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A mole ratio is used to convert from one chemical substance to another!
Unit 9 – Stoichiometry
Stoichiometry is the calculation of
amounts in chemical reactions.
Unit Notes
A balanced chemical equation gives us the same kind of information that a recipe does.
If it takes 2 eggs and 1 cup of flour to make 1 batch of cookies;
How many eggs does it take to make 3 batches of cookies? _____
How many cups of flour would it take to make 3 batches of cookies? _____
If we only have 6 eggs and 4 cups of flour;
How many batches of cookies could we make? _____
What would be left over? _______________
The Law of Conservation of Mass
The mass of the products is equal to the mass of the reactants.
The same number of the same types of atoms that were in the reactants before
the reaction started are present in the products at the end of the reaction.
Mass and Atoms are ALWAYS conserved
in ALL chemical reactions!!!
What does the term “conserved” mean?
Reactants
Products
2 CO + O2
2 CO2
Number of Atoms C:
2
2
Number of Atoms O:
4
4
Mass (from periodic table):
88g
88g
3
2
Balanced Chemical
Equation:
Moles:
The coefficients in a balanced chemical
equation represent the number of moles
of each substance.
Reactants
Products
2 NH3
N2 + 3 H2
Number of Atoms N:
2
2
Number of Atoms H:
6
6
34g
34g
2
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Balanced Chemical
Equation:
Mass (from periodic table):
Moles:
Moles to Moles (One Step)
We can do this first one
together. (It’s not balanced!)
N2(g) +
H2(g) 
NH3(l)
How many moles of hydrogen react completely with 0.60 mol of nitrogen when
producing ammonia (NH3)?
You do these three. Use the following balanced equation to answer questions 1, 2, & 3.
4 Al(s) + 3 O2(g)  2 Al2O3(s)
1.
How many moles of aluminum are needed to form 3.7 mol Al2O3?
Answer:
7.4 mol Al
2.
How many moles of oxygen are required to react completely with 14.8 mol Al?
Answer:
11.1 mol O2
3.
How many moles of Al2O3 are formed when 0.78 mol O2 reacts with aluminum?
Answer:
0.52 mol Al2O3
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Rules that apply to ALL stoichiometric conversions!
Start with the given number and units.
Units must always match _________ and to the _________.
Cross cancel units _________ and to the _________.
You are done when the units left on top match the units that were asked for.
Multiply everything that is on the __________,
Divide by everything that is on the __________!
There are only __2__ types of conversion ratios in stoichiometry!
A mole ratio is used to convert from one chemical to another.
Units are “want over have” and the numbers come from coefficients.
EXAMPLE of a MOLE RATIO
All stoichiometric
conversions must use one
mole ratio to convert from
one substance to another.
2 mol B
5 mol A
A molar mass ratio is the grams in 1 mol of a chemical.
Molar masses are often stated in the question or you can use a
periodic table to add it up.
These have grams in them.
EXAMPLES of MOLAR MASS RATIOS
1 mol H2O
32 g O2
18 g H2O
1 mol O2
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Rules that apply to ALL stoichiometric conversions!
Start with the given number and units.
Units must always match down and to the right.
You are done when the units on top match the units that were asked for.
Multiply everything that is on the top, divide by everything that is on the bottom!
Grams to Grams (Three Step)
Cu +
2 AgNO3
 Cu(NO3)2 +
2 Ag
Example Question:
If 2.25 g of Cu are reacted, how many grams of Ag are produced?
In this question, what is the given (number and units)? ___________________
What units does the question ask for? ____________________
This is the given
It always goes first
This is the mole ratio!
The numbers come from the
coefficients of the equation.
This is a molar mass ratio
(the grams in 1 mole)
In this case, it is the molar mass
of Cu from the periodic table.
It is used to convert from
the given substance to
another substance.
This is another molar mass ratio
(the grams in 1 mole)
This one is the molar mass of Ag
from the periodic table.
2.25 g Cu
1 mol Cu
63.5 g Cu
2 mol Ag
1 mol Cu
107.9 g Ag
= 7.65 g Ag
1 mol Ag
Answer
Cu +
2 AgNO3
 Cu(NO3)2 +
2 Ag
How many grams of AgNO3 (molar mass: 170 g/mol) are consumed when 5.30 g Ag are produced?
Answer:
8.35 g AgNO3
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Yes, this is still
Unit 9
Percent Yield
Words to know:
Actual Yield is the amount of a product that was actually recovered in a real reaction.
Theoretical Yield is the maximum mass (grams) of product that could form in a reaction.
