Transcript 1

1
A Short Review
2
Avogadro’s
Number of
Particles
23
10
6.02 x
Particles
1 MOLE
Mole Weight
3
23
1 mole1 =mole
6.02=x6.02
1023xatoms
molecule
formula
10 ions
units
4
• For calculations of mole-mass-volume
relationships.
– The chemical equation must be balanced.
– The number in front of a formula
represents the number of moles of the
reactant or product.
The equation is balanced.

2 Al + Fe2O3  2 Fe + Al2O3
2 mol
1 mol
2 mol
1 mol
5
Introduction to Stoichiometry:
The Mole-Ratio Method
6
• Stoichiometry: The area of chemistry
that deals with the quantitative
relationships between reactants and
products.
• Examples of questions asked:
• How much product from a specific amount of reactant.
• How much reactant do you start with to get so much
product.
• What if the reaction does not go to completion? What is
the percent yield.
• What if the sample is impure? What is the percent purity.
7
• Mole Ratio: a ratio between the moles
of any two substances involved in a
chemical reaction.
– The coefficients used in mole ratio
expressions are derived from the
coefficients used in the balanced
equation.
8
Examples
9
N2 + 3H2  2NH3
1 mol
3 mol
2 mol
1 mol N 2
3 mol H 2
10
N2 + 3H2  2NH3
1 mol
3 mol
2 mol
3 mol H 2
2 mol NH 3
11
• The mole ratio is used to convert the
number of moles of one substance to
the corresponding number of moles of
another substance in a stoichiometry
problem.
• The mole ratio is used in the solution of
every type of stoichiometry problem.
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The Mole Ratio Method
1. Convert the quantity of starting substance
to moles (if it is not already moles)
2. Convert the moles of starting substance to
moles of desired substance.
3. Convert the moles of desired substance to
the units specified in the problem.
13
Step 1 Determine the number of moles of
starting substance.
Identify the starting substance from the
data given in the problem statement.
Convert the quantity of the starting
substance to moles, if it is not already
done.
 1 mole 
moles = grams 

 mole weight 
14
Step 2 Determine the mole ratio of the
desired substance to the starting
substance.
The number of moles of each substance
in the balanced equation is indicated by
the coefficient in front of each substance.
Use these coefficients to set up the mole
ratio.
moles of desired substance in the equation
mole ratio =
moles of starting substance in the equation
15
In the following reaction how many moles of PbCl2
are formed if 5.000 moles of NaCl react?
2NaCl(aq) + Pb(NO3)2(aq)  PbCl2(s) + 2NaNO3(aq)
 moles of desired substance in the equation 
moles of starting substance  
 = moles of desired substance
moles
of
starting
substance
in
the
equation


Mole Ratio
 1 mole PbCl 2 
5.000 moles NaCl  
  2.500 mole PbCl2
 2 mole NaCl 
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Step 3. Calculate the desired substance in
the units specified in the problem.
• If the answer is to be in moles, the
calculation is complete
• If units other than moles are wanted,
multiply the moles of the desired
substance (from Step 2) by the
appropriate factor to convert moles to
the units required.
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Step 3. Calculate the desired substance in
the units specified in the problem.
 mole weight 
1. To calculate grams: moles x 
  grams
 1 mole 
 18.02 g H 2O 
5.000 mol H 2O  
  90.10 grams H 2O
 1 mol H 2O 
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Step 3. Calculate the desired substance in
the units specified in the problem.
 6.02 x 1023atoms 
2. To calculate atoms: moles  
  atoms
1 mole


 6.02 x 1023 Na atoms 
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5.000 moles Na atoms  

3.011
x
10
Na atoms

 1 mole Na atoms 
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Step 3. Calculate the desired substance in
the units specified in the problem.
3. To calculate molecules:
 6.02 x 1023molecules 
moles x 
  molecules
1 mole


 6.02 x 1023 H 2O molecules 
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5.000 moles H 2O  
  3.011 x 10 molecules H 2O
1 mole H2O


