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The Mole

Chapter 10

1

Objectives

Use the mole and molar mass to make conversions among moles, mass, and number of particles

Determine the percent composition of the components of a compound

Calculate empirical and molecular formulas for compounds

Determine the formulas for hydrates Examine relationship between volume

and amount of gas (Avogadro’s Law –

Ch. 13.2) 2

Dimensional Analysis

Conversion Factors:

– What is 8a*5b/2a=

- Real Example: 1 mile = 5280 ft – Convert 5.5 miles to feet. – 5.5 miles x 5280 ft =

1 mi 20 b

29040 ft

3

The Mole

Mole (mol)

measures the number of particles of a substance

(atom, molecule, formula unit)

Using “mole” is just a shorthand way for saying 6.02 x 1023 particles

6.02 x 1023 particles = Avogadro’s number

4

How much mass is in one

atom of

carbon-12

?

Exactly 12 amu (by definition)

Molar Mass of Atoms

Definition:

12.0000g of Carbon-12 is exactly 1 mol (6.02x1023) of Carbon-12

atoms. Therefore:

(2)

Finding Molar Mass of Other

Elements

All other elements are

determined with respect to

Carbon-12 by mass

spectroscopy.

For example: the He-4 has 4/12

or 1/3 the mass of Carbon-12

7

Finding Molar Mass of Other

Elements

Another element: Fe-56 has 55.935 amu

– So it is 55.935/12 or 4.67 x the mass of C-12

– If we put all the iron isotopes together, the ‘average’ Fe atom is 55.85 amu – Each amu has a mass of 1.66 x 10-24g, so:

8

Finding the Molar Mass of an

Element

9

.

x

.

=

9.27 x 10-23g/atom

.

x

.

=

55.85 g/mol Fe

Mass of Atoms

If you have two 1 g samples of different substances, what do they have in common?

•••••••••••••••••••••••• •••••••••••••••••••••••• •••••••••••••••••••••••• •••••••••••••••••••••••• •••••••••••••••••••••••• ••••••••••••••••••••••••

10

Mass of Atoms

•• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •• •••••• •••••• •••••• •••••• •••••• •••••• •••••• ••••••

The atomic mass of ONE iron

atom is 55.85 AMU (atomic

(3)

The number of grams in a mole

is called the

molar mass

.

13

Why does a mole of iron weigh

more than a mole of carbon?

For the same reason that a dozen bowling balls weighs more than a

dozen golf balls!

Bowling balls are heavier than golf

balls.

14

Mass of Iron on Balance

55.85 g of iron (iron’s molar mass) on scale is:

1 mole of iron (1 mol Fe = 55.85 g Fe)

6.02 x 1023atoms of iron

15

A conversion you already know.

How many eggs are in 14 dozen eggs?

14 dozen eggs x 12 eggs = 168 eggs 1 dozen eggs

How many dozens of eggs would you need

to buy if you were going to feed 504 people 1 egg each?

504 eggs x 1 dozen eggs = 42 dozen eggs 12 eggs

16

Moles and Number of Particles

Examples

If have 2 mole of iron how many atoms of

iron do you have?

2 moles Fe x 6.02x1023atoms = 12.04x1023atoms

1 mole

If have 12.04 x 1023atoms of iron how

many moles of iron have you?

12.04x1023atoms x 1 mole____ = 2.0 moles 6.02x1023atoms

Mass and Moles Examples

What is the mass in grams of 2.00 moles of Cu?

2.00 moles Cu x 63.5 grams = 127 grams Cu 1 mole Cu

How moles is 190. g of Cu? 190. g Cu x 1 mole Cu = 3.00 mol Cu

(4)

Mass

Particles Examples

In order to go from mass to number of particles, you have to go through moles.

How many atoms are in 46.0g of Na?

46.0 g x

.

2.00

2.00 mol Na x .

12.04 10

19

Mass

Particles Examples

What is the mass in grams of 15.05 x 1023 atoms of Al?

15.05 x 1023atoms x

.

2.50

2.50 mol Al x

.

