Moles to Mass & Mass to Moles

Just when we thought we were done with our conversions, here's a lesson with some more converting between moles and mass!
Something to remember though, that I think is handy with ANY of the conversions we learned: always look at the units, and think about what the final unit that you are trying to get is!

Some questions will give you an amount of moles and an equation (which will be balanced or you will have to balance it) and ask you to determine the mass of a compound/element
Converting moles to mass only requires one extra step from moles to moles, and its a step which we've actually learned before!

Here are some lovely examples:
Ex. How many grams of Bauxite (Al2O3) are required to produce 3.5 mol of pure Aluminum?
Step 1: Make an equation for this, and balance it!
2Al2O3 ---> 4Al + 3O2
Step 2: Look at the info you're given. You have the mol of Aluminum, and you want to find out the mass of the Bauxite. Find out the mol of Al2O3 (using the coefficients) so that you can convert it to mass (using molar mass!) :
3.5 mol Al    x    2 mol Al2O3  =   1.75 mol Al2NO3  x    102 g                  =   178.5 g 

                             4 mol Al                                                  1 mol Al2NO3

Step 3: Sig Figs!
178.5 g = 1.8 x 10^2 g

Ex. How many grams of water are produced if 0.84 mol of Phosphoric Acid is completely neutralized by Barium Hydroxide?
2H3PO4 + 3Ba(OH)2 ---> 6H2O + Ba2(PO4)2
0.84 mol H3PO4  x      6 mol H2O  =  2.52 mol   x   18.0 g H2O  =   45 g

                                    2 mol H3PO4                              1 mol

Now here's a different kind of example. Instead of being given the moles of a compound, you are given the mass, and you must find the moles. You have to use the same conversions and all, just in a different order!

Ex. How many moles of Lead II Nitrate are consumed when 4.5 g of Sodium Sulphide completely reacts?
Pb(NO3)2   +  Na2S  ----> PbS  +  2NaNO3

4.5 g  Na2S  x   1 mol              =  0.0576 mol Na2S  x  1 mol Pb(NO3)2  =  0.058 mol

                          78.1 g Na2S                                            1 mol Na2S

Empirical and Molecular Formulas

No, this post isn't overdue at all, no, no. It's just fashionably late :)
This lesson is after Percent Composition, if anyone is keeping track of order!

Off to the lesson now:

Empirical formulas are the SIMPLEST formula of a compound
For example, the molecular formula of Octane is C8H18
The empirical formula of this is C4H9 (because both of the subscripts above are divisible by 2)

Empirical formulas show only the simplest ratios and not the actual number of atoms of compounds, whereas Molecular Formulas give the actual number of atoms. Here are some lovely examples:

Molecular Formulas                               Empirical Formulas
P4O10                                                      P2O5

C6H18O3                                                C2H6O

C5H12O                                                  C5H12O    (cannot be simplified further)

To determine the empirical formula, we need to know the ratio of each element. This is simple when you get the molecular formula, like above, but sometimes you are just given the mass, moles, etc, and so being the incredible chemistry students that we are, we will determine the empirical formulas!:

Example: 

Atoms      Mass        Molar Mass    Moles       Smallest Mole       Ratio
C              50.5g        12 g/mol         4.2 mol            1.33                    4
H              5.26 g        1.0 g/mol       5.26 mol           1.66                   5

N              44.2 g        16 g/mol        3.16 mol            1.0                    3                                              

 Are you asking "Why, how did you do that, you incredible genius?" then here are some general steps:
first, depending on what you are given, you find the mass, molar mass, and moles of each element!
Then, you divide by the smallest amount of moles (in this case, the N because it is 3.16 mol), and that will give you a number, which you will place under the 'smallest mole' column. Now, often this number will be a BEAUTIFUL whole number with no decimals to worry about, in which case you rewrite those numbers under 'ratio' and BAM you've got your equation.
Well this example is different. The Smallest Mole numbers have come out all ugly and have to undergo changes! In this case, since the ending is either .33 or .66, then you multiply all of the smallest mole numbers by 3 to give you a whole number, which gives you the ratio. And so your compount is C4H5N3.
*note: if the ending were .5, you'd multiply by 2, if the ending were .25 or .75 you'd multiply by 4!

IMPORTANT MESSAGE: that rainbow chart took a long time to do. so appreciate the beautiful colors fully. please and thank you

Now what do you do if you're given the empirical formula and you are trying to change it to Molecular Formula?
First of all, you NEED to know the molar mass (in chem 11 it's usually given). Then you find the molar mass of the empirical, which will show you what you need to multiply the subscripts in the Empirical formula by. Here's an incredible example:

The empirical formula for a substance is CH2O and its molar mass is 60.0 g/mol. Determine the molecular formula:
Empirical Formula               Molecular Formula                      
CH2O                                     C2H4O2

30.0g/mol                                60.0 g/mol

In this case, the molar mass of the molecular formula was double the molar mass of the empirical formula, so we just have to multiply the subscripts of the Empirical formula by 2!

Here's a long, but good video on Molecular and Empirical Formulas:

http://www.youtube.com/watch?v=gfBcM3uvWfs

 This is what came up when I looked up "Empirical Formulas". I just had to put this up, cause it just reminds me of Mr Doktor so incredibly much.

Mole to Mole conversions

Mole to mole conversion is basically predicting how many moles will be reacted or produced in a balanced equation. All you have to do is remember the golden rule:
                      What you need over what you have

That's about all you need to know so lets do some examples:

If you have 2moles of water and it decomposes, how many moles of hydrogen are produced?

                          2H2O = 2H+O2
                        
                        2 mol H2O * 2 = 2 mol H
                                            2
 2 moles of hydrogen are produced

If we wanted to know how many moles of oxygen were produced, we would do this

                      2H2O = 2H+O2
 
                     2 mol H2O * 1= 1 mol O2
                                         2

In this case only 1 mole of oxygen would be produced from 2 moles of water.



So now you know how to do Mole to Mole conversions.

Simple right?

So there you have it, Mole to Mole conversions. Simple right?

Percent Composition

Today after a long winters nap, Mr Doktor's E block Chemistry class met once again. The topic of today was percent composition, or what percentage of a compound comes from each element. The basic rule is:

• The percentage by mass of an element in a compound is ALWAYS the same

To find percent by mass, determine the mass of each element present in one mole of the compound.

For example:
NaCl has a total atomic mass of 58.4g/mol(22.9 +35.5)
To find the percent of Na in one mole of NaCl, divide Na's atomic mass by the atomic mass of NaCl
                                  
                                    22.9/58.4= .392= 39% Na

Ex 2:
           You have a 10g sample of H2O. How many grams of Hydrogen are there?

H2O=18g/mol       H2=2g/mol       2/18=.09   (.09)(10)=.9gH

Thats all for today folks and have a............