# Chapter 3 Stoichiometry: Calculations with Chemical Formulas and ... - PowerPoint

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```					          Chemistry, The Central Science, 10th edition
Theodore L. Brown, H. Eugene LeMay, Jr.,
and Bruce E. Bursten

Chapter 3
Stoichiometry:
Calculations with Chemical
Formulas and Equations
John D. Bookstaver
St. Charles Community College
St. Peters, MO                                   Stoichiometry

 2006, Prentice-Hall
Law of Conservation of Mass
“We may lay it down as an
incontestable axiom that, in all
the operations of art and nature,
nothing is created; an equal
amount of matter exists both
before and after the experiment.
Upon this principle, the whole art
of performing chemical
experiments depends.”
--Antoine Lavoisier, 1789
Stoichiometry
Chemical Equations

Concise representations of chemical
reactions

beginning of reaction        end of reaction
Stoichiometry
Anatomy of a Chemical Equation

CH4 (g) + 2 O2 (g)                         CO2 (g) + 2 H2O (g)

In this reaction, the reactants and products are said to be “balanced”
A balanced equation contains the lowest possible whole number coefficients
Stoichiometry
Anatomy of a Chemical Equation

CH4 (g) + 2 O2 (g)            CO2 (g) + 2 H2O (g)

Reactants appear on the
left side of the equation.                 Stoichiometry
Anatomy of a Chemical Equation

CH4 (g) + 2 O2 (g)             CO2 (g) + 2 H2O (g)

Products appear on the
right side of the equation.                 Stoichiometry
Anatomy of a Chemical Equation

CH4 (g) + 2 O2 (g)               CO2 (g) + 2 H2O (g)

The states of the reactants and products       (s) = solid
(l) = liquid
are written in parentheses to the right of           gas
(g) =Stoichiometry
each compound.                               (aq) = aqueous
solution
Anatomy of a Chemical Equation

CH4 (g) + 2 O2 (g)             CO2 (g) + 2 H2O (g)

Coefficients are inserted to
balance the equation.                   Stoichiometry
Subscripts and Coefficients Give
Different Information

• Subscripts tell the number of atoms of
each element in a molecule
Stoichiometry
Subscripts and Coefficients Give
Different Information

• Subscripts tell the number of atoms of
each element in a molecule
• Coefficients tell the number of          Stoichiometry
molecules
Reaction
Types
Stoichiometry
Why Categorize Reaction Types?

• There are three basic reaction types
considered in this chapter
• It is worthwhile recognizing basic
patterns of reactivity so that products of
chemical reactions can be predicted
before reactions are actually carried out

Stoichiometry
Combination Reactions

• Two or more
substances
react to form
one product

• Examples:
N2 (g) + 3 H2 (g)  2 NH3 (g)      Products contain only non-metals
C3H6 (g) + Br2 (l)  C3H6Br2 (l)   (molecular compounds)

2 Mg (s) + O2 (g)  2 MgO (s)                             Stoichiometry
Product is an ionic compound
2 Mg (s) + O2 (g)  2 MgO (s)

In the course of this reaction, the shiny, grey ribbon of Mg reacts with O2(g) in a flame
to produce a white powder

What observations may indicate whether a reaction has occurred or not?
- Evolution of a gas (bubbles)
- Formation of a solid (precipitate) or liquid (harder to detect)        Stoichiometry
- Evolution (or consumption) of heat
- Color change
Decomposition Reactions

• One substance breaks
down into two or more
substances. Usually
brought on by heating.

What would happen to MgCO3(s)
if it were heated?

