Download Stoichiometry - WordPress.com

Lecture Presentation
Chapter 3
Chemical Reactions
and Reaction
Stoichiometry
© 2015 Pearson Education, Inc.
James F. Kirby
Quinnipiac University
Hamden, CT
Stoichiometry
• The study of the mass relationships in
chemistry
• Based on the Law of Conservation of
Mass (Antoine Lavoisier, 1789)
“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
© 2015 Pearson Education, Inc.
Chemical Equations
Chemical equations are concise
representations of chemical reactions.
Stoichiometry
© 2015 Pearson Education, Inc.
What Is in a Chemical Equation?
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
Reactants appear on the left
side of the equation.
Stoichiometry
© 2015 Pearson Education, Inc.
What Is in a Chemical Equation?
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
Products appear on the right
side of the equation.
Stoichiometry
© 2015 Pearson Education, Inc.
What Is in a Chemical Equation?
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
The states of the reactants and products are written
in parentheses to the right of each compound.
(g) = gas; (l) = liquid; (s) = solid;
(aq) = in aqueous solution
© 2015 Pearson Education, Inc.
Stoichiometry
What Is in a Chemical Equation?
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
Coefficients are inserted to balance the equation
to follow the law of conservation of mass.
Stoichiometry
© 2015 Pearson Education, Inc.
Why Do We Add Coefficients Instead of
Changing Subscripts to Balance?
• Hydrogen and oxygen can make water
OR hydrogen peroxide:
 2 H2(g) + O2(g) → 2 H2O(l)
 H2(g) + O2(g) → H2O2(l)
Stoichiometry
© 2015 Pearson Education, Inc.
Three Types of Reactions
• Combination reactions
• Decomposition reactions
• Combustion reactions
Stoichiometry
© 2015 Pearson Education, Inc.
Combination Reactions
• In combination
reactions two or
more substances
react to form one
product.
• Examples:
– 2 Mg(s) + O2(g)
– N2(g) + 3 H2(g)
– C3H6(g) + Br2(l)
© 2015 Pearson Education, Inc.
2 MgO(s)
2 NH3(g)
C3H6Br2(l)
Stoichiometry
Decomposition Reactions
• In a decomposition
reaction one
substance breaks
down into two or
more substances.
• Examples:
– CaCO3(s)
– 2 KClO3(s)
– 2 NaN3(s)
© 2015 Pearson Education, Inc.
CaO(s) + CO2(g)
2 KCl(s) + 3O2(g)
2 Na(s) + 3 N2(g)
Stoichiometry
Combustion Reactions
• Combustion reactions
are generally rapid
reactions that produce
a flame.
• Combustion reactions
most often involve
oxygen in the air as a
reactant.
• Examples:
– CH4(g) + 2 O2(g)
– C3H8(g) + 5 O2(g)
© 2015 Pearson Education, Inc.
CO2(g) + 2 H2O(g)
3 CO2(g) + 4 H2O(g)
Stoichiometry
Formula Weight (FW)
• A formula weight is the sum of the atomic
weights for the atoms in a chemical formula.
• This is the quantitative significance of a
formula.
• The formula weight of calcium chloride,
CaCl2, would be
Ca: 1(40.08 amu)
+ Cl: 2(35.453 amu)
110.99 amu
Stoichiometry
© 2015 Pearson Education, Inc.
Molecular Weight (MW)
• A molecular weight is the sum of the atomic
weights of the atoms in a molecule.
• For the molecule ethane, C2H6, the molecular
weight would be
C: 2(12.011 amu)
+ H: 6(1.00794 amu)
30.070 amu
Stoichiometry
© 2015 Pearson Education, Inc.
Question
• Calculate the molecular weight/formula
weight for each of the following:
– Sulfur trioxide
– Barium phosphate
– Silver nitrate
15
Stoichiometry
© 2015 Pearson Education, Inc.
Ionic Compounds and Formulas
• Remember, ionic compounds exist with
a three-dimensional order of ions.
There is no simple group of atoms to
call a molecule.
• As such, ionic compounds use empirical
formulas and formula weights (not
molecular weights).
Stoichiometry
© 2015 Pearson Education, Inc.
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 =
(FW of the compound)
× 100
Stoichiometry
© 2015 Pearson Education, Inc.
Percent Composition
So the percentage of carbon in ethane is
(2)(12.011 amu)
%C =
=
(30.070 amu)
24.022 amu
30.070 amu
× 100
= 79.887%
Stoichiometry
© 2015 Pearson Education, Inc.
% Composition
• Calculate the percent composition of iron
in a sample of iron (III) oxide
1. Determine formula: Fe2O3
2. Calculate formula
19 mass of compound
• Fe = 2 x 55.85 = 111.70 amu
• O = 3 x 16.00 = 48.00 amu
• Total mass: 111.70 + 48.00 = 159.70 amu
Stoichiometry
© 2015 Pearson Education, Inc.
% Composition
3. Divide total mass of compound by total
mass of element
total mass of element
% by mass 
X 100
total mass of compound
20
111.70amu
% mass Fe 
X 100  69.9%
159.70amu
Stoichiometry
© 2015 Pearson Education, Inc.
