TOPIC III: THE MOLE

LECTURE SLIDES

• Definition and Use

• Avogadro’s Number

• Conversion Problems

• Empirical Formula Calculations

Kotz & Treichel, Chapter 3, Sections 3.6-3.8

From PARTICULATE (“too small to touch”)

to MACROSCOPIC (amounts we can handle):

THE MOLETHE MOLE

Many different items we encounter in our daily lives come packaged in set amounts described by various“counting terms”.

Let’s consider a few of them:

Many different items we encounter in our daily lives come packaged in set amounts described by various“counting terms”.

Let’s consider a few of them:

•shoes and socks and earrings come in pairs (2),

• eggs come in dozens (12),

•pencils are wholesaled by the gross (144), •donuts and sweet corn are often sold as “the baker’s dozen” or “the farmer’s dozen” (13),

•and diet pop and beer by the 6- pack or case (24).....

Chemists deal in atoms, molecules and ions, which need to be counted and measured as well.

BUT:

The mass of one atom of the 19F isotope is 3.156X10-23 g.

The radius of a nucleus is about .001 pm and the radius of an atom about 100 pm. (1012 picometers or 1,000,000,000,000 pm = 1 m)

THEREFORE......

Chemists need their own unit for counting and weighing amounts of substances which come in particle size too tiny to be seen or weighed on any balance.

For convenience in describing amounts of atoms, molecules, and ions , chemists have a unique unit of measure,

THE MOLE

The MOLE

•The chemist’s counting number

•Comes from the Latin meaning “whole heap or pile of”

•SI base unit for measuring amount of substance

•Defined as the number of atoms in exactly 12 grams of 12C, 6.022 X 1023

1 Mole always contains Avogadro’s number of particles or units:

1 Mole = Avogadro’s number of particles

= 6.022136736 X 1023 particles

= 602,213,673,600,000,000,000,000 particles

If one used A’s number to describe macroscopic objects, one would be overwhelmed:

I mole of green peas would cover the entire United States to a depth of 3 miles!

Grams, amu’s, What’s the Difference?

One mole is defined as the number of particlesin exactly 12 g of the 12C isotope of carbon.

Carbon was used as the standard for the amu scale, where the mass of one atom of 12C was defined as 12 amu.

The mole answers the question: “How many atomswould you have if you took the amu scale (which describesmass of one atom) and use it as grams instead?

So, How to Get a Mole:

•We consult the periodic table, obtain the atomic mass of an element in amu’s, the relative mass of one atom.....

•We weigh this amount out in grams.....

•We now have one mole of atoms, A’s number, 6.02 X 10 23 atoms, a convenient “package of atoms”.....

•We have gone into the chemist’s counting system and can deliver not a dozen eggs but a mole of atoms....

This system works because of the relative nature of the atomic mass unit scale, in which all atoms were assigned a mass relative to 12C, the mole standard:

“One mole is the number of atoms in exactly 12 g of 12C”

The mole “pile or heap” of atoms for each element will weigh more or less than the mole “pile” for carbon, depending on whether the individual atoms weigh more or less than carbon.

If one mole of carbon atoms weighs 12.0 g, then one mole of oxygen atoms, which weighs 1.33 times more than carbon, would be: 1.33 X 12.0 g =16.0 g = 1 mol O

Since H atom is 1/12 the mass of a carbon atom, the matching pile of hydrogen atoms would be

1/12 X 12.0 g = 1.00 g = 1 mol H

•For any element, the molar mass, M, is the mass in grams of a mole of atoms, “#g/mol”

•M , molar mass, is NUMERICALLY equal to the mass of one atom in amu’s as given on your PT.

•If, however, one weighs out the molar mass, one has 6.022 x 10 23 atoms every time

The “Molar Mass, M”

MOLES of ATOMS: the MOLAR MASS, M

1 atom 1 mole, A’s # atoms

Li, 6.941 amu Li, 6.941 g

Pb, 207.2 amu Pb, 207.2 g

Zn, 65.39 amu Zn, 65.39 g

Cr

51.9961

24

atomic mass, one atom,atomic mass units,relative to C

molar mass, 6.022 x 1023 atoms, in grams

ONE MOLE

A's NUMBER6.02 x 1023 UNITS

M,molar massin grams

Using this knowledge, the chemist can interconvertgrams, moles, and atoms of any element.