67.5

20

Flowchart

Atoms or

Molecules

Moles

Mass (grams)

Divide by 6.02 X 1023

Multiply by 6.02 X 1023

Multiply by molar mass from periodic table

Divide by molar mass from periodic table

21

Mass, Moles, and Number of

Particles Practice

How many moles are in 25.5 g of Ag? How many grams are 42.60 mol of silicon

(Si)?

How many atoms are in 15.0 mol of Xe? How many moles are in 7.23 x 1024atoms

of Xe?

How many atoms are in 6.50 g of B? How many grams are in 5.53 x 1022atoms

of Mg?

22

Practice

How many moles are in 25.5 g of Ag?

Practice

How many grams are 42.60 mol of silicon (Si)?

(5)

Practice

How many atoms are in 15.0 mol of Xe?

25

Practice

How many moles are in 7.23 x 1024 atoms of Xe?

26

Practice

How many atoms are in 6.50 g of B?

27

Practice

How many grams are in 5.53 x 1022 atoms of Mg?

28

One Mole of

Four

Elements

One mole each of helium, sulfur, copper, and mercury. How many atoms of helium are present? Of sulfur? Of copper? Of mercury?

One-Mole Quantities of Some

Elements & Compounds

(6)

We can also calculate the molar mass

of compounds like

carbon dioxide.

The formula for carbon dioxide is

CO

2

.

That means there is one carbon

atom and two oxygen atoms in

every molecule of CO

2

.

31

We can also calculate the molar mass

of compounds like carbon dioxide.

Carbon weighs 12.01 grams/mole

Oxygen weighs 16.00 grams/mole.

Therefore CO

2

has a molar mass of:

C + O + O

= CO

2

12.01 + 16.00 + 16.00 = 44.01 g 32

Finding Molar Mass of CaCl

2

(assume 1 mole compound)

Atom # mol Atoms in 1 mol of compound

Molar Mass (g/mol)

Total (g) Ca

Cl CaCl2

33

Let’s find out the molar mass of

glucose (C

6

H

12

O

6

).

Use the molar masses from the

periodic table:

Carbon – 12.0 grams/mole

Hydrogen – 1.0 gram/mole

Oxygen – 16.0 grams/mole

34

How many grams are in a mole of

glucose (C

6

H

12

O

6

)?

Carbon: 6 x 12.0 g/mol = 72.0 g Hydrogen: 12 x 1.0 g/mol =12.0 g Oxygen: 6 x 16.0 g/mol = 96.0 g

180.0 g/mol (For consistency, round all molar masses

(elements & compounds) to 0.1 g.)

Moles to Mass of Compound

How many grams of KCl are in 2.30 mol of KCl?

First, find molar mass.

K: 1 K x 39.1 g/mol = 39.1 g Cl: 1 Cl x 35.5 g/mol = 35.5 g

(7)

Moles to Mass of Compound

How many grams of KCl are in 2.30 mol of KCl?

2.30 mol KCl x 74.6 g KCl = mol KCl

=171.6 g KCl = 172 g KCl (proper Sig Figs)

# of moles Molar mass

37

Mass to Moles in a Compound

How many moles of KCl are present in 253.6 grams KCl

?

First, determine the molar mass of KCl.

(Same as before, which is 74.6 g/mol) Then, divide mass by molar mass to get moles.

253.6 g KCl x 1 mol =3.40 mol KCl

74.6 g KCl

 38

Mass to Moles in a Compound

Can determine the number of moles of each of the atoms/ions that make up the compound. Multiply by

ion/compound conversion factor. Conversion factor: mole of specific atom

1 mol of compound Example: 2 mol Cl

-1 mol CaCl2

39

Mass to Moles in a Compound

How many moles of Cl-ions are there 5.5 mol of CaCl2?

5.5 mol CaCl2x 2 mol Cl- = 11.0 mol Cl

-1 mol CaCl2

40

Mass, Moles & Particles

Use the flowchart from the elemental calculations and use them for

compounds.

Moles are still centralGet to mass (g) by multiplying by molar mass Get to # molecules (covalent

compounds) or formula units (ionic compounds) by multiplying by Avogadro’s number.

Mass to Moles in a Compound

How many moles of Na+are there in 561.8 g of Na2CO3?

