Stoichiometry Continued
True Stoichiometry
So far, we have talked about changing
from units of one thing (atoms,
compounds, ions, etc) to different units of
the same thing.
 But stoichiometry is not limited to this. Its
true purpose is to change units of one
thing into units of a different thing.
 In other words, grams of NaCl into grams
of NaOH.

The most important thing
you need to remember is
you will always go through
the mole.
The mole is the key
We have already touched on the
mole and …
Mass (use the molar mass)
 Density (convert it to mass first then pick
it up from there)
 Atoms (Avogadro’s Number)
 Molecules (Avogadro’s Number)
 Ions (Avogadro’s Number)
 Formula Units (Avogadro’s Number)

So, what is molarity and
how do volumes of gases fit
into this?
What is Molarity?
Molarity (M) is the concentration of a
solution expressed as moles of solute
dissolved per Liter of solution.
 In other words, it is moles/Liter
 Its formula is M=moles/volume (in liters)

How do we use this information?
In an experiment, we use a certain volume
of the solution.
 By using the molarity, we will know how
many moles are being used.
 Remember, we usually use volumes in mL in
our labs. So, we need to convert mL to L for
these problems.
 For example, we are going to use 50 mL of 3
M HCl in a lab, how many moles of HCl are
we going to use?

Examples (Remember M=mol/V)
How many moles are in 25 mL of a 15 M
NaCl solution?
 We have a 50 mL solution of HCl that
contains 0.52 moles of HCl. What is the
concentration of this solution? (In other
words, what is the molarity?)
 What volume does 0.34 moles of 12 M HCl
occupy?

What about volumes of gases?
At normal conditions, one mole of any gas
will occupy 22.4 liters.
 In other words, a gas will have a value of
22.4 L/mol at normal conditions.
 If it isn’t normal conditions, a density will
be given so solve for mass.
Examples (both at normal conditions):
 How many moles of H2 (g) occupy 89.76
L.
 What is the volume occupied by 2.5 moles
of O2 gas?

Now, let us practice getting various
things into the mole.
1.
2.
3.
4.
5.
100 mL of 5.6 M of HCl
68.5 L of O2 (g) at normal conditions
89 g of AgCl
9.65 x 1022 ions of Cl75 mL of nitrogen (l) whose density is
0.808 g/mL
So, that is the first step.
Get it into moles.
After that, we need to change
moles to moles.
How do we do that?
That is where a balanced equation comes into
play.
 We can change moles to moles of any two
substances in the chemical equation by using
the coefficients in the balanced equation.
 The conversion factor (also known as the mole
ratio) will be:

molessolving for/molesgiven
For Example
For example, let’s say we have the balanced
equation: 2NaCl  2 Na + Cl2
If we have 5 moles of NaCl, how many moles a
Na and Cl2 will we have after the reaction?
Coefficients
2 moles Na
5 moles NaCl (given) x ______________=
5 moles Na
2 moles NaCl
1 moles Cl2
5 moles NaCl (given) x ______________=2.5
moles Cl2
2 moles NaCl
Then after you have the moles, you can
change it back to whatever you need
Convert the following:
 5.5 moles of NaOH into grams
 10 moles of copper into atoms
 0.8 moles of neon gas at normal conditions to
Liters
 3.5 moles of HCl dissolved in 100 mL of water
is what molarity?
 The volume of 15 moles of gasoline, if the
density of gasoline is 0.77 g/mL

In other words…
1. Is the number they gave me in moles? If not,
how can I change it into moles?
2. If I am changing compounds, I need to change
moles to moles. Use the coefficients from the
balanced equation (the coefficient of the
substance you were given goes on the bottom).
3. Do they want the answer in moles? If not, how
can I change the moles into that unit?
Mass
Summary Chart
In grams
Volume
(Liters of a
gas at STP)
÷
x
22.4
÷
Molar
mass
x
MOLE
Molarity
÷
NA
x
(moles of solute per
liter of solution)
Use the formula:
M = mol/L
Atoms
(or molecules)
6.02 x 10 23 atoms
(molecules) per
mole
Practice Problems

2.00 g H2O (g) = ____________ L H2O (g)

8.00 x 1022 molecules C6H12O6 =
_________ g C6H12O6

1.00 x 1024 atoms Ar = ________ liters Ar
More Practice Problems
Use the following balanced equation to
complete the following:
CaC2 (s) + 2H2O (l)  C2H2 (g) + Ca(OH)2
 How many grams of water are needed to
react with 485 g calcium carbide?
 How many grams of CaC2 are needed to
make 23.6 g C2H2?
 If 55.3 g Ca(OH)2 are formed, how many
grams of water reacted?

Homework

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