Lecture Notes – Ch 9.1 & 9.2 – molmol and
ratios from formulas Std 3e
9.1 – Information from Chemical Equations
The following balanced chemical equation gives us a lot of information. Most importantly, it tells us the ratio of the reactants to the product. This allows us to predict how much product we can make with any amount of reactants.
CO + 2
H_{2} >
CH_{3}OH (molecular equation) 
As you read this equation you can
say: one molecule of CO reacts
with 2 molecules of H_{2} to make 1 molecule of CH_{3}OH. Each number (coefficient) actually
represents how many molecules of each reactant and product is involved in
the reaction.
If we
rewrite the equation with moles of molecules, it will look like this:
6.022
x 10^{23} CO + (2)(6.022 x 10^{23}
) H_{2} > 6.022 x 10^{23} CH_{3}OH
Since 6.022
x 10^{23} = 1 mole, this equation can be rewritten as:
1 mol CO
+ 2 mol H_{2} >
1 mol CH_{3}OH (mole equation) 
As
you read this new equation you say: one
mole of CO reacts with 2 moles of H_{2} to make 1 mole of CH_{3}OH. Each number (coefficient) now
represents how many moles each reanctant and product is involved in the
reaction.
Whether we
have molecules or moles the ratios are the same. The only difference in the second equation is
that we are talking about moles (Avogadro Numbers) of molecules or atoms instead of individual molecules or atoms.
In the
second equation we have the following mole ratios:
1 : 2 : 1
(mole ratio)
CO +
2 H_{2} >
CH_{3}OH (molecular
equation)
1 mol CO + 1 moles H_{2}
> 1 mole CH_{3}OH (mole
equation)
There is one more type of equation, the mass equation. In this type of equation we change the mole
to either gmw (gram molecular weights) or gaw (gram atomic weights). To do a mass equation we first must figure
all of the gmw’s for this equation (note:
this equation has no gaw’s).
For CO we find the mass by:
1
C x
12g = 12 g (gaw for C)
1 O x
16g =
16 g (gaw for O)
Total
= 28 g
total mass CO = 28g (the total gmw for one mole of CO
molecules)
For H_{2} we find the mass by:
total mass 2
H_{2} x 1g = 4 g
(gmw of one mole of H_{2} molecules)
For CH_{3}OH we find the mass by:
1 C x 12g =
12g (gaw of 1 mole of C atoms)
4 H x 1g =
4g (gaw of 4 moles of H atoms)
1 O x 16g =
16g (gaw of 1 mole of O atoms)
Total
= 32 g
total mass
of one mole of CH_{3}OH = 32 g
(gmw for 1 mole CH_{3}OH)
Rewriting the equation to show mass ratios, the equation would
read:
7
: 1 : 8 (simplest mass ratios)
28g CO
+ 4g H_{2} >
32g CH_{3}OH (mass
equation)
The mass
equation is what you need to get the weights for mixing the correct
proportions, because we weigh everything
in grams before mixing them for any reaction.
As you see, you cannot figure out the mass equation without the mole
equation. The mole equation only works
because of molecular equation.
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9.2 MoleMole
Relationships
Example 9.2
(see page 262  263)
Question:
If I have 2
moles of H_{2}O, how many moles of O_{2} can I make? The unbalanced equation for the reaction is
as follows:
(Notice
I’ve changed the starting number of moles from 5.8 to 2 moles of H_{2}O.)
H_{2}O >
O_{2} + H_{2}
Solution:
Here is my way
of working this problem out. First we
must balance the equation so it reads:
2 : 1
: 2 (simplest mole ratios)
2
H_{2}O > O_{2} + 2 H_{2} (molecular
equation)
Next we use
the simplest ratios (the coefficients) as the mole ratios because mole ratios
are the same as molecule ratios. With
this we can come up with several equalities, or ratios:
2
mol H_{2}O = 1 mol O_{2}
2
mol H_{2}O = 2 mol H_{2}
1 mol O_{2} = 2 mol H_{2}
The equality (ratio) we need to use in our grid
is 2 mol H_{2}O = 1
mol O_{2}
Using this
equality in a grid we get:
Given 
Use the equality I chose 

Answer 
2 mol H_{2}O 
1 mol O_{2} 

1 mol O_{2} 
1 
2 mol H_{2} 


Now try working this same equation using these
questions:
1. If you want
to make 3 moles of O_{2} , how
many moles of H_{2}O do you need?
2. If you want
to make 3 moles of O_{2} , how many moles of H_{2} will be made
at the same time?
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Example 9.3 (see page 263 – 264)
Question:
How many
moles of O_{2} are needed to react exactly with 2.15 moles of propane (
C_{3}H_{8} ) in the following reaction?
C_{3}H_{8} + O_{2} >
CO_{2} + H_{2}O
Solution:
First thing
we must do is to balance the equation:
1
: 5 : 3 :
4 (simplest mole ratio)
C_{3}H_{8} + 5
O_{2} > 3 CO_{2} + 4 H_{2}O (balanced molecular equation)
We want to
find the moles of oxygen, so we need an equality to take us from the given 2.15
moles of propane to the moles of oxygen.
So we use the ratio of the coefficients in the molecular equation to
get:
1
mole C_{3}H_{8} = 5 mole O_{2}
Given 
mole ratio 

Answer 
2.15 mole C_{3}H_{8}

5 moles O_{2} 

10.75 moles O_{2} 
1 
1 mole C_{3}H_{8}



Now try
working these problems using the same balanced equation:
1. If you have 2.15 moles C_{3}H_{8}
, how many moles of CO_{2} can
you make?
2. If you have 2.15 moles C_{3}H_{8}
, how many moles of H_{2}O can
you make?