15 The Mole

(1)

The Mole

(2)

Words and Numbers

 1 dozen 1 dozen

 A pairA pair

 A ream (paper)A ream (paper)

 A scoreA score

 One grossOne gross

 A trioA trio

(3)

A mole is:



6.O22 x 10

23 23

entities

 A mole of apples = 6.O22 x 10A mole of apples = 6.O22 x 1023 23 apples_apples

 A mole of dollars = $ 6.O22 x 10A mole of dollars = $ 6.O22 x 1023 23

 A mole of particles: 6.O22 x 10A mole of particles: 6.O22 x 1023 23

particles particles

(4)

A mole is:

 A quantity of substanceA quantity of substance

 Avogadro’ s number of anything!Avogadro’ s number of anything!

 Avogadro’s Number (NAvogadro’s Number (N_A_A)) = 6. 022 x 10 = 6. 022 x 10

(5)

How BIG is Avogadro's number?

 If Avogadro's number of sheets of paper were If Avogadro's number of sheets of paper were

divided into a million equal piles, each pile would be

divided into a million equal piles, each pile would be

so tall that it would stretch from the Earth to the

Sun and beyond.

 A CRAY S-1 super computer with a nominal speed _{A CRAY S-1 super computer with a nominal speed} rating of 1,000 mips (millions of instructions per

rating of 1,000 mips (millions of instructions per

second) would require 1.9 million years to process

Avogadro's number of steps.

(6)

• If we were able to count atoms at the rate of If we were able to count atoms at the rate of

10 million per second, it would take about 2

billion years to count the atoms in one mole

• If you could count 100 particles every If you could count 100 particles every

minute and counted twelve hours every day

and had every person on Earth also

counting, it would take more than four

million years to count a mole of anything.

(7)

Avogadro’s Number

 This number is named in honor of Amedeo Avogadro _{This number is named in honor of Amedeo Avogadro}

(1776 – 1856), who proposed his hypothesis in 1811.

At that time there was no data at all on the number of

particles in a mole, or an agreement on any atomic

weights or the standard.

 Avogadro used other pieces of data (Guy Lussac’s Law _{Avogadro used other pieces of data (Guy Lussac’s Law}

and Dalton’s Atomic theory) to hypothesize that equal

volumes of gases at the same temperature and pressure

contain equal numbers of molecules.

(8)

Avogadro’s Number

 They compared masses They compared masses

using carbon 12 as the

standard!

 The mole is an amount of The mole is an amount of

substance which

contains as many

particles as there are

atoms in 0.012 kilograms

(12 g) of carbon-12.

(9)

The Mole

 1 dozen cookies = 12 cookies_{1 dozen cookies = 12 cookies}

 1 mole of cookies = 6.02 X 101 mole of cookies = 6.02 X 1023 23 cookies_cookies

 1 dozen cars = 12 cars1 dozen cars = 12 cars

 1 mole of cars = 6.02 X 101 mole of cars = 6.02 X 1023 23 cars_cars

 1 dozen Al atoms = 12 Al atoms_{1 dozen Al atoms = 12 Al atoms}

 1 mole of Al atoms = 6.02 X 101 mole of Al atoms = 6.02 X 1023 23 atoms_atoms

Note that the NUMBER is always the same,

Note that the NUMBER is always the same,

but the MASS is very different!

Mole is abbreviated mol (gee, that’s a lot

quicker to write, huh?)

(10)

=

= 6.02 x 106.02 x 102323 C atoms C atoms

=

= 6.02 x 10 6.02 x 102323 H_H 2

2O moleculesO molecules

=

= 6.02 x 106.02 x 102323 NaCl “units”_{NaCl “units”}

(technically, ionics are compounds not

molecules so they are called formula units)

6.02 x 10

6.02 x 102323 Na_Na++ ions and _{ions and}

6.02 x 10

6.02 x 102323 Cl_Cl–– ions_ions

A Mole of Particles

Contains 6.02 x 10

23 23

particles

1 mole C 1 mole H2O

(11)

6.02 x 106.02 x 102323 particles _particles

1 mole1 mole

or

1 mole

6.02 x 10

6.02 x 102323 particles_particles

Note that a particle could be an atom OR a molecule! Note that a particle could be an atom OR a molecule!

