Balancing formulas and other stuff

AS USUAL I GET THE HARD BLOG AGAIN....anyways today we learned about balancing formulas

First of all before you go balancing and telling everyone your a pro at it, we should remember something we learned.  In the equation we have 2 sides.  We HAVE to make it equal or else its entirely bs.  Why you may ask? Because according to the law of conservation matter is never evar destroyed evar.

ok lets get to the learning...or review....mostly review.  Anyway  to balance the equation you must make sure that all the atoms on one side equals the number of atoms on the other side.  There is no cleverly developed way to do this.  The way to do this is by your preference.  I, however like to balance whole groups first (eg OH, SO4, CO2) then balance the atoms in their elemental form (eg O2, K, P4).  Then CHECK CHECK CHECK to make sure both sides are balanced or else you messed up and have to waste over 10 minutes to do it again HAH!  Ms.Chen said i didnt have to go into super tryhard details since this is review so this will have to do, to jog your memories kthxbai.



note how both sides equal to each other.  He added a 2 to H2 on the left side and added a 2 to H20 on the right side



Molar Volume of a Gas at STP

Sup, followers... or follower. Happy New Year and the like.

Pro Tip: When you read a word that is not actually a word and is in all CAPS it is usually an acronym.

SO WHAT COULD STP STAND FOR?

• Scalar triple product
• Sewage treatment plant
• Shielded Twisted Pair
• Shovel test pit
• Signal Transfer Point
• Sodium tripolyphosphate
• South Texas Nuclear Generating Station
• Space Test Program
• Spanning tree protocol
• Steiner tree problem
• Strategic Technology Plan
• Standard Temperature and Pressure

ALL WRONG

except for the last one, that one is correct

Standard Temperature and Pressure are measurements of temperature and pressure that are common in most experiments to allow easier comparison and contrast between different experiments. We want to keep these measurements the same because gases contract and expand as these measurements differentiate.

These measurements are  

Temperature: 0 °C (273.15 K, 32 °F)

Pressure: 1 At the Moment (atm)...............................................................atmosphere

Also: at STP, 1 mole of gas will have a volume of 22.4L

SO WITH ALL OF THAT SAID: MOLE CONVERSION TIME!!!!!!!!!

as with any mole conversion, we need conversion factors!

If 22.4L = 1 mole THEN:

22.4L/1mole OR 1mole/22.4L

Example Question:

Calculate how many moles of NH3 are in 5.6L of gas at STP

(5.6L of gas)(1 mole NH3/22.4L of gas)

L of gas cancel out

(5.6/22.4)moles of NH3

0.25 moles of NH3

dude with some what seems like a Beta version of MS paint teaches you about STP

Diluting Solutions to Prepare Workable Solutions

Concentration is the amount of a substance that exists in a given volume of a solution. And we humans can lessen the concentration of a substance by diluting it. OMG wow so cool.

Wait, what is diluting you ask? This is the process of making a liquid thinner or weaker by adding water or another solvent into it.
Here is a picture of dilution                                            

As more solvent is added to each test tube, the solvent starts to get weaker or thinner.








You might say "We learned this already come on man/woman."

BUT WAIT IM NOT DONE YET MY PUPILS. This leads to the important part of this chapter.

The equation:  moles solute before = moles solute after.

WHATTTT???

In terms of molarity and volume: M1L1 = M2L2

*Note* Subscrite 1 means before and the subscript 2 means after

Here is an example :  Ex.  What volume of 8.00 M of HCl is used to make up 2.00L of 0.500 HCl

REMEMBER TO USE THE EQUATION

L1 = M2 x L2                     L1 = 0.500M x 2.00L
        -----------                  -----------------        =          0.125 L
             M2                               8.00 M

If you still do not understand,check out this website below.
http://www.ausetute.com.au/dilucalc.html

It is happy new year with a touch of MOLARITY

hmm So what is Molarity???? To be honest I don't know :X. Jokes Molarity is the amount of solute dissolved in what ever volume of Solution 

okay? so What is Solute and Solution? HMMM as we were told this was once written into our brains a while ago and since our mind have always being written into since grade 1 most of the stuff that was ever written is erased so it makes perfect sense that no one remembers what solute and solution means. Okay enough random talking lets get to the real stuff so hmm what is solute and solution? Solute is the THING that is BEING DISSOLVED and Solution is THE THING THAT IS DISSOLVING THE SOLUTE (yes solution is all caps because it is cool that way unlike Johnny(FACT)) jokes

The Formula ahh math again...