Question
• Calculate the percent oxygen in a
sample of potassium chlorate.
21
Stoichiometry
© 2015 Pearson Education, Inc.
Avogadro’s Number
• In a lab, we cannot
work with individual
molecules. They are
too small.
• 6.02 × 1023 atoms
or molecules is an
amount that brings
us to lab size. It is
ONE MOLE.
• One mole of 12C has
a mass of 12.000 g.
© 2015 Pearson Education, Inc.
Stoichiometry
Molar Mass
• A molar mass is the mass of
1 mol of a substance (i.e., g/mol).
• The molar mass of an
element is the atomic
weight for the element
from the periodic table.
If it is diatomic, it is twice
that atomic weight.
• The formula weight (in
amu’s) will be the same
number as the molar mass
(in g/mol).
© 2015 Pearson Education, Inc.
Stoichiometry
Using Moles
Moles provide a bridge from the molecular
scale to the real-world scale.
Stoichiometry
© 2015 Pearson Education, Inc.
Mole Relationships
• One mole of atoms, ions, or molecules contains
Avogadro’s number of those particles.
• One mole of molecules or formula units contains
Avogadro’s number times the number of atoms or
ions of each element in the compound.
Stoichiometry
© 2015 Pearson Education, Inc.
Question
1. How many atoms of silver are in 3.50 moles of silver?
2. Determine the number of moles of carbon disulfide in
34.75 grams of carbon disulfide.
3. Determine the number of sulfur atoms in 34.75 grams
of carbon disulfide.
Stoichiometry
© 2015 Pearson Education, Inc.
Question
• How many grams of oxygen are present
in 5.75 moles of aluminum oxide?
Stoichiometry
© 2015 Pearson Education, Inc.
Question
Determine the number of fluorine atoms in
24.24 grams of sulfur hexafluoride.
Stoichiometry
© 2015 Pearson Education, Inc.
Determining Empirical Formulas
One can determine the empirical formula
from the percent composition by following
these three steps.
Stoichiometry
© 2015 Pearson Education, Inc.
Determining Empirical Formulas—
an Example
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
© 2015 Pearson Education, Inc.
Determining Empirical Formulas—
an Example
Assuming 100.00 g of para-aminobenzoic acid,
C:
H:
N:
O:
1 mol
12.01 g
1 mol
5.14 g ×
1.01 g
1 mol
10.21 g ×
14.01 g
1 mol
23.33 g ×
16.00 g
61.31 g ×
= 5.105 mol C
= 5.09 mol H
= 0.7288 mol N
= 1.456 mol O
Stoichiometry
© 2015 Pearson Education, Inc.
Determining Empirical Formulas—
an Example
Calculate the mole ratio by dividing by the smallest number
of moles:
© 2015 Pearson Education, Inc.
C:
5.105 mol
0.7288 mol
= 7.005 ≈ 7
H:
5.09 mol
0.7288 mol
= 6.984 ≈ 7
N:
0.7288 mol
0.7288 mol
= 1.000
O:
1.458 mol
0.7288 mol
= 2.001 ≈ 2
Stoichiometry
Determining Empirical Formulas—
an Example
These are the subscripts for the empirical formula:
C7H7NO2
Stoichiometry
© 2015 Pearson Education, Inc.
Determining a Molecular Formula
• Remember, the number of atoms in a
molecular formula is a multiple of the
number of atoms in an empirical
formula.
• If we find the empirical formula and
know a molar mass (molecular weight)
for the compound, we can find the
molecular formula.
Stoichiometry
© 2015 Pearson Education, Inc.
Determining a Molecular Formula—
an Example
• The empirical formula of a compound
was found to be CH. It has a molar
mass of 78 g/mol. What is its molecular
formula?
• Solution:
Whole-number multiple = 78/13 = 6
The molecular formula is C6H6.
Stoichiometry
© 2015 Pearson Education, Inc.
Combustion Analysis
• Compounds containing C, H, and O are routinely analyzed
through combustion in a chamber like the one shown in
Figure 3.14.
– C is determined from the mass of CO2 produced.
– H is determined from the mass of H2O produced.
– O is determined by the difference after C and H have been determined.
Stoichiometry
© 2015 Pearson Education, Inc.
Combustion Analysis
Suppose that 18.8 grams of glucose was
burned in a combustion train resulting in 27.6
grams of carbon dioxide and 11.3 grams of
water. Calculate the empirical and molecular
formula of glucose. Molar mass = 180 g/mol
37
Combustion Analysis
Steps:
1. Convert 27.6 g CO2 into g of C
2. Convert 11.3 g H2O into g H
3. Calculate g O = g sample - (g C + g H)
4. Find empirical formula as before
(g to moles, mole ratio)
Copyright McGraw-Hill 2009
38
Combustion Analysis
Molecular formula
= molar mass/empirical mass
= 180/30 = 6
Multiply through empirical formula to
obtain new subscripts
Molecular formula = C6H12O6
Copyright McGraw-Hill 2009
39
Question
Combustion of 5.50 g of benzene
produces 18.59 g CO2 and 3.81 g H2O.