• Examples:
CaCO3 (s)  CaO (s) + CO2 (g)
2 KClO3 (s)  2 KCl (s) + 3O2 (g)                        Stoichiometry

2 NaN3 (s)  2 Na (s) + 3 N2 (g)
Combustion Reactions

• Rapid reactions that
produce a flame
• Most often involve
hydrocarbons
reacting with oxygen
in the air

• Examples:
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)     Stoichiometry
Atomic Weights, Formula
Weights, and Molecular
Weights

Stoichiometry
Atomic Weight and Formula Weight
• The weight of an atom of some element (atomic weight) is
equivalent to its periodic table mass, expressed in atomic mass
units (amu)
1 amu = 1.66054 x 10-24g
• Ionic compounds involve huge numbers of ions held together by
electrostatic attraction – the formula of an ionic compound is
represented by the formula unit (example, NaCl represents the
formula unit for sodium chloride)

• So, the “formula weight” (FW) of calcium chloride, NaCl, would
be

Na : 22 .989770 _ amu
Cl : 35 .453 _ amu

 58 .442 4 _ amu
Stoichiometry
Molecular Weight (MW)

• Sum of the atomic weights of the atoms
in a molecule
• For the molecule ethane, C2H6, the
molecular weight would be
C: 2(12.0 amu)
+ H: 6(1.0 amu)
30.0 amu
Often, the text shortens the atomic masses of elements for simplicity. Stoichiometry
You should use the same weights given in your text’s periodic table when
solving problems
Percent Composition

One can find the percentage of the mass
of a compound that comes from each of
the elements in the compound by using
this equation:

(number of atoms)(atomic weight)
% element =                                      x 100
(FW of the compound)

Stoichiometry
Percent Composition

So the percentage of carbon in ethane
(C2H6) is…

(2)(12.0 amu)
%C =                   x 100%
(30.0 amu)
24.0 amu
=              x 100%
30.0 amu
= 80.0%
Stoichiometry
Example – Percentage Composition
• Calculate the percentage carbon by
mass in sodium acetate
Na: 22.989770 amu
C: 2(12.0107 amu)
Formula: NaC2H3O2               H: 3(1.00794 amu)
O: 2(15.9994 amu)
The formula weight for NaC2H3O2 = 82.03379 amu

%C 
2 _ C 12 .0107 amu / C  100 %  29 .2823 %
82 .0337 9 amu
Stoichiometry
Moles

Stoichiometry
Moles and Dozens
• The term “dozen” is used to simplify
descriptions of larger numbers of items (for
example, two dozen apples instead of twenty
four apples)
• Chemists use the term “mole” to describe
large numbers of
atoms/ions/molecules/particles, etc.
• A dozen means twelve objects, a mole means
6.022 x 1023 objects (Avogadro’s number)
• There are 6.022 x 1023 atoms of 12C in a 12g
sample of isotopically pure 12C
Stoichiometry
• How many atoms of 12C are there in
one mole of 12C?
6.022x1023 12C atoms
• How many H2O molecules are there in
one mole of H2O? 6.022x1023 H2O molecules
• How many frosh are there in one mole
of frosh?... 6.022x1023 frosh (see: Lane Hall)

guacomole
Using Avogadro’s Number          Need to know:
number of H2O molecules per mole of H2O
number of O atoms per H2O molecule

• How many oxygen atoms are there in 1
mol of H2O?
 6.022x10 23 _ H 2 O _ molecules  1atom _ O       
1mol _ H 2O
                                                    6.022x10 23 _ O _ atoms
 1H O _ molecule 
        1 _ mol _ H 2 O           2               
• How many hydrogen atoms are there in
1 mol of H2O?
 6.022x10 23 _ H 2 O _ molecules  2atoms _ H 
1mol _ H 2 O 
                                 
 1H O _ molecule   1.204x10 _ H _ atoms

24

        1 _ mol _ H 2 O           2               
• How many H atoms are there in 0.400
mol H2O?
 6.022x1023 _ H 2O _ molecules  2atoms _ H 
0.400mol _ H 2O 
                                                  4.817 x1023 _ H _ atoms
 1H O _ molecule 
        1 _ mol _ H 2O          2                                Stoichiometry

H2O has a “molar mass”
of 18.0 g/mol

 1.66054x10 24 g  6.022x10 23          g
18.0amu
                  
               18.0

      1amu            mol              mol
• 1 atom of 12C has a
mass of 12 amu        This same relationship applies
• 1 mole of 12C has a   to all atoms, ions, molecules, etc.
mass of 12 g
Stoichiometry
• Using the molar mass and dimensional
analysis, it is an easy exercise to Carbon : 6(:12.(0107 amuamu )
Hydrogen 12 1.00794
)

convert masses to moles, and to Oxygen : 6(15.9994 amu )
numbers of atoms/molecules          180 .1558 amu                                        8

