The Mole• Atomic mass provides a means to count

atoms by measuring the mass of a sample• The periodic table on the inside cover of the

text gives atomic masses of the elements• The mass of an atom is called its atomic

mass • When using atomic masses, retain a

sufficient number of significant figures so the atomic mass data contributes only slightly to the uncertainty of the result

• The molar mass allows counting of molecules by mass

• The molar mass is the sum of atomic masses of the atoms in the compounds formula– For example the molar mass of water, H2O, is

twice the mass of hydrogen (1.008) plus the mass of oxygen (15.999) = 18.015

• Strictly speaking, ionic compounds do not have a “molar mass” because they don’t contain molecules

• The mass of the formula unit is called the formula mass

• Formula masses are calculated the same way as molecular masses– For example the formula mass of calcium

oxide, CaO, is the mass of calcium (40.08) plus the mass of oxygen (15.999) = 56.08

• One mole of a substance contains the same number of formula units as the number of atoms in exactly 12 g of carbon-12

• One mole of a substance has a mass in grams numerically equal to its formula mass

• The mass of one mole of a substance is also called its molar mass

• One mole of any substance contains the same number of formula units

• This number is called Avogadro’s number or constant

1 mol formula units = 6.02 x 1023 formula units

• Counting formula units by moles is no different than counting eggs by the dozen (12 eggs) or pens by the gross (144 pens)

• Avogadro’s number is huge because atoms and molecules are so small: a huge number of them are needed to make a lab-sized sample

• Avogadro’s number links moles and atoms, or moles and molecules and provides an easy way to link mass and atoms or molecules

Particles

• Atoms – elements

• Molecules – molecular compounds

• Ions – charged particles and polyatomics

• Formula units – Ionic compounds

• Photons – electromagnetic radiation

• Using water (molar mass 18.015) as an example: 1 mole H2O 6.022 x 1023 molecules H2O

1 mole H2O 18.015 g H2O

18.015 g H2O 6.022 x 1023 molecules H2O

• Within chemical compounds, moles of atoms always combine in the same ratio as the individual atoms themselves so: 1 mole H2O 2 mole H

1 mole H2O 1 mole O

• Stoichiometry is the study of the mass relationships in chemical compounds and reactions

• A common use for stoichiometry is to relate the masses of reactants needed to make a compound

• These calculations can be solved using the factor-label method and equivalence relations relating molecular masses and/or formula masses

• Example: How many grams of iron are in a 15.0 g sample of iron(III) oxide?ANALYSIS: 15.0 g Fe2O3 ? g Fe

LINKS: 1 mol Fe2O3 2 mol Fe

1 mol Fe2O3 159.7 g Fe2O3

1 mol Fe 55.85 g Fe

SOLUTION:

Fe g 5.10

OFe g 0.15

Fe mol 1Fe g 85.55

OFe mol 1Fe mol 2

OFe g 9.159OFe mol 1

32 3232

32

• The usual form for describing the relative masses of the elements in a compound is a list of percentages by mass

• This is called the percentage composition or percentage composition by mass

• The percentage by mass is the number of grams of the element in 100 g of the compound and can be calculated using:

100% element % sample wholeof masselement of mass

• Example: A sample was analyzed and found to contain 0.1417 g nitrogen and 0.4045 g oxygen. What is the percentage composition of this compound?ANALYSIS: Find sample mass and calculate %

LINKS: whole sample = 0.5462 g

SOLUTION:

% 74.06 100% O %

% 25.94 100% N %

sample g 0.5462O g 0.4045

sample g 0.5462N g 0.1417

• Hydrogen peroxide consists of molecules with the formula H2O2

• This is called the molecular formula

• The simplest formula for hydrogen peroxide is HO and is called the empirical formula

• It is possible to calculate the empirical formula for a compound from mass data

• The goal is to produce the simplest whole number mole ratio between atoms

• Example: A 2.012 g sample of a compound contains 0.522 g of nitrogen and 1.490 g of oxygen. Calculate its empirical formulaANALYSIS: We need the simplest whole number

mole ratio between nitrogen and oxygen

SOLUTION:

525.002.002 2.502 1.00

2.501.00

O g 15.999O mol 1

N g 14.01N mol 1

ON ON O N

ON O N

O mol 0.0931 O g .4901

N mol 0.0373 N g 522.0

0.03730.0931

0.03730.0373

• Empirical formulas may also be calculated indirectly

• When a compound made only from carbon, hydrogen, and oxygen burns completely in pure oxygen only carbon dioxide and water are produced

• This is called combustion

• Empirical formulas may be calculated from the analysis of combustion information

