The Mole

Chemical formulas are not simply abbreviations for words. They represent precise quantities.

What 2 “bits” of information do formulas give?

Water for example is H2O.

It indicates that 1 molecule of water consists of 2 hydrogen atoms and 1 oxygen atom.

How could you represent 2 molecules of water?

Use a coefficient

This formula shows 4 hydrogen atoms and 2 oxygen atoms.

•Elements•Number of atoms of each element

2 H2O

In daily life, pieces of matter are often measured either by counting them or by massing them; the choice is determined by convenience. If you buy eggs, you buy them by the dozen—that is, by number. Eggs are easy to count out. So are oranges and lemons.

Other items, though countable, are more conveniently sold by mass. A dozen peanuts is too small a number to buy, and several hundred are too difficult to count. You buy peanuts by the pound or kilogram—that is, by mass.

Chemists, too, are interested in quantity—the quantity of an element or compound, which, like grocery items, can be measured by number or by mass. Although you can easily mass a sample of a substance, yet the number of atoms or molecules in it is much too large to count. Nevertheless, chemists are interested in knowing such numbers.

In the laboratory, when we “run” a reaction we want to know the number of atoms, molecules, or formula units in a substance because these are the entities that react with each other. However, these entities are much too small to count individually, so chemists use a unit called the mole to count by massing them.

The mole was derived from the Latin word moles, meaning “heap” or “pile.” The mole, whose symbol, mol, is the SI base unit for measuring the amount of substance.

A mole is the amount of substance that is equal to the number of carbon atoms in exactly 12 g of the carbon-12 isotope.

One mole always contains the same number of particles, no matter what the substance.

1 mole = 6.022 x 1023 particles

This value is referred to as Avogadro’s number (Amadeo Avogadro) in honor of the man who conceived the basic idea.

One mole each of various substances - Clockwise from top left: 1-octanol (C8H17OH); mercury(II) iodide (HgI2); methanol (CH3OH); sulfur (S8).

The central relationship between the mass of one atom and the mass of one mole of atoms is that the mass of an element is expressed in amu.

The mass of one mole of atoms of an element is the same numeric value but measured in grams.

1 atom of sulfur has a mass of 32.07 amu.

1 mole of sulfur atoms has a mass of 32.07 g.

So,

1 mole S = 6.022 x 1023 atoms S = 32.07 g S

1 mole of Fe2+ = 6.022 x 1023 ions = 55.85 g Fe2+

1 mole H2O = 6.022 x 1023 molecules H2 = 18.02 g H2O

1 mole NaCl = 6.022 x 1023 formula units NaCl = 58.44 g NaCl

•Add up the individual masses of the elements.

•If we are talking about the mass of an atom we call it the atomic mass.

•If a compound is a molecule, we call this the molecular mass.

•If a compound is a formula unit, we call it the formula mass.

How can I get the mass of one unit of a compound?

g-formula mass = 1 MOLE = 6.022 x 1023 “particles”

Atoms Ions Molecules Formula Units

elements charged covalent ionic particles compounds compounds

1 moleof a

substance

= Avogadro’s constant = (6.022 x 1023)

mass of substancein grams

The MoleMatch the representative particle with the appropriate

substance.1. CCl4

2. lead (IV) bromide3. Rb4. carbon monoxide5. Sr2+

6. Ba(NO3)2

7. H2

8. CO32-

9. tungsten10.Al(CN)3

11.How many potassium atoms are contained in one mole of the element?

12.What is the mass of one mole of potassium?

AtomMolecule

IonFormula unit

6.022 x 102339.10 g

A 1-carat diamond has a mass of 0.200 g. How many moles of carbon?

How many atoms of carbon?USE FLM!!!!!!!

Start with your given, identify units of answer, and write conversion factors.

0.200 g C 1 mole C = moles C

12.01 g C

0.200 g C = moles C

0.200 g C=

0.200 g C 1 mole C = 0.0167

moles C12.01 g C

0.200 g C 1 mole C= atoms C

12.01 g C

0.200 g C= atoms C

0.200 g C 1 mole C 6.022 x 1023 atoms C= atoms C

12.01 g C 1 mole C

0.200 g C 1 mole C 6.022 x 1023 atoms C= 1.00 x 1022 atoms C12.01 g C 1 mole C

HINT: If moles of your substance does not appear in your given or in the desired units, then you must do

at least 2 conversion factors.

A ring is constructed out of 3.06 x 1022 atoms Au.

How many grams?

How many moles?

