Transcript Ch 6 LAN 7th Intro Chem Chemical Reactions

Chapter 6 Lecture
Fundamentals of General,
Organic, and Biological
Chemistry
7th Edition
McMurry, Ballantine, Hoeger, Peterson
Chapter Six
Chemical Reactions:
Mole and Mass Relationships
Julie Klare
Gwinnett Technical College
© 2013 Pearson Education, Inc.
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6.1
6.2
6.3
6.4
6.5
The Mole and Avogadro’s Number
Gram–Mole Conversions
Mole Relationships and Chemical Equations
Mass Relationships and Chemical Equations
Limiting Reagent and Percent Yield
Goals
 1. What is the mole, and why is it useful in
chemistry?
 Be able to explain the meaning and uses of the mole and Avogadro’s
number.
 2. How are molar quantities and mass quantities
related?
 Be able to convert between molar and mass quantities of an element or
compound.
 3. What are the limiting reagent, theoretical yield,
and percent yield of a reaction?
 Be able to take the amount of product actually formed in a reaction,
calculate the amount that could form theoretically, and express the
results as a percent yield.
6.1 The Mole and Avogadro’s
Number
How to balance reactions
by counting molecules
Counting molecules: Conceptually
• The reactants in a chemical reaction must be
balanced
• One molecule of hydrogen (H2) reacts with
one molecule of iodine (I2), creating two
hydrogen iodide molecules (HI)
• Of course, we can’t visually count molecules to
achieve this one-to-one ratio
• In fact, the only tool available is an analytical
mass balance (using grams!)
Molecular Weight
• But how do we relate a sample’s mass in grams
to the number of molecules it contains?
• We begin with the concept of ‘molecular
weight’ (molecular mass)
• Let’s start by utilizing the Ch 2 concept of mass
by ‘amu’
– Note: We will be focusing on molecules for a while
• Recall, atomic weight is the average mass of an
element’s isotope atoms (review next slide)
• By analogy, molecular weight (MW) is the
mass of a compound’s molecules
– A molecule’s molecular weight is the sum of the
atomic weights for all the atoms in the molecule
– A salt’s formula weight is the sum of the atomic
weights for all the atoms in the formula unit
fyi: Atomic Weight Calculation
C-12 is 98.89% of all natural carbon
C-13 is 1.11% of all natural carbon
The mass of C-12 is: 12 amu (by definition)
The mass of C-13 is: 13.003355 amu
Atomic weight = [(% isotope abundance) × (isotope mass)]
[98.89% x 12 amu] + [1.11% x 13.0034 amu] =
(0.9889)(12 amu) + (0.0111)(13.0034 amu) = 12.0111
11.8668
0.1443
12.01 amu
fyi: Disambiguation
• For molecules, these terms are identical
– molar weight
– molar mass
– molecular weight
– molecular mass
• For atoms, these terms are identical
– molar weight
– molar mass
– atomic weight
– atomic mass
Molecular Weight (Mass):
Mass in grams of one mole of the substance:
Molar Mass of N = 14.01 g/mol
Molar Mass of N2 = 28.02 g/mol
Molar Mass of H2O = 18.02 g/mol
(2 × 1.008 g) + 16.00 g
Molar Mass of Ba(NO3)2 = 261.35 g/mol
137.33 g + (2 × 14.01 g) + (6 × 16.00 g)
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Molecular weight
• So if the mass (atomic weight) of one atom of
hydrogen (H) is 1.008 amu
– then the mass (molecular weight) of one hydrogen
molecule (H2) is 2.016 amu
• If the atomic weight of one iodine atom (I) is 126.90
amu
– then the molecular weight of one iodine molecule (I2) is
253.80 amu
• Finally, the molecular weight of each HI molecule
must be:
– 1.008 amu + 126.90 amu = 127.91 amu
Counting molecules:
by mass
• Here is where we left our reaction making two
HI molecules from H2 and I2
• We learned in chapter 2 that the iodine atom
(I) is approximately 126 times heavier than the
hydrogen atom (H)
– H = 1.008 amu
– I = 126.90 amu
• So mathematically, it is necessary that the
iodine molecule (I2) is approximately 126 times
heavier than the hydrogen molecule (H2)
– H2 = 2.016 amu
– I2 = 253.80 amu
• Meaning that this reaction is balanced – in amu
units
H2(g) + I2(g) → 2HI(g)
2.016 amu 253.80 amu
1 : 126
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• So by the law of identical ratios, this reaction
must also be balanced – in gram units
H2(g) + I2(g) → 2HI(g)
2.016 grams 253.80 grams
1 : 126
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Now notice
we have gone
from amu to grams
• And again by The Law of Identical Ratios, this
must also be balanced
H2(g) + I2(g) → 2HI(g)
1.00 g
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126 g
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Mass / Number relationship for I2 & H2
• So in order control the number ratio of I2 to H2
molecules at 1:1 we can measure out any of the
following
– so long we keep the mass ratio at 1:126
H2(g) + I2(g) → 2HI(g)
2.016 amu 253.80 amu
2.016 g 253.80 g
1.00 g
126 g
0.0079 g 1.0000 g
But how many molecules is that?