• How many moles of glucose (C6H12O6)
are there in a teaspoon (5.0g)?
So, 1 mol C6H12O6 weighs 180.15588g
 1mol _ C6 H12O6      
5.0 g _ C6 H12O6 
 180.1558 g _ C H O    2.7 7 *102 _ moles _ C6 H12O6
         8     6 12 6 

• How many C-atoms are there in this
mass of glucose?
 6.022x10 23 _ C6 H 12 O6 _ molecules       6 _ C _ atoms   
2.7 7 *10 _ moles _ C6 H 12 O6 
2


 1 _ C H O _ molecule   1.0 x10 C _ atoms

Stoichiometry
23

        1 _ mol _ C6 H 12 O6                6 12 6          
Finding
Empirical
Formulas
An empirical formula is a formula which presents elements in relative numbers.
Example: the molecular formula for hydrogen peroxide is H2O2. The empirical
formula for hydrogen peroxide is HO.
This is obtained from mole ratios of the elements involved.

Stoichiometry
Empirical Formula of Hydrogen
Peroxide
• In an analysis of hydrogen peroxide, it
was found that the sample contained
5.93% hydrogen, and 94.07% oxygen,
by mass. What is the empirical formula
of hydrogen peroxide?
If you could express the percentages of these elements as masses,
you could convert each of them to moles

In 100g of hydrogen peroxide, 5.93g will be H and 94.07g will be O

Knowing the number of moles of each element is the same as knowing
the number of atoms of each element                                       Stoichiometry

Their mole ratio will yield the empirical formula for hydrogen peroxide
Empirical Formula of Hydrogen
Peroxide
• In an analysis of hydrogen peroxide, it was found that
the sample contained 5.93% hydrogen, and 94.07%
oxygen, by mass. What is the empirical formula of
hydrogen peroxide?
Step 1: convert percentages to masses (assume there is 100 g of the sample)

So 5.93 g H, 94.07 g O

Step 2: calculate how many moles of each element there are in the sample

 1mol _ H 
 1.00794g _ H   5.883 mol _ H
5.93g _ H               
              
 1mol _ O 
 15.9994g _ O   5.8795 mol _ O
94.07 g _ O              
Stoichiometry
              
Empirical Formula of Hydrogen
Peroxide
Thus there are 5.887 moles of hydrogen and 5.8795 moles of oxygen.

To calculate the mole ratio, divide each element’s number of moles by the
smallest number of moles:

5.887 mol _ H
mol H:                   1.001
5.8795 mol _ O           For every mole of O in hydrogen peroxide,
there is one mole of H, so the empirical
5.8795 mol _ O           formula is HO
mol O:                   1.000
5.8795 mol _ O

Coefficients in molecular and empirical
Stoichiometry
formulas are whole numbers
Calculating Empirical Formulas

One can calculate the empirical formula from
the percent composition

Stoichiometry
Calculating Empirical Formulas
The compound para-aminobenzoic acid (you may have
seen it listed as PABA on your bottle of sunscreen) is
composed of carbon (61.31%), hydrogen (5.14%),
nitrogen (10.21%), and oxygen (23.33%). Find the
empirical formula of PABA.

Stoichiometry
Calculating Empirical Formulas
Convert % to gram quantities and determine the number of moles of each element

Assuming 100.00 g of para-aminobenzoic acid,

C:      61.31 g C x  1 mol C           = 5.105 mol C
12.01 g C
1 mol H
H:       5.14 g H x                    = 5.09 mol H
1.01 g H
1 mol N
N:      10.21 g N x                    = 0.7288 mol N
14.01 g N
1 mol O
O:      23.33 g O x                    = 1.456 mol O
16.00 g O

Stoichiometry
Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number
of moles:
5.105 mol
C:                  = 7.005  7
0.7288 mol

