Transcript Stoichiometry PP

Chapter 12
Cookies?
 When
baking cookies, a recipe is
usually used, telling the exact
amount of each ingredient
• If you need more, you can double
or triple the amount
 Thus, a recipe is much like a
balanced equation
A. Proportional Relationships
2 1/4 c. flour
1 tsp. baking soda
1 tsp. salt
1 c. butter
3/4 c. sugar

3/4 c. brown sugar
1 tsp vanilla extract
2 eggs
2 c. chocolate chips
Makes 5 dozen cookies.
I have 5 eggs. How many cookies can I
make?
Ratio of eggs to cookies
5 eggs
5 doz.
2 eggs
= 12.5 dozen cookies
Stoichiometry
 Greek
for “measuring elements”
 The calculations of quantities in
chemical reactions based on a
balanced equation.
 We can interpret balanced
chemical equations several ways.
In terms of Particles
 Element-
made of atoms
 Molecular compound (made of only
non- metals) = molecules
 Ionic Compounds (made of a metal
and non-metal parts) = formula
units (ions)
2H2 + O2  2H2O
Two molecules of hydrogen and one
molecule of oxygen form two molecules
of water.
 2 Al2O3 Al + 3O2

2 formula units Al2O3 form 4 atoms Al
and 3 molecules O2
2Na + 2H2O  2NaOH + H2
Looking at it differently
 2H2 +
O2  2H2O
2 dozen molecules of hydrogen and 1
dozen molecules of oxygen form 2 dozen
molecules of water.
 2 x (6.02 x 1023) molecules of hydrogen
and 1 x (6.02 x 1023) molecules of oxygen
form 2 x (6.02 x 1023) molecules of water.
 2 moles of hydrogen and 1 mole of oxygen
form 2 moles of water.

In terms of Moles
Al + 3O2
 2Na + 2H2O  2NaOH + H2
 The coefficients tell us how many
moles of each substance are
needed.
 2 Al2O3
In terms of Mass
The Law of Conservation of Mass
applies
 We can check using moles
 2H2 + O2  2H2O
2.02 g H2
2 moles H2
= 4.04 g H2
1 moles H2

32.00 g O2
1 moles O2
= 32.00 g O2
1 moles O2
36.04 g H2+O2
In terms of Mass
 2H2 +
O2  2H2O
18.02 g H2O
2 moles H2O
= 36.04 g H2O
1 mole H2O
2H2 + O2  2H2O
4.04 g H2 + 32.00 O2 = 36.04 g H2O
A. Proportional Relationships

Stoichiometry
• mass relationships between substances in
a chemical reaction
• based on the mole ratio

Mole Ratio
• indicated by coefficients in a balanced
equation
2 Mg + O2  2 MgO
B. Stoichiometry Steps
1. Write a balanced equation.
2. Identify known & unknown.
3. Line up conversion factors.
•• Mole
molesmoles
moles
Mole ratio
ratio -moles
• Molar mass moles  grams
• Molarity moles  liters soln.
• Molar volume moles  liters gas
Core step in all stoichiometry problems!!
4. Check answer.
Mole to Mole conversions
2 Al2O3 Al + 3O2
 every time we use 2 moles of Al2O3 we
make 3 moles of O2

2 moles Al2O3
3 mole O2
or
3 mole O2
2 moles Al2O3
These are two possible conversion
factors
Mole to Mole conversions
How many moles of O2 are produced
when 3.34 moles of Al2O3 decompose?
 2 Al2O3 Al + 3O2

3.34 mol Al2O3
3 moles O2
= 5.01 moles O2
2 moles Al2O3
Your Turn
2C2H2 + 5 O2  4CO2 + 2 H2O
 If 3.84 moles of C2H2 are burned, how
many moles of O2 are needed? (9.60 mol)
 How many moles of C2H2 are needed
to produce 8.95 mole of H2O? (8.95 mol)
 If 2.47 moles of C2H2 are burned, how
many moles of CO2 are formed? (4.94
mol)
How do you get good at this?
Hydrogen Fuel Cell DVD
Mass in Chemical Reactions
How much do you make?
How much do you need?
Mass-Mass Calculations
We do not measure moles directly, so
what can we do?
 We can convert grams to moles
• Use the Periodic Table for mass
values
 Then do the math with the mole ratio
• Balanced equation gives mole ratio!
 Then turn the moles back to grams
• Use Periodic table values

