Lecture 5
The Mole
Stoichiometry
Limiting Reactant
% Yield
Avogadro’s Number
6.022 x 1023 mol-1
The Mole
1
mol Ne atoms = 6.022 x 1023 Ne atoms = 20.18 g
 1 mol Cl- ions = 6.022 x 1023 Cl- ions = 35.45 g
 The mole is the fundamental unit for counting
particles on a microscopic level.
 It allows us to draw quantitative connections
between macroscopic and atomic level
chemical processes.
Molar Mass of CO2
Molecular Weight = 12.01 amu + 2(16.00 amu)
= 44.01 amu/CO2 molecule
= 44.01 g/mol CO2
 1 mol of CO2 molecules = 6.022 x 1023 CO2 molecules
=44.01 g

Ex) Converting Grams to Moles
 Ex)
How many moles of NaCl are there in
48 g of NaCl?
48 g NaCl x
1 mol
58.44 g
= 0.82 mol NaCl
Ex) Converting Grams to Moles
 Ex)
How many grams of C2H6 are there in
18.7 moles of C2H6?
18.7 mol x
30.08 g
1 mol
= 562 g
Ex) Converting from Grams to
Molecules
 Ex)
How many water molecules are contained
within a 56 g pure sample of water?
56 g x
18.02 g
1 mol
x
6.022x1023
molecules
1 mol
= 1.9 x 1024 molecules of H2O
Ex) Converting from Grams to
Atoms
 Ex)
How many hydrogen atoms are contained
within a 56 g pure sample of water?
56 g x
1 mol
18.02 g
x
6.022x1023
molecules
1 mol
x 2 = 3.74 x 1024 H atoms
Conservation of Atoms in
Chemical Reactions
O
H
H
C
H
1
1
O
H
O
O
2
O
O
C
O
H
H
H
H
CH4 + 2O2  CO2 + 2H2O
2
O
Conservation of Atoms in
Chemical Reactions
O
H
H
C
1
O
H
H
O
O
2
O
C: 12.01 amu = 12.01 amu
H: 4(1.01) amu = 4.04 amu
O: 4(16.00) amu = 64.00 amu
80.05 amu
O
C
O
H
H
H
2
O
H
C: 12.01 amu = 12.01 amu
H: 4(1.01) amu = 4.04 amu
O: 4(16.00) amu = 64.00 amu
80.05 amu
Ex1) Predicting Mass of
Products
 Ex)
What mass of water is produced when
a car burns 246.4 g of methane?
CH4 + 2O2  CO2 + 2H2O
mol CH x 2 mol H O x 18.02 g H O = 553.7 g H O
246.4 g x 116.05
2
1 mol CH
1 mol H O
g CH
4
2
4
2
4
2
Ex2) Predicting Mass of
Products
 How
many grams of CO2 and Fe are produced
when 114 g of carbon monoxide gas is added
to a vessel containing excess hot iron (III) oxide

Step 1. Write a balanced chemical equation
3CO + Fe2O3  3CO2 + 2Fe

Step 2. Find masses of CO2 and Fe
CO2: 114 g x
1 mol CO
28.01 g
CO
x
3 mol CO2
3 mol CO
1 mol CO x 2 mol Fe x
Fe: 114 g x 28.01
g CO 3 mol CO
g CO
x 44.01
1 mol CO2 = 179.12 g CO2
55.85 g Fe
1 mol Fe
2
= 151.5 g Fe
Ex3) Predicting Mass of
Reactants
 What
mass of sodium bicarbonate is needed to
produce 32 g of Na2CO3?
2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g)
32 g Na2CO3
x
1 mol Na2CO3
105.99 g
x
2 mol NaHCO3
1 mol Na2CO3
x
84.01 g
1 mol
NaHCO3
= 51 g NaHCO3
Limiting Reactant
 If
you have set quantities of two different
reactants, one will get used up and some
amount of the other will be left over.
 Limiting Reactant: The reactant that is
used up limits how far the reaction will
proceed.
 Excess Reactant: The reactant that is
leftover when the reaction is complete.
Ex) Limiting Reactant Problem
 Ex)
(a) What is the limiting reactant when 28
g of glucose reacts with 14 g of oxygen gas?
(b) What mass of CO2 is produced?

Step 1. Write a balanced chemical equation
C6H12O6 + 6O2  6CO2 + 6H2O
Ex) Limiting Reactant Problem
(cont.)
 Step
2. Find the mass of CO2 that would be
produced by each reactant
28 g C6H12O6 x
1 mol C6H12O6
180.18 g
C6H12O6
x
6 mol CO2
1 mol
C6H12O6
x
44.01 g CO2
1 mol CO2
= 41 g CO2
Ex) Limiting Reactant Problem
(cont.)
 Step
2. Find the mass of CO2 that would be
produced by each reactant

14 g O2 x
1 mol O2
32 g O2
x
6 mol CO2
6 mol O2
x
44.01 g CO2
1 mol CO2
= 19 g CO2
Ex) Limiting Reactant Problem
(cont.)
 Step
3. Compare the two masses
produced. The reactant that produced
the smallest quantity of product is the
limiting reactant.

19 g < 41 g
 Thus
O2 is the limiting reactant (limits the
amount of product formed)
 19 g of CO2 would be produced in theory
Ex) Percent Yield
 Ex)
Find the percent yield if only 15 grams
of CO2 were produced in the previous
problem.
Actual Yield
 % Yield =
x 100
Theoretical
Yield
Ex) Percent Yield
 Find
the percent yield if only 15 grams of
CO2 were produced in the previous
problem.

15g
19g
x 100 = 79%