May, 2011 - Amines and Amides (Angelo)

Amines:
- Amines are functional groups that have a Nitrogen bonded to Hydrogens or Carbons
- In order to name this, name the carbon chain and add an -amine ending
- Primary Amines have one carbon chain
- Secondary Amines have two carbon chains
- Tertiary Amines have three carbon chains
- The most simple Amine is...

...Methylamine, also known as Aminomethane.


Amides:
- Amides are a functional group with CONO3
- In order to name this, name the carbon chain and add an -amide ending
- The most simple Amide is...

...Ethanamide.

May 5, 2011 - Esters (Angelo)

Esters:
- The functional group...
     O
     ||              |
-----C-----O-----C-----
                    |


...is called an Ester.


- One of the simplest Esters is Ethyl Methanoate.
        O            H     H
        ||             |       |
H-----C-----C-----C-----C-----H
                      |        |
                      H      H


Esterification:
- Esters are formed by the reaction of a Carboxylic Acid and an Alcohol.
        O                                                           O
        ||                                                            ||       
R-----C-----OH     +     OH-R     ----------->     R-----C-----OR     +     H2O


EXAMPLES:


1. Name the following Ester.
                                                                                                       O
                                                                                                       ||
CH3-----CH2-----CH2-----CH2-----CH2-----CH2-----CH2-----CH2-----O-----C-----CH3


Answer: Octyl Ethanoate


2. Name the following Ester.
                                   O
                                   ||
CH3-----CH2-----CH2-----CH2-----O-----CH2-----CH3


Answer: Ethyl Butanoate

Keytones - brian

- a keytone is a hydrocarbon chain with a double bonded oxygen that is not on either end
- follow standard naming rules and add -one to the parent chain

    

Propanone

Example: Draw the structural diagram for the following keytone

                                Heptanone             

Aldehyde's:- an aldehyde is a compound that has a double bonded oxygen at the end of a carbon
- the simplest aldehyde is a methanol (also called formaldehyde)
- follow the standard rules and change the parent chain ending to -al
- be careful when naming aldehyde's and alcohols

Example: draw the structural diagram for the following aldehyde
             
                

 

April 26, 2011 - Halides (Angelo)

Halides:
- Group 7 elements (Flurine, Chlorine, Bromine, Idodine) can bind to a hydrocarbon chain
- Naming follows standard rules with halides using floro-, chloro-, bromo- and iodo-

EXAMPLES:


1. Name the following compound.


Answer: 2 3 Dibromo Pentane

2. Draw the following compound: 1 Chloro 3 Iodo Propane
Answer:

April 26, 2011 - Alcohols (Angelo)

Alcohols:
- an alcohol is a hydrocarbon with a -OH bonded to it
- some naming rules apply but the parent chain ending with -ol
- The following compound is ethanol...


Multiple -OH:
- if a compound has more than one -OH group, number both and add -dio, -trio, et cetera.

EXAMPLES:


1. Name the following compound.


Answer: 1 Propanol

2. Name the following compound.
 

Answer: 1, 3 Propanediol

Alkenes & Alkynes (double & triple bonds)- brian

- carbon can form double & triple bonds with carbon atoms

- when multiple bonds form fewer hydrogen's are attached to the carbon atom

- naming rules are almost the same as with alkanes
-the position of the double/triple bonds always has the lowest number and is put in front of the parent chain

- double bonds(Alkenes) end in ENE

- triple bonds(Alkynes) end in YNE

Example:

Trans & Cis butene
- if two adjacent carbons are bonded by a double bond and have side chains on them two possible compounds are possible
 

          














Multiple double bonds
- more than one double bond can exist in a molecule

- use the same multipliers inside the parent chain

Example:

March 31,11 Polar Molecules

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 is usually balanced
-Molecules can be unsymmetrical in two ways
1)different atoms
2)different number of atoms

http://www.youtube.com/watch?v=LKAjTE7B2x0

March 28, 2011 - Bonding, Bonds and Electronegativity (Angelo)

Types of Bonds:
- there are three main types of bonds
     1. Ionic (metal  -  non-metal)
          - electrons (will now be abbreviated to e.) are transferred from metal to non-metal
     2. Covalent (non-metal  -  non-metal)
          - e. are shared between non-metals
     3. Metallic (metal)
          - holds pure metals together by electrostatic attraction

Electronegativity:
- electronegativity (will now be abbreviated to "en.") is a measurement of an atom's attraction for e. in a bond (NOTE: there are NO UNITS for electronegativity)
     - Fluorine = 4.0
     - Chlorine = 3.0
     - Cesium = 0.8
- atoms with greater en. attract e. more
- Polar Covalent bonds form an unequal sharing of e.
- Non-Polar Covalent bonds form from equal sharing

