Bonds and Electronegativity

There are three main types of bonds, two of which we already learned in grade 10.

• Ionic (Metal-Nonmetal)
• Covalent (Nonmetal-Nonmetal)
• Metallic (Metal-Metal)

In Ionic bonds, electrons(e-) are transferred whereas in Covalent bond, e- are shared between the two atoms. Metallic bonds involve two metals which are held together by electrostatic attraction.

Next we have to talk about electronegativity (EN). EN is a measure of an atoms attraction for e- in a bond. EN ranges from 4 to 0.7 and follows a specific trend on the periodic table.

                                  EN > 1.7    =  Ionic Bond
                                  EN < 1.7    = Polar Covalent Bond
                                  EN = 0       = Nonpolar Covalent Bond


Lets try some examples: 

Ex1.      Ca-P
            1.0-2.19
            1.19 = Polar Covalent Bond

 http://www.youtube.com/watch?v=Kj3o0XvhVqQ





So atoms with greater EN attract more e-s, forming two different kinds of covalent bonds: polar and nonpolar. Polar covalent bonds have an unequal sharing of electrons while nonpolar covlent bonds share electrons equally.

The different types of bonds can be predicted based on the difference of EN of the two atoms:

                                  EN > 1.7    =  Ionic Bond
                                  EN < 1.7    = Polar Covalent Bond
                                  EN = 0       = Nonpolar Covalent Bond


Lets try some examples: 

Ex1.      Ca-P
            1.0-2.19
            1.19 = Polar Covalent Bond

Ex2.       K-Br
              0.82-2.96
              2.14 = Nonpolar Covalent Bond

Ex3.       Cl-Cl
              3.18-3.18
              0 = Ionic Bond

Titrations

Titration is used to find the concentration of known solution. 

However before you start titrating you need to know a few terms.

buret- has known solution

stopcock- valve used to control flow of solution from buret

pipet- measures volume of unknown solution

Erlenmeyer flask- container for unknown solution

indicator- identifies the point of titration

stock solution- the known solution

Now how to do titration

For example find 10mL of  [BaOH] titration with .3 M of HI
first you will have a graph that looks like this...

Now as nice as a graph as this is you need to know what it means before you do anything. After filling up your buret you will write the volume in the initial reading spot. you will then open the stopcock and let the known solution pour into the unknown solution until you have a reaction. After the reaction occurs you will record the final reading and then find the volume used with the difference found between the final and initial reading. You continue this for all the trials and you will end up with a graph that looks something like this.

See how it was done. After this you find the average amount of volume used (however since 9.6 looks like a mistake was made in the process of titration, I will leave it out in order to get a more accurate result). 

Thus the average is 10.2 mL or .0102 L.

You now have the M of HI, the average volume of HI and the volume of BaOH, you can now find the concentration of BaOH.

Here's what the equation looks like

0.3mol/1L x .0102L/1 x 1/1 x 1/.01L =3.06x10^-5 mol/L

Dilutions

When two solutions are mixed, the concentration CHANGES. Dilution is the process of DECREASING the concentration by adding a solvent, usually water. The amount of solute doesn't change. Because the concentration is mol/L, we can write:

C= n/V   n= CV     C1V1=C2V2

example 1
Determine the concentration when 100 mL of 0.10 M HCl is diluted to a final volume of 400 mL.

Because these are not chemical reactions, you do not have to write a balanced equation.
C1V1 = C2V2
(0.1 M) (0.100 L) = C2 (0.400 L)
C2= 0.0250 M

example 2: How much water must be evaporated from 2.00 L of 0.250 M KCl solution fo rthe final concentration to be 2.75 M?
C1V1 = C2V2
(2.00 L)(0.250 M)= V2(2.75)                        2.00 L - 0.18 L = 1.82 L
0.18 L = V2

example 3: 100 mL of 0.250 M Sodium Nitrate is mixed with 200 mL of 0.100 M Sodium Nitrate?
-Determine the # of mol of Sodium Nitrate in the resulting solution
n= CV
n1 + n2 = 0.100 L  x   0.250 mol   +   0.200 L   x  0.100 mol  = 0.250 mol + 0.200 mol

                                         1 L                                        1 L           

                                                                                                     = 0.450 mol

-Determine the final [NaNO3]  (determine the concentration)
0.0450 mol =  0.150 M

  0.300 L

Now what's the point of learning all of this if we're not going to use it?
Well we are going to use it! We're going to be doing a dillution lab! Here's a video that shows how to do some dilution!
Plus, the music is really cool :)

http://www.youtube.com/watch?v=bnQJ2q36d2E&feature=related

More Molarity

Here are some more examples of Molarity, getting trickier and trickier!