The molar mass, “g/mol”, like density, “g/cm3”, is a convenient conversion factor: For any element:

1mole = atomic weight, grams = 6.022 X 101mole = atomic weight, grams = 6.022 X 102323 atoms atoms

Suppose you weighed out 35.89 g of aluminum metal.How many moles and how many atoms of aluminum would be contained in this sample?

35.89 g Al=_ ?____mol Al

#1

Question: 35.89 g Al = ? mol Al = ? atoms Al

Relationships: 1 mol Al = 26.98 g Al = 6.022 X 1023 atoms AlSetup and Solve: g ---> mol

35.89 g Al=_ ?____atoms Al#2

g mol atoms

35.89 g Al

26.98 g Al

1 mol Al=_ ?____mol Al = 1.330 mol Al#1

g mol

ans.

g --------> mol ---------> atoms

35.89 g Al=_ ?____atoms Al#2

26.98 g Al

1 mol Al

1 mol Al

6.022 x 1023 atoms Al

=atoms Al8.011 x 1023

ans

Group Work 3.1

Suppose you weighed out 15.00 g each of V and Mo metal. How many moles and how many atoms of each do you now have?

15.00 g V = ? Mol =? Atoms

15.00 g Mo= ? Mol =? Atoms

What would 9.00 x 1024 atoms of mercury weigh in grams?

Question: 9.00 x 1024 atoms Hg = ? g HgRelationships: 1 mol Hg = 200.59 g Hg = 6.022 X 1023 atoms HgSetup and Solve: atoms -----> mol -----> g

= 9.00 x 1024

atoms Hg g Hg

6.022 x 1023 atoms Hg

=

9.00 x 1024 atoms Hg 1 mol Hg

1 mol Hg

200.59 g Hg

g Hg2998

= 2.998 X 103 = 3.00 x 103 g Hg

atoms mol g

Mercury is a liquid metal with a density of 13.534 g/cm3. If you measured out 75.0 mL of Hg into a graduated cylinder, how many atoms of Hg would be in the sample?

75.0 mL Hg= ? atoms Hg

Question: 75.0 mL Hg = ? Atoms HgRelationships: 13.534 g Hg = 1cm3 or mL Hg 200.59 g Hg = 1 mol Hg 1 mol Hg = 6.022 x 1023 atoms HgSetup and solve: mL---> g ---> mol --->atoms

GROUP WORK 3.2

Molecules, Compounds, and the Mole

Let us now extend the use of molar mass, M, to include all particles chemists need to measure: not just atoms but also especially ions and molecules....

The basic principle is this: whenever you weigh out the“formula weight” of any substance or species in grams, you have A’s number of particles of that species, and the molar mass of that species...

Molar Mass of Molecules

The formula of any molecule describes the number of atoms making up one unit of that molecule:

Br2 The diatomic bromine molecule, as bromine is found in nature: the formula tells us that 2 atoms of bromine are contained in every molecule.

By extension, 2 moles of bromine atoms are contained in every 1 mole of bromine molecules. The calculation of the molar mass of molecular bromine then looks like this:

The atomic weight of Br, from the PT, is 79.904 amu’s.

Therefore:

2 moles of Br = 2 X 79.904 g = 159.808 g

And the molar mass, M, of Br2 is 159.808 g/mol

Now let’s try the molar mass of CH3CH2OH, ethylalcohol:

Molar Mass, M, CH3CH2OH

Element # of atoms M, g/mol total

C 2 12.01 24.02

H 6 1.008 6.048

O 1 16.00 16.00

Total 46.068

MM, CH, CH33CHCH22OH, 46.07 g/molOH, 46.07 g/mol

Molar Mass of Ionic Compounds

The formula of an ionic compound indicates the simplest ratio of ions present in any sample of the compound. It is this “formula unit” that we use for calculating the molar mass.

