Sorry for the late post!
Totally forgot about the lab blog.
The lab we did was Lab 4c - Determining the EmPirical formula of a compound.

In this lab, the materials we had were a pipestem triangle, iron ring, a stand, and a Bunsen burner, and a crucible.

But always remember before you start a lab to put on your lab coat and safety goggles!!

After we heated up the empty crucible for 3 minutes, we weighed it and determined the mass of the crucible + lid.
We then Placed the hydrate into the crucible until it was 1/3 full.

After, we heated the crucible until the bottom of it turned red.
Allow the crucible to cool for 5 minutes, then record the mass of the crucible + lid + contents.

Reheat the crucible so all the water is gone, and check the mass. It should be close or equal to the first mass.

Add a few drops of water to see if there is a change.



Here are some more cool pictures of our experiment!!

Molar volume!

Today we learned how to calculate molar volume!

Some things to know first are
•gases expand and contract (change volume) with changes in temperature and pressure
•we have a standard condition to compare volume of gases called STP ( standard temperature and pressure)
•STP= 1 atmosphere of pressure and a temperature of 0 degree celsius or 273.15k
•at STP, 1 mole gas occupies 22.4 L

This may sound really confusing ( that's what I first thought) but it makes a lot more sense when you learn the formula

Formula= 22.4 L of gas/1 mol of gas
Or 1 mol of gas/22.4 L if gas

These formulas are an add on to the mole map!
But it only attaches to the mole conversion section.

Ex. You have a volume occupied by 6.2g of Sulfur at STP

So...
Step 1- convert grams to moles
6.4g x 1 mol/32g= 0.2moles
Step 2- convert moles to liters using our new formula!
0.2 moles x 22.4L/ 1 mol= 4.48!!

but...

Make sure you remember your sig figs!
4.48 will round up to 4.5!!

Here is a slightly harder example with explanation!

Don't forget that, for our chapter 4 test on monday!

Diluting Solutions to Prepare Workable Solutions

Basically in this lesson we learnt how to dilute solutions to prepare solutions you can actually work with.
We need to be able to make solutions of any concentration.
The equation is actually pretty easy:

Formula: M₁L₁ = M₂L₂

_{What this means is the amount moles of the solute before is equal to how many moles of solute there are after.}

_{The subscript 1 = before}

The subscript 2 = after

Just a little note: Final Volume - Before Volume = Water Added

Example

We have 100 mL of 0.125 M H2SO4. We want to end up with 0.500 M H2SO4. 

HOW DO YOU DO IT?

So remember the formula from above. 

Step 1.

100 mL of 0.125 M --> 0.500 M

100 mL x 0.125 M = mL x 0.500 M

Step 2. 

You have to separate the mL so that it is away from the equation , like you do in algebra.

100 mL x 0.125 M = mL

                                                            0.500 M 

Step 3.

Basically all you have to do is calculate.

Once you've calculated it all you'll get something like this: 

25 = mL.

So you will need 25 mL of H2SO4 to get 0.500 M of H2SO4.

So easy, right?

Key Concepts

• The concentration of a solution is usually given in moles per litre (mol/L). 
• This is also known as molarity.
• Concentration, or Molarity, is given the symbol M. 
• The solute is the substance which dissolves.
• The solvent is the liquid which does the dissolving.
• A solution is prepared by dissolving a solute in a solvent.
• When a solution is diluted, more solvent is added to it.

Can you imagine drinking lemonade without it being diluted first? It'd be pure lemon! Yuck!

Lemonade is made by adding around 4 cups of water to a concentrated solution of lemon.

Without the water it'd just be a really sour syrup! How good can that taste?

If you STILL don't understand, check out this video

How to Dilute a Solution

Back at it!

Well we are back at it! And bringing you today's topic...molarity!

Molarity is used to compare concentrations of solution. The concentration is the amount of solute dissolved in a certain volume of a solution. Molarity is the number of moles of solute in one litre of a solution, and we use the units moles/L.

