Tutorial Mass <----> mole conversions Std 3e

 

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 NOWs.

 

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.

 

 

PROBLEM 1

 

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 Hs 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 its 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 g H2

1 mole H2

1 mol O2

 

 

 

1

2 g H2

2 mol H2

 

 

 

 

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 g H2

1 mole H2

1 mol O2

32 g O2

Answer

64 g O2

1

2 g H2

2 mol H2

1 mole O2

 

 

 

 

 

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 g C3H8

1 mol C3H8

5 mol O2

32 g O2

Answer

348 g O2

1

44 g C3H8

1 mol C3H8

1 mol O2