3.1-3.3
Macroscopic versus Microscopic
I
Chemical symbols
II
Chemical equations
Balanced rxn is lowest ratio
moles
III
Balancing chemical equations
IV
I.
What chemical symbols mean
big, thick wall
H 2O
microscopic
Bond angles
macroscopic
phase
Bond lengths
temperature
density
Dipole moments
Specific heat
pressure
II.
glucose
Chemical Equations:
Symbols
C6 H12O6 s   6 O2 g   6 CO2 g   6 H2Og 
Microscopic viewpoint: 1 molecule of glucose reacts with 6 molecules of oxygen to form 6
molecules of carbon dioxide and six molecules of water.
Macroscopic viewpoint: Solid glucose reacts with gaseous oxygen to form gaseous carbon
dioxide and gaseous water. The mass of all the glucose and oxygen is the same as the mass
of all the carbon dioxide and water
Clearly, a chemical reaction has microscopic scale information (molecule names and ratios),
and macroscopic scale information (phases), but the majority of the important information
(for us) is on the microscopic scale.
II.
Chemical Equations: Conservation of mass
Recall from chapter two that the law of mass conservation is intimately related to chemical
equations. Since chemical equations contain information both macroscopic and
microscopic, let’s see how the law of mass conservation plays on both levels.
Macroscopic
The total mass of the reactants must be
exactly the same as the total mass of
the products. Mass cannot be created
or destroyed.
Microscopic
Every atom of every element in the
reactants must appear in the products.
Atoms cannot be created or destroyed.
This is the origin of
balancing reactions!
link
III.
Moles: the mathematical bridge
While you already know the microscopic meaning of a chemical equation, historically
they didn’t. Their microscopic information was severely limited.
They knew:
•Under some conditions, 12 g carbon reacts with 16 g oxygen (making 28 g of product).
•Under different conditions, 12 g carbon reacts with 32 g oxygen (making 44 g of
product).
•And thousands of similar reactions
So they designed a relative scale (amu) and set one carbon atom at 12 amu and one
oxygen atom at 16 amu. This bridges the macro/micro divide.
But how many carbon atoms in one gram? What’s the relationship between amu and
grams?
III.
Moles: the mathematical bridge
On one level, we don’t care the exact number of amu per gram. We can’t really count all
the atoms and molecules anyway, so a relative scale is good enough—the number of
amu in one gram is real, even if we don’t know it. It still bridges the microscopic and
macroscopic divide.
Because:
1 C atom  12 am u
1 O atom  16 am u
therefore, whatever number of atoms of carbon there
are in 12 g of carbon, there are exactly that many
atoms of oxygen in 16 g of oxygen.
•The mass (in grams) equal to the atomic weight (in amu) is called a molar mass.
•The collection of atoms in one molar mass is called a mole.
•The exact number of atoms in one mole is Avogadro’s number.
6.022 1023
III.
Moles: the mathematical bridge
Because of the law of conservation of mass, we can simply add up the weights of the
individual atoms to come up with the weight of the individual molecule (or one formula
unit, for ionic compounds). And the same reasoning we used to create the molar weights
of the elements works to create the molar weights of compounds as well.
Atomic mass of C = 12.0 amu
Molar mass of C = 12.0 g
1 mole C = 6.022*1023 atoms
Atomic mass of O = 16.0 amu
Molar mass of O = 16.0 g
1 mole O = 6.022*1023 atoms
Molecular mass of
CO2 = 44.0 amu
Molar mass of CO2 = 44.0 g
1 mole CO2 = 6.022*1023 molecules
Formula mass of
CaO = 56.1 amu
Molar mass of CaO = 56.1 g
1 mole
CaO = 6.022*1023 formula units
Microscopic units
Macroscopic units
Since we will often use molar
mass to convert between
microscopic and macroscopic
viewpoints, we will usually
represent molar mass with
units of grams per mole,
link
II. & III.
Chemical reactions & moles
A new view of the chemical equation can now be given:
The “mole” unit is very much like the “dozen” unit, except, umm, a whole lot bigger.
Remember that the coefficients are only ratios. If you double the moles of hydrogen
and nitrogen, then you double the moles of ammonia. This will be VERY important
in stoichiometry.
IV.
Balancing chemical reactions
The new, “bridging” view of the chemical reaction demands correct chemical
coefficients. We need to know how to make them. This is called “balancing the
reaction”.
IV.
Balancing chemical reactions
IV.
Balancing chemical reactions:
table of chemical lovin’
KClO3  C12 H22O11  KCl  CO2  H2O
1
1
12
22
14
K
Cl
C
H
O
1
1
1
2
3
Its usually easiest to start with elements other than C, H, and O. But they’re already
balanced, so let’s start with C.
IV.
Balancing chemical reactions:
table of chemical lovin’
KClO3  C12 H22O11  KCl  12 CO2  H2O
1
1
12
22
14
K
Cl
C
H
O
1
1
12
2
25
We could try to fix the O, but it doesn’t look like a whole number coefficient helps. Let’s
fix hydrogen instead, and hope the oxygen problem goes away.
IV.
Balancing chemical reactions:
table of chemical lovin’
KClO3  C12 H22O11  KCl  12 CO2  11 H2O
1
1
12
22
14
K
Cl
C
H
O
1
1
12
22
35
Now we have to face oxygen. We need to increase oxygen on the reactant side, where
it appears in two places. In both cases the coefficient will affect two other elements.
But in the first compound, the other two elements only appear together on the product
side. Whereas in the second compound, the other two elements are in different
products; that sounds more complicated.
IV.
Balancing chemical reactions:
table of chemical lovin’
8 KClO3  C12 H22O11  KCl  12 CO2  11 H2O
8
8
12
22
35
K
Cl
C
H
O
1
1
12
22
35
We’re almost done. We just need to fix the KCl and we’re through.
Balancing chemical reactions:
table of chemical lovin’
IV.
8 KClO3  C12 H22O11  8 KCl  12 CO2  11 H2O
8
8
12
22
35
Boo ya!
K
Cl
C
H
O
8
8
12
22
35
3.4 – 3.6
Stoichiometry
II.
Inputs
II.
I.
Two Hearts of
Stoichiometry
Outputs
I.
Stoichiometry Defined
Stoichiometry is the mathematics of chemicals and chemical reactions.
One “heart” of stoichiometry is the repeated bridging of the
microscopic/macroscopic divide. We have only one mathematical concept that
achieves this: the mole. Thus, a large part of stoichiometry is the manipulation of
moles.
The second “heart” of stoichiometry involves reactions only. Because a balanced
reaction is a statement of ratios of the substances involved, we often develop
microscopic knowledge of one component of the reaction and leverage it to find out
something about a different component of the reaction.
If in doubt, convert to moles.
I.
Converting to/from moles
We have two “facts” about moles: Avogadro’s number and molar masses. One relates
to numbers of particles and the other relates to masses.
Converting from numbers to moles:
423eggs 1 dozen   35.3 dozen
 12 eggs
6.5110
19