Lets make sure you know the basics. A 10.00 moles/L solution is more concentrated than a 2.00 moles/L. Yes or no?
Hopefully you said yes. If you didn't, stop taking chemistry. Just kidding. But keep it in mind.
As always, we need a formula. Here's the one for molarity!

Molarity = moles of solute (mol)
                 ________________
                volume of solution (L)

or more simply if you need it:
M = mol
      -----
         L

Rearrange to solve for mol and L
mol = m x l and L = mol/M

Now onto some real questions.
Calculate the molar concentration of a solution that has 0.510 moles of NaOH in 1.400 L of solution.

Plug in the information given into the equation, and don't forget about the sig figs! We learned them for a reason people!

so M NaOH = 0.510 mol NaOH / 1.400 L
With the calculations, it comes out to M NaOH = 0.364286 M NaOH
But is that right? No!!!
The correct answer is 0.364 M NaOH.

There are three other types of questions we will cover:

1) calculating the number of grams
2) how many moles are contained in ___ L of ___ mole/L
3) What volume of ____ mole/L contains ___g of ___?

Lets start with calculating grams.
What is the number of grams of CaOH in 1.30 L of a 0.75 M Ca(OH)2 solution?
moles Ca(OH) 2 = 0.75 M Ca(OH) 2 x 1.30 L

We have to convert now, since it isn't in grams. So convert to grams.
Molar Mass = 40.1 + 17.0 x 2 = 74.1 g/mol
0.975 moles Ca(OH)2 x 74.1 g Ca(OH)2
                                     ------------------
                                     1 mole Ca(OH)2
So the answer is 72.2475 g Ca(OH)2
Don't forget the sig figs though! The answer is 72 g Ca (OH)2

Now, how many moles of Ca(OH) 2 are contained in 0.25 L of 0.80 mole/L Ca(OH)2?
0.8 x 0.25 = 0.20 mole Ca(OH)2

And thirdly, What volume of 0.50 mole/L Ca(OH)2 contains 0.10 g of Ca(OH)2?
volume required = moles
                            ------
                            concentration

1.7 x 10^-3 mole Ca(OH)2 / 0.50 mole/L Ca(OH)2
=0.0033 L

There you go! Now you can take some information and do four different things with it. Cool, right?
Well I know you have just been skimming this anyways waiting for the awesome videos, well they are coming at you now!

Basic Molarity Instructions
Need more help?
Molarity Rap (YOU MUST WATCH THIS!!)

QUIZ DAY!

Okay so cool, we finished our quiz on Molecular and Empirical Forumla, along with Percent Composition!
there isnt alot to say about this because we didnt have an actual lesson before or after the quiz...

I found that the quiz was quite straight forward! there was nothing on that quiz that Ms. Chen didnt
teach us. (im not being a suck up here)

Hopefully its not one of those quizzes where i thought i did well, but when i get it back... i didnt do so great.
If your still not getting it here are some VIDEOS:

Molecular and Empirical Formula: http://www.youtube.com/watch?v=gfBcM3uvWfs
and of course... A SONG, pretty intense: http://www.youtube.com/watch?v=slO9kN1XCnE
.... ok so they arent much good at singing, but if u look at the lyrics its about moles and molecular formula!

Percent Composition: http://www.youtube.com/watch?v=xbEeyT8nK84

ENJOY!

Review for Quiz!

Next class is quiz day!! so today we did a lot of review!