 

100 mL of 0.250 M Iron (II) Chloride reacts with excess copper. How many grams of Iron are produced?

Now, what do we ALWAYS do first?
Free body diagram!
Just kidding. Balanced Equation!
FeCl2 + Cu -> CuCl2  + Fe
 

0.250 mol   x   0.100 L  x  1   x   55.8 g  =  1.40 g
      L                                    1        1 mol    

Use the numbers given to you: the Molarity and volume, and then use the mole ratio and molar mass to get the end product! Remember, what you need over what you have!

ex 2. A 3.00 g piece of Iron is added to a beaker containing 100 mL of 0.750 M AgNO3. Determine the LR
Fe + 2AgNO3 -> Fe(NO3)2 + 2Ag

0.100 L   x   0.750 mol   x   1  x   55.8 g  =  2.09 g of Fe needed

                          1 L               2       1 mol 

Remember to convert mL to L and be consistent in your units!
To find the LR, as taught before, just convert one of your reactants' numbers to the number (it can be either mass, volume, moles) of your other reactant, to see if you have enough or not!
In this case, the AgNO3 is the LR

ex. 3 A beaker contains 100 mL  of 1.5 M HCl. Excess Zinc is added to the beaker. Determine how many litres of Hydrogen gas should be produced.

2HCl + Zn -> ZnCl2 + H2

0.100 L   x   1.5 mol   x   1   x  22.4 L  =  1.68 L

                      1 L              2        1 mol

Simply use mole ratio to convert to Hydrogen gas, and remember, in this case, because Hydrogen is a gas, you can use the volume at STP (22.4 L)!

Here's a link to website that explains it quite well, too!

http://dl.clackamas.cc.or.us/ch105-04/molarity.htm

Titration of Vinegar!!

Now you know all about titration, lets try it out to see if it really works!!!

For this lab we used Vinegar (acetic acid or CH3COOH) and Sodium Hydroxide (NaOH).
We also used:

Phenolphthalein Indicator
Pipet
Lab Stand
Buret
Buret Clamp
Funnel
Erlenmayer Flask

The NaOH had a concentration of 1.0M and we had to find the volume of NaOh needed to neutralize 10.0mL of Acetic Acid.
The balanced equation is:

NaOH + CH3COOH = NaCH3COOH + HOH(water)

Next we set up the Buret with NaOH up to the 0.3mL line and measured 10.0mL of vinegar into an Erlenmayer Flask placed underneath. Then we put a drop of Phenolphthalein into the vinegar.

We then added NaOH into the vinegar b it by bit until the solution in the Erlenmayer Flask turned pink. We then recorded the amount of NaOH we had used and started over. We did this test 4 times to be sure of our accuracy and then found the average of all of the tests.

Our average of mLs used was 8.8mL and we used these results to determine the concentration of acetic acid.

0.0088L x 1mol x1 x 1           = 0.88M CH3COOH
                  L       1   0.010L

The acepted value of Acetic acid is 0.85M.

We determined our percent error with the equation:

Theoretical - Experimental     x100 = percent error
         Theoretical


  0.85-0.88 x100 =  3.5% error
       0.85


And thats all for now folks!!

Limiting Reactants

Obviously in chemical reactions not all the chemicals will be fully used up, usually there is one chemical that will be used up first, stopping the reaction. The chemical that will be used up first is called the Limiting
Reactant (LR).

Of course one needs to know how to find the limiting reactant before they can do much, which is what I will show you next.

For example
Find the LR when 2.2 moles of Sodium reacts with .19 moles of Oxygen.
First write the balanced chemical equation  4Na + O₂ -> 2Na₂O
Next you will take one of the chemicals and figure out how many moles of the second equation you need to fully react with the first equation I will use Sodium to start  2.2/1 x 1/4 = .55 mol (don't forget your significant digits)
Now how to interpret the answer the .55 means that you need .55 mol of Oxygen to fully react with the 2.2 mol of Sodium, of course you only have .19 mol of Oxygen, therefore Oxygen is the limiting reactant.

This method can also be expanded with using grams or liters etc. instead of moles but the basic equation is still the same.
If you're still not sure here's another example.
You have 1.40 L of  Calcium find the LR when calcium reacts with 2.00 mol of Nitrogen find the LR.
4Ca + 3N₂ ->  2Ca₂N₃ 
1.4/1 x 1/22.4 x 3/4 = .0469 mol
Therefore Calcium is the LR

And here's a question for you to solve, if you have 2 mol of Calvin and the monster under his bed can only eat 22 L will Calvin survive the night?