Mass-Mass Conversion
If 10.1 g of Fe+3 are added to a solution
of Copper (II) Sulfate, how much solid
copper would form?
 Fe + CuSO4  Fe2(SO4)3 + Cu
 2Fe + 3CuSO4  Fe2(SO4)3 + 3Cu

10.1 g Fe
1 mol Fe
55.85 g Fe
= 0.181 mol Fe
2Fe + 3CuSO4  Fe2(SO4)3 + 3Cu
3 mol Cu
0.181 mol Fe
= 0.272 mol Cu
2 mol Fe
63.55 g Cu
0.272 mol Cu
= 17.2 g Cu
1 mol Cu
All Together It Looks Like
10.1 g Fe 1 mol Fe 3 mol Cu 63.55 g Cu
55.85 g Fe 2 mol Fe 1 mol Cu
= 17.2 g Cu
Examples
To make silicon for computer chips they
use this reaction
 SiCl4 + 2Mg  2MgCl2 + Si
 How many grams of Mg are needed to
make 9.3 g of Si?
 How many grams of SiCl4 are needed
to make 9.3 g of Si?
 How many grams of MgCl2 are
produced along with 9.3 g of silicon?

Gases and Reactions
We Can Also Change
Movie
 Liters of a gas to moles
If we are at STP, Standard
Temperature and Pressure, which
is 0ºC and 1 atmosphere pressure
At STP
 22.4 L of a gas = 1 mole of any gas
molecules

Molar Volume at STP
1 mol of a gas=22.4 L
at STP
Standard Temperature
&
0°C and 1 atm.
Pressure
LITERS
OF GAS
AT STP
Molar Volume
(22.4 L/mol)
MASS
IN
GRAMS
Molar Mass
(g/mol)
6.02 
MOLES
1023
particles/mol
Molarity (mol/L)
LITERS
OF
SOLUTION
NUMBER
OF
PARTICLES
Volume-Volume Calculations
How many liters of CH4 at STP are
required to completely react with 17.5 L
of O2 ?
 CH4 + 2O2  CO2 + 2H2O

1 mol O2 1 mol CH4 22.4 L CH4
17.5 L O2
22.4 L O2 2 mol O2 1 mol CH4
= 8.75 L CH4
Gas Stoichiometry Problems

How many grams of KClO3 are req’d to produce
9.00 L of O2 at STP?
2KClO3  2KCl + 3O2
?g
9.00 L
9.00 L
O2
1 mol
O2
2 mol
KClO3
122.55
g KClO3
22.4 L
O2
3 mol
O2
1 mol
KClO3
= 32.8 g
KClO3
Stoichiometry in the Real
World
Limiting Reactants and Yield
Limiting Reactants

Available Ingredients
• 4 slices of bread
• 1 jar of peanut butter
• 1/2 jar of jelly

Limiting Reactant
• bread

Excess Reactants
• peanut butter and jelly
Limiting Reactants

Limiting Reactant
• used up in a reaction
• determines the amount of product

Excess Reactant
• Not completely used up
• added to ensure that the other
reactant is completely used up
• cheaper & easier to recycle
Limiting Reactants
1. Write a balanced equation.
2. For each reactant, calculate the
amount of product formed.
3. Smaller answer indicates,
• the limiting reactant
• and the amount of product formed.
Limiting Reactants

79.1 g of zinc react with 0.90 L of
2.5M HCl. Identify the limiting and
excess reactants. How many liters of
hydrogen are formed at STP?
Zn + 2HCl 
0.90 L
79.1 g
2.5M
ZnCl2 + H2
?L
Limiting Reactants
Zn + 2HCl
79.1 g
0.90 L
2.5M
79.1
g Zn
1 mol
Zn
65.39
g Zn

ZnCl2 + H2
?L
1 mol
H2
22.4 L
H2
1 mol
Zn
1 mol
H2
= 27.1 L
H2
Limiting Reactants
Zn + 2HCl
79.1 g
0.90 L
2.5M
0.90
L
2.5 mol
HCl
1L

1 mol
H2
2 mol
HCl
ZnCl2 + H2
?L
22.4
L H2
= 25 L
1 mol
H2
H2
Limiting Reactants
Zn: 27.1 L H2
HCl: 25 L H2
Limiting reactant: HCl
Excess reactant: Zn
Product Formed: 25 L H2
left over zinc
Another Example
1. Do two stoichiometry problems.
2. The one that makes the least
product is the limiting reagent.