Determine the empirical formula and the
molecular formula of benzene given that
its molar mass is 78 g/mol.
Empirical formula – CH
Molecular Formula – C6H6
40
Quantitative Relationships
• The coefficients in the balanced equation show
 relative numbers of molecules of reactants and
products.
 relative numbers of moles of reactants and
products, which can be converted to mass. Stoichiometry
© 2015 Pearson Education, Inc.
Stoichiometric Calculations
We have already seen in this chapter how to
convert from grams to moles or moles to
grams. The NEW calculation is how to
compare two DIFFERENT materials, using
the MOLE RATIO from the balanced
equation!
Stoichiometry
© 2015 Pearson Education, Inc.
An Example of a Stoichiometric Calculation
• How many grams of water can be
produced from 1.00 g of glucose?
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
• There is 1.00 g of glucose to start.
• The first step is to convert it to moles.
Stoichiometry
© 2015 Pearson Education, Inc.
An Example of a Stoichiometric Calculation
• The NEW calculation is to convert
moles of one substance in the equation
to moles of another substance.
• The MOLE RATIO comes from the
balanced equation.
• C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
Stoichiometry
© 2015 Pearson Education, Inc.
An Example of a Stoichiometric Calculation
• The last step is to convert the moles of
water to grams of water.
Stoichiometry
© 2015 Pearson Education, Inc.
Limiting Reactants
• 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, the H2).
Stoichiometry
© 2015 Pearson Education, Inc.
Limiting Reactants
In the example below, the O2 would be the
excess reagent.
Stoichiometry
© 2015 Pearson Education, Inc.
Limiting Reactants
• The limiting reactant is used in all stoichiometry
calculations to determine amounts of products
and amounts of any other reactant(s) used in a
reaction.
Stoichiometry
© 2015 Pearson Education, Inc.
Practice Problem
• Given the reaction:
2Mg(s) + O2(g)
2MgO(s)
• If we have 42.5 g of Mg and 33.8 g of O2,
what is the limiting reagent?
Stoichiometry
© 2015 Pearson Education, Inc.
2Mg(s) + O2(g)
Given:
Find:
Concept Plan:
2MgO(s)
42.5 g Mg, 33.8 g O2
Limiting reagent
g Mg
mol Mg
1 mol Mg
24.31 g
g O2
2 mol MgO
2 mol Mg
mol O2
1 mol o2
32.00 g o2
mol MgO
mol MgO
2 mol MgO
1 mol o2
Relationships: Molar mass Mg = 24.31 g and 2 mol Mg: 2 mol MgO
Molar mass O2 = 32.00 g and 1 mol O2 : 2 mol MgO
Stoichiometry
© 2015 Pearson Education, Inc.
Solution:
2Mg(s) + O2(g)
2MgO(s)
1 mol Mg
2 mol MgO
42.5 g Mg 

 1.75 mol MgO
24.31 g Mg
2 mol Mg
1 mol o2
2 mol MgO
33.8 g o2 

 2.11mol MgO
32.00 g o2 1 mol o2
Check:
© 2015 Pearson Education, Inc.
•Compare the number of moles of MgO made by
each reactant.
•Mg gives the least number of moles of MgO.
Stoichiometry
•Mg is the limiting reagent.
Question
• When 35.50 grams of nitrogen react with
25.75 grams of hydrogen, how many grams
of ammonia are produced?
• How many grams of excess reagent remain
in the reaction vessel?
Stoichiometry
© 2015 Pearson Education, Inc.
Theoretical Yield
• The theoretical yield is the maximum
amount of product that can be made.
– In other words, it’s the amount of product
possible as calculated through the
stoichiometry problem.
• This is different from the actual yield,
which is the amount one actually
produces and measures.
Stoichiometry
© 2015 Pearson Education, Inc.
Percent Yield
One finds the percent yield by
comparing the amount actually obtained
(actual yield) to the amount it was
possible to make (theoretical yield):
Percent yield =
actual yield
theoretical yield
× 100
Stoichiometry
© 2015 Pearson Education, Inc.
Question
• What is the percent yield for the reaction
PCl3(g) + Cl2(g) → PCl5(g)
if 119.3 g of PCl are formed when 61.3 g
of Cl2 react with excess PCl3?
Stoichiometry
© 2015 Pearson Education, Inc.
Chapter 3 Post-Test – show all calculations!
When a 2.00-g strip of zinc metal is placed in an aqueous
solution containing 2.50 g of silver nitrate, the reaction is
Zinc (s) + Silver Nitrate (aq)
Silver(s) + Zinc Nitrate (aq)
(a) Write and balance the above equation.
(b) Which reactant is limiting?
(c) How many grams of silver form?
(d) How many grams of zinc nitrate form?
(e) How many grams of the excess reactant are left at
the end of the reaction?
Stoichiometry
© 2015 Pearson Education, Inc.