Esters

• carbon attached to a double bonded oxygen and a single bonded oxygen
• ends with -oate

http://www.youtube.com/watch?v=NKSd2RJolNA

Watch this video on how to write the functional group esters

Janine R

Nitro, Amide, & Amine

Nitro

• nitrogen has a double bond with an oxygen and a single bond with oxygen
• located at the end of a parent chain

Nitro Pentane

3 Nitro Benzoic Acid

Amide

• oxygen and NH2 bond
• ends in -amide

Propanamide

Amine

• nitrogen part of the parent chain
• ends as Amine

pentyl amine

Janine R

Aldehydes and Caroxylic Acid

Aldehydes

-double bonded O --> end of your cahin
- simplest aldehyde is methanol
-  endings : al

Example:

Carboxylic Acid

-from by the functional group


- uses standard rules
- parent cahin ends : oic acid

Nicole H

Alicylics and Aromatics

- carbon chains can form 2 types of closed loops
- Alicylics are loops usually made with single bonds
- if parent chain is a loop standard naming rules apply
        - with one addition: "cyclo" is added in front of the parent chain

3 different ways to draw organic compunds:

1. Complete Structural Diagram

2. Condensed Structural Diagram

3. Line Diagram

- numbering can go anywhere starting with clockwise or counter clockwise on the loop but the side chains must be the lowest possible

BENZENE:

Nicole H

Alcohols, Halides, Ketone and Ethers

ALCOHOLS:
-contain functional groups OH (hydroxol)
one OH : Ends in ol
two OH : Ends in diol
three OH : Ends in triol

Example : Ethanol



HALIDES:
- it only includes group 17 from the periodic table
ending : O
Flourine: Floro
Chlorine: Cloro
Bromine: Bromo
Iodine: Ioda

Example: 1 cloroethane



KETONE:
- double bonded oxygen
- ends in anone
-double bond cannot be in the beginning or the end

Example: butanone



ETHERS:
-only contains oxygen
- always between two carbons
- numerical order ( not alphabetical order like the others)
-ends in ether

Example: dimethyl ether

Nicole H

Alkenes and Alkynes (double&triple bonds)

• carbon can form double and triple bonds with carbon atoms
• naming rule  --the position of the double/triple bonds always have the lowest number and is out in front of the parent chain
• double bonds (alkenes) end in -ene
• triple bonds (alkynes) end in -yne

EX.

5, 5 diethyl 2,6,7 trimethyl 3 octene

Ethene

4 methyl 2 pentyne

Multiple Double Bonds

• more than one double bond can exist in a molecule
• you can use the same multipliers inside the parent chain

Janine R

Organic Chemistry----Nomenclature

CH4

• organic chemistry is the study of carbon compounds(carbon forms multiple covalent bonds)
• carbon compounds can form chains, rings or branches
• simplest organic compounds are made of carbon and hydrogen
• saturated compounds have no double or  triple bonds
• compounds with only a single bonds are called Alkanes and always end in -ane

Nomenclature

There are 3 types of Organic compounds:

1. straight chains
2. cyclic chains
3. aromatics

Straight chain rules

1. Circle the largest continuous chain and name them as the base chain (ex. meth, eth, prop)
2. Number the base chain so side chains have the lowest possible numbers
3. Name each side chain using the -yl ending
4. Give each side chain the appropriate number (if there is more than one identical side chain numbers, labels are slightly different)
5. List side chains alphabetically

EX.

             

3-ethyl 2,4 dimethyl pentane

Janine R

Polar and Non-Polar Solvents Lab

Objective of this lab: to determine if Glycerin is Polar or Non-Polar

We set up 6 clean, dry test tubes and divided them into 3 sets. 3 test tubes were half filled with water and the other 3 were filled with paint thinner. The first 3 tubes with water, we added table salt, sugar and iodine to each one. All dissolved in each of the solvents. We then added the same 3 substances in each of the test tubes filled with paint thinner. Non of the substances dissolved in the paint thinner. The iodine turned the paint thinner a magenta colour. At the end of the lab we combined water, glycerin and the paint thinner into one tube. It didnt dissolve and the glycerin and paint thinner stayed at the top, creating a jelly look. (like the picture below)

                                       (water, paint thinner and glycerin combined in one test tube)



Janine R

Intramolecular Bonds

Intramolecular: bonds that exist within molecules
     -ionic and covalent
Intermolecular: bonds that exist between molecules
     - stronger the intermolecular nonds the higher boiling point or melting point
     - 2 types : Vander waals and Hydrogen bonds


Vander Waal Bonds

• based on electron distribution
• two categories:   1. Dipole 2. Bonas
• if a molecule is polar the positive end of one molecule will be attracted to the negative end of another molecule