Avogadro’s Number as

Conversion Factor

(12)

1. Number of atoms in 0.500 mole of Al

1. Number of atoms in 0.500 mole of Al

a) 500 Al atoms

b) 6.02 x 10b) 6.02 x 102323 Al atoms_{Al atoms} c) 3.01 x 10

c) 3.01 x 1023 23 Al_Alatoms_atoms

2.Number of moles of S in 1.8 x 10

2.Number of moles of S in 1.8 x 102424 S atoms_{S atoms} a) 1.0 mole S atoms

a) 1.0 mole S atoms

b) 3.0 mole S atoms

c) 1.1 x 10

c) 1.1 x 104848_{mole S atoms}_{mole S atoms}

Learning Check

(13)

 The Mass of 1 mole (in grams)The Mass of 1 mole (in grams)

 Equal to the numerical value of the average Equal to the numerical value of the average

atomic mass (get from periodic table)

atomic mass (get from periodic table)

1 mole

1 mole of C atoms of C atoms = = 12.0 g12.0 g 1 mole

1 mole of Mg atoms of Mg atoms == 24.3 g24.3 g 1 mole

1 mole of Cu atoms of Cu atoms == 63.5 g63.5 g

Molar Mass

(14)

The mole and the Periodic

Table

 Mass on the periodic table represents:Mass on the periodic table represents:



Mass of 1 atom in atomic mass

units

(u)(u)



Mass of 1 mole of the atom in

grams

(15)

Avogadro’s number

 12g of C12g of C1212 = 6.022 x10_{= 6.022 x10}23 23 C atoms = 1 mole_{C atoms = 1 mole}

2.02g of H

2.02g of H₂₂ = 6.022 x10 = 6.022 x1023 23 molecules of H_{molecules of H} 2

2= 1 = 1

mole mole

65.39g of Zn = 6.022 x10

65.39g of Zn = 6.022 x1023 23 Zn atoms = 1 mole_{Zn atoms = 1 mole}

32.00g of O

32.00g of O₂₂ = 6.022 x10 = 6.022 x1023 23 molecules of O_{molecules of O} 2

2= 1 = 1

(16)

Find the molar mass

Find the molar mass

(usually we round to the tenths place)

Learning Check!

A. 1 mole of Br atoms

B. 1 mole of Sn atoms

= 79.9 g/mole

(17)

Mass in grams of 1 mole equal numerically to

the sum of the atomic masses

1 mole of CaCl

1 mole of CaCl₂₂ = 111.1 g/mol = 111.1 g/mol

1 mole Ca1 mole Ca x 40.1 g/mol x 40.1 g/mol +

+ 2 moles Cl2 moles Cl x 35.5 g/mol = 111.1 g/mol CaCl x 35.5 g/mol = 111.1 g/mol CaCl₂₂

1 mole of N

1 mole of N₂₂OO₄₄ = 92.0 g/mol= 92.0 g/mol

Molar Mass of Molecules and

Compounds

(18)

A.

Molar Mass

of K

₂₂

O = ? Grams/mole

B.

Molar Mass

of antacid Al(OH)

₃₃

= ?

Grams/mole

Learning Check!

(19)

Prozac, C

Prozac, C₁₇₁₇HH₁₈₁₈FF₃₃NO, is a widely used NO, is a widely used antidepressant that inhibits the uptake

antidepressant that inhibits the uptake

of serotonin by the brain. Find its molar

mass.

Learning Check

(20)

Other Names Related to Molar

Mass

 Molecular Mass/Molecular Weight:_{Molecular Mass/Molecular Weight:} If you have a single _{If you have a single}

molecule, mass is measured in amu’s instead of grams. But,

the molecular mass/weight is the

the molecular mass/weight is the same numerical valuesame numerical value as 1 as 1 mole of molecules. Only the units are different. (This is the

mole of molecules. Only the units are different. (This is the

beauty of Avogadro’s Number!)