Copper reacts with sulfur to form copper
(I) sulfide. If 10.6 g of copper reacts with
3.83 g S how many grams of product
will be formed?
If 10.6 g of copper reacts with 3.83 g S.
How many grams of product will be
formed?
Cu is
 2Cu + S  Cu2S
Limiting

1 mol
Cu2S 159.16 g Cu2S
1
mol
Cu
Reagent
10.6 g Cu
63.55g Cu 2 mol Cu
1 mol Cu2S
= 13.3 g Cu2S
1
mol
S
3.83 g S
32.06g S
1 mol Cu2S 159.16 g Cu2S
1 mol S
1 mol Cu2S
= 19.0 g Cu2S
Limiting reactants
How many grams of H2O is produced if 20.0 grams of H2 react
with 20.0 grams of O2? Which reactant is limiting? Which
reactant is excess? How much excess remains?
2H2 + O2  2H2O
20. 0 g H2
Start with one
of your givens
20.0 g O2
Start with your
other given
1 mol H2
2.02 g H2
2 mol H2O
1 mol O2
2 mol H2O
32.00 g O2
2 mol H2
1 mol O2
18.02 g H2O
= 178 g H2O
1 mol H2O
18.02 g H2O
= 22.5 g H2O
1 mol H2O
To find the answer for how many grams of H2O produced compare
both of your answers, the lowest number is the correct answer.
Limiting Reactants: How to find excess
To find out how much excess is left over you take what you produce
and determine excess reactant actually used.
22.5 g H2O
1 mol H2O
2 mol H2
18.02g H2O
2 mol H2O
2.02 g H2
1 mol H2
= 2.52 g H2
This is how many
Based on limiting
reactant
grams you used up.
Now take the answer and subtract from your given amount of
excess. The answer that you find is how much excess is left over.
20.0g – 2.52g = 17.5g H2
Your Turn
If 10.1 g of magnesium and 2.87 L of
HCl gas are reacted, how many liters of
gas will be produced?
 How many grams of solid will be
produced?
 How many atoms of Cl are in the solid?
 How much excess reagent remains?

Your Turn II
If 10.3 g of aluminum are reacted with
51.7 g of CuSO4 how much copper will
be produced?
 How much excess reagent will remain?

Yield
The amount of product made in a
chemical reaction.
 There are three types
 Actual yield- what you get in the lab
when the chemicals are mixed
 Theoretical yield- what the balanced
equation tells you you should make.
 Percent yield
Percent Yield
measured in lab
actual yield
% yield 
 100
theoretical yield
calculated on paper
Percent Yield

When 45.8 g of K2CO3 react with
excess HCl, 46.3 g of KCl are formed.
Calculate the theoretical and % yields
of KCl.
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
?g
actual: 46.3 g
Percent Yield
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
?g
actual: 46.3 g
Theoretical Yield:
45.8 g
K2CO3
1 mol
K2CO3
138.21 g
K2CO3
2 mol
KCl
74.55
g KCl
1 mol
K2CO3
= 49.4
1 mol
g KCl
KCl
Percent Yield
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
49.4 g
actual: 46.3 g
Theoretical Yield = 49.4 g KCl
% Yield =
46.3 g
49.4 g
 100 = 93.7%
Details
 Percent
yield tells us how “efficient”
a reaction is.
 Percent yield can not be bigger
than 100 %.
Another Example
 6.78
g of copper is produced when
3.92 g of Al are reacted with excess
copper (II) sulfate.
 2Al + 3 CuSO4  Al2(SO4)3 + 3Cu
 What is the actual yield?
6.78 g Cu
 What is the theoretical yield? 13.9 g Cu
48.8 %
 What is the percent yield?
Chapter 16
Energy in Chemical Reactions
How Much?
In or Out?
Energy
Energy is measured in Joules or calories
 Every reaction has an energy change
associated with it
 Exothermic reactions release energy,
usually in the form of heat.
 Endothermic reactions absorb energy
 Energy is stored in bonds between atoms

