Transcript Chapter 3.6-3.10 Student Notes Stoichiometry Chapter 3 Table of Contents 3.1 Counting by Weighing 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass 3.5 Percent Composition of Compounds SEE PREVIOUS LESSONS ON OTHER SLIDE.

Chapter 3.6-3.10
Student Notes
Stoichiometry
Chapter 3
Table of Contents
3.1
Counting by Weighing
3.2
Atomic Masses
3.3
The Mole
3.4
Molar Mass
3.5
Percent Composition of Compounds
SEE PREVIOUS LESSONS ON OTHER SLIDE SHOW
3.6
Determining the Formula of a Compound
3.7
Chemical Equations
3.8
Balancing Chemical Equations
3.9
Stoichiometric Calculations: Amounts of Reactants
and Products
3.10 The Concept of Limiting Reagent
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2
Chapter 3
TUESDAY - B DAY - SEPT. 17, 2013
Table of Contents
•
•
•
•
Pick up calculator, notes packet, and pre-lab packet.
CW: Notes 3.1 to 3.5
CW: Finish Test part 2 ch.1-2 if not done yet
CW/HW: Read lab and complete Pre-lab questions.
Note that the reading helps with the pre-lab questions
due Thursday for lab. Try to be here by 7:45 for setup to
give us time to discuss pre-lab and complete lab.
• HW: Section 3.1-3.5 pg. 117 #21, 26, 27, 29, 30, 31, 33,
35, 37, 39b,45b, 47b, 49b, 52, 57a-g, 61, 63 due
Monday 9/23/13 to go over.
3
Section 3.6
Assignments Monday 9/23/13
Determining the Formula of a Compound
• Pick up handouts for use later & have out ch. 3 notes
and HW ch. 3
• CW: Notes 3.5-3.6 finished together
• CW: Empirical and Molecular Formula Race Game
• CW/HW: Notes 3.7-3.8 with computers.
• HW: Formal Lab report due on Wednesday.
• Be sure you completed homework 3.1-3.5 previously
assigned and turn in by Wednesday.
• HW: Empirical Formula and Balancing Equations
w/sheets due on Friday.
• HW: Be reading chapter 3. The next test is over ch.3-4
and is tentatively scheduled for October 16th. Ch. 4 is
mostly new material not covered in Gen. Chemistry.
4
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Section 3.5
Percent Composition of Compounds
ASSIGNMENTS Ch. 3 Homework - Tuesday - 9/17/13
• HW: Section 3.1-3.5 pg. 117 #21, 26, 27, 29, 30, 31, 33,
35, 37, 39b,45b, 47b, 49b, 52, 57a-g, 61, 63 due
Monday 9/23/13 to go over.
• You must show your work for most of these with
calculations and units. No work for math problems will
receive no points.
• CW: Finish test part 2 if you haven’t.
• HW: Read and complete Pre-lab Determination of the
Molecular Weight of an Acid for lab on Thursday.
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5
Section 3.6
Determining the Formula of a Compound
Formulas
•
Empirical formula = CH
 Simplest whole-number ratio
• Molecular formula = (empirical formula)n
[n = integer]
• Molecular formula = C6H6 = (CH)6
 Actual formula of the compound
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22
Formulas
Empirical formula: the lowest whole number
ratio of atoms in a compound.
Molecular formula: the true number of
atoms of each element in the formula of a
compound.
 molecular formula = (empirical
formula)n [n = integer]
 molecular formula = C6H6 = (CH)6
 empirical formula = CH
Formulas (continued)
Formulas for ionic compounds are ALWAYS
empirical (lowest whole number ratio).
Examples:
NaCl
MgCl2
Al2(SO4)3
K2CO3
Formulas (continued)
Formulas for molecular compounds MIGHT be
empirical (lowest whole number ratio).
Molecular:
H2O
C6H12O6
C12H22O11
Empirical:
H2O
CH2O
C12H22O11
10
Chemical Formulas from Mass
Percent Composition
• We can “reverse” the process of finding
percentage composition.
• First we use the percentage or mass of each
element to find moles of each element.
• Then we can obtain the empirical formula by
finding the smallest whole-number ratio of
moles.
– Find the whole-number ratio by dividing each
number of moles by the smallest number of
moles.
Empirical Formula Determination
1. Base calculation on 100 grams of
compound. (If given masses, go to #2.)
