THE MOLE
Chapter 11
Writing and balancing chemical equations
• Chemical equation: identities and
quantities of substances involved in
chemical/physical change
– Balance using Law of Conservation of
Mass and Law of Definite Composition
• same # of atoms on each side
• fixed ratio of elements in compound
Writing and balancing chemical equations
1. translate statement: reactants  products
2. balance atoms using stoichiometric coefficients
3. Adjust stoichiometric coefficients (if necessary)
– Smallest whole # preferred
4. Check
5. Specify states of matter
Samples
• Stoichiometry: study of quantitative aspects of
chemical formulas/reactions
• Mole: unit chemists use to count chemical entities by
weighing them
11.1 Measuring Matter
• The Mole
– SI Unit for the amount of a substance
– The number of particles equal to the number of
atoms in exactly 12.0 grams of carbon-12
– Also called Avogadro’s number
• 1 mol = 6.02 x 1023 particles
The mole
mole - amount of substance that contains same # of
entities as atoms in 12g of carbon-12.
1 mol contains 6.022x1023 entities
Avogadro’s number (N)
1 mole H2O contains 6.022 x 1023 H2O molecules
1 mole KNO3 contains 6.022x1023 KNO3 formula units
1 mole Hg contains 6.022x1023 Hg atoms
Mole represents large
quantity of microscopic particles.
The mole
•Molar mass (M) (gmw)- mass of 1 mole of entities
–M (g/mol) numerically equal to formula weight (amu)
CH4 = 1(12.10 g/mol) + 4(1.008 g/mol) = 16.04 g/mol
Substance
CaCO3
O2
H2O
Copper
mass of 1
atom (molecule)
100.09 amu
32.00 amu
18.02 amu
63.55 amu
mass of 1
mole of atoms (molecules)
100.09 g
32.00 g
6.022 x 1023 entities
18.02 g
63.55 g
Can weigh out grams
using scale
Use mass to ‘count’
entities.
Relating moles to chemical formulas
Glucose C6H12O6 ( M = 180.16 g/mol)
Table 3.2
Carbon (C)
Hydrogen (H)
Oxygen (O)
Atoms/molecule
of compound
6 atoms
12 atoms
6 atoms
Moles of atoms/
mole of compound
6 moles
of atoms
12 moles
of atoms
6 moles
of atoms
Atoms/mole of
compound
6(6.022 x 1023)
atoms
12(6.022 x 1023)
atoms
6(6.022 x 1023)
atoms
Mass/molecule
of compound
6(12.01 amu)
= 72.06 amu
12(1.008 amu)
= 12.10 amu
6(16.00 amu)
= 96.00 amu
12.10 g
96.00 g
Mass/mole of
compound
72.06 g
180.16 g/mole
Interconverting Moles, Mass, and # of Chemical Entities
gmw
Mass (g) = no. of moles x
1 mol
No. of entities = no. of moles x
6.022x1023 entities
1 mol
Mass Percent from Chemical Formula
Mass % of element X =
moles X in formula (molar mass of X)
1 mol compound
x 100%
molar mass of compound
i.e. Mass % of H in H2O = 2 mol H
1.008 g H
1 mol H2O
1 mol H
18.02 g H2O
1 mol H2O
= 11.19% H by mass
x 100%
Particles
• Atoms
– Single elements
• Formula units
– Ionically bonded compounds
• Molecules
– Covalent bonded compounds
(6.02 x 1023
molecules)
(6.02 x 1023
molecules)
macro
micro
Figure 3.6
Avogadro’s # takes us to/from
macroscopic/microscopic
Law of Conservation of Mass
(1.20 x 1024
molecules)
Particle  Mole Problems
• REMEMBER
– The number of particles in 1 mole of ANY substance is
ALWAYS the same (6.02 x 1023)
1 mol = 6.02 x 1023 particles
– How many molecules are in 2.2 moles of water?
1 mol H2O = 6.02 x 1023 particles
Use this as your conversion factor!!!
1 mol H2O
6.02 x 1023 molecules
OR
6.02 x 1023 molecules
1 mol H2O
Calculating amounts of reactant and product
• Balanced equation needed for stoichiometric
calculations
– Ratios of reactants/products to calculate
amounts of reactants/products
Particle  Mole Practice Continued
Take your given value
and put it over ONE.
Multiply by the conversion factor that
allows you to cancel out the top unit.
Leaving you with the unit you WANTED.
2.2 mol H2O
6.02 x 1023 molecules
24
=
1.3244
x
10
X
1
1 mol H2O
ANSWER: 1.3 x 1024 molecules of H2O
How many moles of sodium carbonate contain
7.9 x 1024 formula units?