 Formula Mass/Formula Weight:Formula Mass/Formula Weight: Same goes for Same goes for

compounds. But again,

compounds. But again, the numerical value is the samethe numerical value is the same. . Only the units are different.

Only the units are different.

(21)

Mole Calculations

 Example: If there are 9.50 x10Example: If there are 9.50 x103030

molecules of CO

molecules of CO₂₂, then calculate the , then calculate the moles of CO

moles of CO_2._2.

 n = N/Nn = N/N

A A

 n = n = 9.50 x109.50 x1030 30 molecules_molecules/6.022x 10 _{/6.022x 10}2323 mol_mol-1-1

(22)

Aluminum is often used for the structure Aluminum is often used for the structure

of light-weight bicycle frames. How

many grams of Al are in 3.00 moles of

Al?

3.00 moles Al ? g Al

Converting Moles and Grams

(23)

1. Molar mass of Al

1. Molar mass of Al 1 mole Al = 27.0 g Al1 mole Al = 27.0 g Al

2. Conversion factors for Al

27.0g Al

27.0g Al or or 1 mol Al 1 mol Al

1 mol Al 27.0 g Al1 mol Al 27.0 g Al

3. Setup

3. Setup 3.00 moles Al x 3.00 moles Al x 27.0 g Al 27.0 g Al 1 mole Al

1 mole Al

Answer

(24)

Mole Calculations

 n = # of molesn = # of moles

 NN

A

A = Avogadro’s number 6.022 x10 = Avogadro’s number 6.022 x10 2323

 N = number of particlesN = number of particles



N = n * N

A A

(25)

Molar Mass

 Calculate the molar mass of Al(NOCalculate the molar mass of Al(NO₃₃))₃₃  MM= 1Al+ 3N +9OMM= 1Al+ 3N +9O

 MM= (1 x 26.98) + (3 x 14.007) + (9 x MM= (1 x 26.98) + (3 x 14.007) + (9 x

16.00) = 213.00 g/mol

 213.00 grams is the mass of one mole of 213.00 grams is the mass of one mole of

aluminum nitrate.

 213.00 grams of aluminum nitrate 213.00 grams of aluminum nitrate

contains 6.022 x 10

contains 6.022 x 102323 entities of Al(NO_{entities of Al(NO} 3

(26)

(27)

Moles to Mass and Back!

 mass (m): units gramsmass (m): units grams

 Molar mass (MM) : units g/molMolar mass (MM) : units g/mol

 Moles (n) : units molMoles (n) : units mol

 Mass = moles * molar massMass = moles * molar mass

(28)

Atoms/Molecules and Grams

 Since 6.02 X 10_{Since 6.02 X 10}2323 particles = 1 mole _{particles = 1 mole}

AND

1 mole = molar mass (grams)

1 mole = molar mass (grams)

 You can convert atoms/molecules to _{You can convert atoms/molecules to}

moles and then moles to grams! (Two step

process)

 You can’t go directly from atoms to _{You can’t go directly from atoms to}

grams!!!! You MUST go through MOLES.

grams!!!! You MUST go through MOLES.

 That’s like asking 2 dozen cookies weigh _{That’s like asking 2 dozen cookies weigh}

how many ounces if 1 cookie weighs 4 oz?

You have to convert to dozen first!

(29)

÷÷ molar mass molar mass x x by 6 x 10by 6 x 102323

Grams

Grams MolesMoles particles particles

xx molar mass molar mass ÷ ÷ by 6 x 10by 6 x 102323

Everything must go through

Moles!!!

Calculations

(30)

Atoms/Molecules and Grams

How many atoms of Cu are present

How many atoms of Cu are present

in 35.4 g of Cu?

35.4 g Cu 1 mol Cu 6.02 X 1023 atoms Cu 63.5 g Cu 1 mol Cu

(31)

Moles to Mass and Back!

 1. Calculate the mass of 0.500 moles 1. Calculate the mass of 0.500 moles of calcium chloride.

of calcium chloride.  n = 0.500moln = 0.500mol

 MM=110.98g/mol MM=110.98g/mol

 m= n* MM = 0.500mol*110.98g/molm= n* MM = 0.500mol*110.98g/mol

(32)

Moles, Mass and

Particles



m = n* MM

and

N = n * N

A A

Combine the two equations

(33)

Moles, Mass and

Particles



Find the number of atoms in 8.30g of

oxygen gas (O

₂₂

).