Percent Yield is a measure of how complete a reaction was.
FORMULA FOR PERCENT YIELD:
% Yield =
Actual Yield
x 100
Theoretical Yield
Cu + 2 AgNO3  Cu(NO3)2 + 2 Ag
Enough copper and silver nitrate are reacted to produce a theoretical yield of 32.6 g silver, Ag.
If 26.3 g of silver is actually recovered, what is the percent yield of silver in this reaction?
Actual Yield: ____________
Theoretical Yield: ____________
x 100 = ______________ %
C8H18(g) + 25 O2(g)
16 CO2(g) + 18 H2O(l)
Enough octane (C8H18) and oxygen are combusted to produce a theoretical yield of 5.35 g water.
If 4.85 g of water is actually recovered, what is the percent yield of water in this reaction?
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Percent Yield (when theoretical yield is not given)
Theoretical Yield is the maximum mass (grams) of product
that can form in a reaction!
So, determining theoretical yield is a grams to grams conversion.
2 Pb(NO3)2  2 PbO + 4 NO2 + O2
An actual yield of 2.53 g PbO (223.2 g/mol) product is recovered by reacting 5.60 g Pb(NO3)2
(331.2 g/mol), what is the percent yield of PbO in the reaction?
This has a 2-step solution:
First find the theoretical yield (in grams) of PbO in the reaction…
5.60 g Pb(NO3)2
1 mol Pb(NO3)2
2 mol PbO
223.2 g PbO
331.2 g Pb(NO3)2
2 mol Pb(NO3)2
1 mol PbO
Then find percent yield…
% Yield =
= 3.77 g PbO
This is the
Theoretical Yield
Actual Yield
Theoretical Yield
x 100
Actual Yield: _2.53 g PbO _
Theoretical Yield: _3.77 g PbO _
x 100 = ______________ %
SiO2(s) + 3 C(s)  SiC(s) + 2 CO(g)
If 50.0 g of SiO2 (60.1 g/mol) is reacted with an excess of carbon, an actual yield of 27.9 g of CO
(28.0 g/mol) is produced. What is the percent yield of CO in this reaction?
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Stoichiometry - Limiting Reactant
In a chemical reaction, the limiting reactant is the one that is completely consumed when the
reaction is completed (none of it is left over).
The other reactant is called the excess reactant because it is left over when the reaction is
completed.
The limiting reactant is used to determine the maximum amount of product that can be
produced.
Identifying the Limiting Reactant:
To identify the limiting reactant in a chemical reaction
you must find how many moles of one of the products
can be produced by each reactant.
The reactant that produces LESS of a product is the limiting reactant!
Example:
CaC2(s) + 2 H2O(l)  C2H2(g) + Ca(OH)2(aq)
Identify the limiting reactant when 3.27 mol of CaC2 is added to 5.03 g of H2O.
Find the moles of one of the products produced (just pick one) for each of the reactants:
3.27 mol CaC2
5.03 g H2O
1 mol C2H2
1 mol CaC2
1 mol H2O
18.0 g H2O
= 3.27 mol C2H2
1 mol C2H2
2 mol H2O
= 0.140 mol C2H2
H2O is the Limiting Reactant!
It produces less.
H2O is the limiting reactant because reacting all 5.03 g H2O completely will produce less C2H2
than reacting all of the CaC2 .
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Theoretical Yield and Limiting Reactant
Theoretical Yield is the maximum mass (grams) of product that can form in a reaction!
When you are asked to find theoretical yield and you are given masses for all of the reactants,
you must find the theoretical yield (in grams) for each of the reactants.
The LOWEST theoretical yield (in grams) is the true theoretical yield!
Example:
P4(s) + 2 O2(g)
4 PO(g)
What is the theoretical yield of PO (47.0 g/mol) that can be formed when 5.20 g P4 reacts with
3.15 g O2?
5.20 g P4
1 mol P4
124 g P4
3.15 g O2
1 mol O2
32.0 g O2
4 mol PO
1 mol P4
4 mol PO
2 mol O2
47.0 g PO
1 mol PO
47.0 g PO
1 mol PO
= 7.88 g PO
= 9.25 g PO
7.88 g PO is the theoretical yield, because it is the smaller of the two!
Even though P4 had the larger given mass it had the smaller theoretical yield.
The mole ratio MUST be used to convert from one chemical substance to another!
MASS and ATOMS are ALWAYS conserved in ALL chemical reactions!
The mass of the products is equal to the mass of the reactants!
Stoichiometry! Frustrating chemistry students since 1792!
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