3.06 x 1022 atoms Au

1 mole Au

= moles Au6.022 x 1023 atoms

Au

3.06 x 1022 atoms Au

= moles Au

3.06 x 1022 atoms Au

1 mole Au 196.97 grams Au

= grams Au6.022 x 1023 atoms

Au1 mole Au

3.06 x 1022 atoms Au

1 mole Au= 0.0508

moles Au6.022 x 1023 atoms

Au

3.06 x 1022 atoms Au

1 mole Au

= grams Au6.022 x 1023 atoms

Au

3.06 x 1022 atoms Au

1 mole Au 196.97 grams Au =

10.0grams Au6.022 x 1023 atoms

Au1 mole Au

1 molecule of H2O2 atoms of H1 atom of O

1 mole of H2O2 moles of H1 mole of O

Counting Atoms and MoleculesA raindrop contains

about 0.050 g of water.

How many molecules of water?

How many moles of O?

How many atoms of H?

= 1.7 x 1021 molecules H2O

How many molecules of water?

How many moles of O?

= 0.0028 moles O

0.050 grams H2O

1 mole H2O

18.02 grams H2O

6.022 x 1023 molecules H2O =

1 mole H2O

0.050 grams H2O =

1 mole H2O

=18.02 grams H2O

1 mole O

=1 mole H2O

How many atoms of H?

0.050 grams H2O

1 mole H2O 2 mole H 6.022 x 1023 atoms H

=18.02 grams

H2O1 mole

H2O1 mole H

0.050 grams H2O

1 mole H2O6.022 x 1023

molecules H2O2 atoms H

=18.02 grams

H2O1 mole H2O 1 molecule

H2O

= 3.3 x 1021 atoms H

OR

YOU TRY THIS ONE

Moles of What?!?!One teaspoon of table sugar

(sucrose) is approximately 15 g of C12H22O11.

How many moles of sugar is this?

15 g C12H22O11 1 mole C12H22O11

=342.34 g C12H22O11

0.044 mole C12H22O11

How many atoms of carbon are in this 15 g sample?15 g

C12H22O11

1 mole C12H22O11 6.022 x 1023

molecules C12H22O11

12 atoms C

342.34 g C12H22O11

1 mole C12H22O11 1 molecule C12H22O11 = 3.2 x 1023 atoms

C

Moles of What?!?!What mass of CaCl2 would be needed to furnish

1.00 mole of Cl1- ions?

1.00 mole Cl1-

ions1 mole CaCl2 110.98 g CaCl2

=2 moles Cl1- ions 1 mole CaCl2

= 55.5 g CaCl2

What are we looking for as an answer?!?!

Percent CompositionCompare the mass of each element present in one mole

of a compound to the total mass of one mole of a compound.

% Composition

=Mass of Element

Mass of CompoundEXAMPLE: Find the % composition of Al2(SO4)3 Al = 2 x 26.98 =

53.96

S = 3 x 32.07 = 96.21

O = 12 x 16.00 = 192.00

342.17

= 342.17 g Al2(SO4)3

Al = x =

S = x =

O = x =

Al = 2 x =

S = 3 x =

O = 12 x =

Al = 2 x 26.98 =

S = 3 x 32.07 =

O = 12 x 16.00 =

Al = 2 x 26.98 = 53.96

S = 3 x 32.07 = 96.21

O = 12 x 16.00 = 192.00

Check the sum of the %’s. They should sum to very near

Percent CompositionEXAMPLE: Find the % composition of Al2(SO4)3Al = 2 x 26.98 =

53.96

S = 3 x 32.07 = 96.21

O = 12 x 16.00 = 192.00

342.17

% Al =

x 100%

=

% S =

x 100%

=

% O =

x 100%

=

53.96

342.1715.77 % Al

96.2128.12 % S

342.17

192.0056.11 % O

342.17

100.00 %

EXAMPLE: Find the mass percent (% composition) of Ferrous nitrate

Fe = x =

N = x =

O = x =

% Fe =

x 100%

=

% N =

x 100%

=

% O =

x 100%

=

55.85

179.8731.05 % Fe

28.0215.58 % N

179.87

96.0053.37% O

179.87100.00 %

What is the formula for ferrous nitrate?Fe(NO3)2

Fe = 1 x =

N = 2 x =

O = 6 x =

Fe = x 55.85 =

N = x 14.01 =

O = x 16.00 =

55.85

28.02

96.00

179.87

= 179.87 g Fe(NO3)2

If you had a 100.0 g sample of ferrous nitrate, how many grams of iron are there?

31.05 g Fe

Nitrogen? 15.58 g NIf you had a 540.0 g sample of ferrous nitrate, how many grams of iron are there? 540.0 x (.3105 % Fe)

= 167.7 g Fe540.0 g

Fe(NO3)2 =31.05 g Fe

100 g Fe(NO3)2

Nitrogen? 540.0 x (.1558 % N) =

540.0 g Fe(NO3)2 =

15.58 g N

100 g Fe(NO3)2

84.13 g N

Which toothpaste component

has a higher % by mass of fluoride?CaF2 NaF

Calcium fluoride 48.67% F, while sodium fluoride 45.25% F

CaF2

Empirical FormulasWe can also do the reverse and use percent composition data to determine empirical and molecular formulae.

