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To understand how to use stoichiometry in scientific terms, you will have to
understand how to use it in…
EVERYDAY EQUATIONS!!
• For example, when baking food, a recipe is given for you to
follow. If you need to make a larger number of the item, you
could double or even triple the quantity of ingredients you
use.
• “A balanced chemical equation provides the same kind of
quantitative information that a recipe does.”
• In a recipe you can imagine the ingredients as reactants
and the cookies as the products.
Stoichiometry
• Balanced chemical equations tell you what
amounts of reactants to mix and what
amount of products to expect from the
reaction
– Chemist use balanced chemical equations as
a basis to calculate how much reactant is
needed or product is formed in a reaction
– The calculation of quantities in chemical
reactions is the subject of chemistry called
stoichiometry
• Stoichiometric calculations allow chemist to keep
track of reactants and products in a chemical
reaction
Interpreting Chemical Equations
• Consider the following equation
N2(g) + 3H2(g)  2NH3(g)
• The balanced chemical equation can be interpreted in
terms of different quantities such as: numbers of atoms,
molecules, moles, mass, and volume
– Examples:
• Atoms – In formation of ammonia two atoms of nitrogen react with 6
atoms of hydrogen. These eight atoms are recombined in the
product
• Molecules – One molecule of nitrogen reacts with 3 molecules of
hydrogen to form two molecules of ammonia
• Moles – One mole of nitrogen gas reacts with 3 moles of hydrogen
gas to form 2 moles of ammonia gas
• Mass – The mass of1mole of N2 (28 grams) plus the mass of 3
moles of H2 (6.0 grams) equals the mass of 2 moles of NH3 (34
grams)
• Volume –22.4L of N2 gas reacts with 67.2L H2 (3 x 22.4L)gas to
form 44.8L of NH3 (2 x 22.4L) gas (Remember that the volume of
one mole of gas = 22.4L)
Mass Conservation in Chemical Reactions
• Mass and atoms are conserved in every
chemical reaction
Writing and using mole ratios
N2(g) + 3H2(g)  2NH3(g)
• A mole ratio is a conversion factor derived from the
coefficients of a balanced chemical equation interpreted
in terms of moles
• Example:
1 mol N2
3 mol H2
2 mol NH3
1 mol N2
3 mol H2
2 mol NH3
• In chemical calculations, mole ratios are used to convert
between moles of reactant and moles of product,
between moles of reactants, or between moles of
products
Mole-Mole Calculations
• How many moles of ammonia are produced
when 0.6 moles of nitrogen reacts with
hydrogen?
1 mol N2
N2(g) + 3H2(g)  2NH3(g)
3 mol H
2
x mole G
Given
0.6 moles N2
b mole W
=xb/a mol W
a mole G
Mole Ratio
2 mol NH3
1 mol N2
Calculated
2 mol NH3
= 1.2 moles NH3
1 mol N2
Entered into calculator 0.6 X 2 = 1.2
3 mol H2
2 mol NH3
Mole-Mole Calculations
• How many moles of excess hydrogen are consumed
when 0.6 moles of nitrogen reacts with hydrogen to form
ammonia?
1 mol N2
3 mol H2
N2(g) + 3H2(g)  2NH3(g)
x mole GX
Given
0.6 moles N2 X
b mole W
=xb/a mol W
a mole G
Mole Ratio
3 mol H2
1 mol N2
Calculated
= 1.8 moles H2
Entered into calculator 0.6 X 3 = 1.8
Flip mole
ratio so N2
cancel each
other
2 mol NH3
1 mol N2
3 mol H2
2 mol NH3
Chemical Calculations:
• In chemical calculations, mole ratios are used to convert
between moles of reactants and moles of product, between
moles of reactants, or between moles of products.
• A mole ratio is a conversion factor derived from the
coefficients of a balanced chemical equation interpreted in
terms of moles.