Bonds:
- the type of bond formed can be produced by looking at the difference in en. of elements
     - if (en. > 1.7), it is an Ionic bond
     - if (en. < 1.7), it is a Polar Covalent bond
     - if (en. = 0), it is a Non-Polar Covalent bond


EXAMPLES:


1. Predict the type of bond formed.

                         F  -  F

Identify the electronegativity of each element: The electronegativity of Fluorine is 3.98.
Find the difference by forming and solving an equation: 3.98 - 3.98 = 0.00
State what type of bond was formed: Because the electronegavitity of this bond is 0.00, we may conclude that this is a Non-Polar Covalent bond.

2. Predict the type of bond formed.

                         Mg  -  S


Identify the electronegativity of each element: The electronegativity of Magnesium is 1.31, and the electronegativity of Sulphur is 2.58.
Find the difference by forming and solving an equation: 1.31 - 2.58 = -1.27 = 1.27 (NOTE: find and use the absolute value)
State what type of bond was formed: Because the electronegativity of this bond is 1.27, we may conclude that this is a Polar Covalent bond

3. Predict what time of bond was formed. Then identify the positive and negative sides of the following bond.

                         Ca  -  Se


Identify the electronegativity of each element: The electronegativity of Calcium is 1, and the electronegativity of Selenium is 2.55.
Find the difference by forming and solving an equation: 1 - 2.55 = -1.55 = 1.55
State what type of bond was formed: Because the electronegativity of this bond is 1.55, we may conclude that this is a Polar Covalent bond.
State which side is negative, and which side is positive: The electronegativity of Selenium is greater than the electronegativity of Calcium. From this, we may conclude that Selenium is the negative side and Calcium is the positive side because the electron will be attracted to Selenium.
Represent this with a diagram: δ + Ca  -  Se - δ

Acid-Base Reactions (brian)

strong acids & strong bases:
-strong acids(SA) dissociate to produce H+ ions
-stong bases(SB) dissociate to produce OH- ions
-when a SA and SB mix they form water and an ionic salt
-total volume changes
NaOH + HCl > NaCl + H2O

pH & pOH:
-pH is a measure of the hydrogen ions present in a solution
pH = -log[H+]
-pOH is a measure of hydroxide ions present in a solution
pOH = -log[OH-]

 

 

Ion Concentrations (brian)

Dissociation:
-ionic compounds are made up of 2 parts
 a) cation - positively charged particle
 b) anion - negatively charged particle
-when ionic compounds are dissolved in water the cation and anion separate from each other
-this process is called dissociation
-when writing dissociation equation the atoms and charges must balance
-the dissociation of sodium chloride is
NaCl > Na+ + Cl-
-if the volume does not change then the concentration of individual ions depends on the balanced coefficients in the dissociation equation

Example: BaSO4 > Ba2+ + SO4 2-
Example: 0.250M solution of KOH
               KOH >     K+      +  OH-
             0.250M  0.250M  0.250M

Dilutions (brian)

Diluting solutions:-when two solutions are mixed the concentration changes
-dilutions is the process of decreasing the concentration by adding a solvent (usally water)
-the amount of solute does not change
n1=n2-because concentration is mol/L we can write C=n1Vand
n=CV
so C1V1=C2V2

Example: determine the concentration when 200ml of 0.20M HCL is diluted to a final volume of 400ml
               (0.20mol/l)(0.200L)=C2(0.400L)
               C2=0.1M

Titrations (Zac)

-a titration is an experimental technique used to determine the concentration of an unknown solution.
Terms & Equipment
-buret - contains the known solution. used to measure how much is added
-stop cock- valve used to control the flow of solution from the buret
-pipet - used to accurately measure the volume of unknown solution
-erlenmeyer flask - container for unknown solution
-indicator - used to identify the endpoint of the titration
-stock solution - known solution
Eg. Zac completed a titration .330 M NaOH with 15.00 mL samples of HI of unknown concentration.  The data he gathered is below.  Determine the concentration of HI.

