You have a great web site in my web helps to give you this same information in a slightly different way. Take a look at it now by clicking on Mole <---> grams conversion tutorial . Here I will go through what we have done in DO NOW’s.
I am going to add something to the diagram you find at that website. So the complete diagram will give you everything you need for solving the problems below. Remember it takes avocadoes to make guacamole from atoms or molecules.
Avogadro gaw or atoms
or molecules
<----------------> MOLES
<--------------->
grams 6.022 x 1023 (of
first | atom or gmw | molecule) | |
(mole ratio from)
** balanced | equation ** | | | Avogadro | gaw or atoms or
molecules
<----------------> MOLES
<--------------->
grams 6.022 x 1023
(of 2nd
atom or gmw molecule)
Why the MOLES two times? If you start with moles of one kind of molecule or atom and need to change to moles of a different molecule or atom, you use the mole ratio from the chemical equation to make the change.
If you are give
8 grams of H2 , how many grams of O2 will you need
to burn up all 8 grams of H2 for the following balanced chemical
equation:
2
H2 + O2
-----> 2 H2O
Solution:
First look at the balanced equation and the mole
ratios:
2
H2 + O2
-----> 2 H2O
mole ratios 2
: 1 : 2
This
tells you how many moles of
O2 will be
needed for each mole of H2
you have. But, we were given
grams of H2 so we have to change from grams to moles before we can
use the mole ratio. So we use the gmw (gram molecular weight) of H2
to do that. The gmw of H2 can be retrieved from the
periodic table
2 H’s x 1g
= 2 g for the gmw
of H2 because there are two
atoms in the molecule.
|
8 g H2 |
1 mole H2 |
|
|
|
|
|
1 |
2 g H2 |
|
|
|
|
With this
step, we went from grams to MOLES on our diagram. But we need the moles of O2 not H2 . This
is the step that requires the extra MOLES I put into the diagram above. Now it’s time to cancel the units we can and also
change moles of H2
to moles of O2 . For
that we need to use the mole ratio for
the balanced equation.
|
8 |
1 |
1 mol O2 |
|
|
|
|
1 |
2 |
2 |
|
|
|
Now that we are in moles of O2 we need to take the last step and
convert to grams of O2 . For
that we use the gmw of O2 which is 2 O x 16g = 32g which is also equal to 1 mole.
|
8 |
1 |
1 |
32 g O2 |
Answer |
64 g O2 |
|
1 |
2 |
2 |
1 |
|
|
PROBLEM 2
If you are given 96.1 g C3H8 . You are asked to find out how many grams of O2 will be needed to burn up all of the 96.1 g
of C3H8 , given
the following unbalanced chemical
equation:
C3H8 + O2 ----->
CO2 + H2O
Solution:
Your first
task is to balance the equation. If you do not balance the equation, you will
not know the mole ratios of the reactants and products and will not be able to
convert from moles of
C3H8
to moles moles of O2
, which is the 3rd
step in the solution of this problem.
C3H8 + 5
O2 -----> 3 CO2 + 4
H2O
mole ratio 1 :
5 : 3 :
4
We will
need the gmw (gram molecular
weight) of both C3H8 and
O2
3 C x
12g = 36 g 2
O x 16g = 32 g/mole of O2
8 H x 1 g = 8 g
--------
44
g/mole C3H8
The last
piece to our puzzle is the mole ratio of
C3H8 : O2
given to us by the balanced formula:
1 mole C3H8 =
5 mole O2
|
96.1 g C3H8 |
1 mol C3H8 |
5 mol O2 |
32 g O2 |
|
|
|
1 |
44 g C3H8 |
1 mol C3H8 |
1 mol O2 |
|
|
Now do the unit
cancellation, multiply all the numbers left on the top, multiply all the
numbers together from the 2nd row and divide the 2nd row
into the top row number
|
96.1 |
1 |
5 |
32 g O2 |
Answer |
348 g O2 |
|
1 |
44 |
1 |
1 |
|
|