• Example: The combustion of a 5.217 g sample of a compound of C, H, and O gave 7.406 g CO2 and 4.512 g of H2O. Calculate the empirical formula of the compound. ANALYSIS: This is a multi-step problem. The

mass of oxygen is obtained by difference:

g O = 5.217 g sample – ( g C + g H )

The masses of the elements may then by used to calculate the empirical formula of the compound

SOLUTION:

OCHOHCOHC

O mol 0.1681 O g 2.690 :O

H mol 0.5009 H g 0.5049 :H

C mol 0.1683 C g 2.022 :C

g 2.690 2.527g - g 5.217 O mass

g 2.473 H and C of mass total

H g 0.5049 OH g 4.512

C g 2.022 CO g 7.406

31.0002.9801.001

O g 15.999O mol 1

H g 1.008H mol 1

C g 12.011C mol 1

OH g 18.015H g 2.0158

2

CO g 44.009C g 011.12

2

0.16810.1681

0.16810.5009

0.16810.1683

2

2

• The formula for ionic compounds is the same as the empirical formula

• For molecules, the molecular formula and empirical are usually different

• If the experimental molecular mass is available, the empirical formula can be converted into the molecular

• The molecular formula will be a common multiplier times all the coefficients in the empirical formula

• Example: The empirical formula of hydrazine is NH2, and its molecular mass is 32.0. What is its molecular formula?ANALYSIS: The molecular mass (32.0) is some

simple multiple of the mass calculated from the empirical formula (16.03)

SOLUTION:

422.0022.001

16.0332.0

HN HN

thenis formulacorrect The

2.00 multiplier

• The coefficients of a balanced chemical equation provide the mole-to-mole ratios for the substances involved in the reaction

• Whenever a problem asks you to convert between different substances, the calculation usually involves a mole-to-mole relationship

• How to detect unbalanced equations will be covered shortly

• Example: If 0.575 mole of CO2 is produced by the combustion of propane, C3H8, how many moles of oxygen are consumed? The balanced equation is:C3H8 + 5 O2 3 CO2 + 4 H2O

ANALYSIS: Relating two compounds usually requires a mole-to-mole ratio

SOLUTION:

2HC mol 1O mol 5

83 O mol 2.88 HC mol 575.083

2

• In many problems you will also need to perform one or more mole-to-mass conversions

• These types of stoichiometry problems are summarized with the flowchart:

• Example: How many grams of Al2O3 are produced when 41.5 g Al react?2Al(s) + Fe2O3(s) Al2O3(s) + 2 Fe(l)

ANALYSIS: 41.5 g Al ? g Al2O3

SOLUTION:

32OAl mol 1OAl g 102.0

Al mol 2OAl mol 1

Al g 26.98Al mol 1

3232

OAl g 78.45 Al g 41.5

OAl gramsOAl moles Al moles Al grams

32

3232

• Chemical equations provide quantitative descriptions of chemical reactions

• Conservation of mass is the basis for balancing equations

• To balance an equation:1) Write the unbalanced equation

2) Adjust the coefficients to get equal numbers of each kind of atom on both sides of the arrow

• Guidelines for Balancing Equations:1) Balance elements other than H and O first

2) Balance as a group any polyatomic ions that appears unchanged on both sides of the arrow

3) Balance separately those elements that appear somewhere by themselves

• As a general rule you should use the smallest whole-number coefficients when writing balanced chemical equations

• All reactions eventually use up a reactant and stop

• The reactant that is consumed first is called the limiting reactant because it limits the amount of product that can form

• Any reagent that is not completely consumed during the reactions is said to be in excess and is called an excess reactant

• The computed amount of product is always based on the limiting reagent

• Example: How many grams of NO can form when 30.0 g NH3 and 40.0 g O2 react according to:4 NH3 + 5 O2 4 NO + 6 H2O

ANALYSIS: This is a limiting reactant problem

SOLUTION:

form NO g 30.01 andreagent limiting theis O

NO g 30.01O g 40.0

NO g 9.52NH g 0.30

2

NO mol 1NO g 01.30

O mol 5NO mol 4

O g 00.32O mol 1

2

NO mol 1NO g 01.30

NH mol 4NO mol 4

NH g 03.17NH mol 1

3

22

2

33

3

• The amount of product isolated from a chemical reactions is almost always less than the calculated, or maximum, amount

• The actual yield is the amount of the desired product isolated

• The theoretical yield is the amount that would be recovered if no loss occurred (the calculated, maximum amount)

• The percentage yield is the actual yield as a percentage of the theoretical yield

• When working with percentage yield:– Remember they involve a measured (actual

yield) and calculated (theoretical yield) quantity– The calculation may be done in either grams or

moles – The result can never be a number larger than

100%

%100 yield percentage yield ltheoreticayield actual