2. Determine moles of each element in 100
grams of compound. (or sample given)
3. Divide each value of moles by the
smallest of the values.
4. Multiply each number by an integer to
obtain all whole numbers.
Empirical Formula Determination
Adipic acid contains 49.32% C, 43.84% O, and
6.85% H by mass. What is the empirical formula of
adipic acid?
1. Base calculation on 100 grams of compound.
In 100 g sample, we would have
49.32 g Carbon
43.84 g Oxygen
6.85 g Hydrogen
Empirical Formula Determination
(part 2)
2. Determine moles of each element in
100 grams of compound.
Carbon: 49.32 g / 12.01 g/mol = 4.107 mol
Oxygen: 43.84 g / 16.00 g/mol = 2.740 mol
Hydrogen: 6.85 g / 1.01 g/mol = 6.78 mol
Empirical Formula Determination
(part 3)
3. Divide each value of moles by the
smallest of the values.
Carbon: 4.107 mol /2.74 mol = 1.50
Oxygen: 2.74 mol /2.74 mol = 1.00
Hydrogen: 6.78 mol/2.74 mol = 2.47
Empirical Formula Determination
(part 4)
4. Multiply each number by an integer
to obtain all whole numbers.
Carbon: 1.50 x 2 = 3.0
Oxygen: 1.0 x 2 = 2.0
Hydrogen: 2.47 = 2.5 x 2 = 5
Empirical formula: C3H5O2
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular formula
of adipic acid?
1. Find the formula mass of C3H5O2
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular formula
of adipic acid?
2. Divide the molecular mass by the
mass given by the emipirical formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
Finding the Molecular Formula
The empirical formula for adipic acid is
C3H5O2. The molecular mass of adipic acid
is 146 g/mol. What is the molecular formula
of adipic acid?
3. Multiply the empirical formula by this
number to get the molecular formula.
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
(C3H5O2) x 2 = C6H10O4
19
Relating Molecular Formulas
to Empirical Formulas
• A molecular formula is a simple integer multiple
of the empirical formula.
• That is, an empirical formula of CH2 means that
the molecular formula is CH2, or C2H4, or C3H6,
or C4H8, etc.
• So: we find the molecular formula by:
molecular formula mass
empirical formula mass
= integer (nearly)
We then multiply each subscript in the empirical formula by the
integer.
Section 3.6
Determining the Formula of a Compound
Analyzing for Carbon and Hydrogen
•
Device used to determine the mass
percent of each element in a compound.
Combustion Analysis Method for
determining molecular mass
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Section 3.6
Determining the Formula of a Compound
Empirical & Molecular Formula Partner Team Race
Activity
• Working with a partner, you will solve one problem at a
time. Both of you will solve the problem on your paper
and then decide if you agree. Consider the correct
number of significant figures as well and race the
answer to me when you think you have a consensus of
the final answer. If you are correct, I will say “yes” and
you will proceed to the next problem. If I say “no”, you
will need to try again. If I say “no” Sig. Fig., your answer
is right to the wrong significant figures. Use the AP
periodic table values which are mostly to the hundredths
for atomic mass.
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21
Section 3.6
Determining the Formula of a Compound
Assignments Monday
•
•
•
•
•
CW: Notes 3.5-3.6 finished together
CW: Empirical and Molecular Formula Race Game
CW/HW: Notes 3.7-3.8 with computers.
HW: Formal Lab report due on Wednesday.
Be sure you completed homework 3.1-3.5 previously
assigned and turn in by Wednesday.
• HW: Empirical Formula and Balancing Equations
w/sheets due on Friday.
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22
Combustion Analysis Method for
determining molecular mass
• Menthol, the substance we can smell in
mentholated cough drops, is composed of
C, H, and O. A 0.1005 g sample of menthol
is combusted, producing 0.2829 g of CO2
and 0.1159 g of H2O. What is the empirical
formula for menthol?
Combustion Analysis Method for
determining molecular mass (cont)
Combustion Analysis Method for
determining molecular mass (cont.)
More complicated example:
You do not have to record but do look over it.
Nasty example - you don’t
have to record but look over.
Notes 3.7
Writing Chemical Equations
• A chemical equation is a shorthand
description of a chemical reaction, using
symbols and formulas to represent the
elements and compounds involved.
