Lecture Notes – Ch 9.1 & 9.2 – mol-mol 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 H2  ---->  CH3OH     (molecular equation)

As you read this equation you can say:   one molecule of CO reacts with 2 molecules of H2 to make 1 molecule of CH3OH.  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 1023 CO  +  (2)(6.022 x 1023 ) H2  ---->  6.022 x 1023 CH3OH

Since 6.022 x 1023 = 1 mole, this equation can be rewritten as:

 1 mol CO    +    2 mol H2  ---->  1 mol CH3OH     (mole equation)

As you read this new equation you say:   one mole of CO reacts with 2 moles of H2 to make 1 mole of CH3OH.  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 H2       ------>           CH3OH     (molecular equation)

1 mol CO       +   1 moles H2      ---->      1 mole CH3OH    (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 H2 we find the mass by:

total mass 2 H2 x 1g  =  4 g  (gmw of one mole of H2 molecules)

For CH3OH 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 CH3OH = 32 g  (gmw for 1 mole CH3OH)

Rewriting the equation to show mass ratios, the equation would read:

7           :     1           :        8                   (simplest mass ratios)

28g CO  +  4g H2   ---->  32g CH3OH         (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  Mole-Mole Relationships

Example 9.2 (see page 262 - 263)

Question:

If I have 2 moles of H2O, how many moles of O2 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 H2O.)

H2O  ---->  O2  +  H2

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 H2O   ---->   O2  +  2 H2        (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 H2O  =  1 mol O2

2 mol H2O  =  2 mol H2

1 mol O2     =  2 mol H2

The equality (ratio) we need to use in our grid is    2 mol H2O  =  1 mol O2

Using this equality in a grid we get:

# Given

Use the equality I chose

# Answer

2 mol H2O

1 mol O2

1 mol O2

1

2 mol H2

Now try working this same equation using these questions:

1.   If you want to make 3 moles of O2 ,  how many moles of H2O do you need?

2.   If you want to make 3 moles of O2 , how many moles of H2 will be made at the same time?

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Example 9.3  (see page 263 – 264)

Question:

How many moles of O2 are needed to react exactly with 2.15 moles of propane ( C3H8 ) in the following reaction?

C3H8   +   O2    ---->    CO2  +  H2O

Solution:

First thing we must do is to balance the equation:

1       :      5          :          3       :     4 (simplest mole ratio)

C3H8   +   5 O2    ---->    3 CO2  +  4 H2O    (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 C3H8  =  5 mole O2

mole ratio

# Answer

2.15 mole C3H8

5 moles O2

10.75 moles O2

1

1 mole C3H8

Now try working these problems using the same balanced equation:

1.  If you have 2.15 moles C3H8 ,  how many moles of CO2 can you make?

2.  If you have 2.15 moles C3H8 ,  how many moles of H2O can you make?