Trial                                1                     2                     3                      4
Final Reading (mL)        11.9                22.8                34.2                43.1
Initial reading (mL)          .7                  11.9                22.8                34.2 
Volume Used (mL)        11.2               10.9                11.4                 8.9*
*insignificant due to inaccuracy
NaOH + HI = NaI + HOH

Solutions & Molarity (Zac)

-Solutions are homogeneous mixtures composed of a solute and a solvent
-solute is the chemical present in the lesser amount (whatever is dissolved)
-solvent is the hemical present in the greater amount (whatever does the dissolving)
-Chemicals dissolved in water are aqueous
eg. NaCl (aq)
-Concentration can be expressed in many different ways (g/L, mL/L, % by volume, % by mass, mol/L)
-the most common way is mol/L which is also called Molarity
-mol/L = M
-[HCl] = concentration of HCl
Eg. Determine the concentration of 16.0 g of C6H12O6 in 0.750 L of water

16.0g x 1mol/180g x 1/.75L = .119 M

February 4th, 2010: Energy & Percent Yield (Zac)

-Enthalpy is the energy stored in chemical bonds
-symbol of Enthalpy is H
-Change in enthalpy is ΔH
-In exothermic rxns enthalpy
-In endothermic rxns enthalpy increases










Calorimetry
- To experimentally determine the heat released we need to know 3 things
1. Temperature change (ΔT)
2. Mass (m)
3. Specific Heat Capacity (C)

These are related by the equation:
ΔH=mCΔT

Example:
Calculate the heat required to warm a cup of 400g of water (C=4.19J/g°C) from 20.0°C to 50.0 °C.
Δ  H = (400)(4.18)(30)
Δ  H = 50160 °C

Percent Yield
-the theoretical yield of a reaction is the amount of products that should be formed
-the actual amount depends on the experiment
-the percent yield is like a measure of success (how close is the actual amount to the predicted amount?
% Yield = Actual/theoretical x 100


Example
-In a double replacement reaction Silver Nitrate is reacted with Barium Sulphate.  If 2.0 mol of Silver nitrate react and 468g of silver sulphate are produced, determine the theoretical yield of Ag2SO4, and determine the percent yield.
 
First, balance the equation: 2AgNO3 + BaSO4 = Ag2SO4 + Ba(NO3)2
Then, determine the theoretical yield of Silver sulphate: 2.0 mol x 1/2 x 312g/1mol = 312g of Ag2SO4
then divide actual by theoretical and multiply by 100: 468/312 x 100: 150 % yield

Feb.8/2011:Limiting Reactants (Brian)

In chemical reactions, usually one chemical gets used up before the other
-The chemical used up first is called the limiting reactant
-Once it is used up the reaction stops
-L.R determines the quantity of products formed
-To find the L.R assume one reactant is used up, determine how
  much of this reactant is required

example:
Find the L.R when 2.8mol of H2 reacts with 1.25mol of O2
2H2+O2 -----> 2H2O
2.8 x 1 = 1.4 mol L.R or 1.25mol x 2 = 2.5
         2                                      1

Feb.8/2011:Percent Yield (Brian)

-The theoretical yield of a reaction is the amount of products that should be formed
-The actual amount depends on the experiment
-The percent yield is like a measure of success

* % yield = Actual      multiplied by 100
                Theoretical  example:
in a double replacement reaction silver nitrate is reacted with barium sulphate. If 2.0mol of silver nitrate react and 468g of silver sulphate are produced.
-determine the theoretical yield of Ag2SO4
-determine the percent yield
2AgNO3+BaSO4 ----> Ba(NO3)2+Ag2SO42.0mol x 1 x 311.9g = 311.9g                          468g x 100 = 150%
               2    1mol                                          311.9g

February 2, 2011: Other Conversions, Volume and Heat (Angelo)

Volume and Heat:
- volume  at STP can be found using the conversion factor 22.4 L/mol (litres/mol)
- heat can be included as a separate term in chemical reactions (this is called ENTHALPHY)
- reactions that release heat are exothermic
- reactions that absorb heat are endothermic
- both can be used in stoichiometry

EXAMPLES:


1. If 5.0 grams of Potassium chlorate decompose according to the reaction below, what volume of Oxygen gas (at STP) is produced?

Create your balanced equation: 2KClO3 -----> 2KCl + 3O2
Create your equation: 5.0 grams * (1 mol / 122.5 g.) * (3 mols / 2 mols) * (22.4 L. / 1 mol) = ?
Calculate: 5.0 grams * (1 mol / 122.5 g.) * (3 mols / 2 mols) * (22.4 L. / 1 mol) = 1.4 L.

January 27: Mass to Mass Conversions (Angelo)

Mass to Mass Conversions:
- mass to mass problems involved one additional conversion
- this chart is a visual example of how to perform mass to mass conversions
GRAMS OF A -------> MOLES OF A -------> MOLES OF B -------> GRAMS OF B


EXAMPLES:


1. Lead (IV) Nitrate reacts with 5.0 grams of Potassium Iodide. How many grams of Lead (IV) nitrate are required for a complete reaction?