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Mole-Mole
Calculations
21
Phosphoric Acid
• Phosphoric acid (H3PO4) is one of the
most widely produced industrial
chemicals in the world.
• Most of the world’s phosphoric acid is
produced by the wet process which
involves the reaction of phosphate
rock, Ca5(PO4)3F, with sulfuric acid
(H2SO4).
Ca5(PO4)3F(s) + 5H2SO4  3H3PO4 + HF + 5CaSO
4
22
Calculate the number of moles of phosphoric acid
(H3PO4) formed by the reaction of 10 moles of sulfuric
acid (H2SO4).
Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4
1 mol
5 mol
3 mol
1 mol
5 mol
Step 1 Moles starting substance: 10.0 mol H2SO4
Step 2 The conversion needed is
moles H2SO4  moles H3PO4
Mole Ratio
3 mol H 3PO4
10 mol H 2SO4 x
= 6 mol H 3PO4
5 mol H 2SO4
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Calculate the number of moles of sulfuric acid
(H2SO4) that react when 10 moles of Ca5(PO4)3F react.
Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4
1 mol
5 mol
3 mol
1 mol
5 mol
Step 1 The starting substance is 10.0 mol Ca5(PO4)3F
Step 2 The conversion needed is
moles Ca5(PO4)3F  moles H2SO4
Mole Ratio
5 mol H 2SO4
10 mol Ca 5 (PO 4 )3F x
= 50 mol H 2SO4
1 mol Ca 5 (PO4 )3F
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Mole-Mass
Calculations
25
1. The object of this type of problem is
to calculate the mass of one
substance that reacts with or is
produced from a given number of
moles of another substance in a
chemical reaction.
2. If the mass of the starting substance
is given, we need to convert it to
moles.
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3. We use the mole ratio to convert
moles of starting substance to moles
of desired substance.
4. We can then change moles of desired
substance to mass of desired substance
if called for by the problem.
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Examples
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Calculate the number of moles of H2SO4 necessary to
yield 784 g of H3PO4
Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4
The conversion needed is
grams H3PO4  moles H3PO4  moles H2SO4
Mole Ratio
 1 mol H 3PO4 
784 g H 3PO4 

 98.0 g H 3PO4 
 5 mol H 2SO4 

 = 13.3 mol H 2SO4
 3mol H 3PO4 
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Calculate the number of grams of H2 required to form
12.0 moles of NH3.
N2 + 3H2  2NH3
The conversion needed is
moles NH3  moles H2  grams H2
Mole Ratio
 3 mol H 2   2.02 g H 2 
12.0 mol NH 3 
  1mol H  = 36.4 g H2
2 
 2 mol NH 3  
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Mass-Mass
Calculations
31
Solving mass-mass stoichiometry problems
requires all the steps of the mole-ratio
method.
1. The mass of starting substance is converted to
moles.
2. The mole ratio is then used to determine moles
of desired substance.
3. The moles of desired substance are converted
to mass of desired substance.
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Diagram to Successful Calculations
Chemical
mole wt
moles of A
grams of A
of A
mole wt
moles of B
equation
grams of B
of B
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Calculate the number of grams of NH3 formed by the
reaction of 112 grams of H2.
N2 + 3H2  2NH3
grams H2  moles H2  moles NH3  grams NH3
 1 mol H 2   2 mol NH 3   17.0 g NH 3 
112 g H 2 
 = 628 g NH3


 2.02 g H 2   3 mol H 2   1 mol NH 3 
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Limiting-Reactant and
Yield Calculations
35
Limiting Reagent
36
• The limiting reagent is one of the
reactants in a chemical reaction.
• It is called the
limiting reagent
because the amount of it present is
insufficient to react with the amounts
of other reactants that are present.
• The limiting reagent limits the amount
of product that can be formed.
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How many bicycles
can be assembled
from the parts
shown?
From
From
three
eight
fourwheels
pedal
framesfour
four
assemblies
bikes
bikes
bikes
can
can
bethree
be
constructed.
constructed.
can be constructed.
The limiting part is the
number of pedal
assemblies.
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First Balance Your Assembly
2 Wheels + 1 Frame + 1 Pedal→ 1 Bike
1 Bike 

8 Wheels 
  4 Bikes
 2 Wheels 
1 Bike 

4 Frames 
  4 Bikes
 1 Frame 
1 Bike 

3 Pedals 

 1 Pedal 
3 Bikes
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Steps Used to Determine the
Limiting Reagent
40
1. Calculate the amount of product (moles or
grams, as needed) formed from each reactant.
2. Determine which reactant is limiting. (The
reactant that gives the least amount of
product is the limiting reagent; the other
reactant is in excess.
3. Calculate the amount of the other reactant
required to react with the limiting reagent,
then subtract this amount from the starting
quantity of the reactant. This gives the
amount of the substance that remains
unreacted.
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Examples
42
How many moles of HCl can be produced by reacting
4.0 mol H2 and 3.5 mol Cl2? Which compound is the
limiting reagent?
H2 + Cl2 → 2HCl
Step 1 Calculate the moles of HCl that can form
from each reactant.


2
mol
HCl
4.0 mol H 2 
  8.0 mol HCl
 1 mol H 2 
3.5 mol Cl2  2 mol HCl   7.0 mol HCl
 1 mol Cl 2 
Step 2 Determine the limiting reagent.
The limiting reagent is Cl2 because it
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produces less HCl than H2.
How many grams of silver bromide (AgBr) can be
formed when solutions containing 50.0 g of MgBr2
and 100.0 g of AgNO3 are mixed together? How
many grams of the excess reactant remain unreacted?
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 1 Calculate the grams of AgBr that can form
from each reactant.
The conversion needed is
g reactant → mol reactant → mol AgBr → g AgBr
 1 mol MgBr2   2 mol AgBr   187.8 g AgBr 
50.0 g MgBr2  
 
  1 mol AgBr   102 g AgBr

 184.1 g MgBr2   1 mol MgBr2  
 1 mol AgNO3   2 mol AgBr  187.8 g AgBr 
110.5 g AgBr
100.0 g AgNO3 