Section 3.7
Chemical Equations
•
A representation of a chemical reaction:
C2H5OH + 3O2  2CO2 + 3H2O
reactants
•
products
Reactants are only placed on the left side
of the arrow, products are only placed on
the right side of the arrow.
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25
Section 3.7
Chemical Equations
C2H5OH + 3O2  2CO2 + 3H2O
•
•
•
The equation is balanced.
All atoms present in the reactants are
accounted for in the products.
1 mole of ethanol reacts with 3 moles of
oxygen to produce 2 moles of carbon
dioxide and 3 moles of water.
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26
Section 3.7
Chemical Equations
•
The balanced equation represents an
overall ratio of reactants and products,
not what actually “happens” during a
reaction.
•
Use the coefficients in the balanced
equation to decide the amount of each
reactant that is used, and the amount of
each product that is formed.
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32
Writing Chemical Equations
• Sometimes
additional
information
about the
reaction is
conveyed in
the equation.
Chapter 8
Section 1 Describing Chemical
Reactions
Symbols Used in Chemical Equations
Chapter menu
Resources
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Section 3.8
Balancing Chemical Equations
Writing and Balancing the Equation for a Chemical Reaction
1. Determine what reaction is occurring.
What are the reactants, the products, and
the physical states involved?
2. Write the unbalanced equation that
summarizes the reaction described in
step 1.
3. Balance the equation by inspection,
starting with the most complicated
molecule(s). The same number of each
type of atom needs to appear on both
reactant and product sides.
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28
Balancing Chemical Equations
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Click here for visualization.
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3–
36
Guidelines for Balancing
Chemical Equations
• If an element is present in just one compound on each
side of the equation, try balancing that element first.
• Balance any reactants or products that exist as the
free element last.
• In some reactions, certain groupings of atoms (such
as polyatomic ions) remain unchanged. In such cases,
treat these groupings as a unit.
• At times, an equation can be balanced by first using a
fractional coefficient(s). The fraction is then cleared
by multiplying each coefficient by a common factor.
Section 3.8
Balancing Chemical Equations
Exercise
Which of the following correctly balances the
chemical equation given below? There may be more
than one correct balanced equation. If a balanced
equation is incorrect, explain what is incorrect about
it.
CaO + C  CaC2 + CO2
I.
II.
III.
IV.
CaO2 + 3C  CaC2 + CO2
2CaO + 5C  2CaC2 + CO2
CaO + (2.5)C  CaC2 + (0.5)CO2
4CaO + 10C  4CaC2 + 2CO2
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30
Section 3.8
Balancing Chemical Equations
Concept Check
Which of the following are true concerning balanced
chemical equations? There may be more than one true
statement.
I. The number of molecules is conserved.
II. The coefficients tell you how much of each
substance you have.
III. Atoms are neither created nor destroyed.
IV. The coefficients indicate the mass ratios of the
substances used.
V.The sum of the coefficients on the reactant side
equals the sum of the coefficients on the product
side.
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31
Section 3.8
Balancing Chemical Equations
Notice
•
•
•
•
The number of atoms of each type of
element must be the same on both sides
of a balanced equation.
Subscripts must not be changed to
balance an equation.
A balanced equation tells us the ratio of
the number of molecules which react and
are produced in a chemical reaction.
Coefficients can be fractions, although
they are usually given as lowest integer
multiples.
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32
Table 3.2 Information Conveyed by the
Balanced Equation for the Combustion
of Methane
Copyright © Houghton Mifflin Company. All rights reserved.
3–
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Stoichiometric Calculations
•
Chemical equations can be used to relate the
masses of reacting chemicals.
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42
Stoichiometric Equivalence
and Reaction Stoichiometry
• A stoichiometric factor or mole ratio is a
conversion factor obtained from the stoichiometric
coefficients in a chemical equation.
• In the equation: CO(g) + 2 H2(g)  CH3OH(l)
– 1 mol CO is chemically equivalent to 2 mol H2
– 1 mol CO is chemically equivalent to 1 mol
CH3OH
2 mol H2 is chemically
1 mol
1 mol–CO
1 mol CO equivalent to
2 mol
H2
–––––––––
–––––––––––––
CH3OH –––––––––––––
2 mol H2
1 mol CH3OH
1 mol CH3OH
43
Concept of Stoichiometric Equivalence
One car may be equivalent
to either 25 feet or 10 feet,
depending on the method of
parking.