Create your balanced equation: Pb(NO3)4 + 4KI -----> 4KNO3 + PbI4
Create your equation: 5.0 grams (KI) * (1 mol / 166 g. (KI)) * (1 mol (Pb(NO3)4) / 4 mols (KI)) * (455.2 g. (Pb(NO3)4) / 1 mol) = ?
Calculate: 5.0 grams (KI) * (1 mol / 166 g. (KI)) * (1 mol (Pb(NO3)4) / 4 mols (KI)) * (455.2 g. (Pb(NO3)4) / 1 mol) = 3.4 grams (Pb(NO3)4)

January 25, 2011: Moles to Mass and Mass to Moles (Angelo)

Moles to Mass and Mass to Moles:
- some questions will give you an amount of moles and ask you to determine the mass
- converting moles to mass only requires one additional step

EXAMPLE:


1. How many grams of Bauxite (Al2O3) are required to produce 3.5 mols of pure Aluminum.

Create your balanced equation: 2NO3 -----> 4Al + 3O2
Create your equation: 3.5 mol (Al) * (2 mols (Al2O3) / 4 mols (Al)) * (102 grams (Al2O3) / 1 mol (Al2O3)) = ?
Calculate: 3.5 mol (Al) * (2 mols (Al2O3) / 4 mols (Al)) * (102 grams (Al2O3) / 1 mol (Al2O3)) = 1.8 x 10^2 grams


2. How many moles of Lead (II) Nitrate are consumed when 4.5 grams of Sodium Sulphide completely react?

Create your balanced equation: Pb(NO3)2 + Na2S -----> PbS + 2NaNO3
Create your equation: 4.5 grams * (1 mol / 78.1 g.) * (1 mol / 1 mol) = ?
Calculate: 4.5 grams * (1 mol / 78.1 g.) * (1 mol / 1 mol) = 0.058 grams

January 21, 2011: Mole to Mole Conversions (Angelo)

Mole to Mole Conversions:
- coefficients in balanced equations tell us the number of moles reacted or produced
- they can also be used as conversion factor:
3X + Y = 2Z
- WHAT YOU NEED over WHAT YOU HAVE


EXAMPLES:


1. If 0.20 mol of Methane are consumed in a combustion reaction, how many moles of CO2 are produced?

Create your balanced equation: CH4 + 2O2 -----> CO2 + H2O
Create your equation:  0.20 mol (CH4 ) * (1 mol (CO2) / 1 mol (CH4)) = ?
Calculate: 0.20 mol (CH4 ) * 1 mol (CO2) / (1 mol (CH4) = 0.2 mol (CH4))

2. When 2.0 mol of Copper (II) react with Iron (II) Chloride, how many moles of iron should be producted?

Create your balanced equation: Cu + FeCl2 -----> CuCl2 + Fe
Create your equation: 2.0 mol * (1 mol / 1 mol) = ?
Calculate: 2.0 mol

NOTE: When you are comfortable, you may remove what type of element, compound, etc. you are using.

Jan 18th, 2010 - Stoichiometry (Zac)

-Stoichiometry is a branch of chemistry that deals with the quantitive analysis of chemical  reactions
-It is a generalization of mole conversions to chemical reaction
-Understanding the 6 types of chemical reactions is the foundation of stoichiometry
SYNTHESIS
A + B = AB
-Usually elements that form compounds
-The formation reactions must be balanced
eg.  2Al + 3F2 = 2AlF3
eg.  2SO2 + O2 = 2SO3
DECOMPOSITION
AB = A + B
-Reverse of Synthesis
-Always assume that the compounds decompose into single elements during the reaction
eg.  Mn(C2O4)2 = Mn + 4C +4O2
eg.  2C12H22O11 = 24C + 22H2 + 11O2
SINGLE REPLACEMENT
A + BC = B + AC
eg. 3Mg + 2(Al(NO3)3 = 2Al + 3Mg(NO3)2
DOUBLE REPLACEMENT
AB + CD = AD + BC
eg. 2Al(NO3)3 +3SrCO3 = Al2(CO3)3 + Sr(NO3)3
eg. AgNO3 + NaCl = NaNO3 + AgCl
NEUTRALIZATION
-A reaction between an acid and a base (acids usually include hydrogen & bases have hydroxide)
eg. 3Ca(OH)2 +2H3PO4 = Ca3(PO4)2 + 6HOH
COMBUSTION
-Reactions of something (usually hydrocarbon) with air
-Hydrocarbon compustion always produces carbon dioxide and water.
eg. CH4 + 4O2 = CO2 + 2H2O
eg. 4Al + 3O2 = 2Al2O3