• We know that:
– 2.016 amu of H2 contains one molecule, and
– 253.80 amu of I2 contains one molecule
• Our final question is, how many molecules are
contained in:
– 2.016 g of H2
– 253.80 g of I2
That amount we call The Mole
• A mole is the amount of substance whose
mass in grams is numerically equal to its
molecular weight
– 2.016 g of H2 contains one mole of H2 molecules
– 253.80 g of I2 contains one mole of I2 molecules
• The mole (NA) has now been measured
– One mole of any molecule contains 6.022 × 1023
molecules
– One mole of any salt contains 6.022 × 1023
formula units
Getting from amu to grams
• The number ‘126.904,’
typically placed as shown,
functions BOTH:
– as the mass in amu per
atom
– and the mass in grams per
mole of atoms
• So I has a mass of
126.904 grams per mole
of I atoms
– I = 126.904 g/mol
The technical definition of a mole
How many silicon atoms are in 0.532 moles of
silicon?
a.
b.
c.
d.
8.83 × 10−25 atoms
3.81 × 1023 atoms
2.6 × 1023 atoms
3.8 × 1023 atoms
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6.2 Gram–Mole Conversions
fyi: Synonyms
• The terms Molecular Weight (section 6.1) and
Molar Mass (this section – 6.2) are identical
– Also, the term Molecular Mass is allowed
– Also, the term Molar Weight is allowed
• When we defined Molecular Weight in Section
6.1, we had not yet define the Mole
– so you couldn’t have understood the more
common term “Molar” Mass
• The molar mass of water is 18.02 g/mol
• So how many moles of water are there in 27
grams?
• Most importantly, molar mass serves as a
conversion factor between numbers of
moles and mass in grams for use in
dimensional analysis
Worked Example 6.3
Molar Mass: Mole to Gram Conversion
Ibuprofen is a pain reliever used in Advil. Its molecular weight
is 206.3 g/mol.
If a bottle of Advil contain 0.082 mole of ibuprofen, how many
grams of ibuprofen does it contain?
WORKED EXAMPLE 6.3
Molar Mass: Mole to Gram Conversion (Continued)
We now have the tools to convert:
• grams → moles → number of atoms
• grams → moles → number of molecules
• grams → moles → number of formula units
• And we can go backwards
33
What is the mass in grams of 3.2 × 1022
molecules of water?
a.
b.
c.
d.
0.0029 g
339 g
0.90 g
0.96 g
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6.3 Mole Relationships and Chemical Equations
Molar coefficients:
using Mole ratios as
conversion factors
• Coefficients in a balanced chemical
equation tell us the necessary ratio of
moles of reactants
– and how many moles of each product are
formed
• Especially useful, coefficients can be put
in the form of mole ratios
– which act as conversion factors when
setting up dimensional analysis calculations
Mole to mole
This reaction represents the reaction of A2 (red) with B2
(blue)
a) Write a balanced equation for the reaction
b) How many moles of product can be made from 1.0
mole of A2? From 1.0 mole of B2?
How many moles of NH3 can be produced from
5.00 moles of H2 according to the following
equation?
N2 + 3 H2  2 NH3
a.
b.
c.
d.
0.67 mol
2.00 mol
3.33 mol
7.50 mol
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6.4 Mass Relationships and Chemical Equations
Putting it all
together
• The actual amounts of substances used in the
laboratory must be weighed out in grams
• Furthermore, customers generally want the
quantity of product reported in terms of grams
We already have the necessary tools
• Mass to mole conversions
• Mole to mole conversions
• Mole to mass conversions
• Mole to mole conversions (section 6.3) are carried
out using mole ratios as conversion factors
If you have 9.0 moles of H2, how many moles of NH3 can you make?