LDF London Dispersion Focus

• is present in all molecules
• -creates weakest bond
• - if a substance is nonpolar dipole the focus doesnt exist
• electrons are free to move around and wil randomly be grouped on one side of the molecule
• this creates a temporarily dipole and can cause a weak bond to form
• the more electrons in the molecule the stronger the LDF can be

Hydrogen Bonding

• if Hydrogen is bonded to certain elements (F, O , or N) the bond is highly polar
• thsi forms a very strong intermolecular bond

Nicole H

Polar Molecules

- polar molecules have an overall charge seperation
- unsymmetrical molecules are usually polar
molecule dipoles: are resulted of unequal sharing of electrons in a moleculE

Predicting Polarity

- If a molecule s symmetrical the pull of electrons are usually balanced
- molecules can be un-symetrical in 2 ways:
1. different atoms
2. different number of atoms

EXAMPLE:
Molecule                  Diagram                                              Polarity

CO2     Non polar (because it is symmetrical)

Nicole H

Types of Bonds

There are three main types of bonds:
1. IONIC (metal - non metal)
      - electrons are transferred from metal to non-metal
2. COVALENT (non-metal - metal)
      - electrons are shared between non metals
3. METALLIC (metal)
      - holds pure metals together by electrostatic attraction

Electronegativity(en)
- en is a measure of an atoms attraction for electrons in a bond
Flourine = 4.0
- atoms with greater en attract electrons more

polar covalent: bonds form from an un-equal sharing of electrons
non-plolar covalent: bonds form from equal sharing

BONDS:
- types of bonds formed can be predicted by looking at the difference in electronegativity of the elements
en > 1.7 = ionic bond
en< 1.7 = polar covalent bonds
en = 0   = non-polar covalent

for examples of electronegativity and bonding please watch the link below :)
http://www.youtube.com/watch?v=Kj3o0XvhVqQ

Nicole H

Ion Concentrations

• when ionic compounds are dissolved in water they contain cation and anion seperate from each other.
• this process is called DISSOCIATION
• Dissociation: is the process in which this happens

Dissociation
ionic compunds are made up of two parts:
-cation - positively charged particles
- anion - negatively charged particles

• when writing dissociation equations the atoms and charges must balance

                ex : Fe(OH)2 ----> Fe 2+ + 2 OH -

EXAMPLE :

Determine [Na+] and [PO4   3-] in a 1.5 M solution of Na3PO4

Na3PO4 ---> 3 Na +    +        PO4 3-

1.5 M  x   3     =  4.5 M = [Na+]
                1

Nicole H

Dilution Lab

We did a lab where we had to figure out which solution contained 0.05 M.  Mr. Doktor did not help us, he only gave hints. Also, we could only use the supplies given at our lab stations



Nicole H

Dilutions

Dilutions - when 2 solutions are mixed the concentration changes

                - the process of decreasing the concentration by adding a solvent(usually water)

                - the amount of solute does not change, only the volume

                - because concentration is mol/L we use the following equations: 

                                       C=n/v and n=Cv  so that    C1V1=C2V2

*always remember to convert mL into L

EX.

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

V1=100mL-->.10L               C1V1=C2V2

C1=0.10M                            

C2=400mL                           C1V1 = V2    ---->  (10.0)(10) = 200mL         changeV=200-10=190mL

                                               C2                             (0.50)

EX.

If a 75.0mL sample of 0.500M KCl is added to 100mL of 0.250M KCl what will the resulting solutions concentrations be?

0.075L x 0.500M = 0. 0375mol               

                   L

0.100L x 0.250M = 0.025mol

                   L

Then add both mols together,

0.0375+0.025= 0.0625mol

After divide the mols by adding both amounts of litres together,

0.0625M = 0.357M

0.175L

Janine R

Lab-Titration

We did a titration lab today, click on the following link for an example:)
http://www.youtube.com/watch?v=8UiuE7Xx5l8



Janine R

Titrations

Titration is an experimental technique used to determine the concentration of an unknown solution.

•  buret - contains the known solution, used to measure  how much is added
• stopcock(tap) - valve used to control the flow of solution from the buret
• pipet - used to accurately measure the volume of a unknown solution
• erlenmeyer flask - container for unknown solution
• indicator - used to identify the end point of the titration
• stock solution - known solution

Janine R

Solution Stoichemetry Question Examples

A beaker contains 100mL or 1.5M HCl. Excess zinc is added to the beaker. Determine how many litres@STP of hydrogen gas should be produced.


2HCl + Zn --> ZnCl2 + H2

1.5mol x 0.100L x 1 x 22.4L = 1.7L
     L                       2     mol

Determine the number of mles of H2O produced when 0.250L of 0.100M NaOH is neutralized by H2SO4.