Empirical formulas simplest whole number ratio of atoms in a compound.

Molecular formulas show the actual numbers of atoms of each element in a compound.

REMEMBER THE TUNE:Percent to mass

Mass to mole

Divide by the smallest

Multiply ‘til whole

There are two ways to determine empirical formulas.

Experimental Data % composition data

Need to know mass of Assume the amounteach element in the to be 100 glaboratory sample

EXAMPLE:

40.0% C 6.71% H 53.3% O

1. Assume 100 g sample

40.0 g C 6.71 g H 53.3 g O

2. Convert to moles40.0 g C 1 mole C

=12.01 g C

6.71 g H

1 mole H=

1.01 g H

53.3 g O

1 mole O=

16.00 g O

= 3.33 moles C

= 6.64356 moles H

= 3.33 moles O

= 3.33 moles C

= 6.64356 moles H

= 3.33 moles O

3.Find the mole ratio (divide by the smallest number of

moles.)

3.33 moles C =

3.33 moles

6.64356 moles H =

3.33 moles

3.33 moles O =

3.33 moles = 1 C = 2 H = 1 O

If the mole ratio ends in

.20 (x5) .25 (x4) .33 or .66 (x3) .50 (2)

Otherwise round to one of the above decimals or to the nearest whole number.4. Empirical Formula CH2O or CHOH

EXAMPLE: A binary compound consisting only of

nitrogen and oxygen with a mass percent of 37.00% N

37.00 g N

1 mole N=

14.01 g N

63.00g O

1 mole O=

16.00 g O

2.6409707 mole N=

3.9375 mole O=

2.6409707 mole N=

2.6409707 mole

3.9375 mole O=

2.6409707 mole

= 1 N = 1.490929

O

= 1 N = 1.490929

O multiply by 2 to eliminate and make whole number ratios.

N2O3

EXAMPLE: 37.70% Na, 22.95% Si, 39.34% O

37.70 g Na

1 mole Na

=22.99 g

Na

22.95 g Si

1 mole Si=

28.09 g Si

39.34g O

1 mole O=

16.00 g O1.63984 mol

Na =0.8170167 mol

Si =2.45875 mol

O =

= 2.007 Na = 1 Si = 3.009 O

1.63984 mol Na

=0.8170167

mol

0.8170167 mol Si =

0.8170167 mol

2.45875 mol O

=0.8170167

mol

Na2SiO3

Molecular FormulasMolecular formulas show the actual number of atoms of each element in a molecule, as well as, the ratio of atoms.

In order to solve these types of problems, 3 pieces of information are needed.

• molecular mass of the compound

• empirical mass of the compound• number of empirical units present in the molecular formula

EXAMPLE: What is the formula of an unknown

substance with a percent composition

of 5.90% H and 94.10% O. It also has a

molecular mass of 34.00 g/mol.5.90% H and 94.10% Omeans5.90 g H and 94.10 g O

5.90 g H

1 mole H=

1.01 g H

94.10 g O

1 mole O=

16.00 g O5.841584 mole H = 5.88125 mole O =

5.841584 mole H=

5.841584 mole

5.88125 mole O=

5.841584 mole

= 1.00 H = 1.006790 O

Empirical Formula is: HO

The empirical mass is 17.01 g (1.01 + 16.00)

We need to determine how much bigger the molecular formula is compared to the empirical.

34.00=

17.01

Molecular mass is given in the problem

34.00= 2

17.01

Molecular formula is 2 times the empirical

Empirical Formula is: HOMolecular Formula is: H2O2

HydratesHydrates are crystals that contain water molecules in their crystal structure.

The crystal has crystallized from a water solution with molecules of water adhering to the particles of the crystal.

EXAMPLE: NiSO3 . 6 H2O

1 crystal of NiSO3 with 6 molecules of H2O

To calculate the formula of a hydrate, we need:

• mass of the hydrated sample

• mass of dry sample (anhydrous, w/o water) after the H2O has been driven off.

• Mole ratio of the anhydrous compound to the H2O driven off.

Finding the formula of a hydrate is very similar to finding empirical formulas.

EXAMPLE: We have a 10.407 g sample ofhydrated barium iodide. The

sample is heated to drive off the water. The

dry sample has a mass of 9.520 g. What is the formula of the

hydrate?The difference between the initial mass and that of the dry sample is the mass of water that was driven off.

 Mass of hydrate 10.407 g

Mass of dry sample

- 9.520 g

Mass of water 0.887 g

9.520 g BaI2

1 mole BaI2

=391.13 g

BaI2

0.887 g H2O

1 mole H2O

=18.02 g

H2O0.02434 moles

BaI2

= 0.0492 moles H2O =

0.02434 moles BaI2 =

0.02434 moles

0.0492 moles H2O =0.02434 moles

= 1 BaI2= 2 H2O

Formula: BaI2 . 2 H2O