One mole of CO may be
equivalent to one mole of
CH3OH, or to one mole of CO2,
or to two moles of CH3OH,
depending on the reaction(s).
44
Outline of Simple Reaction Stoichiometry
Note: preliminary and/or follow-up calculations
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Calculating Masses of Reactants and Products in Reactions
1. Balance the equation for the reaction.
2. Convert the known mass of the reactant
or product to moles of that substance.
3. Use the balanced equation to set up the
appropriate mole ratios.
4. Use the appropriate mole ratios to
calculate the number of moles of desired
reactant or product.
5. Convert from moles back to grams if
required by the problem.
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46
Outline of Stoichiometry Involving Mass
To our simple
stoichiometry
scheme …
… and a
conversion to
mass at the end.
… we’ve added a
conversion from
mass at the
beginning …
Substances A and B
may be two reactants,
two products, or
reactant and product.
Think: If we are given
moles of substance A
initially, do we need to
convert A to grams?
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Calculating Masses of Reactants and Products in Reactions
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35
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
WEDNESDAY - SEPT 25, 2013
• CW: Notes 3.9
• Turn in Formal Lab Report from last Thursday’s titration
lab AND HW 3.1 to 3.5 - go over any questions
• CW: Limiting Reactants problem w/sheet handout - HW
if not finished in class due FRI
• HW: Balancing Equations practice quick w/sheet due
FRI
• HW: Complete the empirical vs. molecular formula and
% composition w/sheet from Monday - due Friday
• CW/HW: pre-lab moles vs. molar mass graphical with
limiting reactants - using graphing to determine
unknown.
48
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Stoichiometry
“In solving a problem of this sort, the
grand thing is to be able to reason
backward. This is a very useful
accomplishment, and a very easy one,
but people do not practice it much.”
Sherlock Holmes, in Sir Arthur Conan
Doyle’s A Study in Scarlet
Review: Chemical Equations
Chemical change involves a reorganization of
the atoms in one or more substances.
C2H5OH + 3O2 → 2CO2 + 3H2O
reactants
products
When the equation is balanced it has
quantitative significance:
1 mole of ethanol reacts with 3 moles of
oxygen to produce 2 moles of carbon dioxide
and 3 moles of water
Calculating Masses of Reactants
and Products
1. Balance the equation.
2. Convert mass or volume to moles, if
necessary.
3. Set up mole ratios.
4. Use mole ratios to calculate moles of
desired substituent.
5. Convert moles to mass or volume, if
necessary.
Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of
oxygen. How many grams of aluminum oxide are
formed.
1. Identify reactants and products and write the
balanced equation.
4 Al
+ 3 O2
2 Al2O3
a. Every reaction needs a yield sign!
b. What are the reactants?
c. What are the products?
d. What are the balanced coefficients?
Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of
oxygen. How many grams of aluminum oxide are
formed?
4 Al + 3 O2 2Al2O3
6.50 g Al
1 mol Al
2 mol Al2O3 101.96 g Al2O3
26.98 g Al
4 mol Al
6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 =
1 mol Al2O3
= ? g Al2O3
12.3 g Al2O3
Working a Stoichiometry Problem
Question- How many grams of water can be produced
from 10. grams of H2 with excess O2?
You try.
=
Working a Stoichiometry Problem
Question- How many grams of water can be produced
from 10. grams of H2 with excess O2?
Answer: 2H2 + O2 --> 2H2O
10 g H2
1 mol H2
2 mol H2O
2.02 g H2
2 mol H2
18.02 g H2O
1 mol H2O
= 89.2=89 g
Limiting Reactant
The limiting reactant is the reactant
that is consumed first, limiting the
amounts of products formed.
The reactant left over is in EXCESS.
(remember nuts and bolts mini-lab)
57
Limiting Reactants
• Many reactions are carried out with a limited
amount of one reactant and a plentiful amount of
the other(s).
• The reactant that is completely consumed in the
reaction limits the amounts of products and is
called the limiting reactant, or limiting reagent.
• The limiting reactant is not necessarily the one
present in smallest amount.
58
Limiting Reactant Analogy
If we have 10
sandwiches, 18
cookies, and 12
oranges …
… how many
packaged meals
can we make?
59
Limiting Reactant Analogy
There would only be 9 because
you need 20 cookies to use all of
the sandwiches 1:2 ratio and you
only need 10 oranges with 10
sandwiches so oranges and
sandwiches are in excess.
