Polar Molecules

-Polar molecules have an overall charge separation
-unsymmetrical molecules are usually polar
-molecular dipoles are the result of unequal sharing of electrons in a molecule

Predicting Polarity
-if a molecule is symmetrical the pull of e-(electrons) usually balance
-Molecules can be symmetrical in two ways
  *different atoms
  *different number of atoms

These are examples of molecules with polar and non-polar bonds

 

Bonding - Bonds and Electonegativity

There are 3 types of bonds. There is an ionic bond, which is the bond of a metal and a non metal. In ionic bonds, electrons are transferred from metal to non-metal. There is a covalent bond, which is the bond of a non metal and a non metal. In covalent bonds, electrons are shared between non metals. And there is a metallic bond, which is the bond of metals. A description of a metallic bond is: holds pure metal together by electrostatic attraction. 


After learning about bonds, we learned about electronegativity. Electronegativity (en) is a measure of an atoms attraction for electrons in a bond. For example:

• Flourine has an en. of 4.0
• Chlorine has an en. of 3.0
• Cesium has an en. of 0.9

You may be asking how we know the electronegativity of each. Well you can see by the table shown below:


Atoms with greater electronegativity attract electrons more. Polar covalent bonds form from an unequal sharing of electrons. Non-polar covalent bonds form from equal sharing of electrons.

We then went into more detail about bonds. The type of bond formed can be predicted by looking at the difference in electronegativity. For example:

• Electronegativity (en.) of the elements.
• en > 1.7 : ionic bond
• en < 1.7 : polar covalent bond
• en = 0 : non-polar covalent bond

Examples: Predict the type of bond formed

• O - O     3.44 - 3.44 = 0 non polar covalent bond
• F - Cl     3.99 - 3.18 = 0.81 polar covalent bond
• Ba - I     0.89 - 2.66 = 1.77 ionic bond

Solution Stoichiometry - Dillutions

When two solutions are mixed together the molarity changes. Dillution is the process of decreasing the concentration by adding a solvent (usually H20). The amount of solute does not change as n1 = n2. And because concentration is mol/L we can write..

• C = n/V and..
• n = CV so..
• C1V1 = C2V2

Example:

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

     C1V1 = C2V2

     (0.1M)(0.100L) = C2(0.400L)

     C2 = 0.025 M

• How much water must be added to 10.0mL of 10.0M Na2SO4 to give a solution with a concentration of 0.50M?

     C1V1 = C2V2

     (10.0M)(0.010L) = (0.50M)V2

     V2 = 0.200L

              200mL - 10mL = 190mL

     

Solution Stoichiometry - Titrations

A titration is an experimental technique used to determine the molarity of an unknown solution. When doing a titration, some special equipment is needed:

• Buret - contains the known solution, used to determine how much solution is added
• Stopcock - valve used to control the flow of solution from the buret
• Pipet - used to accurately determine volume of unknown solution
• Erlenmeyer Flask - container for unknown solution
• Indicator - used to identify the end point of titration
• Stock Solution - known solution

Example:

1. Jerome wants to determine the molarity of a Sodium Hydroxide sample so he does a titration with Hydrochloric Acid (HCl). He gathers the following data. 

Trial	1	2	3	4
Final Reading (mL)	13.3	26.0	38.8	13.4
Init. Reading (mL)	0.2	13.3	26.0	0.6
Volume Used (mL)	13.1	12.7	12.8	12.8

NaOH samples = 10.00mL          [HCl] = 0.75M   ( NaOH + HCl -> NaCl + HOH)

The first thing to do is to find the average volume used. Now, since trials 2,3 and 4 are relatively close to each other, we can exclude trial 1 in the averaging process. The average volume used is 12.8 mL. Now we used this in our conversion. 


0.0128L x 0.75mol x 1 x 1mol = 0.96M NaOH 
                     1L        1   0.010L

Buret and Stopcock

Erlenmeyer Flask
Pipet