Actually, we needn’t ask what kind of compound we are getting the M for; we simply calculate for all atoms found in the formula of any species!

Cl2 (2 Cl) Fe(CN)2 (1Fe 2C 2N) (NH4)2CO3 (2N 8H 1C 3O)

M, CH3CH2OH, 46.07 g/mol, use in problems:

Given a Given a massmass, or , or volumevolume and and densitydensity, , solve for:solve for:

a) moles of compound or individual atoms

b) grams of individual atoms

c) number of molecules or atoms

M, CH3CH2OH, 46.07 g/mol, use in problems:

Given a Given a massmass, or , or volumevolume and and densitydensity, , solve for:solve for:

a) moles of compound or individual atoms

b) grams of individual atoms

c) number of molecules or atoms

How many moles of ethyl alcohol are contained in a sample that weighs 33.95 g? (CH3CH2OH, 46.07 g/mol).

Question: 33.95 g CH3CH2OH = ? mol CH3CH2OH

Relationship: 46.07 g CH3CH2OH = 1 mol

Setup and Solve: ( g ---> mol)

33.95 g CH3CH2OH= ? mol CH3CH2OH

g ----------> mol

33.95 g CH3CH2OH = ? mol CH3CH2OH

46.07 g

1 mol

= .7369 mol CH3CH2OH

1 molecule CH3CH2OH,

2 C atoms

contains:

6 H atoms

1 O atom

C C

H

H

H H

H

O

H

1 mole CH3CH2OH,

2 mole C atoms

contains:

6 moles H atoms 1 mole O atoms

6.022x1023 molecules of CH3CH2OH

2 X 6.022x1023

Carbon atoms

6 X 6.022x1023

Hydrogen atoms

1 X 6.022x1023

Oxygen atoms

How many moles of hydrogen are contained in33.95 g CH3CH2OH?

Question: 33.95 g CH3CH2OH = ? mol H

Relationship: 46.07 g CH3CH2OH = 1 mol CH3CH2OH 1 mol CH3CH2OH = 6 mol H

Setup and Solve: ( g ---> mol CH3CH2OH ---> mol H)

33.95 g CH3CH2OH= ? mol H

33.95 g CH3CH2OH = ? mol H

46.07 g CH 3CH 2OH

CH 3CH 2OH

CH 3CH 2OH

1 mol 6 mol

1 mol

H

= 4.422 mol H

g mol alcohol mol H

How many grams of hydrogen are contained in33.95 g CH3CH2OH?

Question: 33.95 g CH3CH2OH = ? g H

Relationship: 46.07 g CH3CH2OH = 1 mol CH3CH2OH 1 mol CH3CH2OH = 6 mol H 1 mol H = 1.008 g HSetup and Solve: ( g CH3CH2OH ---> mol CH3CH2OH ---> mol H -----> g H)

33.95 g CH3CH2OH = ? g H

g alcohol mol alcohol mol H

33.95 g CH3CH2OH

= ? g H

46.07 g CH 3CH 2OH

CH 3CH 2OH

CH 3CH 2OH

1 mol 6 mol

1 mol

H

= 4.457 g H

g H

1 mol H

1.008 g H

How many grams of carbon are contained in 33.95 g CH3CH2OH?

Question: 33.95 g CH3CH2OH = ? g C

Relationship: 46.07 g CH3CH2OH = 1 mol CH3CH2OH 1 mol CH3CH2OH = 2 mol C 1 mol C = 12.01 g C

Setup and Solve: ( g CH3CH2OH ---> mol CH3CH2OH ---> mol C -----> g C)

Group Work 3.3

How many atoms of carbon are contained in 33.95 g CH3CH2OH?

Question: 33.95 g CH3CH2OH = ? atoms C

Relationship: 46.07 g CH3CH2OH = 1 mol CH3CH2OH 1 mol CH3CH2OH = 2 mol C 1 mol C = 6.02 x 1023 atoms C

Setup and Solve: ( g CH3CH2OH ---> mol CH3CH2OH ---> mol C -----> atoms C)

33.95 g CH3CH2OH = ? atoms C

g alcohol mol alcohol mol C atoms C

g alcohol mol alcohol mol C

33.95 g CH3CH2OH

= ? atoms C

46.07 g CH 3CH 2OH

CH 3CH 2OH

CH 3CH 2OH

1 mol 2 mol

1 mol

C

= 8.873 x 1023 atoms C

atoms C

1 mol C

6.02x1023 atoms C

What mass of ethyl alcohol, CH3CH2OH, would contain2.06 X 1024 atoms of carbon?