• 2) Mole-to-mass and mass-to-mole
conversions (from section 6.2) are carried out
using molar mass as a conversion factor
• 3) Mass to mass conversions cannot be
carried out directly
•
•
•
•
If you know the mass
of A and need to find
the mass of B
first convert the mass
of A into moles of A
then carry out a mole
to mole conversion to
find moles of B
then convert moles of
B into the mass of B
• So after collecting the fundamental data, always
follow this three step process for mass to mass
• 1) convert grams of A to moles via molar mass
• 2) convert moles of A to B via mole ratio
• 3) convert mole of B to grams via molar mass
Let’s go back to our hydrogen / iodine
reaction giving hydrogen iodide
47
10.0 grams of HI requires how many grams of H2
I2 = 253.80 g/mol
H2 = 2.016 g/mol
HI = 127.90 g/mol
48
How many grams of oxygen are needed to react
with 25.0 g of K according to the following
equation?
4 K(s) + O2(g)  2 K2O(s)
a.
b.
c.
d.
10.2 g O2
2.56 g O2
8.66 g O2
5.12 g O2
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6.5 Limiting Reagent and Percent Yield
Vocabulary:
Reactant = Reagent
limiting reactant (reagent)
•
Reactants are not always perfectly balanced
•
When running a chemical reaction, we
generally ‘overcharge’ one of the reactants
As our mechanic has ‘overcharged’ tires
limiting reactant (reagent)
•
The limiting reactant is the reactant that runs
out first
•
The reactant that never runs out is called the
excess reactant
So which is ‘limiting here?’ Tires or car bodies.
limiting reactant (reagent)
•
In order to make this ‘reaction’ balanced
stoichiometrically, how many tires should our
mechanic actually have on hand?
32 tires would perfectly ‘balance’ this ‘reaction’
• Back to chemistry
A stoichiometric mixture of reactants
contains the relative amounts (in moles) of
reactants that matches the coefficients in the
balanced equation
N2(g) + 3H2(g)  2NH3(g)
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A stoichiometric mixture of reactants
• Contains the relative amounts (in moles) of
reactants that matches the coefficients in the
balanced equation
N2(g) + 3H2(g)  2NH3(g)
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A limiting reactant mixture
on the other hand, contains an excess of one
of the reactants
The other reactant is thus limiting
N2(g) + 3H2(g)  2NH3(g)
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A limiting reactant mixture
on the other hand, contains an excess of one
of the reactants
The other reactant is thus limiting
N2(g) + 3H2(g)  2NH3(g)
Which reactant is in
excess in this
reaction mixture?
Which reactant is
limiting?
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Definition: Limiting Reactant
N2(g) + 3H2(g)  2NH3(g)
• The limiting reactant (reagent) is the
reactant that runs out first and thus
constrains (limits) the amounts of
product(s) that can form
 H2 in the previous example is limiting
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• But the limiting reactant must be
identified before the correct amount of
product can be calculated
2H2 + O2  H2O
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Example
10.0 g of Chemical A are reacted with 10.0 g of B
• What information do you need to calculate the mass
of the product (C) that will be produced?
• The mole ratios between A, B, and C
– ie, the balanced reaction equation
• The molar masses of A, B, and C
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Example
10.0 g of Chemical A are reacted with 10.0 g of B
• What information do you need to calculate the mass
of the product (C) that will be produced?
A + 3B 2C
• molar masses:
– A is 10.0 g/mol
– B is 20.0 g/mol
– C is 25.0 g/mol?
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Example
10.0 g of Chemical A are reacted with 10.0 g of B
A + 3B 2C
• 1) Convert known masses of reactants to moles
1 mol A
10.0 g A ´
= 1.00 mol A
10.0 g A
1 mol B
10.0 g B ´
= 0.500 mol B
20.0 g B
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Example
10.0 g of Chemical A are reacted with 10.0 g of B
A + 3B 2C
• 2) Convert to the number of moles of product
2molC
1.00molA ´
= 2.00molC
1molA
2molC
0.500molB ´
= 0.333molC
3molB
So B is limiting reactant
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Example
10.0 g of Chemical A are reacted with 10.0 g of B
A + 3B 2C
• Choose the least number of moles of product
formed as limiting reactant: B is limiting
2molC
0.500molB ´
= 0.333molC
3molB
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Example
10.0 g of Chemical A are reacted with 10.0 g of B
A + 3B 2C
• 3) Convert moles of C to grams of C using the molar
mass
25.0 g C
0.333 mol C ´
= 8.33 g C
1 mol C
66
shortcut to limiting reactant
• How do we determine quickly which reactant
is limiting
– and by deduction, which is in excess?
