0.125 0.125

Write ratio as subscripts: NiBr2

Solution

71

Converting Decimals to Whole

Numbers

When the number of moles for an element is a

decimal, all the moles are multiplied by a small

integer to obtain whole number.


Table 6.4

72

Aspirin is 60.0% C, 4.5 % H and 35.5 % O.

Calculate its empirical (simplest) formula.

Learning Check

73

STEP 1. Calculate the moles of each element in

100 g.

100 g aspirin contains 60.0% C or 60.0 g C,

4.5% H or 4.5 g H, and 35.5% O or 35.5 g O.

60.0 g C x 1 mol C = 5.00 mol C

12.01 g C

4.5 g H x 1 mol H = 4.5 mol H

1.008 g H

35.5 g O x 1mol O = 2.22 mol O

16.00 g O

Solution

74

Solution (continued)

STEP 2. Divide by the smallest number of mol.

5.00 mol C = 2.25 mol C (decimal)

2.22

4.5 mol H = 2.0 mol H

2.22

2.22 mol O = 1.00 mole O

2.22

75


3. Use the lowest whole number ratio as subscripts

When the moles are not whole numbers, multiply

by a factor to give whole numbers, in this case x 4.

C: 2.25 mol C x 4 = 9 mol C

H: 2.0 mol H x 4 = 8 mol H

O: 1.00 mol O x 4 = 4 mol O

Using these whole numbers as subscripts the

simplest formula is

C9H8O4

76

Chapter 6 Chemical

Quantities

6.6

Molecular Formulas



77

A molecular formula

• Is a multiple (or equal) of its empirical formula.

• Has a molar mass that is the empirical formula

mass multiplied by a whole number.

molar mass = a whole number

empirical mass

• Is obtained by multiplying the empirical formula by

a whole number.

Relating Molecular and

Empirical Formulas

78

Diagram of Molecular and

Empirical Formulas

A small integer links

• A molecular formula and its empirical

formula.

• A molar mass and its empirical formula

mass.


79

Determine the molecular formula of compound that

has a molar mass of 78.11 g and an empirical

formula of CH.

STEP 1. Empirical formula mass of CH = 13.02 g

STEP 2. Divide the molar mass by the empirical

mass.

78.11 g = 5.999 ~ 6

13.02 g

STEP 3. Multiply each subscript in C1H1 by 6.

molecular formula = C1x 6 H1 x 6 = C6H6

Finding the Molecular Formula

80

Some Compounds with

Empirical Formula CH2O


Table 6.5

81

A compound has a molar mass of 176.1g and an

empirical formula of C3H4O3. What is the

molecular formula?

1) C3H4O3

2) C6H8O6

3) C9H12O9

Learning Check

82

A compound has a formula mass of 176.1 and an

empirical formula of C3H4O3. What is the

molecular formula?

2) C6H8O6

C3H4O3 = 88.06 g/EF

176.1 g (molar mass) = 2.00

88.06 g (empirical mass)

Molecular formula = 2 x empirical formula

C3 x 2H4 x 2O3 x 2 = C6H8O6

Solution

83

A compound contains C 24.27%, H 4.07%, and Cl

71.65%. The molar mass is about 99 g. What are

the empirical and molecular formulas?

STEP 1. Calculate the empirical formula.

Write the mass percents as the grams in a

100.00-g sample of the compound.

C 24.27 g H 4.07 g Cl 71.65 g

Molecular Formula

84


(Continued)

Calculate the number of moles of each element.

24.27 g C x 1 mol C = 2.021 mol C

12.01 g C

4.07 g H x 1 mol H = 4.04 mol H

1.008 g H

71.65 g Cl x 1 mol Cl = 2.021 mol Cl

35.45 g Cl

85


(Continued) Divide by the smallest number of moles:

2.021 mol C = 1 mol C

2.021

4.04 mol H = 2 mol H

2.021

2.02 mol Cl = 1 mol Cl

2.021

Empirical formula = C1H2Cl1 = CH2Cl

Calculate empirical mass (EM) CH2Cl = 49.48 g

86


(Continued)

STEP 2. Divide molar mass by empirical mass.

Molar mass = 99 g = 2

Empirical mass 49.48 g

STEP 3. Multiply the empirical formula by the small

integer to determine the molecular formula.

2 x (CH2Cl)

C1 x2 H2 x 2 Cl1x 2 = C2H4Cl2

87

A compound is 27.4% S, 12.0% N and 60.6

% Cl. If the compound has a molar mass of

351 g, what is the molecular formula?

Learning Check

88

In 100. g, there are 27.4 g S, 12.0 g N, and 60.6 g Cl.

27.4 g S x 1 mol S = 0.854 mol S

32.07 g S

12.0 g N x 1 mol N = 0.857 mol N

14.01 g N

60.6 g Cl x 1mol Cl = 1.71 mol Cl

35.45 g Cl

Solution

89

Divide by the smallest number of moles

0.854 mol S /0.854 = 1.00 mol S

0.857 mol N/0.854 = 1.00 mol N

1.71 mol Cl/0.854 = 2.00 mol Cl

empirical formula = SNCl2 = 116.98 g

Molar Mass/ Empirical mass

351 g = 3

116.98 g

molecular formula = (SNCl2)3 = S3N3Cl6


6.1 atomic mass and formula mass6...45 molar mass conversion factors • are written from molar...

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