11.2 Mass and the Mole
• Atomic mass
– The mass of an atom relative to the mass assigned to carbon-12
• Molar mass
– The mass in grams of one mole of any pure substance
– Use the average atomic mass off the periodic table
• Molar mass of an element
– Atomic mass in grams per mole (g/mol)
• Molar mass of oxygen
= 16.00 g/moL
• Molar mass of helium
= 4.00 g/moL
Calculating amounts of reactant and product
• Calculate moles of O2 consumed when 10 mol of
H2O are produced (using balanced equation
from Table 3.5)?
– C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
• Calculate mass of CO2 produced burning 1.00 g
butane (C4H10).
– 2C4H10(l) + 13O2(g)  8CO2(g) + 10H2O(l)
• Practice, practice, practice!!!
Mass & Mole Problems
• How many moles are in 82.2g of aluminum?
1 mole of Al = 26.98g Al
Mass & Mole Problems
• How many grams are in 3.5 mol of neon?
1 mole of Ne = 20.18g Ne
11.3 Moles of Compounds
• Formula mass – the sum of the atomic
masses of all the atoms in a compound
H2O
H: 2 x 1.01amu = 2.02 amu
O: 1 x 16.00amu = 16.00 amu
18.02 amu
Chemical rxts. that involve limiting reagents
• Limiting reagent- reactant that forms fewer
moles of product
– Limits amount of product formed (chair
analogy)
– To decide which reagent is limiting reagent:
1. If information given in g, convert to moles
2. Use ratio from balanced equation to find
moles of final product possibly produced
3. Reagent that produces least possible
moles
of product is limiting reagent
What mass of NH3 is produced from the rxt. of 1.00 g H2(g) w/
1.00 g N2(g)?
(Info on both reactants given)
3H2(g) + N2(g)  2NH3(g)
Chemical rxts. that involve limiting reagents
Microscopic Picture
• Is H2 or O2 the limiting reagent?
– H2
Moles of Compounds Continued
• Molar mass
– The mass in grams of 1 mole of a substance
– Formula mass of H2O = 18.02 amu
– Molar mass of H2O = 18.02g/mol
1 mol of H2O = 18.02 g = 6.02 x 1023 molecules
Mass  Mole Problems
•
Changing the mass to moles or vice versa using
the molar mass
•
1.
How many moles are in 11.2g of NaCl?
Determine the molar mass.
- Na: 1 x 23.00 = 23.00g
- Cl: 1 x 35.45 = 35.45g
= 58.45g/mol
Therefore, 1 mol NaCl = 58.45g
Mass  Mole Problems
2. Convert between the molar mass and the moles.
1 mol of NaCl = 58.45g
1 mol of NaCl
58.45g NaCl
Multiply by the conversion factor that
allows you to cancel out the top unit.
Leaving you with the unit you WANTED.
Take your given value
and put it over ONE.
11.2g NaCl
1
OR
58.45g NaCl
1 mol of NaCl
X
1 mol NaCl
58.45g NaCl
=0.1916167665 mol NaCl
ANSWER: 0.192 mol NaCl
More Practice
• What is the mass of 2.50 mol of NaCl?
– Find the molar mass…
1 mol of NaCl = 58.45g
1 mol of NaCl
58.45g NaCl
Take your given value
and put it over ONE.
2.50 mol NaCl
X
1
OR
58.45g NaCl
1 mol of NaCl
Multiply by the conversion factor that
allows you to cancel out the top unit.
Leaving you with the unit you WANTED.
58.45g NaCl
1 mol NaCl
= 146.125g NaCl
ANSWER: 146.13g NaCl
Multi-Step Conversions
• Mass-Particle
g  mol  particles
• Particle-Mass
Particles mol  g
• What is the mass of 8.2 x 1022 atoms of calcium?
1 mol Ca = 6.02 x 1023 atoms Ca
1 mol Ca = 40.08g Ca
Mole Volume Problems
• Equal volumes of gases at the same
temperature and pressure contain the same
number of particles.
• Molar volume
– The volume of 1 mol of gas at standard conditions (STP)
– STP
• standard temperature and pressure: 0oC and 1 atm
– 1 mol = 22.4 liters
Mole  Volume Practice Problem
• What is the volume of 0.35 moles of helium
gas at STP?
Take your given value
and put it over ONE.
0.35 mol He
X
1
Multiply by the conversion factor that
allows you to cancel out the top unit.
Leaving you with the unit you WANTED.
22.4 L
1 mol He
= 7.84L He
ANSWER: 7.84L He
Percent Composition
• Percent Composition
– The percent by mass of each element in a
compound
mass of element
mass of compound
X 100 = % composition
Percent Composition Problem
• Calculate the percent composition of hydrogen in
water
H2O
H: 2 x 1.01amu = 2.02 amu
O: 1 x 16.00amu = 16.00 amu
18.02 amu