Stoichiometry Readings
Reading 1: What is stoichiometry?
The word itself is probably seems a little strange. It is derived from two Greek words
which when put together mean “element measure”. In this unit we will connect the
concept of equations (the element portion) with the mole concept (the measure portion).
These problems will command a comfort with writing formulas and chemical equations,
as well as calculating mole conversions.
A typical stoichiometry problem centers around a balanced equation such as the
following:
2 AgNO3 + Cu(OH)2  Cu(NO3)2 + 2 AgOH
The questions posed may be something like: If you have 20 grams of silver nitrate, how
many grams of silver hydroxide could be produced?
Notice the complexity of this problem. The information you are given if related to silver
nitrate, but the information you are asked to solve for is related to sodium hydroxide.
Now don’t just read that and keep going – take a minute to go back and look at the
equation. Circle the compound that you are given information about. Thank you, now
put a box around the compound you are asked to solve for. Can you see a way to
related these two different compounds? Stoichiometry will be about learning how.
Mole Ratios
Back when we learned balancing equations, we talked about the coefficients in
balanced equations. We learned that these numbers can represent the number of
molecules OR the number of moles. It will be the mole perspective we are considering
here. To illustrate, take another look at the equation above. The equation is saying
this:
“two moles of silver nitrate combine with one mole of copper hydroxide to form one mole
of copper nitrate and two moles of silver hydroxide”
For stoichiometry, we do not need to consider all of the compounds present in the
equation, we only need to consider the ones related to our problem. Each pair of
compounds can be seen as a mole ratio. Consider all the following possibilities:
2 moles of AgNO3 combines with 1 mole of Cu(OH)2
1 mole of Cu(OH)2 will create 1 mole of Cu(NO3)2
1 mole of Cu(NO3)2 will create 2 moles of AgOH
2 moles of AgNO3 will create 2 moles of AgOH
1 moles of Cu(NO3)2 will be created with 2 moles of AgOH
Take a moment to see that all of these relationships (ratios) come from the coefficients
in the balanced equation above. Make sure that you understand where each of these
numbers comes from.
Mole Ratios as Conversion Factors
It will be useful in this chapter to see these ratios in conversion factor form. That is to
say, we write them as fractions much as we do for mole conversions or generic factor
label problems. Using our problem again from above, we have the following examples :
2moles AgNO3 1mole Cu (OH )2
or
1mole Cu (OH )2
2moles AgNO3
1mole Cu (OH )2
or
1mole Cu ( NO3)2
1mole Cu ( NO3)2
1mole Cu (OH )2
So we understand from the balanced equation that 2 moles of AgNO 3 will create 1 mole
of Cu(NO3)2. But what if we don’t have 2 moles of AgNO3? What if we only have 1
mole of AgNO3? How much Cu(NO3)2 can we produce? You should see that ½ as
much AgNO3 (1 mole) will produce ½ as much Cu(NO3)2, or 0.5 moles. Let’s look at this
using factor label:
1 mole of AgNO3 X
1 mole of Cu(NO3)2
= 0.5 moles of Cu(NO3)2
2molesofAgNO3
Take a minute to look over that equation and make sure you understand where each of
these numbers came from.
1 mole of AgNO3 X
Given Stated in
the
question
1 mole of Cu(NO3)2
= 0.5 moles of Cu(NO3)2
2molesofAgNO3
Mole ratio
from the
balanced
equation
Calculation
Question: How many moles of Cu(OH)2 are needed to react with 0.35 moles of AgNO3?
What is the given? 0.35 moles
What is the appropriate mole ratio? The question is asking about copper
hydroxide and silver nitrate. The appropriate mole ratio is 1 mole of 2 moles of
AgNO3 for every 1 mole of Cu(OH)2.
Now set up the factor label problem:
0.35 moles AgNO3 X
1 mole of Cu(OH)2
= 0.175 moles Cu(NO3)2
2molesofAgNO3
Notice that in each of the problems, you start your problem with information
about one compound and then finish the problem with information regarding a
different compound. Make sure that you understanding where mole ratios come
from and how they can be used to convert between compounds this way.
Reading 2: Stoichiometry and Mass
At the beginning of this reading we showed the combination of measurement and
balanced equations. As we know, our lab measurements are not made in moles, but in
grams. Our final step in learning stoichiometry problems then is to be able to convert
from grams to moles or vice versa. Remember that in the last example, the quantity
given was 0.35 moles of silver nitrate. If we were in the lab, instead of moles we would
be measuring out grams of silver nitrate. So, how many moles of copper hydroxide are
used with 179 grams of silver nitrate? How would we go about solving this problem? It
might be very tempting to simply write the following:
79 grams AgNO3 x
1 mole of Cu(OH)2
= 39.5 moles of Cu(OH)2
2molesofAgNO3
Go back and look at that carefully. Do the units cancel out? No. The ratio of Cu(OH)2
to AgNO3 is a MOLE ratio not a GRAM ratio and as a result, this problem will not work
as it is written.
Now look at the correct way to solve the problem:
79 grams AgNO3 x
1 mole AgNO3
= 0.46 moles AgNO3
170 g AgNO3
0.46 moles AgNO3 X
1 mole of Cu(OH)2
= 0.23 moles Cu(OH)2
2molesofAgNO3
Think about the following questions:
 What step did we have to take before we could use the mole ratio?
 If we needed to solve for grams of Cu(OH)2 instead of moles of Cu(OH)2, what
would you do differently/additional?
Reading 3: Stoichiometry and Limiting Reactants
Let’s say you wanted some peanut butter cookies with the Hershey’s kisses in the
middle. You would need a recipe. Here is what you might see as a recipe, as well as
the ingredients you have on hand:
Recipe:
In my kitchen:
½ C brown sugar
½ C white sugar
2 eggs
1 tsp vanilla
2 ½ C of flour
1 t baking soda
1 t salt
24 Hershey kisses
5 C brown sugar
3 C white sugar
1 dozen eggs
1 bottle of vanilla
8 1/3 C of flour
full box of baking soda
10 t salt
30 Hershey kisses
Yield: 24 cookies
How many cookies could you make? Could you make three batches? Take a minute to
see.
You should see that you are only able to make 1 batch. You are limited to only 1 batch.
The fact that you only have 30 Hershey’s kisses was limiting the number of cookies you
could make.
A chemical reaction is like a recipe. How much you can produce is limited by the
amount of reactant you have available. Let’s look at the following example:
2 KI + Pb(NO3)2 PbI2 + 2 KNO3
Let’s imagine that we have a great need to make some lead iodide. So we go to the
chemistry cupboard to find out how much potassium iodide and how much lead nitrate
we have. We find 250 grams of KI and 325 grams of Pb(NO3)2. Now how do we figure
out how much PbI2 we can make?
We need to consider how much product (PbI2) each of the reactants would make. We
must do each one separately.
Let’s start with KI. In the space provided here, use stoichiometry to calculate how many
moles of PbI2 would be produced from our given 250 grams if KI.
Now let’s look at Pb(NO3)2. Again, use stoichiometry to determine how many moles of
PbI2 would be produced from our given 325 grams of Pb(NO3)2.
Now compare these two figures. Which reactant is producing less product? This will be
your limiting reactant.