1 m ole
4

atom s

1
.
08

10
m oles
23

 6.02  10 atom s
Converting from moles to numbers:
18.3 dozen 12 eggs  2.20  102 eggs
 1 dozen 
 6.02  1023 m olecules
  1.79  1024 m olecules
2.98 m oles
1 m ole


I.
Converting to/from moles
8.28 m oles NaCl 58.4 g NaCl   484 g NaCl
 1 m ole NaCl 
 1 m ole NaCl 
  45.9 nm oleNaCl
2.68 g NaCl
 58.4 g NaCl 
I.
Mole relationships within a compound
• In one mole of H2O, there are one mole of oxygen
atoms (one O atom per molecule)
• In one mole of H2O, there are two moles of hydrogen
atoms (two H atoms per molecule)
For example:
 3 m ole H 
  58.2 m oles H atom s
19.4 m oles NH 3 
 1 m oles NH 3 
I.
Moles and reactions
But remember, the coefficients represent ratios. So then if, instead of 3 moles of
hydrogen, I actually have 19.4 moles of hydrogen. Then the reaction would consume
6.47 moles of nitrogen because:
 1 m ole N 2 
  6.47 m oles N 2
19.4 m oles H 2 
 3 m oles H 2 
I.
Quick list
•
•
•
•
•
Massmoles
Molesnumbers
Numbersmass
Numbersmoles
Moles Amoles B
(in balanced rxn)
If in doubt, convert something to moles!
Therefore we can also do any of
these in combination. To go
from mass to numbers, convert
mass to moles and then moles to
numbers.
But notice, pretty much any
combination you devise has to go
through moles. Moles are the
linchpin to stoichiometry.
II.
Given the balanced reaction:
2 NaOH(aq) + Cl2(g)  NaOCl(aq) + NaCl(aq) + H2O(l)
How many grams of NaOH are needed to react with 25.0 g of Cl2?

 2 m ol NaOH  40.0 g NaOH 


  28.2 g NaOH
70
.
9
g
Cl
1
m
ol
Cl
1
m
ol
NaOH

2 
2


25.0 g Cl2  1 m ol Cl2
How many molecules of NaOCl are produced from the above
reaction?