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 169.9 g AgNO3   2 mol AgNO3  1 mol AgBr 
How many moles of silver bromide (AgBr) can be
formed when solutions containing 50.0 g of MgBr2
and 100.0 g of AgNO3 are mixed together? How
many grams of the excess reactant remain unreacted?
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 2 Determine the limiting reagent.
The limiting reactant is MgBr2 because it
forms less AgBr.
 1 mol MgBr2   2 mol AgBr   187.8 g AgBr 
50.0 g MgBr2  
 
  1 mol AgBr   102 g AgBr

 184.1 g MgBr2   1 mol MgBr2  
 1 mol AgNO3   2 mol AgBr  187.8 g AgBr 
110.5 g AgBr
100.0 g AgNO3 



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 169.9 g AgNO3   2 mol AgNO3  1 mol AgBr 
How many grams of the excess reactant (AgNO3)
remain unreacted?
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 3 Calculate the grams of unreacted AgNO3.
First calculate the number of grams of
AgNO3 that will react with 50.0 g of MgBr2.
The conversion needed is
g MgBr2 → mol MgBr2 → mol AgNO3 → g AgNO3
 1 mol MgBr2  2 mol AgNO3   169.9 g AgNO3 
50.0 g MgBr2  
  92.3 g AgNO3


 184.1 g MgBr2  1 mol MgBr2   1 mol AgNO3 
The amount of AgNO3 that remains is
100.0 g AgNO3 - 92.3 g AgNO3 = 7.7 g AgNO463
Percent Yield
47
The quantities of products calculated
from equations represent the maximum
yield (100%) of product according to the
reaction represented by the equation.
48
Many reactions fail to give
a 100% yield of product.
This occurs because of side reactions
and the fact that many reactions are
reversible.
49
• The theoretical yield of a reaction is
the calculated amount of product that
can be obtained from a given amount
of reactant.
• The actual yield is the amount of
product finally obtained from a given
amount of reactant.
50
• The percent yield of a reaction is the
ratio of the actual yield to the theoretical
yield multiplied by 100.
actual yield
x 100 = percent yield
theoretical yield
51
Silver bromide was prepared by reacting 200.0 g of
magnesium bromide and an adequate amount of silver
nitrate. Calculate the percent yield if 375.0 g of silver
bromide was obtained from the reaction:
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 1 Determine the theoretical yield by
calculating the grams of AgBr that can be
formed.
The conversion needed is
g MgBr2 → mol MgBr2 → mol AgBr → g AgBr
 1 mol MgBr2 
200.0
g
MgBr


2 
184.1
g
MgBr
2 

 2 mol AgBr   187.8 g AgBr 
408.0 g AgBr



 1 mol MgBr2   1 mol AgBr 
52
Silver bromide was prepared by reacting 200.0 g of
magnesium bromide and an adequate amount of silver
nitrate. Calculate the percent yield if 375.0 g of silver
bromide was obtained from the reaction:
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 2 Calculate the percent yield.
must have same units
actual yield
percent yield =
x 100
theoretical yield
must have same units
375.0 g AgBr
x 100 = 91.9%
percent yield =
408.0 g AgBr
53
Percent Purity
54
• The percent purity of a reaction is the
ratio of the theoretical yield (calculated)
to the sample weight multiplied by 100.
theoretical yield
x 100 = percent purity
sample weight
55
A 5.00 gram sample of impure CaCO3 (limestone)
produces 1.89 grams of CO2 when heated. Calculate
the percent purity of CaCO3 according to:
CaCO3 (s)  CaO (s) + CO2
Step 1 Determine the actual grams (theoretical )
of CaCO3 in the original sample.
The conversion needed is
g CO2 → mol CO2 → mol CaCO3 → g CaCO3
1.89 g CO2 
 1 mol CO2   1 mol CaCO3 

 

44.0
g
CO
1
mol
CO
2  
2 

 100. g CaCO3 


 1 mol CaCO3 
4.30 g CaCO3
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A 5.00 gram sample of CaCO3 produces 1.89 grams of
CO2 when heated. Calculate the percent purity
according to:
CaCO3 (s)  CaO (s) + CO2
Step 2 Calculate the percent purity.
theoretical yield
percent purity =
x 100
sample weight
4.30 g CaCO3
x 100 = 86.0%
percent purity =
5.00 g CaCO3
57
484 gram of copper ore is oxidized by HNO3 to yield
1000. grams of Cu(NO3)2. Find the percent purity of
copper in the ore according to:
3 Cu(s) + 8 HNO3(aq)  3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)
The conversion needed is
g Cu(NO3)2 → mol Cu(NO3)2 → mol Cu → g Cu
3mole Cu
1mole Cu(NO3 )2
63.5 g Cu

 339g Cu

1000.g Cu(NO3 )2 
3mole
Cu(NO
)
1mole Cu
187.5g Cu(NO3 )2
3 2
339g
%purity 
 100  70.0%
484 g
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