Question: 2.06 X 1024 atoms C = ? g CH3CH2OH

Relationship: 6.02 x 1023 atoms C = 1 mol C 2 mol C = 1 mol CH3CH2OH 1 mol CH3CH2OH = 46.07 g CH3CH2OH

Setup and Solve: (atoms C-----> mol C ---> mol CH3CH2OH ---> g CH3CH2OH )

GROUP WORK 3.4

New TOPIC:% Composition from Formula of the Compound

The formula of a compound can be used to determinethe mass % of each element by using the molar relationships we have learned...

Consider the following:

What is the % by mass of Carbon, Hydrogen and Oxygenin a sample of ethyl alcohol, CH3CH2OH?

Mass % of Elements in a Compound

This is done using the formula weight calculations andthe approach:

% by mass = total mass of one element (the part) X 100 % total mass of compound (the whole)

based on one mole of the compound

% Composition of Ethyl Alcohol, CH3CH2OH

Molar Mass Calculation:2 C = 2 X 12.011 = 24.022 g C6 H = 6 X 1.008 = 6.048 g H 1 O = 1 X 15.999 = 15.999 g O 46.069 g/mol CH3CH2OH

%C = 24.022 g C X 100 = 52.144% C 46.069 g CH3CH2OH

%H = 6.048 g H X 100 = 13.13% H 46.069 g CH3CH2OH

%O = 15.999 g O X 100 = 34.728% O 46.069 g CH3CH2OH

CHECKING.....

52.144% C 13.13 % H 34.728% O 100.002 % CH3CH2OH

= 100.00% (2 digits allowed after decimal)

Formula from % Composition

If one can go from formula to % composition,one should be able to go from % composition to the formula of the compound.

This is quite true (almost):

We can take % composition back to the “empirical formula”, which describes the simplest mole ratio ofatoms in the formula... That is not always the same thing...

Empirical Vs Molecular Formulas

Name MolecularFormula

EmpiricalFormula

“n”

Acetylene C2H2 (CH)n 2

Benzene C6H6 (CH)n 4

VinylAcetylene

C4H4 (CH)n 6

For “organic” or “molecular carbon containing compounds”, there exist a list of compounds which share almost every conceivable empirical formula. To determine which compound one has fromanalytical data, one needs the molar mass aswell, as we will see...

For most ionic compounds and simple molecularcompounds, the empirical and molecular formulasare identical.

Empirical Formula from % Composition:

Let’s take our ethyl alcohol compound back to itsformula from its percent composition.The trick is to use 100 grams of sample whenever one is calculating from mass %. Then we have:

52.144% C = 52.14 g carbon 13.13 % H = 13.13 g hydrogen 34.728% O = 34.73 g Oxygen

Our procedure will be:% element ---> g element ---> mol element ---> simplest mole ratio

Method of choice: make a chart:

element grams Molarmass

#moles Simplestmole ratio

List eachseparately

Use basedon 100 gwhen %given

Use massof 1 molof atoms

#g X 1 mol #g

(calculate and insert)

Divide allmoles bysmallest #of moles

Exp dataifprovided

“atomicweightonly”

Chart out your information in this fashion to determine formula:

grams Molarmass

Moles:#g/ M

Simplestmole ratio

C 52.14 g 12.01 4.341 4.341/2.171 =2.000

H 13.13 g 1.008 13.03 13.03/2.171= 6.002

O 34.73 g 16.00 2.171 2.171/2.171= 1.000

Moles: 52.14 g C X 1 mol C = 4.341 mol C 12.01 g C

Your empirical formula is the simplest ratio of molesof the elements in the formula, for which you have determined that:

For every 1.00 mole of O atoms, you have 6.002 mole of H atoms and 2.000 mole of C atoms

Empirical formula: C2H6O

GROUP WORK 3.5

Cumene is an organic hydrocarbon containing only C and H. It is 89.94% C and 10.06% H. What is its empirical formula?