• To change one units to another, you will need a map to
ensure that the units as well as their labels are correct.
Map on next page
Stoichiometry Calculator:
Coefficient
Volume A
22.4 L
mol
Mass A
mw
mol
Mole A
Volume B
22.4 L
mol
Coefficient
6.02×1023 molec
mol
Molecule A
Mole B
mw
mol
6.02×1023 molec
mol
Coefficient
Molecule B
Mass B
How To Use The Map:
• In order to use this map to convert units of a substance, you
need to start by drawing a t-like outline and put the given
information in the first empty space and nothing in the space
beneath it. For example, if you have 20 mol H, and you want
to know
howAmany L of O there are in the equation 2H + O
20 mol
H2O , this is how you would start :
• Then, you need to follow the map and convert 20 mol A to
liters by adding 22.4 L/mol into your outline like so:
20 mol A 22.4 L A
mol A
• Last, you will need to convert your liters of H to liters of O
by following the map once again and adding the coefficient
to your outline. (make sure that units match up with the
information diagonal to it.) (coefficient is the number in front
Continued on next page
of the element. It will look like this now:
20 mol A 22.4 L A 1 L B
mol A 2 L A
• Now that you are done creating your outline, you will need
to multiply the top and bottom numbers by each other and
divide them. This is what you will end up with:
448
224
answer!!
2
• When you are done, make sure to add your units of your substa
224 Liters of O
final answer
Another example using “the map”:
• How many molecules of oxygen are produced when 29.2 g
of water is decomposed by electrolysis according to this
electricity
balanced 2H
equation:
2O (L)
2H2 (g) + O 2 (g)
1. List the known:
mass of water= 29.2 g
2. Calculate:
1 mol H2O
29.2 g H2O 18.0 g H2O
23
1 mol O2 6.02 10 molec.
1 mol O2
2 mol H2O
4.88 1023 molecules O2
3. Check answer:
Yes, the answer is correct
Mass-Mass Calculations
• Calculate the number of grams of NH3 produced by the
reaction of 5.40 grams of hydrogen with excess nitrogen
N2(g) + 3H2(g)  2NH3(g)
1. Convert 5.40 grams H2 into moles
2. Change the moles of hydrogen to moles of ammonia
3. Change the moles of ammonia to grams of ammonia
5.40 g H2 X
1 mol H2
2.0 g H2
Step 1
X
2 mol NH3 17.0 g NH3
X
= 31g NH3
3 mol H2
1 mol NH3
Step 2
Step 3
Calculator: 5.4 ÷ 2 X 2 ÷ 3 X 17 = 31 g
1 mol N2
3 mol H2
2 mol NH3
1 mol N2
3 mol H2
2 mol NH3
Mole-Mole Calculations
Practice Problems
• How many moles of ammonia are produced
when 3 moles of nitrogen reacts with hydrogen?
N2(g) + 3H2(g)  2NH3(g)
• How many grams of ammonia are produced
when 84 grams of nitrogen reacts with
hydrogen?
Other Stoichiometric Calculations
•
In a typical stoichiometric problem, the given quantity is first converted to
moles. Then the mole ratio from the balanced equation is used to calculate
the number of moles of the wanted substance. Finally, the moles are
converted to any other unit of measurement related to the unit mole, as the
problem requires.
– How many molecules of NH3 are produced when 5.4 grams of hydrogen
react with excess nitrogen?
N2(g) + 3H2(g)  2NH3(g)
1. Convert 5.40 grams H2 into moles
2. Change the moles of hydrogen to moles of ammonia
3. Change the moles of ammonia to molecules of ammonia
5.40 g H2 X
1 mol H2
2.0 g H2
Step 1
X
2 mol NH3
X
3 mol H2
Step 2
6.02x1023 molecules NH3
1 mol NH3
= 1.08x1024 NH3molecules
Step 3
Calculator: 5.4 ÷ 2 X 2 ÷ 3 X 6.02x1023 = 1.08x1024
Limiting and Excess Reagents:
•
In a chemical reaction, an insufficient quantity of any of
the reactants will limit the amount of product that forms.
