MOLARITY!

Hi everyone! So last class we learned about the molar concentration aka molarity of solutions.

Recall:
- solution = a homogenous mixture where one substance is dissolved in another
- solvent = larger quantity; dissolves the solute
- solute = smaller quanitity; it's the amount of solute being dissolved in a certain amount of solution



Why do we need to know molarity?
ANSWER: To compare the amount of solute dissolved in a certain volume of solution!

"Molar Concentration" (aka molarity) is the number of moles of a solute in 1 litre of solution

• measures the concentration of solute in the solution
• "M" is the symbol we use to represent molarity
• "mol/L" is the unit

There are 3 different formulas (but as long as you know one, you can find the others):

1) MOLARITY = Moles/Litres
2) LITRES = Moles/ Molarity
3) MOLES = Molarity x Litres

Let's see some examples! 

1) Calculate the MOLARITY of a 1.5 L solution that contains 0.24 moles of HCL.

    Formula: M = mol/L
                  M = 0.24mol/1.5L
                  M = 0.16M of HCL

2) Calculate the concentration of 60 mL solution that contains 0.075 moles of NH4Cl. 

    Formulas: M = mol/L
                   M = 0.075mol/(0.060L)
                   M = 1 M  

3) What is the mass of solute for 125mL of 0.25 M Ba(NO3)2?

    Formula: mol = L x M
                  mol = 125mL x 1L/1000mL = 0.125 L
                  mol = 0.125L x 0.25 M
                  mol = 0.03125 mol Ba(No3)2 
                  0.031 mol x 261.3g of Ba(NO3)2/ 1mol = 8.2g of Ba(NO3)2 

 4) What volume in ml of a 0.25 mol/L NaOH solution is needed for 0.050mol of NaOH?


   Formula: L = mol/M
                 L = 0.050mol/0.25M
                 L = 0.20 L x 1000 = 2.0 x 10^2 mL



Here is an informative video on molarity to further educate you guys!

to be hydrated or ANhydrated? (LAB 4C)

So, over the course of two classes, we've once again did a lab!! This time, it was all about finding the empirical formula of a hydrate!

Now to give you a better understanding of this lab, here are some background information:

• A hydrate is a compound that contains a definite number of water molecules (H2O) 
• Without containing water, it becomes an anhydrous (without water) salt
• The formula of a hydrate is AB•xH2O where AB is the anhydrous salt and x is the number of moles of water per mole of the anhydrous salt

**always use crucible tongs and never touch or eat the substance!**


Here is the hydrate that we used:

Now in order to find the empirical formula, we have to find the mass of water that's in the hydrate by finding how much water was given off by burning the compound.
BUT WAIT! Before that, we must:
1. Find mass of empty crucible
2. Find mass of crucible + hydrate (before it's burned)
3. Subtract the mass in step 2 from the mass in step 1 to get mass of HYDRATE
Then, we heat the hydrate in the crucible until the substance changes to a white-ish colour. After letting it cool (SAFETY FIRST OF COURSE! - don't want to get your fingers burned) we weigh the crucible + anhydrous salt (as the water has been dried off). The mass should now be lighter than it was before. We do this process twice to make sure the water is completely gone! (the two masses should be within 0.03g)

(anhydrous salt of copper sulphate)

Next:
4. Using lowest mass of the crucible + anhydrous, subtract the mass of step 1 from this to get the mass of ANHYDROUS salt
5. Subtract the mass of the anhydrous salt from the mass of the hydrate to get mass of water that WAS in the compound


Given that the molar mass of the anhydrous salt (CuSO4) is 159.6g/mol, we can find the # of moles of the anhydrous.
mass of anhydrous salt x 1mol/159.6g = moles of anhydrous salt

Recall (from our quiz) that we have to find the ratio of moles of water to moles of anhydrous. Therefore, we find the moles of water from the molar mass of H2O:
mass of water given off x 1mol/18.0g = moles of H2O


Then, divide the two moles by the smallest molar amount (does this sound familiar now? :D)  and VOILA! You should get a whole-number ratio.


Thus, the empirical formula of this hydrate is CuSO4•5H2O (copper (II) sulphate pentahydrate)

Also, if you try adding water into the crucible with the anhydrous, an exothermic reaction will occur along with the changing of the color of the substance to the original hydrate color. Neat huh!

(teehee, goodluck!)

Organic Compounds - Formulablahblah

HAAII.

procrastinating for the win.

cats make me laugh so hard...

so last class we learned about empirical formulas, and this class we progressed to empirical formulas of ORGANIC compounds.  heres where things get tricky... :(

so a reminder: organic compounds MUST CONTAIN CARBON
-examples:  C
                    CH
                    CHO
CO is not one, because in an organic compound the hydrogen attaches to the carbon, and then the oxygen attaches to the hydrogen. NOT THE CARBON.

so to find the empirical formula of an organic compound, you need to first BURN the compound
second, you must collect and find the mass of the products (and convert to moles)

so say you have a mystery (organic) substance containing carbon and hydrogen, and when you burn it you end up with 10.0g of CO2 and 10.0g of H2O. Calculate the empirical formula.

STEPS:

1. Find the moles of the 2 compounds.

2. Divide each mole with the smallest molar mass.

3. If the resulting decimals are not even/close, multiply both substances until they are

(e.g. if you have 3.66mol H and 1 mol C, multiply both by 3 so you will have 11 mol H and 3 mol C)

4. If you are given the mass of the compound before it was burned, you should check the mass of the compound in the empirical formula to make sure it matches.  If not, then there might have been some oxygen in the original compound.

EXAMPLE

First Step: convert to moles

10.0g CO2 x 1/44 = .227mol

10.0g H20 x 1/18 = 0.556mol

Second Step: convert moles of compounds to C and H

0.227mol CO2 x 1mol C/1mol CO2 = 0.227mol C

0.556mol H2O x 2mol H/1mol H2O = 1.11mol H

and since there are 2H, then we multiply this by 2 to get a grand total of 1.11mol

Step 3: Now divide each by the smallest # of moles...

.227mol CO2 / .227 = 1

1.11mol H20 / .227 = 4.9, which rounds to 5 :)

so our empirical formula becomes CH5 

SUCCESS

Also remember to check your masses with the mass in the question (if stated) to check your answer! To do this, convert your moles of C and H back to grams and add them together. 

Total mass = grams of C + grams of H

If it's not equal (UH OH).. don't worry! There's just an extra step. Oxygen must be present, so the remainder of the grams is the grams of oxygen. Mass of O = mass of compound - mass of (C + H)

From there, convert the grams to moles and do step 3 again! Phew...

http://www.youtube.com/watch?v=9qGo9YTdnMg

^this is a video i found (and for the life of me, cant figure out how to insert...) and its kind of helpful especially if you want to isolate each element

as a reward, i has un autre chatton amusant <3

HAHAHAHAHAHAHAHAHAHAHAHAHAHAH TOOOO GOOOOOOOD

<3 Heather

Self Portrait:

sleep with one eye open...