 First, find Molar Mass of Na2CO3:

Na: 2 x 23.0 g/mol = 46.0 g C: 1 x 12.0 g/mol = 12.0 g O: 3 x 16.0 g/mol = 48.0 g

(8)

Mass to Moles in a Compound

How many moles of Na+are there in 561.8 g of Na2CO3?

Next, find the number of moles of Na2CO3.

561.8 g x 1 mol = 5.300 mol Na2CO3

106.0 g

Use conversion of moles of ion/mole of

compound.

5.300 mol Na2CO3x 2 mol Na+ = 10.6 mol Na+

1 mol Na2CO3

43

Mass to Number of Particles

How many Na+ ions are in 561.8 g of Na2CO3?

Note: Looking for individual # of ions. Will be a very large number. First, find # of moles of the ion or

element you are interested in. In this case it was 10.6 mol Na+.

44

Mass to Number of Particles

Finally, multiply # of mol by Avogadro’s Constant.

10.6 mol Na+x 6.02 x 1023 ions 1 mol

= 6.38 x 1024 Na+ions

45

Mass to Number of Particles

How many formula units of Na2CO3are there in the 5.30 mol of Na2CO3?

5.30 mol Na2CO3 x 6.02 x 1023 Frm Unts 1 mol of Na2CO3 = 31.9 x 1023 formula Units

46

Practice Problems

Determine the # of formula units, the number of moles of each ion, and the number of each ion in:

a) 2.50 mol ZnCl2 b) 623.7 g of Fe2S3

Determine the mass in grams of 2.11x1024 formula units of Na

2S.

Practice Problems

Determine the # of formula units in 2.50 mole of ZnCl2.

(9)

Practice Problems

Determine the # of moles Zn2+ ions in 2.50 mol of ZnCl2.

49

Practice Problems

Determine the # of moles Cl- ions in 2.50 mol of ZnCl2.

50

Practice Problems

Determine the # of Zn2+ions in 2.50 mol of ZnCl2.

51

Practice Problems

Determine the # of Cl-ions in 2.50 mol of ZnCl2.

52

Practice Problems

Determine the # of formula units in in 623.7g of Fe2S3.

Practice Problems

Determine the # of formula units in in 623.7g of Fe2S3.

(10)

Practice Problems

Determine the # of moles of Fe3+ions in 623.7g of Fe2S3.

55

Practice Problems

Determine the # of moles of S2-ions in 623.7g of Fe2S3.

56

Practice Problems

Determine the # Fe3+ions ions in 623.7g of Fe2S3.

57

Practice Problems

Determine the # S2-ions in 623.7g of Fe2S3.

58

Practice Problems

Determine the # Fe3+ions and S2-ions in 623.7g of Fe2S3.

Practice Problems

Determine the mass in grams of 2.11x1024 formula units of Na

2S.

(11)

Practice Problems

Determine the mass in grams of 2.11x1024 formula units of Na

2S.

61

Ch. 10.4 - Percent Composition

Percent composition is the

percent, by mass, of each element in a compound.

In general, it’s the mass of the element/mass of the formula:

Mass of element x 100 = %mass Mass of compound of element

62

Percent Composition

Example: If a 50.0 g sample of H2O contains 5.6 g of H and 44.4 g of O the percent composition is:

 5.6 g H x 100% = 11.1% H 50.0 g H2O

 44.5 g O x 100% = 88.9% O 50 g H2O

63

Percent Composition

You can calculate the percent composition by finding the mass of each element in 1 mole of a

compound.

Example: Water’s formula is H2O which means there are 2 mol of hydrogen and 1 mol oxygen in one mol of water.

64

Percent Composition

So, if you have 1 mol of water, you have 18.0 g of water (molar mass) The 18.0 g of water is made up of 2

mol Hydrogen (2 g) and 1 mol oxygen (16 g).

H: 2 g x 100% = 11.1% 18.0 g

O: 16 g x 100% = 88.9% 18 g

Percent Composition Example

Calculate the percent composition of each element in Ca(OH)2.

