Transcript Stoichiometry - AP Chemistry

Lecture Presentation
Chapter 3
Chemical Reactions
and Reaction
Stoichiometry
© 2015 Pearson Education, Inc.
James F. Kirby
Quinnipiac University
Hamden, CT
Edited by Brian Fain
Warm Up
• Balance the following equation. Why must
we do so?
__CH4(g) + __O2(g)
___CO2(g)+__H2O(g)
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Warm Up
• Balance the following equation. Why must
we do so?
__CH4(g) + 2 O2(g)
___CO2(g)+ 2H2O(g)
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Warm Up 10/29
1. Which has more mass, a mole of water
(H2O) or a mole of glucose (C6H12O6)?
2. Which contains more molecules, a mole of
water or a mole of glucose?
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Warm Up 11/3
The percentage by mass of
phosphorus in Na3PO4 is
a) 44.0 b) 11.7 c) 26.7 d)18.9.
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Warm Up 11/3
The percentage by mass of
phosphorus in Na3PO4 is
a) 44.0 b) 11.7 c) 26.7 d)18.9.
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Agenda 11/3
1. Lecture Notes: 30 mins
2. Presentations: 30 mins
3. Work Time on Lab Report –
Due Thursday
4. HW Assignments: Finish
Chapter 3: 3.7 – end
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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
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Chemical Equations
Chemical equations are concise
representations of chemical reactions.
Stoichiometry
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What Is in a Chemical Equation?
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
Reactants appear on the left
side of the equation.
Stoichiometry
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What Is in a Chemical Equation?
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
Products appear on the right
side of the equation.
Stoichiometry
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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
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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
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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
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Three Types of Reactions
• Combination reactions
• Decomposition reactions
• Combustion reactions
Stoichiometry
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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)
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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)
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CaO(s) + CO2(g)
2 KCl(s) + O2(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)
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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
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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
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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
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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
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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
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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.
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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).
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Stoichiometry
Using Moles
Moles provide a bridge from the molecular
scale to the real-world scale.
Stoichiometry
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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
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Determining Empirical Formulas
One can determine the empirical formula
from the percent composition by following
these three steps.
Stoichiometry
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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
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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
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Determining Empirical Formulas—
an Example
Calculate the mole ratio by dividing by the smallest number
of moles:
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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
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Ethanol contains 52.2% carbon,
13.0% hydrogen, and 34.8% oxygen
by mass. The empirical formula of
ethanol is
a.
b.
c.
d.
C 2H 5O 2.
C2H6O.
C 2H 6O 2.
C 3H 4O 2.
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Stoichiometry
Ethanol contains 52.2% carbon,
13.0% hydrogen, and 34.8% oxygen
by mass. The empirical formula of
ethanol is
a.
b.
c.
d.
C 2H 5O 2.
C2H6O.
C 2H 6O 2.
C 3H 4O 2.
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Stoichiometry
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
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Determining a Molecular Formula: Example
Determine the molecular formula of a
compound with an empirical formula of
CF2 and a molar mass of 200.04 g/mol
Stoichiometry
© 2015 Pearson Education, Inc.
Determining a Molecular Formula: Example
Determine the molecular formula of a
compound with an empirical formula of
CF2 and a molar mass of 200.04 g/mol
1. Molar mass of Empirical Formula
Stoichiometry
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Determining a Molecular Formula: Example
Determine the molecular formula of a
compound with an empirical formula of
CF2 and a molar mass of 200.04 g/mol
1. Molar mass of Empirical Formula
2. Determine molar masses of different
multiples for a match
Stoichiometry
© 2015 Pearson Education, Inc.
Determining a Molecular Formula: Example
Determine the molecular formula of a
compound with an empirical formula of
CF2 and a molar mass of 200.04 g/mol
1. Molar mass of Empirical Formula:
C = 12.01 g/mol x 1 = 12.01
F = 19.00 g/mol x 2 = 38.00
TOTAL  50.01 g/mol
Stoichiometry
© 2015 Pearson Education, Inc.
Determining a Molecular Formula: Example
Determine the molecular formula of a
compound with an empirical formula of
CF2 and a molar mass of 200.04 g/mol
2. Molar masses of multiples to find a
match…
Multiple
Formula Mass
x2
C2F4
100.02
Stoichiometry
© 2015 Pearson Education, Inc.
Determining a Molecular Formula: Example
Determine the molecular formula of a
compound with an empirical formula of
CF2 and a molar mass of 200.04 g/mol
2. Molar masses of multiples to find a
match…
Multiple
Formula Mass
x2
C2F4
100.02
x3
C3F6
150.03
Stoichiometry
© 2015 Pearson Education, Inc.
Determining a Molecular Formula: Example
Determine the molecular formula of a
compound with an empirical formula of
CF2 and a molar mass of 200.04 g/mol
2. Molar masses of multiples to find a
match…
Multiple
Formula Mass
x2
C2F4
100.02
x3
C3F6
150.03
x4
C4F8
200.04
Stoichiometry
© 2015 Pearson Education, Inc.
Determining a Molecular Formula: Example
Determine the molecular formula of a
compound with an empirical formula of
CF2 and a molar mass of 200.04 g/mol
2. Molar masses of multiples to find a
match…
Multiple
Formula Mass
x2
C2F4
100.02
x3
C3F6
150.03
x4
C4F8
200.04
Stoichiometry
© 2015 Pearson Education, Inc.