• An example of excess and limiting reagents is making
muffins. You have more than enough flour, butter, water,
and eggs. However, you only have one cup of sugar. The
number of muffins you can make will be limited by how
much sugar you have on hand. Therefore, the sugar is
the limiting ingredient in this recipe and the other
ingredients are the excess ingredients. A chemist often
has the same situation,
with balanced
• Abut
chemist’s
recipe ischemical
a balanced
equations.
equation such as:
N2 (g) + 3H2 (g)
2NH3 (g)
Limiting Reagent:
• To show how to find the limiting and excess reagent in an
actual equation, we will use the formula for ammonia:
N2 (g) + 3H2 (g)
2NH3 (g)
• Before this reaction takes place, nitrogen and hydrogen
are in a 2:3 ratio. One molecule (mole) of N2 reacts with
three molecules (moles) of H2 to make two molecules of
NH3. When this happens, all of the H has been used up,
and the reaction stops. One molecule of leftover nitrogen is
• Therefore, in this reaction, only the hydrogen molecule is
left.
used up, and consequently, it is the limiting reagent, or the
reagent that determines the amount of product that can be
made in the reaction.
More on next page
Limiting and Excess Reagents
• In a chemical reaction, an insufficient quantity of any of
the reactants will limit the amount of product formed.
– Limiting reagent is the reactant that determines the amount of
product that can be formed by a reaction
– Excess reagent is the reactant that is not completely used up in
a chemical reaction
2Na + Cl2  2NaCl
Cl Cl
Cl Cl
Na
Cl Cl
Na
Cl Cl
Na
Cl Cl
Na
Cl Cl
Na
Na
Cl Cl
Na
Cl Cl
Na
Na
Na
Identify the limiting reagent and the excess reagent
from the above reaction about to occur
Before reaction
Cl Cl
Cl Cl
Cl Cl
Cl Cl
Cl Cl
Cl Cl
Cl Cl
Cl Cl
+
2Na + Cl2  2NaCl
Na
Na
Na
Na
Na
Na
Na
Na
Na
Na
After reaction
In this reaction, sodium (Na) is the limiting reagent because it is entirely
used up in the reaction.
Chlorine Gas (Cl2) is the excess reagent because some chlorine gas is left
over after the reaction
Limiting Reagent Problem
•
Copper reacts with sulfur to form copper(I) sulfide according to the
following balanced equation.
2Cu(s) + S(s)  Cu2S(s)
•
What is the limiting reagent when 80.0 grams of copper react with
25.0 grams of sulfur?
1.
Convert mass to moles:
1 mole Cu
= 1.26 mole Cu
63.5g Cu
1 mole S
= 0.779 mole S
25.0g S X
32.1g S
80.0g Cu X
2. Determine how many moles of sulfur you need to react with 1.26
moles of Cu
1.26 mole Cu X
1 mole S
= 0.630 mole S
2 mole Cu
3. Because 0.630 moles of S is less than the 0.779 moles of sulfur
present in the reaction, sulfur is in excess and Cu is limiting
Limiting Reagent Problem
•
Copper reacts with sulfur to form copper(I) sulfide according to the
following balanced equation.
2Cu(s) + S(s)  Cu2S(s)
•
What is the maximum number of grams of Cu2S that can be formed
when 80.0 grams of Cu react with 25.0 grams of S?
1.
2.
Convert mass to moles:
1 mole Cu
80.0g Cu X
63.5g Cu
1 mole S
25.0g S
X
32.1g S
= 0.779 mole S
X
1 mole S
2 mole Cu
= 0.630 mole S
Determine the limiting and excess reagents
•
4.
= 1.26 mole Cu
Determine how many moles of sulfur you need to react with 1.26 moles of Cu
1.26 mole Cu
3.