58
Energy
C + O2  CO2+ 395 kJ
C + O2
395kJ
C + O2
Reactants

Products
59
In terms of bonds
C
O
O
O
C
O
Breaking this bond will require energy
O
C
O C O
O
Making these bonds gives you energy
In this case making the bonds gives you
more energy than breaking them
60
Exothermic
The products are lower in energy than
the reactants
 Releases energy

61
Energy
CaCO
 CaO
CaCO
CaO
+ CO+2 CO2
3 + 176
3 kJ
CaO + CO2
176 kJ
CaCO3
Reactants

Products
62
Endothermic
The products are higher in energy than
the reactants
 Absorbs energy

63
Chemistry Happens in
 MOLES
An equation that includes energy is
called a thermochemical equation
 CH4 + 2 O2  CO2 + 2 H2O + 802.2 kJ
 1 mole of CH4 makes 802.2 kJ of
energy.
 When you make 802.2 kJ you make 2
moles of water

64
CH4 + 2 O2  CO2 + 2 H2O + 802.2 kJ

If 10. 3 grams of CH4 are burned
completely, how much heat will be
produced?
10. 3 g CH4
1 mol CH4
16.05 g CH4
802.2 kJ
1 mol CH4
=514 kJ
65
CH4 + 2 O2  CO2 + 2 H2O + 802.2 kJ
How many liters of O2 at STP would be
required to produce 23 kJ of heat?
 How many grams of water would be
produced with 506 kJ of heat?

66
Heats of Reaction
Enthalpy
The heat content a substance has at a
given temperature and pressure
 Can’t be measured directly because
there is no set starting point
 The reactants start with a heat content
 The products end up with a heat content
 So we can measure how much enthalpy
changes

68
Enthalpy
Symbol is H
 Change in enthalpy is DH
 delta H
 If heat is released the heat content of
the products is lower
 DH is negative (exothermic)
 If heat is absorbed the heat content of
the products is higher
 DH is negative (endothermic)

69
Energy
Change is down
DH is <0
Reactants

Products
70
Energy
Change is up
DH is > 0
Reactants

Products
71
Heat of Reaction
The heat that is released or absorbed in a
chemical reaction
 Equivalent to DH
 C + O2(g)  CO2(g) +393.5 kJ
 C + O2(g)  CO2(g)
DH = -393.5 kJ
 In thermochemical equation it is important
to say what state
 H2(g) + 1/2O2 (g) H2O(g) DH = -241.8 kJ
 H2(g) + 1/2O2 (g) H2O(l) DH = -285.8 kJ

72
Heat of Combustion
The heat from the reaction that completely
burns 1 mole of a substance
 C2H4 + 3 O2  2 CO2 + 2 H2O
 C2H6 + O2  CO2 + H2O
 2 C2H6 + 5 O2  2 CO2 + 6 H2O
 C2H6 + (5/2) O2  CO2 + 3 H2O

73
Standard Heat of Formation
The DH for a reaction that produces 1
mol of a compound from its elements at
standard conditions
 Standard conditions 25°C and 1 atm.
0
 Symbol is DH
f

The
standard heat of formation of an
element is 0
This
includes the diatomics
74
What good are they?
There are tables (pg. 190) of heats of
formations
 The heat of a reaction can be calculated
by subtracting the heats of formation of
the reactants from the products

DH =
0
DH f (reactants)
0
- DH f (products)
75
Examples

CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
0
DH f CH4 (g) = -74.86 kJ
0
DH f O2(g) = 0 kJ
0
DH f CO2(g) = -393.5 kJ
0
DH f H2O(g) = -241.8 kJ
DH= [-393.5 + 2(-241.8)]-[-74.68 +2 (0)]
 DH= 802.4 kJ

76
Examples

2 SO3(g)  2SO2(g) + O2(g)
77
Why Does It Work?
If H2(g) + 1/2 O2(g) H2O(g) DH=-285.5 kJ
 then
H2O(g) H2(g) + 1/2 O2(g) DH =+285.5 kJ
 If you turn an equation around, you change
the sign
 2 H2O(g) H2(g) + O2(g) DH =+571.0 kJ
 If you multiply the equation by a number,
you multiply the heat by that number.

78
Why does it work?
You make the products, so you need
there heats of formation
 You “unmake” the products so you have
to subtract their heats.
 How do you get good at this

79