2 atoms of oxygen per molecule



m= 8.30g n= m/MM N = n*N

A

*2



N= (8.30g/32.00g)*

6.022 x 10

2323

mol

-1-1

* 2

(34)

Learning Check!

How many atoms of K are present in 78.4

How many atoms of K are present in 78.4

g of K?

(35)

Learning Check!

What is the mass (in grams) of 1.20 X 10

What is the mass (in grams) of 1.20 X 102424 molecules of glucose (C

(36)

Learning Check!

How many

How many atomsatoms of O are present in 78.1 g of O are present in 78.1 g of oxygen?

of oxygen?

78.1 g O₂ 1 mol O₂ 6.02 X 1023 molecules O

2 2 atoms O

(37)

Moles, Mass and

Particles

 Pg 186 # 29Pg 186 # 29

 n = 55.6 mol of watern = 55.6 mol of water

 MM= 18.02g/molMM= 18.02g/mol

 Mass= n* MM= 18.02g/mol * Mass= n* MM= 18.02g/mol * 55.6mol

55.6mol

(38)

Moles, Mass and

Particles

 Pg 186 # 30Pg 186 # 30

 MM of styrene = 104.16g/molMM of styrene = 104.16g/mol

 n= 255moln= 255mol

 m= 255mol*104.16g/mol= 2.66 m= 255mol*104.16g/mol= 2.66 x10

x1044g_g

 How many atoms of C are there in How many atoms of C are there in 0.800kg of styrene?

(39)

Moles, Mass and

Particles

 How many atoms of C are there in How many atoms of C are there in 0.800kg of styrene?

0.800kg of styrene?  N = m/MM * NN = m/MM * N_A_A *8 *8

 N= (800g/104.16g/mol)*NN= (800g/104.16g/mol)*N

A

A*8*8

(40)

Law of Definite

Proportions

 The Law of Definite Composition The Law of Definite Composition

states: The masses of the elements states: The masses of the elements

are always present in the same are always present in the same

proportion by mass in a compound. proportion by mass in a compound.

(41)

Molecular and Empirical

Formulae

 Molecular Formula-Molecular Formula-_{represents the}_{represents the}

actual # of atoms of elements in a unit of actual # of atoms of elements in a unit of

the compound the compound

 E.g HE.g H₂₂OO₂₂- 2 hydrogen, 2 oxygen atoms- 2 hydrogen, 2 oxygen atoms

 Empirical Formula-Empirical Formula-_{the simplest whole}_{the simplest whole}

number ratio of elements in a compound number ratio of elements in a compound

(42)

Molecular and Empirical Formulae

Compound

Compound Molecular Molecular Formula Formula Empirical Empirical Formula Formula Glucose

Glucose CC₆₆HH₁₂₁₂OO₆₆ CHCH₂₂OO

Octane

Octane CC₈₈HH₁₈₁₈ CC₄₄HH₉₉

Copper(ii)

chloride

CuCl

(43)

Calculating Empirical

formulas

A) Using % composition

 1. Convert % element into mass of element 1. Convert % element into mass of element

by assuming a 100 g sample mass by assuming a 100 g sample mass

 2. Find the # of moles of each element.2. Find the # of moles of each element.  3. Express the mole ratio between the 3. Express the mole ratio between the

elements in terms of small whole numbers. elements in terms of small whole numbers.