• Three step program
1 Calculate the number of moles of each reactant
2 divide the number of moles of each reactant by
its coefficient from the balanced equation
3 the reactant with the smaller result is limiting
If 56 g of K is reacted with 56 g of oxygen gas
according to the equation below, indicate the mass of
product that can be made and identify the limiting
reactant.
4 K(s) + O2(g)  2 K2O(s) K2O = 94.2g/mol
a.
b.
c.
d.
67g K2O; K is the limiting reactant.
270 g K2O; K is the limiting reactant.
270 g K2O; O2 is the limiting reactant.
67g K2O; O2 is the limiting reactant.
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Theoretical vs actual yield
• Theoretical Yield
– The maximum amount of a given product that can
be formed once a limiting reactant has been
completely consumed – it is a calculation
• Actual yield
– The amount of product that is actually produced
in a reaction – it is a measurement
– It is usually less than the maximum expected
(theoretical yield)
FYI
• Theoretical yield is found by using the amount
of limiting reactant calculated in a mass-tomass calculation
• For the chemist in the lab, the actual yield is
found by weighing the amount of product
obtained
• Theoretical Yield
– For our auto mechanic, the theoretical yield is
8 finished cars
• Actual yield
– If one car body was damaged beyond repair,
the actual yield would have been 7 finished
cars
• Percent yield is the percent of the theoretical
yield actually obtained from a chemical reaction
actual yield
Percent yield =
´100
theoretical yield
• For our car mechanic:
WORKED EXAMPLE 6.8 Percent Yield
The combustion of acetylene gas (C2H2) produces carbon
dioxide and water as indicated in the following reaction
2 C2H2 (g) + 5 O2 (g) → 4 CO2 (g) + 2 H2O (g)
When 26.0 g of acetylene is burned in sufficient oxygen for
complete reaction, the theoretical yield of CO2 is 88.0 g
Calculate the percent yield for this reaction if the actual yield
is only 72.4 g CO2
Analysis—The percent yield is calculated by dividing the actual
yield by the theoretical yield and multiplying by 100
Percent yield =
actual yield
´100
theoretical yield
WORKED EXAMPLE 6.8 Percent Yield (finished)
actual yield
Percent yield =
´100
theoretical yield
72.4 g CO 2
´100 = 82.3%
88.0 g CO 2
WORKED EXAMPLE 6.9 Mass to Mole Conversions:
Limiting Reagent and Theoretical Yield
The element boron is produced commercially by the reaction of
boric oxide with magnesium at high temperature.
B2O3 (l) + 3 Mg (s) → 2 B (s) + 3 MgO (s)
What is the theoretical yield of boron when 2350 g of boric
oxide is reacted with 3580 g of magnesium? The molar
masses of boric oxide and magnesium are 69.6 g/mol and
24.3 g/mol, respectively
WORKED EXAMPLE 6.9 Mass to Mole Conversions:
Limiting Reagent and Theoretical Yield (Continued)
If 28.56 g of K2O is produced when 25.00 g K is
reacted according to the following equation, what
is the percent yield of the reaction?
4 K(s) + O2(g)  2 K2O(s)
a.
b.
c.
d.
87.54%
95%
94.82%
88%
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Chapter Summary
1. What is the mole, and why is it useful in
chemistry?
• A mole refers to Avogadro’s number 6.022 ×
1023 formula units of a substance.
• One mole of any substance has a mass (a
molar mass) equal to the molecular or formula
weight of the substance in grams.
• Because equal numbers of moles contain
equal numbers of formula units, molar masses
act as conversion factors between numbers of
molecules and masses in grams.
2. How are molar quantities and mass
quantities related?
• The coefficients in a balanced chemical
equation represent the numbers of moles of
reactants and products in a reaction.
• The ratios of coefficients act as mole ratios that
relate amounts of reactants and/or products.
• By using molar masses and mole ratios in
factor-label calculations, unknown masses or
molar amounts can be found from known
masses or molar amounts.
3. What are the limiting reagent, theoretical
yield, and percent yield of a reaction?
• The limiting reagent is the reactant that runs
out first.
• The theoretical yield is the amount of product
that would be formed based on the amount of
the limiting reagent.
• The actual yield of a reaction is the amount of
product obtained.
• The percent yield is the amount of product
obtained divided by the amount theoretically
possible and multiplied by 100%.
The official name for what we are doing
is Stoichiometry
amu to grams
• Recall: in Ch 2 we calculated the masses in amu
needed to balance this reaction
H2(g) + I2(g) → 2HI(g)
2.016 amu 253.80 amu
– So the mass ratio of I2 to H2 needed is actually: 125.90
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