… how many
packaged meals
can we make?
If we have 10
sandwiches, 18
cookies, and 12
oranges …
60
Molecular View of the Limiting Reactant Concept
1. Why is ethylene left
over, when we started
with more bromine
than ethylene? (Hint:
count the molecules.)
2. What mass of ethylene
is left over after
reaction is complete?
(Hint: it’s an easy
calculation; why?)
When 28 g (1.0
mol) ethylene
reacts with …
… 128 g (0.80 mol)
bromine, we get …
… 150 g of 1,2dibromoethane, and
leftover ethylene!
Recognizing and Solving
Limiting Reactant Problems
•
•
•
We recognize limiting reactant problems by the
fact that amounts of two (or more) reactants are
given.
One way to solve them is to perform a normal
stoichiometric calculation of the amount of
product obtained, starting with each reactant.
The reactant that produces the smallest amount
of product is the limiting reactant.
61
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Exercise
Consider the following reaction:
P4 (s) + 5 O2 (g)
2P2O5 (s)
If 6.25 g of phosphorus is burned, what mass
of oxygen does it combine with?
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Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Exercise
Consider the following reaction:
P4 (s) + 5 O2 (g)
2P2O5 (s)
If 6.25 g of phosphorus is burned, what mass
of oxygen does it combine with?
6.25 g / 124 g/mol = .0504 mol 1:5 ratio so
.0504... x 5 =0.252 mol x 32 g/mol = 8.06 g O2
8.07 g
if you do everything and then round at the end
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36
Section 3.10
The Concept of Limiting Reagent
Concept Check
Which of the following reaction mixtures could
produce the greatest amount of product? Each
involves the reaction symbolized by the equation:
2H2 + O2  2H2O
Identify limiting reactants & excess reactant
a)
b)
c)
d)
e)
2 moles of H2 and 2 moles of O2
2 moles of H2 and 3 moles of O2
2 moles of H2 and 1 mole of O2
3 moles of H2 and 1 mole of O2
Each produce the same amount of product.
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46
Section 3.10
The Concept of Limiting Reagent
Notice
•
We cannot simply add the total moles of
all the reactants to decide which reactant
mixture makes the most product. We
must always think about how much
product can be formed by using what we
are given, and the ratio in the balanced
equation.
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47
Section 3.10
The Concept of Limiting Reagent
Concept Check
• You know that chemical A reacts with
chemical B. You react 10.0 g of A with
10.0 g of B.
 What information do you need to
know in order to determine the mass
of product that will be produced?
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48
Section 3.10
The Concept of Limiting Reagent
Let’s Think About It
•
Where are we going?

•
To determine the mass of product that will be
produced when you react 10.0 g of A with 10.0 g
of B.
How do we get there?

We need to know:
• The mole ratio between A, B, and the product
they form. In other words, we need to know
the balanced reaction equation.
• The molar masses of A, B, and the product
they form.
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49
Section 3.10
The Concept of Limiting Reagent
Exercise
You react 10.0 g of A with 10.0 g of B. What
mass of product will be produced given that
the molar mass of A is 10.0 g/mol, B is
20.0 g/mol, and C is 25.0 g/mol? They react
according to the equation:
A + 3B  2C
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50
Section 3.10
The Concept of Limiting Reagent
Percent Yield
•
An important indicator of the efficiency of
a particular laboratory or industrial
reaction.
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51
Section 3.10
The Concept of Limiting Reagent
Exercise
Consider the following reaction:
P4(s) + 6F2(g) ? 4PF3(g)85.0 g
?g
 What mass of P4 is needed to produce 85.0
g of PF3 if the reaction has a 64.9% yield?
 % yield = actual / theoretical
0.649 = 85.0 g / t
t = 130.97 = 131 g
131 g PF3 / 88 g/mol = 1.49 mol (1.5 mol)
ratio 4 mol PF3 to 1 mol P4 1.49/4=.3725 mol P4
.3725 mol x 124 g/mol = 46.19 = 46.2 g P4
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52
Section 3.10
The Concept of Limiting Reagent
Exercise
Consider the following reaction:
P4(s) + 6F2(g)  4PF3(g)
 What mass of P4 is needed to produce
85.0 g of PF3 if the reaction has a 64.9%
yield?