1. Determine the mass of each element present in 1 mol of cmpd.

Ca: 1 mol x 40.1 g/mol = 40.1 g O: 2 mol x 16.0 g/mol = 32.0 g H: 2 mol x 1.0 g/mol = 2.0 g 2. Determine mass in g of one mole

(12)

Percent Composition Example

(Continued)

3. Calculate percentage of each element

Ca: 40.1 g/74.1 g * 100 = 54.1% H: 2.0 g/74.1 g * 100 = 2.7% O: 32.0 g/74.1 g * 100 = 43.2%

67

Percent Composition Practice

Determine the %

Comp of each element in: 1. KBr 2. Fe2O3

3. Barium nitrate

68

Empirical Formulas

An empirical formula for a compound is the formula of a substance written with the smallest integer subscripts. In other words, the simplest mole ratios. You can use the percent composition of a

compound to determine is empirical formula.

69

Empirical Formula

Steps to determine formula:

1. Consider 100g of compound

2. Convert percentages of elements to grams

3. Divide each element’s respective mass by its molar mass to obtain moles

4. Divide each mole value by the smallest mole value. This give the mole ratio. 5. Multiply by appropriate number to get

whole number subscripts.

70

Empirical Formula Example

Determining the Empirical Formula from the Masses of Elements.

We have determined the mass percentage composition of calcium chloride: 36.0% Ca and 64.0% Cl. What is the empirical formula of calcium chloride?

Empirical formula Calcium Chloride

(36% Calcium; 64% Chlorine)

Atom Mass % In grams Molar Mass

Moles Ratio Ca

Cl

(13)

Empirical Formula Practice

Determine the empirical formula of a compound that is 36.8% nitrogen and 63.2% oxygen.

73

More Empirical Formula Practice

Determining The Empirical Formula from Percentage Composition. (General)

Benzene is a widely used industrial solvent. This compound has been analyzed and found to contain 92.26% carbon and 7.74% hydrogen by mass. What is its empirical formula? Hint: Consider a 100 g sample.

The empirical formula is:

74

Compounds with different molecular formulas can have the same empirical formula, and such substances will have the same percentage composition.

Remember that the molecular formula has the actual number of atoms of each element that make one molecule of that compound.

Eg. Compound A = C2H2 Compound B = C6H6 both have the empirical formula =

Molecular Formulas

75

Molecular Formula from Empirical

Formula

The molecular formula of a compound is a multiple of its empirical formula.

Molecular mass = n x empirical formula mass where n = number of empirical formula units in the molecule.

76

Determining the Molecular Formula from the Percent Composition and Molar Mass. We have already determined the mass

composition and empirical formula of benzene (CH). In a separate experiment, the molar mass of benzene was determined to be 78.1. What is the molecular formula of benzene

Mass of empirical formula =

Molecular Formula Example

Molar mass benzene _ = Mass of empirical formula

Molecular Formula Example

What is the molecular formula of a compound that has an empirical formula of CH3and a molar mass of 30.0 g/mol.

What is the mass of each empirical unit? C: 1 x 12.0 = 12.0

H: 3 x 1.0 = 3.0 15.0

How many times units? 30/15 = 2 So molecular formula is…C H

(14)

Molecular Formula Practice

1. Empirical Formula is

NO2; molar mass is

92.0 g/mol

2. A compound contains 26.76% C, 2.21%H, 71.17% O and has a molar mass of 90.04 g/mol. Determine its molecular formula.

79

Ch. 10.5 - Salt Hydrates

 Water molecules bound to a compound

For example, CaCl2•2H2O

– Each molecule of calcium chloride has two water molecules bound to it

See “conceptual” picture next slide

80

CaCl

2

•2H

2

O

Ca+2 Cl

-Cl- Cl

-Water

Water 

-

-81

Example of Percent

Composition of a Hydrate

Determine the percent salt and percent water (by mass) in 1 mol of CaCl2•2H2O.

1. Find the mass of 1 mole of the compound.

Ca: 1 x 40.1 = 40.1

Cl: 2 x 35.5 = 71.0 = 147.1 g/mol H2O: 2 x 18.0 = 36.0

82

Example of Percent

Composition of a Hydrate

2. Find the percent of salt.

111.1 g CaCl2 = 75.5% CaCl2

147.1 g CaCl2•2H2O 3. Find the percent of water.

