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

The Mole

Chapter 101

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)

5

Molar Mass of Atoms

Definition: 12.0000g of Carbon-12 is exactly

1 mol (6.02x1023) of Carbon-12 atoms.

Therefore:

-2423

12 g C-12 1 mol 1 atomx x 1.66x10 g/amu

1 mol 6.02 10 atoms 12 amux

6

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 10-24 g, so:

8

Finding the Molar Mass of an Element

9

x= 9.27 x 10-23 g/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

•• •• •• •• •• •• •• •• •• •• •• ••

•• •• •• •• •• •• •• •• •• •• •• ••

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

11

The atomic mass of ONE iron atom is 55.85 AMU (atomic mass units).

The molar mass of (a mole of) iron is 55.85 grams. 12

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 1023 atoms of iron

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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 6.02x1023 atoms = 12.04x1023 atoms 1 mole

If have 12.04 x 1023 atoms of iron how many moles of iron have you?

12.04x1023 atoms x 1 mole____ = 2.0 moles

6.02x1023 atoms

17

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

63.5 g

18

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 mol Na x

19

Mass ↔ Particles Examples

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

15.05 x 1023 atoms x

2.50 mol Al x

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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 1024 atoms

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

of Mg?

22

Practice

How many moles are in 25.5 g of Ag?

23

Practice

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

24

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?

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One-Mole Quantities of Some Elements & Compounds

30

We can also calculate the molar mass of compounds likecarbon dioxide.

The formula for carbon dioxide is CO2.

That means there is one carbon atom and two oxygen atoms in every molecule of CO2.

31

We can also calculate the molar mass of compounds like carbon dioxide.

Carbon weighs 12.01 grams/moleOxygen weighs 16.00 grams/mole.

Therefore CO2 has a molar mass of:

C + O + O = CO2

12.01 + 16.00 + 16.00 = 44.01 g 32

Finding Molar Mass of CaCl2(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 (C6H12O6).

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 (C6H12O6)?

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.)

35

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 74.6 g/mol

36

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

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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 compoundExample: 2 mol Cl-

1 mol CaCl239


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

5.5 mol CaCl2 x 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.

41


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

Molar Mass = 106.0 g/mol

42


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 Na2CO3 x 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 Na2CO3 are

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 Na2S.

47

Practice Problems

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

48

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.

53

Practice Problems

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

54

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.

59

Practice Problems


Hint: First find Molar Mass.

60

Practice Problems


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

65

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 g2. Determine mass in g of one mole of

compound. 74.1 g66

Percent Composition Example (Continued)

3. Calculate percentage of each elementCa: 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 compound2. Convert percentages of elements to

grams3. 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 ExampleDetermining 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?

71

Empirical formula Calcium Chloride(36% Calcium; 64% Chlorine)

Atom Mass % In grams MolarMass

Moles Ratio

Ca

Cl

Ratio Ca : Cl Empirical Formula

72

Empirical Formula Practice Determine the empirical formula of a

compound that is 36.8% nitrogen and 63.2% oxygen.

73

More Empirical Formula PracticeDetermining 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

77

Molecular Formula Example What is the molecular formula of a

compound that has an empirical formula of CH3 and 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 15.0How many times units? 30/15 = 2So molecular formula is…C2H6 78

Molecular Formula Practice1. 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

CaCl2•2H2O

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 = 40.1Cl: 2 x 35.5 = 71.0 = 147.1 g/molH2O: 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% CaCl2 = 24.5% H2O.

83

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

60.6 g of Ni(CN)2

84


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

85


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 α n; hold P & T constantV = k x n

89

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 pressure (STP)

90

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 L1 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 LawV 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/POr more commonly

PV = nRTthe IDEAL GAS EQUATION

95

Ideal Gas Law

The most common value of R is 0.0821 L-atm

Mol-KThere 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 depends on what unit (mostly pressure) you are using.

96

Relationship to Other Gas Laws

P1V1 = P2V2

n1T1 n2T2Both 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 M stand 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

101

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…Ptotal = P1 + P2 + P3…Pn

102


John Dalton1766-1844

103


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 N2 in a vessel at STP. The partial pressure of CO2 is 0.70 atm and the partial pressure of N2 is 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 atm105

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


Ratio of pressure of component to total gas pressure:

= RT & V are constants so…

=

107


= This means that the partial pressure

of a gas is equal to its mole fraction multiplied by the total pressure:

Pa = na * Ptotal

ntotal

Mole Fraction

108

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 = = 8.6 atm For O2: PO2 = = 2.7 atm For He: PHe = = 1.0 atm The sum of the pressures is 12.3 atm.

110

Mole Fraction Example 2A 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?

So am I.113

Documents

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