2NaOH +H2SO4 --> Na2SO4 + 2H2O


0.100mol x 0.250L x 2 = 0.025mol
      L                          2





Janine R

Solution Stoichimetry

Solutions: homogenous mixtures composed of a solute and a slovent
-- solute : chemical present in the lesser amount
                             (whatever is dissolved)
--solvent : chemical present in the greater amount
                            (whatever does the dissolving)

- chemicals that dissolved in water are aqueous


MOLARITY

-concentrations can be expressed in many difderent ways
     --> g/L , mL/L, % by volume, % by mass, mol/L
- most common one we will be doing is mol/L

watch this video to learn how to do molarity practice problems and learn the molarity volume

http://www.youtube.com/watch?v=Ed1S0fLBqk4

Nicole H

TEST NEXT CLASS !

We will be having a test next class on the following conversions:

mole--->mole  given mole x need = resultant mole
                                             have

mole--->mass  given mole x need x mm = resultant mass
                                             have   1mol

mass--->mole  given mass x 1mole x need = resultant mole
                                              mm       have

mass--->mass  given mass x 1mole x need x mm = resultant mass
                                               mm      have   1mol



Janine R

Limiting Reactants

In chemical reactions, usually one chemical gets used up before the other. The chemical used up first is the limiting reactant. Once the limiting reactant is used up the reaction is stopped.
- Limitng reactant 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 :

2 H2 + O2 --> 2 H2O
8 mol x 2  = 16 mol
            1
Hydrogen = limiting factors

please watch this video for more examples of limiting reactants.. thanks :)
http://www.youtube.com/watch?v=fQyS7cZIJ1Q

Nicole H

Energy and Percent Yield

• enthalpy is the energy stored in chemical bonds
• symbol of enthalpy is H
• unit of Joules is J
• change in enthalpy is delta H
• in exothermic reactions enthalpy increases
• in endothermic reaction enthalpy increases

Calorimetry:

-to experimentally determine the heat released we need to know 3 things

1. temperature change deltaT

2. Mass. (m)

3. specific heat capacity (C) --> ability to hold heat

equation: H = mCT

Example:

- Calculate the heat required to warm a cup of 400g of water (c=4.8J/g celsius) from 20 to 50 degrees

H= mCT

H= (4.oo)(4.18)(30)

H= 50160 J --> 50.2 kJ

PERCENT YIELD

• the theoretical yield of a reaction is the amount of products that should be form
• the actual amount depends on the experiment
• the percent yiels is like a measure of success

- how close is the actual amount to the predicted amount?

    %yield = actual              x 100

                   theoretical

Nicole H

Other Conversions

• Volume at STP can be found using hte conversion factor 22.4L/mol
• Heat can be included as a seperate term in chemical reaction called enthalpy
• reaction that releases heat are exothermic
• reaction that absorbs heat are endothermic
• both can be used in stoichiometry

Examples:

- if 5.og of potassium chlorate decompose according to the reaction below, what volume of oxygen gas @STP is produced?
                                                          2 KClO3 --> 2 KCl + O2
5.0g x 1mol x 3  = 0.0612 mol x 22.4L =  1.4 L
          120.6   2                           1 mol

-When zinc reacts with hydrochloric acid exactly 1.00 L of hydrogen gas is produced at STP what mass of zinc was reacted?
                                   Zn + 2 HCl --> H2 + ZnCl2
1.00 L x 1 mol x 1 x 65.4  = 2.92 g
              22.4L   1    1mol

Nicole H

Stoichiometry Investigation

Problem : Does stoichemistry accuratley predict the mass of products produced in chemical reaction?

In this experiment we have to figure out how many moles of percipitate were formed when 2.00g of Stronium nitrate is disslved in 50mL of water and then reacted with the excess copper sulphate.

1) carefully measure about 3.oog of copper(ii) sulphate.
2) crush the copper(ii) sulphae into a fine powder using a mortar and pestle.
3) dissolve the copper sulphate in 50 mL
4) measure 2.00g of stronium nitrate and sissolvr iy on 50 mL of watter
5) slowly pour the two solutions
6) stir the mixture to complete the reaction
7) write your group name on  a piece of filter paper
8) find and record the mass of the filter paper
9)using a funnel and a n erlenmeyer flask, place the filter paper in funnel. Slowly pur the mixture into the funnel
10) pour the filtrate into the waste colection bottle
11) place the filter paper in the drying oven and record the mass when its dry

After following the procedure we discoved the formed moles of percipitate was 0.00953 mol

Nicole H

Mole to Mass and Mass to Mole

•  some questions will give you an amount of moles and ask you to determine the the mass. 
•  converting moles to mass only requires one additional step

                                        MOLES ------------- MOLES
                                             I                                  I
                                             I                                  I
                                             I                                  I
                                        MASS                         MASS

EXAMPLES:

--How many grams of (Al2O3) are required to produce 3.5 mol of pure aluminium.