Determining a Molecular Formula: Example
Determine the molecular formula of a
compound with an empirical formula of
CF2 and a molar mass of 200.04 g/mol
Is there a simpler way?
Stoichiometry
© 2015 Pearson Education, Inc.
Determining a Molecular Formula: Example
Determine the molecular formula of a
compound with an empirical formula of
CF2 and a molar mass of 200.04 g/mol
Is there a simpler way?
YES!
Stoichiometry
© 2015 Pearson Education, Inc.
Determining a Molecular Formula: Example
Determine the molecular formula of a
compound with an empirical formula of
CF2 and a molar mass of 200.04 g/mol
What we multiply by:
Molar mass of molecular formula = 4
Molar mass of empirical formula
Stoichiometry
© 2015 Pearson Education, Inc.
Determining a Molecular Formula: Example
Determine the molecular formula of a
compound with an empirical formula of CF2 and
a molar mass of 200.04 g/mol
What we multiply by:
Molar mass of molecular formula = 4 = 200.04
Molar mass of empirical formula
50.01
50.01
CF2
x4
200.04
x4
C4F8
Stoichiometry
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Ribose has a molecular weight of 150
grams per mole and the empirical
formula CH2O. The molecular formula
of ribose is
a.
b.
c.
d.
C 4H 8O 4.
C5H10O5.
C6H14O4.
C6H12O6.
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Stoichiometry
Ribose has a molecular weight of 150
grams per mole and the empirical
formula CH2O. The molecular formula
of ribose is
a.
b.
c.
d.
C 4H 8O 4.
C5H10O5.
C6H14O4.
C6H12O6.
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Stoichiometry
Warm Up 11/5
A bicycle manufacturing company has 4813
wheels, 2304 frames, and 2254 handlebars.
a. How many bicycles can be manufactured
using these parts?
b. How many parts of each kind are left over?
c. Which part limits the production of
bicycles?
Stoichiometry
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Warm Up 11/5
A bicycle manufacturing company has 4802
wheels, 2299 frames, and 2249 handlebars.
a. How many bicycles can be manufactured
using these parts? 2250
b. How many parts of each kind are left over?
304 wheels, 50 frames, 0 handlebars
c. Which part limits the production of
bicycles? Handlebars!
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Stoichiometry
Plan for Today
1.
2.
3.
4.
Debrief Lab Report
Limiting Reactant Activity
Lecture Notes
Homework: Read New Lab and Do Pre-Lab
Stoichiometry
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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
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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
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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
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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
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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.
Stoichiometry
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An Example of a Stoichiometric Calculation
• There is 18.00 g of water for every 1
mole of water.
• We can use this to cancel moles of
water and get our answer in grams of
water!.
Stoichiometry
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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
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Limiting Reactants
In the example below, the O2 would be the
excess reagent.
Stoichiometry
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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
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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
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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
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Warm Up 11/12
When 3.14 g of Compound X is
completely combusted, 6.91 g of CO2
and 2.26 g of H2O form. The
molecular formula of Compound X is
a. C7H16.
c. C5H8O2.
b. C6H12O.
d. C4H4O3.
Stoichiometry
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When 3.14 g of Compound X is
completely combusted, 6.91 g of CO2
and 2.26 g of H2O form. The
molecular formula of Compound X is
a. C7H16.
c. C5H8O2.
b. C6H12O.
d. C4H4O3.
Stoichiometry
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C6H6 + 2 Br2  C6H4Br2 + 2 HBr
When 10.0 g of C6H6 and 30.0 g of Br2
react as shown above, the limiting
reactant is
a.
b.
c.
d.
Br2.
C 6H 6.
HBr.
C6H4Br2.
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Stoichiometry
C6H6 + 2 Br2  C6H4Br2 + 2 HBr
When 10.0 g of C6H6 and 30.0 g of Br2
react as shown above, the limiting
reactant is
a.
b.
c.
d.
Br2.
C 6H 6.
HBr.
C6H4Br2.
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Stoichiometry
2 Fe + 3 Cl2  2 FeCl3
When 10.0 g of iron and 20.0 g of
chlorine react as shown, the
theoretical yield of FeCl3 is
a.
b.
c.
d.
10.0 g.
20.0 g.
29.0 g.
30.0 g.
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Stoichiometry
2 Fe + 3 Cl2  2 FeCl3
When 10.0 g of iron and 20.0 g of
chlorine react as shown, the
theoretical yield of FeCl3 is
a.
b.
c.
d.
10.0 g.
20.0 g.
29.0 g.
30.0 g.
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Stoichiometry
C3H4O4 + 2 C2H6O 
C7H12O2 + 2 H2O
When 15.0 g of each reactant was
mixed, 15.0 g of C7H12O2 formed. The
percentage yield of this product is
a. 100%.
c. 65%.
b. 75%.
d. 50%.
Stoichiometry
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C3H4O4 + 2 C2H6O 
C7H12O2 + 2 H2O
When 15.0 g of each reactant was
mixed, 15.0 g of C7H12O2 formed. The
percentage yield of this product is
a. 100%.
c. 65%.
b. 75%.
d. 50%.
Stoichiometry
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