 Nothing out of the normal 

Molarity = Moles of Solute(mole)/Volume of Solution (L)

M = mol/L  SEE WHY mole isn't M but the lame abbreviation of mol since Molarity decided to steal it..

okay the fun examples now 

4 grams (C12H22O11) is dissolved in a 350 ml teacup. What is the molarity of the sugar solution?

First step

C12H22O11= C(12)(12) + H(1)(22) + O(16)(11)
C12H22O11= 144 + 22+ 176
C12H22O11= 342 g/mol

4 g/(342 g/mol) = 0.0117 mol

350 ml x (1L/1000 ml) = 0.350 L

and the final step 

0.0117/0.35 = 0.033 mol/L

ahh that was lots of typing T T

Okay second question

How many moles of salt are contained
in 300 mL of a 0.40 M NaCl solution?

0.300 L x 0.40 moles NaCl
               1.00 L                 = 0.12 moles NaCl

Third question

15grams of salt with a molarity of 0.458M what is the volume of salt?

Atomic mass of NaCl is 58.5g/mole

15G x 1 mole/ 58.5g = 0.256410 moles

0.256410/0.458 = 0.56 Ml

okay a little game that i found using Molarity

• Materials Needed:
• computer
• projector
• molarity slides
• periodic table
• notes
• wordbank

• Procedure:
• Opener:
1. Review with students the formulas for molar mass (atomic mass = 1 mole), molecules (6.02 x 10^23 molecules = 1 mole), molarity (mole/liter), and toxicity (weight in kg * LD50).
• Development:
1. Divide class into 2 teams.
2. Show the 5 categories. Explain that higher point questions are harder.
3. Explain that each team must elect a captain, who will be the person to give the final answer.
4. Explain that team 1 will choose a question to answer. Team 2 must also prepare to answer the question in case team 1 gives an incorrect answer. The team with the most points wins!
• Closure:
1. Suggestion: offer extra credit or free homework grade to the winners (although no incentive was necessary for a heated competition!).

Lab 4C Day a.k.a time consuming heating stuff up lab

    Today we had another lab which featured finding and calculating the mass of various objects such as anhydrous salt, empty crucible, water given off, etc.  To begin the lab we had to wear the awesome looking safety goggles.  Then ms. chen previewed over what we should do, and everyone i mean everyone ran to get all the supplies.  Each group of two acquired a pipestem triangle, iron ring, stand, and bunsen burner.  We set up the equipment so that the bunsen burner is directly underneath the crucible which is on the  pipestem triangle that is held midair by the iron ring locked onto the stand.  Before starting the lab we had to turn on the bunsen burner by connecting it to gas and lighting it up in order to dry out the crucible that is placed on the triangle.  Once dry we weigh the crucible and record it.  Shortly after we obtain some anhydrous salt equal to 1/3 of the max volume the crucible can hold and weighed and recorded them down again.  Then comes the fun part........heating stuff up



anhydrous salt





    Now we had to heat the anhydrous salt for 5 min.  During the heating you can see that the anhydrous salt which WAS BLUE TURNED GREENISH WHITE.  After the 5min and seeing cool colour changing, we let it cool and recorded its mass.  This process was repeated to see if the mass were the same.  Then we had to do the last step in our procedure, add water to the dehydrated salt.  As soon as we added a few drops of water.......IT TURNED BACK TOO NORMAL OMGWTH!! NO WAI!!!  Then we recorded down this amazing discovery and dumped it in the garbage.  Then we filled in the rest of the information we needed to fill in with the data from the lab.  As soon as we were done we handed our lab filled with all our data and observation to ms.chen.  It was fun

The Empirical Formula of an Organic Compound

What is ORGANIC COMPOUND?

Does that mean that the compound is twice as expensive and has absolutely no difference whatsoever? (distastefully biased plug on organic foods) No, it means that the compound contains carbon. BUT it is noteworthy that certain compounds containing carbon are indeed inorganic, like carbides and carbonates. In this lesson we learn how to calculate the empirical formula of such compounds with chemistry's favorite furry little critter, MOLES.