5.09 mol
H:                  = 6.984  7
0.7288 mol

0.7288 mol
N:                  = 1.000
0.7288 mol

1.458 mol
O:                  = 2.001  2
0.7288 mol                         Stoichiometry
Calculating Empirical Formulas

These are the subscripts for the empirical formula:

C7H7NO2                     N
C

O

H         Stoichiometry
Combustion Analysis

• Compounds containing C, H and O are routinely
analyzed through combustion in a chamber like this
– C is determined from the mass of CO2 produced
– H is determined from the mass of H2O produced
– O is determined by difference after the C and H have been
determined
organic compound + O2  CO2 + H2O
Stoichiometry
Determining the moles of C, H, and O enables the
empirical formula to be calculated
Combustion Analysis Problem (CxHy)
• Combustion of toluene, an organic
compound that contains only carbon
and hydrogen, produces 5.86 mg of
CO2 and 1.37 mg H2O. What is the
empirical formula for toluene?
CxHy+ O2  CO2 + H2O
Strategy:
1) Find the number of moles of C (from moles of CO2)
and moles of H (from moles of H2O)
2) Divide by the smallest number to get the empirical formula
Stoichiometry
Combustion Analysis Problem (CxHy)
• massCO2 and massH2O  mol C and mol H:
C 5.86 _ mgCO 2   1g   1mol _ CO   1mol __ C   1.33 x10 mol _ C

                 
 44.0095g _ CO  
1mol
2

CO                  1
4

 1000mg                 2          2

1g  1mol _ H 2O  2mol _ H 
H      1.37 _ mgH2O

         18.0152 g _ H O  1mol _ H O   1.520 x10 mol _ H
                             
4

 1000mg         8     2           2  

• Divide by smallest number of moles to get
subscript for empirical formula
1.52 0 x10 4 mol _ H                       Is the answer C1H1.14? No
H:               4
 1.14
1.331 x10 mol _ C                           Is the answer C1H1? No
1.331 x10 4 mol _ C                When is a number close enough to
C:              4
1                                         Stoichiometry
1.331 x10 mol _ C                   round off?
(should be no more than +/- 2% away)
Combustion Analysis Problem (CxHy)
• Multiply each of these numbers by
whole numbers until both are whole
numbers
this is 0.25%
Multiple 1    2     3          7          different than 8

H: 1.14 2.28 3.42 … 7.98

C:    1    2      3   …     7

So empirical formula is C7H8
Stoichiometry
 8.0  7.98 
            100 %  0.25 %
    8.0     
A Combustion Analysis Problem
(CxHyOz)
• A 0.1005 g sample of menthol, a
compound consisting of carbon,
hydrogen, and oxygen, is combusted,
producing 0.2829 g CO2 and 0.1159 g
H2O. What is the empirical formula of
menthol?
CxHyOz+ O2  CO2 + H2O
Strategy:
1) Convert moles of CO2 to moles of C; moles of H2O to moles of H
2) Find masses of C and H from moles of C and moles of H
3) Use these to find mass of O  find moles of O
Stoichiometry
4) Use mole ratios to get empirical formula
   1mol _ CO2  1mol _ C 
1                 0.2829 _ gCO2 
                   
              6.4281 x103 mol _ C

 44.0095g _ CO2  1mol _ CO2 
moles C, H                           1mol _ H 2O  2mol _ H 
0.1159 _ gH 2O 
                     
               1.2866 x102 mol _ H

 18.01528 g _ H 2O  1mol _ H 2O 

 12.0107g _ C 
2                 6.428 x10
1
3
 1mol _ C   0.077205 g _ C
mol _ C               
              
grams C, H
 1.00794g _ H 
1.286   6               
x10 2 mol _ H 
 1mol _ H   0.012969 g _ H

              
3                 sample mass = mass of C + mass of H + mass of O
grams O
0.1005 g _ sample  0.077205 g _ C  0.012969 g _ H
 0.0103 2 g _ O
 1mol _ O 
moles O       0.01032 g _ O                6.453 x10 4 mol _ O
 15.9994g _ O 
              