(i.e. Divide by the smallest mole value) (i.e. Divide by the smallest mole value)

 Pg 209 #9-12, pg 210#13-16, Pg 209 #9-12, pg 210#13-16,

(44)

Calculating Empirical

formulas

 The chlorofluorocarbon Freon-12 is The chlorofluorocarbon Freon-12 is 9.90% carbon, 58.6% chlorine, and 9.90% carbon, 58.6% chlorine, and

31.5% fluorine by mass. What is the 31.5% fluorine by mass. What is the

empirical formula of freon-12? empirical formula of freon-12?  Step 1: Step 1:

 C: 9.90% X 100.0 g = 9.90 g C: 9.90% X 100.0 g = 9.90 g

 Cl: 58.6% X 100.0 g = 58.6 gCl: 58.6% X 100.0 g = 58.6 g

(45)

Step 2: Find # moles

 Step 2: Divide by the mole mass of each Step 2: Divide by the mole mass of each

element

 C: 9.90 g / (12.0 g/mol) = 0.825 moles of C: 9.90 g / (12.0 g/mol) = 0.825 moles of

C

 Cl: 58.6 g/(35.5 g/mol) = 1.65 moles of Cl: 58.6 g/(35.5 g/mol) = 1.65 moles of

Cl

(46)

Step 3: Find the mole

ratio

 C: 0.825 mol/0.825 mol = 1.00 C: 0.825 mol/0.825 mol = 1.00

 Cl: 1.65 mol/ 0.825 mol = 2.00Cl: 1.65 mol/ 0.825 mol = 2.00

 F: 1.66 mol/ 0.825 mol = 2.01 F: 1.66 mol/ 0.825 mol = 2.01

Ratio: 1C: 2Cl:1F Ratio: 1C: 2Cl:1F

(47)

Calculating the

Molecular Formula

 1. Convert % to mass1. Convert % to mass  2. Find # moles 2. Find # moles

 3. Find the empirical formula3. Find the empirical formula

 4. Find the molar mass of the empirical formula_{4. Find the molar mass of the empirical formula}  5. Find the multiplier _{5. Find the multiplier}

Multiplier = molecular mass/empirical massMultiplier = molecular mass/empirical mass

(48)

Molecular Formulae of

Hydrates

 Find the # moles of the anhydrous saltFind the # moles of the anhydrous salt

 Find the # of moles of waterFind the # of moles of water

 Find the ratio of the moles of salt to Find the ratio of the moles of salt to moles of water

moles of water

 Ratio: moles of water/moles of saltRatio: moles of water/moles of salt

 The ratio number is the amount of The ratio number is the amount of water

water

(49)

Molecular Formulae of Hydrates

 A hydrated compound contains 76.2 % LaI_{A hydrated compound contains 76.2 % LaI}

3

and 23.8% water. Find the molecular formula

of the compound

of the compound  76.2g of LaI_{76.2g of LaI}

3

3 and 23.8g water and 23.8g water

 Mol LaIMol LaI

3

3 = 76.2g/520 = 76.2g/520 g/molg/mol = 0.147mol = 0.147mol

 Mol water = 23.8g/18.02Mol water = 23.8g/18.02_g/mol_g/mol=1.32mol=1.32mol

 Ratio: 0.147/0.147= 1 1.32/0.147= 9 water Ratio: 0.147/0.147= 1 1.32/0.147= 9 water



LaI

(50)

Molecular Formulae of Hydrates

 This is your experimental data:This is your experimental data:

 Mass of watch glass and sample: 65.0gMass of watch glass and sample: 65.0g  Mass of watch glass = 15.0gMass of watch glass = 15.0g

 Mass of sample after heating = 42.2gMass of sample after heating = 42.2g  Your sample contained a hydrate of Your sample contained a hydrate of

barium hydroxide- find the molecular

formula of the hydrate.

(51)

Molecular Formulae of Hydrates

 1) Sample mass before heating1) Sample mass before heating

 Mass of hydrate= 65.0g-15.0g= 50.0gMass of hydrate= 65.0g-15.0g= 50.0g  Mass anhydrous salt= 42.2-15.0= 27.2gMass anhydrous salt= 42.2-15.0= 27.2g  Mass water=50.0g-27.2g= 22.8gMass water=50.0g-27.2g= 22.8g

 Mol water= 22.8/18.02Mol water= 22.8/18.02_g/mol_g/mol = 1.2653 mol = 1.2653 mol  Mol Ba(OH)Mol Ba(OH)₂₂ = 27.2/171.35 = 27.2/171.35g/molg/mol= 0.1587= 0.1587