Quiz topics

• percent compositions
• empirical formula
• molecular formula
• calculating empirical formula of organic compounds

Percent compositions

ex. CO2
Total MM = 44          %compositions = mass of elements  x 100%
MM of C = 12                                       mass of comp
MM of O = 32

% of C = 12g/mol x 100% = 27.3%
               44g/mol

% of O = 32g/mol x 100% = 72.7%
               44g/mol

talk through of CO2 composition

Empirical Formula

- lowest term ratio of atoms in formula
* All ionic compounds are empirical formula


We have 10.87 g of Fe and 4.66g of O, whats the empirical formula
#1 convert to moles

Fe  10.87g x 1mol  =  0.195 moles
                     55.8g
O    4.66g x 1 mol  =  0.291 moles
                    16g
#2 divide both by smallest molar amount
Fe = 1
O = 1.5
#3 scale to whole numbers
Fe = 2
O = 3

Molecular Formula

empirical formula of C2H5 molar mass of 58g/mol
MM C2H5 = 29g/molE
n = 58g/mol = 2
      29 g
MF= 2( C2H5) = C4H10

really helpful video


New Stuff!!!

Empirical Formula of an organic compound


Found by

• burning wood (reacting with CO20
• collecting and weighing the products
• from mass of a product the moles of each element in the original organic compound can be calculated

5g of a carbon and hydrogen , 15g of CO2 and 8.18g of H2O

15.0g = 1 mol = 0.341
             44g

8.18g = 1 mol = 0.454
             18g

Molecules of C = 0.341 mol CO2 x 1mol C = 0.341 molecules 0.314 = 1 x 3 =3
                                                        1mol CO2                          0.314

Molecules of H = 0.454 mol H2O x 2mol H = 0.908 molecules 0.908 = 2.66 x 3 = 8
                                                        1mol H2O                          0.314

C3H8

Checking Mass

0.314 x 12g  = 3.77
             1mol                                3.77 +  0.908 = 4.68g
0.908 x 1g     = 0.908
             1 mol


Study, Study, Study, Quiz next day!!

Empirical + Molecular Formula

Today's lesson is on empirical and molecular formula.

Empirical Formula = gives the lowest term ratio of atoms (or moles of course) in the formula.
Ionic compounds are empirical formulas as well!

And here we have our example:

Lets say we have 10.87 g of Fe and 4.66 g of O. What is the empirical formula?
Well, first we convert grams to moles.

we want to convert the 10.87 grams so we will put that first. We want to eliminate grams so be sure to put the molar mass of the element in grams on the bottom.
10.87 g x 1 mole
                ------
                55.8 g = 0.195 moles
4.66 g x 1 mole
              -----
              16 g = 0.291 moles

Now the smaller one of the two is the first one, so be use that one and divide by that amount.
0.195/0.195 = 1
0.291/0.195 = 1.5
Now we have to scale ratios to whole numbers. We need to get rid of that 1.5. We do that by multiplying both by two, because then the 1.5 turns into a 3!
Fe 1 x2 --> 2
O  1.5 x2 ---> 3
So the answer would be Fe2O3

Here is a problem for you to try for practice:
A compound used for sweetening is 57.14% C, 6.16% H, 9.52% N, and 27.18% O.  Calculate the empirical formula of the sweetener and find the molecular formula.  (The molar mass of the compound is 294.30 g/mol)
Remember to follow the steps!
1) convert to moles
2) divide both by the smallest molar amount
3) scale ratios

Now onto molecular formula:

Lets use this formula to help answer the questions:

n = molar mass of compound
      --------------------------
      molar mass of the empirical formula

The empirical formula of a gas is CH2. What is the molecular formula if the molar mass is 42 g/mol?

First, lets calculate the molar mass of the compound, because we need it in the equation.
The mass of carbon is 12, and the mass of hydrogen is 1, but there is 2 o them. So the molar mass of the compound is 14.

Now lets plug that into the equation!

n = 42 g/mol
      --------
      14 g/mol
which equals 3 g/mol.
Now write the answer 3(CH2)
make sure to distribute the 3.
so it is actually C3H6 = MF propane
Make sure you write the MF! It stands for molecular formula. You can remember it by thinking that it is Melanie's initals!

If you want a little extra help, I've got videos galore for you!
Molecular and Empirical Formula
A helpful chemistry tutorial
Extra help!
A super cool 16 minute tutorial! woah!