4
mole ratios    mol _ C
 9.96
mol _ H
 19 .93
mol _ O
1
Stoichiometry
mol _ O                mol _ O              mol _ O
C10H20O
I’ve got the empirical formula…now what?
• We are usually more interested in a compound’s
molecular formula than its empirical formula – how do
we get the molecular formula once we know the
empirical formula?
• If the molar mass of the substance is known, then the
molecular formula can be determined as follows:
                              
 empirical _ formula _ weight   # _ of _ empirical _ formula _ units _ in _ molecular _ formula
molar _ mass
                              
                              

For example, benzene has an empirical formula of CH,
and a molar mass of ~78 g.mol-1. What is the molecular                   There are 6 CH
formula for benzene?                                                     units in benzene
         g     
     78                                     H
        mol    6                      HC
C
CH
            g                                             Stoichiometry
Benzene: C6H6
 13.01864                              HC       CH
           mol                              C
H
Molecular formula
Stoichiometric Calculations

Stoichiometry refers to the relationships that exist between the
numbers of reactant and product molecules in the balanced
chemical equation.

Let Ω mean “is stoichiometrically equivalent to”
2 mol H2 Ω 1 mol O2 Ω 2 mol H2O
Stoichiometry
Stoichiometric Relationships
• In the combustion problem we looked at earlier, we
converted a certain number of grams of CO2 to moles
of carbon (and g H2O to mol H). In doing this, we used
a stoichiometric relationship: 1 atom of C per 1
molecule of CO2

1 mol CO2 Ω 1 mol C

• The same relationships allow us to indirectly relate
masses of reactants and products together in some
balanced equation (or reactants and reactants,
products and products) – example, what mass of water
can be made from 1.00 g of C6H12O6 and enough O2
to ensure complete reaction?
Stoichiometry
C6H12O6 + 6 O2  6 CO2 + 6 H2O
Stoichiometric Calculations – What mass of H2O
can be made from 1.00 g C6H12O6?
C6H12O6 + 6 O2  6 CO2 + 6 H2O

Starting with 1.00 g of C6H12O6…
we calculate the moles of C6H12O6…
use the coefficients to find the moles of H2O…   Stoichiometry
and then turn the moles of water to grams
Stoichiometric Calculations
From the mass of
substance A you can
use the ratio of the
coefficients of A and B
to calculate the mass
of substance B formed
(if it’s a product) or
used (if it’s a reactant)

Stoichiometry
Stoichiometric Calculations
C6H12O6 + 6 O2  6 CO2 + 6 H2O

• Or, do it all in one calculation:

 1mol _ C 6 H 12 O6     6mol _ H 2 O      18.0 g _ H 2 O 
1.00 g _ C 6 H 12 O6 
 180.0 g _ C H O       
 1mol _ C H O     
 1mol _ H O   0.600g _ H 2 O

            6   12 6            6 12 6              2    

Stoichiometry
Stoichiometric Calculations
C6H12O6 + 6 O2  6 CO2 + 6 H2O

• We can also use the stoichiometric              use these for
relationships in the balanced                dimensional analysis
equation to calculate other things            conversion factors

– What mass of O2 is required to react
6 mol O2  1 mol C6H12O6
with 1.00g C6H12O6?
Ans = 1.07g O2
– What mass of CO2 will be produced? 1 mol C6H12O6  6 mol CO2

– What mass of C6H12O6 is required to
6 mol H2O  1 mol C6H12O6
produce 1.00g H2O?
Stoichiometry
Limiting Reactants
If there were only 0.50 g O2 available for the combustion of 1.00g C6H12O6,
it would not be possible to obtain 0.600g of H2O product, because O2 would
run out before all of the C6H12O6 reacts

rem: it takes at least 1.07 g O2 for all of the C6H12O6 to react

C6H12O6 + 6 O2  6 CO2 + 6 H2O

Because the amount of H2O that can be produced depends
on the amount of O2 present initially, we say that O2 is the
limiting reactant.
Stoichiometry
How Many Cookies Can I Make?