36.0 g H2O = 24.5% H2O

147.1 g CaCl2•2H2O

Or you can say 100% – 75.5% CaCl

Determination of Formula

Example

A nickel(II) cyanide hydrate,

Ni(CN)2•XH2O, contains 39.4% water by mass. What is the formula of the hydrated compound?

1. Assume 100 g of the compound. If it is 39.4% water, it is 100 - 39.4 or 60.6% Ni(CN)2. So there are

(15)

Determination of Formula

Example

2. Determine the number of moles of the salt and the water.

60.6 g Ni(CN)2 = 0.547mol Ni(CN)2 110.7 g/mol

39.4 g H2O = 2.19 mol H2O 18.0 g/mol

Ni: 1 x 58.7 = 58.7 C: 2 x 12.0 = 24.0 N: 2 x 14.0 = 28.0 110.7

85

Determination of Formula

Example

3. Determine the mol ratio of the water to the salt.

2.19 mol H2O 0.547 mol Ni(CN)2

= 4 mol H2O/mol Ni(CN)2

So the formula is Ni(CN)2•4H2O 86

Moles and Gases

(Ch. 13.1-13.2)

Remember Boyle’s Law, Charles’s Law and Combined Gas Law? Boyle: P1V1=P2V2

Charles: V1/T1=V2/T2 Combined: P1V1 = P2V2

T1 T2

There is another law that relates volume to moles.

87

The Gas Laws – Avogadro’s Law

Amedeo

Avogadro

(1776 – 1856)

studied the

relationship

between

volume and

amount of gas.

88

The Gas Laws – Avogadro’s Law

Avogadro’s Principle: Equal volumes of

gas at the same P & T contain equal numbers of particles (molecules or atoms).

Avogadro’s Law – The volume of a gas

at constant P & T is directly proportional to the number of moles of gas.

Volume αn; hold P & T constant

V = k x n

Avogadro’s Principle

As an extension of Avogadro’s Principle, 1 mol of a gas at 0˚C (273 K) and 1 atm of pressure occupies a volume of 22.4 Liters (Molar

Volume)

0˚C (273 K) and 1 atm of pressure are standard temperature and

(16)

Avogadro’s Principle

Since volume is related to moles, if you know the volume a gas occupies at STP, you know the # of moles.

22.4 L 1 mol

Example: If you have 11.2 L of a gas at STP, how many moles are there?  11.2 L x 1 mol = 0.500 mole

22.4 L

91

Avogadro’s Principle Practice

Given the conditions

at STP find:

Volume of 0.881 mol

of gas

# of moles of N2(g) in

a 2.0 L flask

# of atoms of Kr in

28.5 L

92

Ch. 13.2 - Ideal Gas Law

The gas laws that we covered all have one thing in common: they relate the volume of the gas to one of the other variables.

Boyle: V 1/P Charles: V  T Avogadro: V  n

We can put them all together to get…

93

Ideal Gas Law

V

nT

P

To change from a

proportionality,

, to an

equation, we introduce, R,

the proportionality constant

or

Universal Gas Constant.

94

Ideal Gas Law

This makes the previous proportionality relationship into

V = R*nT/P Or more commonly

PV = nRT

the IDEAL GAS

EQUATION

Ideal Gas Law

 The most common value of R is 0.0821 L-atm

Mol-K

There are other values of R with different units that are listed in your book, but we’ll use the one above. Note: they all are equivalent, it just

(17)

Relationship to Other Gas Laws

P

1

V

1

= P

2

V

2

n

1

T

1

n

2

T

2

Both equations are equal to a constant, which is…R (the gas

constant)

97

Ideal Gas Equation Example

A deodorant can has a volume of 0.175 L and a pressure of 3.80 atm at 22ºC. How many moles of gas are contained in the can?

Given: P=3.80 atm; V = 0.175 L; T=22+273=295 K; R = 0.0821 L-atm/mol-K

PV=nRTLooking for n (moles) So… n = PV = (3.80 atm)(0.175 L)

RT (0.0821L*atm/mol*K)*295K

=0.0275 mol 98

Ideal Gas Equation Practice

Calculate the volume that a 0.323 mol sample of gas will occupy at 265 K and 0.900 atm.