Italicized
numbers are
molar masses
Because 0.630 moles of S is less than the 0.779 moles of sulfur present in the
reaction, sulfur is in excess and Cu is limiting
Calculate the maximum amount of product formed using the limiting reagent Cu
1.26 mol Cu X
1 mole Cu2S
2 mole Cu
X
159.1 g Cu2S
1 mole Cu2S
= 100 grams Cu2S
Percent Yield:
• In your classroom, not everyone gets 100% on everything.
The grade you receive on a quiz or test is a ratio of the
number of questions answered correctly and the number of
total questions on the exam. Chemists use similar
calculations when the product from a chemical reaction is
•less
Thethan
theoretical
expected
yield
based
is one
on tool
the equation.
used in finding the percent
yield of the equation. The theoretical yield is the maximum
amount of product that could be formed from given amounts
of
reactants.
•Another
tool is the amount of product that actually forms in
your lab when the reaction is carried out that is called the
actual yield.
• The percent yield is the ratio of the actual yield to the
theoretical yield shown as a percent.
Continued on next page
Percent Yield Continued:
• The formula to find percent yield is:
Percent yield= actual yield
theoretical yield
100
• The actual yield of the chemical reaction is mostly less than the
yield and therefore the yield is less than 100% for the most part.
•
The percent yield is a measure of the efficiency of a
reaction carried out in the laboratory. (basically the
measure of how messy the experiment was in the lab)
• In order to perform this procedure, remember that you MUST
have both the theoretical and actual yields.
Percent Yield Example:
• What is the percent yield if 13.1 g CaO is actually produced
when 24.8g CaCO3 is heated?:
CaCO3 (S)
CaCO (S) + CO2 (g)
1. State your knowns:
actual yield= 13.1 g CaO
theoretical yield= 13.9 g CaO
2. Calculate:
percent yield =13.1 g CaO
13.9 g CaO
=
3. Check your answer
94.2%
100
Theoretical yield example:
• Calcium carbonate, which is found in seashells, is
decomposed by heating. The balanced equation for this
reactionCaCO
is: 3 (S)
CaO (S) + CO2 (g)
What is the theoretical yield?
1. State your knowns:
Mass of CaCO3= 24.8 g
2. Calculate:
1 mol CaCO3 1 mol CaO 56.1 g CaO
24.8 g CaCO3100.1 g CaCO31 mol CaCO3 1 mol CaO
=
3. Check your answer
13.9g CaO
Percent Yield
• No process is one hundred percent efficient
• Because no process is one hundred percent efficient,
calculated yields from a chemical reaction are always
greater than the actual yield
• The actual yield of a chemical reaction divided by the
calculated yield (theoretical yield) multiplied by 100% is
the percent yield of a chemical reaction
% Yield = Actual Yield ÷ Theoretical Yield X 100%
• The percent yield is a measure of the efficiency of a
reaction carried out in the laboratory.
Percent Yield Problem
•
Copper reacts with sulfur to form copper(I) sulfide according to the
following balanced equation.
2Cu(s) + S(s)  Cu2S(s)
•
If 95 grams of Cu2S forms from the reaction of 80.0 grams of Cu with
25.0 grams of S, what was the percent yield of the reaction?
1.
Convert mass to moles:
2.
Determine how many moles of sulfur you need to react with 1.26 moles
of Cu
3.
Determine the limiting and excess reagents
4.
Calculate the maximum amount of product formed (Theoretical yield)
using the limiting reagent (Cu)
1.26 mol Cu X
5.
1 mole Cu2S
2 mole Cu
Italicized
159.1 g Cu2S
numbers are
X
= 100 grams Cu2S
molar masses
1 mole Cu2S
Divide the actual yield by the theoretical yield and multiply by 100%
95 grams Cu2S
100 grams Cu2S X 100% = 95%