46.1 g P4
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52
Yields of Chemical Reactions
• The theoretical yield of a chemical reaction is the
calculated quantity of product in the reaction.
• The actual yield is the amount you actually get
when you carry out the reaction.
• Actual yield will be less than the theoretical yield,
for many reasons … can you name some?
actual yield
Percent yield = ––––––––––––– × 100
theoretical yield
Example - % Yield
• 2CO(g) + O2(g) --> 2CO2(g)
Calculate the % yield if 69.1g of CO
combines with excess O2 to form an
experimental yield of 48.3L of CO2 @STP
Example - % Yield
• 2CO(g) + O2(g) --> 2CO2(g)
Calculate the % yield if 69.1g of CO
combines with excess O2 to form an
experimental yield of 48.3L of CO2 @STP
69.1 g / 28 g⋆mol-1 = 2.47 mol CO thus
creates same amount of moles of CO2. For
gases only, 22.4 L = 1 mol of any gas, so
• 2CO(g) + O2(g) --> 2CO2(g)
Calculate the % yield if 69.1g of CO
combines with excess O2 to form an
experimental yield of 48.3L of CO2 @STP
4
4
2.47 mol
4
4
8
87.3% yield
76
Actual Yield of ZnS Is Less than
the Theoretical Yield
Solutions and
Solution Stoichiometry
• Solute: the substance being dissolved.
• Solvent: the substance doing the dissolving.
• Concentration of a solution: the quantity of a
solute in a given quantity of solution (or solvent).
– A concentrated solution contains a relatively
large amount of solute vs. the solvent (or
solution).
– A dilute solution contains a relatively small
concentration of solute vs. the solvent (or
solution).
– “Concentrated” and “dilute” aren’t very
77
78
Molar Concentration
Molarity (M), or molar concentration, is the
amount of solute, in moles, per liter of solution:
moles of solute
Molarity = ––––––––––––––
liters of solution
• A solution that is 0.35 M sucrose contains 0.35
moles of sucrose in each liter of solution.
• Keep in mind that molarity signifies moles of
solute per liter of solution, not liters of solvent.
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Preparing 0.01000 M KMnO4
Weigh 0.01000
mol (1.580 g)
KMnO4.
Dissolve in water. How much
water? Doesn’t matter, as long as
we don’t go over a liter.
Add more water
to reach the 1.000
liter mark.
80
Example 3.23
What is the molarity of a solution in which 333 g potassium
hydrogen carbonate is dissolved in enough water to make 10.0 L
of solution?
Example 3.24
We want to prepare a 6.68 molar solution of NaOH (6.68 M
NaOH).
(a) How many moles of NaOH are required to prepare 0.500 L of
6.68 M NaOH?
(b) How many liters of 6.68 M NaOH can we prepare with 2.35 kg
NaOH?
Example 3.25
The label of a stock bottle of aqueous ammonia indicates that the
solution is 28.0% NH3 by mass and has a density of 0.898 g/mL.
Calculate the molarity of the solution.
81
Dilution of Solutions
• Dilution is the process of preparing a more dilute
solution by adding solvent to a more concentrated
one.
• Addition of solvent does not change the amount of
solute in a solution but does change the solution
concentration.
• It is very common to prepare a concentrated stock
solution of a solute, then dilute it to other
concentrations as needed.
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Dilution Calculations …
•
•
•
•
… couldn’t be easier.
Moles of solute does not change on dilution.
Moles of solute = M × V
Therefore …
Mconc × Vconc = Mdil × Vdil
Visualizing the Dilution of a Solution
We start and end
with the same
amount of solute.
83
Addition of
solvent has
decreased the
concentration.
Example 3.26
How many milliliters of a 2.00 M CuSO4 stock solution are
needed to prepare 0.250 L of 0.400 M CuSO4?
84
Solutions in Chemical
Reactions
• Molarity provides an additional tool in
stoichiometric calculations based on
chemical equations.
• Molarity provides factors for converting
between moles of solute (either reactant or
product) and liters of solution.
85
86
Adding to the previous stoichiometry scheme …
If substance A is a
solution of known
concentration …
… we can start with
molarity of A times
volume (liters) of the
solution of A to get
here.
If substance B is in
solution, then …
… we can go from
moles of substance B to
either volume of B or
molarity of B. How?
Limiting Reactants
Click here to watch visualization.
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