The smell of vinegar is caused by acetic acid (CH3COOH). It is an organic compound

We know that the empirical formula will contain the elements C and H because they are common in all organic compounds. What we do not know, are the amounts of each. If we let x represent the number of carbon and y represent the number of hydrogen, we can use a method to solve for each. THERE ARE 5 STEPS THAT ONE MUST TAKE TO ACHIEVE THIS GLORY. Actually 6, but the 6th is situaltional. It is also noteworthy that when an organic compound combusts, the compounds carbon dioxide and water are ALWAYS formed.

1st STEP:

Convert the mass of the CO2 and H2O 

2nd STEP:

Find the moles of C and H from the moles of CO2 and H2O

There is 1 carbon in CO2 and 2 hydrogens and H2O so basically multiply the value of H2O in moles obtained from the previous step by 2

3rd STEP:

Divide both your values by the smallest molar amount

4th STEP:

Change the ratios until a whole number for both is reached by multiplying by 1 (2/2, 3/3, 4/4) ETC

The empirical formula will be represented by the fraction obtained

5th STEP:

Convert moles back to grams and add them to Check your work

Confused? EXAMPLES HO!

A 3.79 g sample of an organic compound is burned to yield 6.61 g of CO2 and 3.59 g of H2O. Find the empirical formula.

STEP 1:

(6.61g CO2)(1 mol CO2/44.0g CO2) = 0.150 mol CO2

(3.59g H2O)(1 mol H2O/18.02g H2O) = 0.199 mol H2O

STEP 2:

(0.150 mol CO2)(1 mol C/1 mol CO2) = 0.150 mol C

(0.199 mol H2O)(2 mol H/1 mol H2O) = 0.398 mol H

STEP 3:

0.150 mol C/0.150 mol C = 1

0.398 mol H/0.150 mol C = 2.65

STEP 4:

(2.65/1)(3/3) = 8/3 ----> 8 = y 3=x

C3H8

STEP 5:

(0.150 mol C)(12 g/1 mol)= 1.80g C

(0.398 mol H)(1 g/1 mol) = 0.398g H

1.80 + 0.398 = 2.198

BUT WAIT! 2.198 != 3.97.

EXPLAINED: Some organic compounds may contain oxygen aswell. THUS, the sample mass - mass of C and H = mass of oxygen

Dude, most likely jewish, teaching this:

THE TOTALLY AWESOME!!! CLASS(EMPIRICAL+MOLECULAR FORMULA)

As the title suggested this class was more than just cool IT WAS AWESOME!! You won't be surprised because every single class of ours is super awesome because we are the awesome JOHNNY!!! (the Johnny part was a joke). okay back to our class~. Sooo what's did we learn today? Is that avocado lover with his not so cute moles planning for the next destructive weapon again??? The question remains to be answered but the important part is that we learned something in CHEM AGAIN~~~.

Okay the serious stuff is comming
Empirical Formula mean The simplest whole number ratio of atoms of each element present in a compound. (WOW DOESN"T THAT SOUND COOL?)
 Because the Pure awesomeness of Ionic compounds. They are given in Empirical formula already. (please don't go and ask the atoms why, they are just that awesome).  The Covalent compounds are the strange ones they come in the right ratio but NOT SIMPLIFIED in math class that would be 1 mark off... so if a ionic compound went and did a math test with covalent the covalent would just barely pass while ionic gets perfect omg><.

Formula = n = molar mass of the compound/molar mass of the empirical formula
Okay now lets have some examples
A compound have 47.25% copper and 52.75% chlorine.
Find the empirical formula for this compound.
 Cu 47.25g Atomic mass 63.6
47.25/63.6 = 0.74
Cl 52.75g Atomic mass 35.5
52.75/35.5 = 1.49
Cu 0.74/0.74 = 1
Cl 1.49/0.74= 2.01 = 2
So the empirical formula for Cu Cl
is CuCl2

Ex 2~ what is the molar formula? molecular mass = 132.16 , empirical formula = C2H4O

n = 132.16/42 = 3.2
MM = 42g/mol
MF= 3(C2H4O) = C6H12O3

Ex3  empirical formula of a compound is CH and the molar mass is 104 g/mol, calculate the molecular formula.
mass of C  =  12.0 g/mol
mass of H  =   1.01 g/mol
empirical formula mass  =   13.0 g/mol
 CH =  (104 g/mol)(1 mol/13.0 g)  =   8.00

MF  =   8(CH) or C₈H₈