Empirical Formula using Experimental Data

A new compound weighing 0.678 g, containing xenon and fluorine, was made from a mixture of the gases in sunlight. If the xenon showed a mass of 0.526 g, what is the empirical formula of the compound?

Note:0.678 g compound - 0.526 g Xe = .152 g F2

Although the element itself is diatomic, when combined into a compound it exists as individual atoms, and the atomic weight of F, 19.0 g/mol, is used for formula calculation, not 38.0 g/mol F2.

Grams M Moles: #g/ M Simplest mole ratio

Xe .526 131.29 .00401 .00401/.00401 = 1

F .152 19.00 .00800 .00800/.00401 = 2

EMPIRICAL FORMULA: XeF2

Molecular and Empirical Formula

Nicotine, a poisonous compound found in tobaccoleaves, is 74.0% C, 8.65% H, and 17.35% N. Its molar mass is 162 g/mol.

What are the empirical and molecular formulas of this compound?

Note: When the molar mass is included in the problem, the exact molecular formula can be determined...

grams Molar mass Moles Simplest mole ratio

C 74.0 g 12.01 6.16 6.16/1.238 = 4.98 H 8.65 g 1.008 8.58 8.58/1.238 = 6.93N 17.35 g 14.01 1.238 1.238/1.238= 1.000

Empirical Formula: C5H7N

In order to obtain the molecular formula, you must divide the molar mass (mass of molecule) by the empirical formula mass (the mass of the simplest ratio of atoms...)

Empirical Formula Mass:

5C = 5 X 12.01 = 60.057H = 7 X 1.008 = 7.0561N = 1 X 14.01 = 14.01 81.12 g/mol

Molar mass = 162 = 2.00 = “n”Emp. Form. mass 81.12

(C5H7N)n = (C5H7N)2 : C10H14N2

Hydrated Compounds

Crystalline ionic solids (salts!) are frequently found innature or are produced from aqueous solutions with a specific number of water molecules associated with each set of formula ions:

CuSO4. 5H2O NiCl2.6H2O CaSO4.2H2O

Frequently the color of the salt depends on the presence of these “waters of hydration.”

Hydrated Compounds

Crystalline ionic solids (salts!) are frequently found innature or are produced from aqueous solutions with a specific number of water molecules associated with each set of formula ions:

CuSO4. 5H2O NiCl2.6H2O CaSO4.2H2O

Frequently the color of the salt depends on the presence of these “waters of hydration.”

The number of water molecules associated with a particular salt is characteristic but not easy to predict: therefore the value is determined experimentally...

The hydrated salt is weighed, heated carefully to drive off the water, and reweighed.

The mass of the water driven off is calculated, converted to moles and compared to moles of the parent, anhydrous salt to determine theformula of the hydrate...

Naturally occurring hydrated copper(II) chlorideis called eriochalcite. If 0.235 g of CuCl2.xH2O is heatedto drive off the water, 0.185 g residue remains. What is the value of x?

0.235 g hydrate - 0.185 g parent salt = 0.050 g H2O

1Cu= 1 X 63.55 = 63.552Cl = 2 X 35.45 =70.90 134.45 g/mol CuCl2

2H= 2 x 1.008= 2.0161O= 1 x 16.00= 16.00 18.02 g/mol H2O

grams Molar mass Moles Simplest mole ratio

CuCl2 0.185 g 134.45g/mol .00138 .00138/.00138= 1H2O 0.050 g 18.02 g/mol .00278 .00278/.00138= 2

Formula of eriochalcite: CuCl2. 2H2O

Note: The correct answer will always be a whole number, with one mole of parent compound and(usually) several moles of water....

GROUP WORK 3.6

If 1.023 g of a hydrated compound, CuSO4. xH2Oshows a mass of 0.654 g when dehydrated, what is the formula of the compound? (CuSO4, 159.6 g/mol;H2O, 18.02 g/mol).