A 47.3 L container containing 1.62 mol of He is heated until the pressure reaches 1.85 atm. What is the

temperature in ⁰C?

Kr gas in a 18.5 L cylinder exerts a pressure of 8.61 atm at 24.8ºC What is the mass of Kr?

99

Gas Density and Molar Mass

Let Mstand for molar mass (g/mol) Where n = m/M ;m = mass in grams, n

is moles

And we know PV=nRT So then PV= (m/M)RT Rearranging gives: M = mRT

V P Finally we get: M = dRT/P

100 Mass/volume =

Density

Gas Density & Molar Mass

Example 1

What is molar mass of a gas that has a density of 1.40 g/L at STP?

Given: T=273; P=1.00 atm; d=1.40 g/L; R=0.0821 L-atm/mol-K

Know M = DRT/P so…

=(1.40 g/L)(0.0821 L-atm/mol-K)(273K) 1.0 atm

= 31.4 g/mol

Dalton’s Law of Partial

Pressures

Dalton’s Law of Partial Pressure

(end of Ch. 12.1) states that the total pressure of a mixture of gases is equal to the sum of the pressures of all the gases in the mixture. In other words…

(18)

Dalton’s Law of Partial

Pressures

John Dalton 1766-1844

103

Dalton’s Law of Partial

Pressures

partial pressure depends only on:

– number of moles of gas – container volume – temperature of the gas.

It does not depend on the identity of the gas.

Because gas molecules are so far apart, they don’t interact with each other.

104

Dalton’s Law Example 1

We have a mixture of O2, CO2, and N2in a vessel at STP. The partial

pressure of CO2 is 0.70 atm and the partial pressure of N2is 0.12 atm. What is the partial pressure of O2?

Answer: What is total pressure? 1 atm, so…

1.00 = 0.70 + 0.12 + P(O2)

P(O2) = 1.00 – 0.70 -0.12 = 0.18 atm

105

Partial Pressure & Mole

Fraction

The partial pressure of a gas

component in a mixture is dependent on how much (moles) are there: Pa = nART/V

The total pressure of a gas mixture is Ptotal = ntotalRT/V

106

Partial Pressure & Mole

Fraction

 Ratio of pressure of component to total gas pressure:

=

/

/

RT & V are constants so…

=

Partial Pressure & Mole

Fraction

=

 This means that the partial

pressure of a gas is equal to its mole

fraction multiplied by the total

pressure:

 Pa= na * Ptotal

(19)

Mole Fraction Example 1

A mixture of O2, N2, and He gases are in an enclosed tank. The total pressure in the tank is 12.3 atm. If there are 10.6 mol of N2, 3.3 mol O2 and 1.2 mol He in the tank, what is the partial pressure of each gas?

109

Mole Fraction Example 1

Find total Number of moles:

10.6 + 3.3 + 1.2 = 15.1 mol gas total. For N2: PN2= .

.

∗ 12.3

= 8.6 atm

For O2: PO2= .

.

∗ 12.3

= 2.7 atm

For He: PHe= .

.

∗ 12.3

= 1.0 atm

The sum of the pressures is 12.3 atm.110

Mole Fraction Example 2

A gas mixture containing 0.538 mol He, 0.315 mol Ne and 0.103 mol Ar is confined to a 7.00 L vessel at 25°C.

a) Calculate the partial pressure of each gas in the mixture

b) Calculate the total pressure of the mixture. P(He) = 0.538 mol(0.0821 L-atm/mol-K)(298K)/7.0 L

= 1.88 atm

P(Ne) = 0.315 mol(0.0821 L-atm/mol-K)(298K)/7.0 L = 1.10 atm

P(Ar) = 0.103 mol(0.0821 L-atm/mol-K)(298K)/7.0 L = 0.36 atm

111

Mole Fraction Example 2

Can do previous problem because the gases are at same T & V. Therefore, their mole ratios will be

proportional.

Greater moles means greater partial pressure!

112

Tired of moles?

References

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