 1 m ol NaOCl 

 
70
.
9
g
Cl
1
m
ol
Cl
2 
2


25.0 g Cl2  1 m ol Cl2
 6.02  1023 m oleculesNaOCl 

  2.12  1023 m oleculesNaOCl
1 m ol NaOCl


Reaction Yields
The reaction yields you calculate from the balanced reaction are called theoretical
yields. It is the amount you are supposed to get. It is not usually the amount you
actually get. You are clumsy (admit it). And there are usually other reactions going
on that use up your reactants making other things.
The actual yield is not something you get from theory. It is an experimental fact and
can only be learned from doing calculations with numbers from the experiment.
actual yield
Percent yield 
theoretical yield
Often written as a percent
Normal stoichiometry calculation
Limiting Reagents (I)
molecular picture
Limiting Reagents (II)
Balanced rxn
To determine which reagent is the limiting reagent, divide the moles of each reagent
by its stoichiometric coefficient. I call this the limited reagent lovin’ index (LRLI).
Whichever LRLI is smallest is the limiting reagent.
2 C8H18 + 27 O2  16 CO2 + 18 H2O
3 moles
16 moles
3 moles octane / 2 = 1.5
16 moles oxygen / 27 = 0.59
smallest
Once you have the limiting reagent, do two things:
1. Base all stoichiometric calculations off of your limiting reagent. It determines your
theoretical yields, etc.
2. Throw away all the LRLI. You won’t use it again.
Stoichiometry Possibilities
% yield
Volume A
grams B
grams A
Moles A
atoms/molecules A
Moles B
atoms/molecules A
Limiting
Reagent
% yield
Volume A
grams B
grams A
Balanced rxn
Moles A
atoms/molecules A
Moles B
atoms/molecules A
S
Solutions
• Solutions are homogeneous mixtures (think
salt dissolved in water).
• Mixtures do not have molecular weights, so
“converting to moles” is different.
• Molarity (M) measures the moles of solute
(the salt) per liter of solvent (the water)
• Molarity is a lot like a density. Instead of mass
per volume its moles per volume.
S
Solutions (II)
• Molarity serves the same stoich fxn as molar weights. That is,
it is another way to convert to moles.
% yield
Volume A
grams B
grams A
Balanced rxn
Liters sol’n
Moles A
atoms/molecules A
Moles B
Molarity B
atoms/molecules A
S
How many mL of 0.300 M sulfuric acid is to be added to 50.0 mL of 0.250 M barium
hydroxide solution to complete the following balanced reaction:
BaOH 2 (aq)  H2 SO4 (aq)  BaSO4 ( s)  2 H2O(l )
A. Convert to moles
B. Moles B  Moles A
C. Moles A  volume A
.050 L sol' n. 0.250 m oles BaOH 2  1 m ole H 2 SO4  1 L H 2 SO4 sol' n. 
 1 L BaOH 2 sol' n.  1 m ole BaOH 2  0.3 m ole H 2 SO4 


=0.042 L
= 42 mL


S
Solutions (III)
• While the solute is primarily what we use in solution
stoichiometry, the amount of solvent matters too.
• Increasing the amount of solvent is called “diluting”
the solution.
• The mathematics of dilution is quite simple:
Mi Vi  M f V f
“i” means inital
“f” means final
molarity
volume
• What volume of 18.0 M H2SO4 is required to prepare 250.0 mL
of 0.500 M aqueous H2SO4?
Mi Vi  M f V f
Re-solve equation for the unknown and substitute in the values:
Vi 
M f V f
Mi
0.500 M  250 mL

 6.94 mL
18.0 M
Describe how you would prepare 2.10 L of a 0.360 M solution
of glacial acetic acid starting with concentrated sulfuric acid
that is 12.5 M.
Mi Vi  M f V f
Vi 
M f V f
Mi
0.360 M  2.10 L

 0.0605L
12.5 M
After extracting the 60.5 mL of glacial acetic acid, you would then dilute
until you have 2.10 L, in other words you would add approximately 2.04 L.
Titration
• Lab technique—details don’t matter for us
• Just stoichiometry—nothing new for us
% yield
Volume A
grams B
grams A
Balanced rxn
Liters sol’n
Moles A
atoms/molecules A
Moles B
Molarity B
atoms/molecules A
Titration example
• What is the molarity of a sulfuric acid solution if a 25.0 mL
sample is titrated to equivalence with 50.0 mL of 0.150 M
sodium hydroxide solution?
Titration-speak for “reacted with”
H2SO4(aq) + 2 NaOH(aq)  Na2SO4(aq) + 2 H2O(l)


1
m ol H 2 SO4
  0.150
L
 2 m ol NaOH  0.025 L H 2 SO4 sol' n. 
.050 L NaOH sol' n. 0.150 m ol NaOH  1 m ol H 2 SO4 