1. Balance the equation                                          Al2O3 -->    Al  +     O2
                                                                         2  Al2O3 --> 4 Al  +  3   O2
2.Find the molar mass of (Al2O3) = 102g
3. Solve the problem                                  3.5mol Al x 2 Al2O3 x 102g  = 178.5g --> 1.8x 10 power of 2
                                                                                       4 Al        1 mol

-- How many grams of water are produced of 0.84 mol of (H3PO4) is completely neutralized barium hydroxide?

2H3PO4 + 3Ba(OH)2  --> 6H(OH) + Ba3 (PO4)2

0.84 mol x 6 x 1.8g  = 45.36g --> 45g
                 2   1mol
Nicole H

Classifying Chemical Reactions

Direct Combination Reaction: 2 or more reactants come together to form a single product
EX.  A + B ---> AB
        2Na + Cl2 ---->2NaCl
        S + O2 ---->SO2

Decomposition: a single compound is broken down into 2 or more smaller compounds or elements
EX.  AB ---> A + B
        2 H2O ----> 2H2 + O2
        CaCo3 ----> CaO + CO2

Single-Replacement: an uncombined element displaces an element that is part of a compound
EX.  A + Bx ---> Ax + B
         Mg + CuSO4 ----> MgSO4 + Cu
         Cl2 + 2KI ----> FeSO4 + Cu

Double-Replacement: atoms or ions from 2 different compounds replace eachother
EX.  Ax + By ----> Ay + Bx
        CaCO3 + 2HCl ----> CaCl2 + H2CO3

Janine R

Molecular Formulas

- If you know an empirical formula to find the molecular formula you need the molar mass.

Example :
-- The empirical formula for the substance CH2O and its molar mass is 60.0 g/mol . Determine the molecular formula.

                                                        EMPIRICAL                      MOLECULAR
                                                             CH2O             x2                  C2H4O2
M.M.=12.0+2(1.0) +16.0
         = 30.0                                       30.0g/mol           x2                  60.0g/mol


*watch this video to help you solve empirical and molecular formulas*

http://www.youtube.com/watch?v=AFqwtY7m2PI

Nicole H

Empirical Formula

Empirical formula: are the simplest formula a compound
  - shows the simplest ratios, not the actual atoms

EXAMPLE:

 this is the difference between an empirical formula and a molecule formula

- to determine the empirical formula we need to know the ratio of each element

- to determine the ratio we use this table to help us solve the problem:

ATOM   MASS(g)    MOLAR MASS   MOLES             MOLE                        RATIO

                                                               (g/mol)       SMALLEST MOLE

please refer to this link on how to solve empirical probblems:

http://video.about.com/chemistry/Empirical-Formula.htm 

Nicole H

December 12-Density and Moles/Density of Gas

Density and Moles

These are the equations for density.
-density is measured in g/L or g/mL


EX.
How many moles are in 11.5mL of water? (H2O=18)

11.5g x 1mol = 0.639g<---don't forget about
             18 g                       significant digits




EX.
A small sample of silver contains 1.23x10(23) silver atoms. If the density is 10.5g/mL what is the volume of the silver sample?
-this is a multi-step conversion so you will have to go from atoms->moles->mass->density

1.23x10(23) atoms x      1mol      x 107.9g x 1mL = 2.1mL
                                 6.02x10(23)    1mol      10.5g

Density of Gas

-@STP we can find density by:
   mm                molar mass-->g/mol
22.4L/mol        molar volume-->L/mol

EX.
Determine the density of methane(CH4=16g) at STP.

16g
1mol  = 0.71g
22.4L
mol

Janine Roldan

December 3 - Moles of Iron and Copper Lab

Today we were assigned to do a lab. In this lab we had to determine the number of moles of copper produced in the reaction of iron and copper (ii) chloride, the number of moles of iron used up in the reaction of iron and copper (ii) chloride, the number of atoms and formula units involved in the reaction and, the ratio of moles or iron to moles of copper. 

This is our procedure:

• First of all we weighed a dry 250 mL to find its mass.
• Then we added about 8 grams of copper (II) chloride crystals into the beaker and weighed it.
• After, we added 50 mL of water to the beaker
• Then weighed the 2 nails and placed them into the copper (II) solution
• Then saw a formation happening to the nails and after a couple of minutes we carefully used tongs to pick up the nails and removed any of the remaining copper from it.
• After we dried the nails and removed the copper, we then weighed the nails again to find its new mass.
• The excess solution was poured into the beaker carefully through a funnel and filter paper.
• At last we placed the copper in a drying oven and then find the new mass of the beaker and the copper 

1. Janine Roldan

Multi-Step Conversion Review