• You can make cookies
until you run out of one
of the ingredients
• Once this family runs
out of sugar, they will
(at least any cookies
you would want to eat)

Stoichiometry
How Many Cookies Can I Make?

• In this example the
sugar would be the
limiting reactant,
because it will limit the
amount of cookies you
can make

Stoichiometry
Limiting Reactants
2H2(g) + O2(g)  2H2O(g)           10H2  10H2O
7O2  14H2O

• The limiting reactant is the reactant present in
the smallest stoichiometric amount
– In other words, it’s the reactant you’ll run out of first (in
this case, H2)

Stoichiometry
Limiting Reactants
2H2(g) + O2(g)  2H2O(g)

In the example below, the O2 would be the excess
reagent (i.e. there would be some O2 left when the
reaction has finished). H2 is the limiting reagent.

Stoichiometry
Theoretical Yield
• The theoretical yield is the amount of product
that can be made upon complete reaction of
the limiting reactant
– In other words it’s the amount of product possible
as calculated through the stoichiometry problem
• The theoretical yield is calculated from the
number of moles of limiting reactant
• This is different from the actual yield, the
amount one actually produces and measures
Br   S   N            S   Sn(C4H9)3    Pd(PPh3)2Cl2    S   S       N
+    2
N   S   Br                       DMF, 150oC, 5h       N       S     S   Stoichiometry
+
by-products
Theoretical Yield
• Q: The most important commercial
process for the production of ammonia
(NH3) is the Haber process:
3H2(g) + N2(g)  2NH3(g)
• How many moles of NH3 can be formed
from reaction of 3.0 mol N2 with 6.0 mol
H 2?

Strategy: find out which reactant is limiting and then
use the number of moles of it to figure out the theoretical yield
Stoichiometry
3H2(g) + N2(g)  2NH3(g)
•     Which reactant is limiting? Three common
ways to check this:
Solve for the amount of NH3 produced by each amount
1)      of reactant. The reactant that produces a smaller
amount of NH3 must be limiting.
–    In this case, we would convert the number of
moles of each reactant to moles of NH3:
 2mol _ NH 3 
H2 is limiting reactant                 1mol _ N   6.0mol _ NH 3
3.0mol _ N 2                              this method also
          2 
gives you the
 2mol _ NH 3                 theoretical yield
 3mol _ H   4.0mol _ NH 3
6.0mol _ H 2              
           2 

–    Thus, since the H2 initially present produces a
smaller amount of NH3, it must be limiting (if we
wanted to produce 6.0 mol NH3 from the N2 Stoichiometry
present, we’d have to add more H2)
3H2(g) + N2(g)  2NH3(g)
2) –   Find how many moles of H2 are needed to react
with this much N2:

 3mol _ H 2   
3.0mol _ N 2 
 1mol _ N       9.0mol _ H 2

          2   
•   Thus, it would take 9.0 mol H2 to react
completely with 3.0 mol N2. There is not
that much H2 present initially (rem: there are
only 6.0 mol H2), so H2 must be limiting.
Stoichiometry
H2 is limiting reactant
3H2(g) + N2(g)  2NH3(g)
–     Take the number of moles of each reactant
present and divide it by the coefficient that
corresponds to it in the balanced chemical
equation. The smallest result indicates the
limiting reactant:
The numbers                       3.0mol _ N 2         
 1mol _ N ( from _ equation)   3.0
                             
H2 is limiting reactant
calculated here                     2                  
only tell us which
         6.0mol _ H 2        
reactant is limiting
 3mol _ H ( from _ equation)   2.0
                             
           2                 

•      Thus H2 is limiting (since 2.0 is smaller than
3.0).

Stoichiometry
Percent Yield
A comparison of the amount actually
obtained to the amount it was possible
to make
Actual Yield
Percent Yield =               x 100%
Theoretical Yield
In the last problem, the theoretical yield was 4.0 mol NH3. If after carrying
out this reaction, you only obtained 3.3 moles of NH3, the % yield would be

 3.3mol _ NH 3 
 4.0mol _ NH  x100%  82.5%  82%
% _ yield                 
             3 
Stoichiometry

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