1L
Convert NaOH to moles
Convert to molarity (check it out)
Moles NaOH to moles sulf. acid
The reaction between sulfuric acid and potassium hydroxide can be described
by the equation Matt will write on a side board. A 25.00 mL sample of sulfuric
acid requires 42.61 mL of of a 0.1500 M potassium hydroxide solution to titrate
to its end point. Calculate the concentration of the sulfuric acid solution.
Percent Composition
• Percent Composition: Identifies the elements present
in a compound as a mass percent of the total
compound mass.
• The mass percent is obtained by dividing the mass of
each element by the total mass of a compound and
converting to percentage.
Percent Composition example
Saccharin: C7H5NO3S
FW = 183.18
g
7
m ole
%C 
 45.89%
g
183.18
m ole
12.01
g
1
m ole  7.648%
%N 
g
183.18
m ole
14.01
g
5
m ole  2.751%
%H 
g
183.18
m ole
1.008
g
3
m
ole
%O 
 26.20%
g
183.18
m ole
16.00
g
1
m ole  17.50%
%S 
g
183.18
m ole
32.06
Percent Composition Backwards
• Start with percents, and work back to mole ratios
• Percents are independent of mass, so choose a convenient
mass (100 g is a good choice)
A compound has the following mass percents: 27.931% Fe, 24.057% S, and 48.012%
O. What is its formula?
Assume 100 g of total mass, then
percents of each element turn into
masses of each element
Now convert to moles to get the mole ratio
 1 m ol Fe 
  0.5001m ol Fe
27.931 g Fe
 55.847 g Fe 
 1 m ol S 
  0.7502 m ol S
24.057 g S 
 32.066 g S 
 1 m ol O 
  3.000 m ol O
48.012 g O 
 15.999 g O 
The lowest whole
number ratios emerging
from these three
numbers are:
2 Fe, 3 S, and 12 O. The
formula from this is
Fe2S3O12, which can be
written more
informatively as
Fe2(SO4)3.
Mass Spectrometer
Mass Spectrograph
Formulas
•
•
•
The percent composition technique just given finds the lowest whole number ratio
of atoms in the formula. This formula is called the empirical formula.
The following compounds have the same empirical formula (CH): acetylene,
benzene. They have different molecular formulas.
Mass spectrometry can find a molar weight (but gives no info on what elements
are in the compound)
So percent composition and mass spectrometry combined can find the molecular formula
A compound with 92.3% C and 7.74% H was subjected to mass spectrometry and
found to have a molecular weight of 26.038 amu. What is its molecular formula?
1.
2.
3.
Use percent comp. to get moles of each.
Construct empirical formula
Use mass spec. number to get molecular
formula.
92.3 g C 
1 m ol C 
  7.68 m ol C
 12.011 g C 
7.74 g C  1 m ol H   7.68 m ol H
 1.008 g C 
This is a 1:1 ratio. The empirical
formula CH has a molecular mass of
13.019 amu.
26.038amu/13.019amu = 2
C2 H 2
The compound most responsible for the odor of garlic has the empirical formula
C3H5S2. What is its percent composition?
Mass spectrographic analysis shows its molar weight to be 146 g/mol. What is
its molecular formula?
Combustion Analysis
•
Combustion analysis is one of the most common methods for
determining empirical formulas.
•
A weighed compound is burned in oxygen and its products
analyzed by a gas chromatogram.
•
It is particularly useful for analysis of hydrocarbons. (C, H, O)
How do you get oxygen?
Subtract the mass of carbon and hydrogen from the
sample mass. The rest of the mass is oxygen.
The CO2 and water will weigh more together than the
sample, because some of the oxygen came from the air,
adding to the mass!
Combustion Analysis Example
Ascorbic acid is known to contain only C, H, and O. A 6.49 mg sample was subjected to
combustion analysis which produced 9.74 mg CO2 and 2.64 mg H2O. What is the
empirical formula of ascorbic acid?
1. Get the masses of carbon and
hydrogen
2. Get the mass of oxygen
3. Get the moles of each component
4. Construct an empirical formula.
 12.01 g C 
  0.00266g C
44
.
01
g
CO
2 



0.00264g H 2O  2.016 g H   0.000295g H
 18.02 g H 2O 
0.00974g CO2 
6.94 mg ascorbicacid  2.66 mg C  0.295mg H   3.53 mg O
 2.66  103 g C 


 12.011 g C 
 2.21 104 m olesC
Putz around to find the ratio. In
this case the C:H:O ratio is:
 0.295 103 g H 


 1.008 g H 
 2.93  104 m olesC
4
1: :1
3
 3.53  103 g O 


 16.00 g O 
 2.21 104 m olesO
This gives an empirical formula of C3H4O3.
Combustion analysis of an unknown compound containing carbon and hydrogen
produced 4.544 g of carbon dioxide and 2.322 g of water. What is the empirical
formula of the compound?
C2H5