(TG): Ethers!

ETHERS!

• An ether consists of 2 chains that are separated by a single O in the middle
• Name both chains with the -yl ending and then end it with ether
• So, methyl ethyl ether
• or, propyl ethly ether

EXAMPLES: Name these!

propyl methyl ether

    dimethyl ether


DRAW THESE! 


diethyl ether: 











butyl methyl ether

 


 good job!

(TG): The Lab!

The Lab

Today in class we did a lab on esters!

The lab consisted of mixing a carboxylic acid, and an alchohol and in the end we would get an ester that smelled like spearmint! (the gun) 

The first step  was to mix the acid and the alcohol, then what we had to do was put the mixture in a hot water mixture so that the 2 mixtures could bond.

After waiting for 15 minutes, we took the  mixture out and left it in a cold water beaker for 2 minutes.

Finally, the resulting mixture smelled like spearmint!

The other smells we could have done were raspberry, or orange but when we tried them, the smell was very hard to distinguish!

(TG) Carboxylic Acids, Alcohols and Esters!

CARBOXYLIC ACIDS

•  A functional group formed by a doubled bonded O and an OH at the end of a Carbon chain
• Standard rules apply to naming, but change the parent chain ending to -oic acid
• The simplest carboxylic acid is methanoic acid
• 

Examples! Name these compounds: 

            (ethanoic acid, or commonly known as acetic acid.)




     (butanoic, or butyric acid)

Now, draw THESE compounds:

Propanoic acid:  









Pentanoic acid:

Now heres an extra tricky carboxylic acid that every chemist should know:

Benzenoic Acid:






 Great job! Now you're a pro... So lets move on to alcohols.... (dont get TOO excited high school students)

ALCOHOLS

• Alcohols are the functional group that contain OH's as side chains
• There can be OH's on any part of the side chain
• There can be multiple OH's coming off the main chain, and when this happens you use the prefixes di tri tetra etc. (example) methadiol, butatriol etc
• All standard naming rules apply!

Methanol:













Now here are some examples for you try out!: NAME THESE ALCOHOLS:

    Ethanol



                   Ethanediol



DRAW THESE:


propanetriol: 




butanediol:





Here again is a super tricky one when benzene becomes an alcohol!

PHENOL:








great work... now lets move on to our final topic ESTERS!

ESTERS!

• Esters first off, are not to be confused with ethers.
• An ester has 2 chains separated by an O, and then one chain right next to the O has a double bonded O on the first C
• Esters are formed when you add carboxylic acids and alcohols together but we'll get into that after we do the basics
• Name the chain without the double bonded O with an -yl ending, and the one with the double O with an -oate ending.

Methyl Methanoate:

Now that you've seen what it looks like, go ahead and name these:

      methyl ethanoate


  ethyl ethanoate



Draw these!

propyl ethanoate:




ethyl propanoate:




\








and finally, the finale. ESTERFICATION: when a carboxylic acid and an alcohol mix!








Here is a basic explanation. The 2 H's and O form water, and the remainder form the Ester!
Now do these esterfication processes!






   GOOD JOB GUYS!

Amines & Amides

• Amines are funtion groups that contain a Nitrogen compound bonded to either Hydrogens or Carbons
  
  
  - Primary amines have 1 carbon chain
  - Secondary Amines have 2 carbon chains
  - Tertiary amines have 3 carbon chains

EXAMPLES:

 Methylamine

Pentylamine

Amides:

• Amides are functional groups with CONH3
  
  
• Name the alkyl (carbon) chain and add -amide ending
• The simplest amide is ethanamide
  
  Ethanamide

Alicyclics & Aromatics

• Carbon chains can form two types of closed loops
• Alicyclics are loops usually made with single bonds
• If the parent chain is a loop standard naming rules apply with one addition

Cyclopentane

• There are three different ways to draw organic compounds

Condensed Structural Diagram

Complete Structural Diagram

Line Diagram

• Numbering can start anywhere and go clockwise and counterclockwise on the loop but side chain numbers must be the lowest possible

EXAMPLES:
 1,1 dimethyl cyclobutane

1,3 diethyl cyclohexane

• Loops can also be a sidechain
• Same rules apply but the side chain is given a cyclo-prefix
  
  2 cyclobutane octane
  or
  2 methyl 3 cyclopropyl 1 pentene

Aromatics:

• Benzene (C6H6) is a cyclic hydrocarbon with unique bonds between the carbon atoms
  Structurally it can be drawn with alternating double bonds
  
  
  
  or with a circle in the middle
  
  

(TG) ALKANES

ORGANIC CHEMISTRY - ALKANES

• Organic chemistry is the study of carbon compounds
• Carbon will form multiple covalent bonds
• Carbon compounds can form chain rings or branches
• There are less than 100000 non organic compounds
• Organic compounds number more than 17 000 000
• The simplest organic compounds are made of Carbon and hydrogen

 When naming an organic compound take the longest chain of carbon that are connected.

 C - C - C - C - C    5 carbon chain - PENTANE
\
Use prefixes to name the carbon chains:

1 - meth
2 - eth
3- pro
4 - but
5 - pent
6 - hex
7 - hept
8 - oct
9 - non
10 - dec

when dealing with Alkanes the ending will always be a preffix + ane.   (ie. propane, decane, methane etc)


EXAMPLES!


Here are some very basic alkanes, can you name them?

 C - C  = Ethane
C - C - C - C  = Butane
C = Methane

Now as mentioned before, organic chemistry deals with Carbons AND HYDROGENS. So whenever you see an open valence electron on a carbon, pair it with a hyrdrogen. (Carbon can make 4 bonds)

EXAMPLES!

Methane - CH(4)

Butane - CH(3) - CH(2) - CH(2) - CH(4)

Now moving on to the final piece of information dealing with organic compounds. When a compound has a normal chain, with several smaller ones sticking out of it, you name the side chain with the same prefixes but then add the ending - yl. Also you number the chains from left to right, and put the number of the side chain next to the prefix, AND, your side chains have to add up to be the lowest possible number.


EXAMPLES!

 C - C - C - C - C     2 methyl  pentane
                   C



Also, when there is multiple chains of the same number use prefixes such as:

di
tri
tetra
etc

Examples:

C - C - C -C            2, 3 dimethyl butane
      C   C


C - C - C - C -C      2 ethyl pentane
      C
      C

(NR) KETONES

- A ketone is a hydrocarbon chain with a double bonded oxygen that is NOT on either end.

- still follow the standard rules and add "-ONE" to the parent chain.

EX:
draw the following organic chemicals!

4,4 dibromo 6 floro 2,3 dimethyl 3 hexanone

3 ethyl 5,1 difloro 2 phenyl octone

name the following organic chemicals!

 

ALDEHYDES

- an aldehyde is a compound that has a double bonded oxygen at the end of the chain.

- the simplest aldehyde is methanal.

- follow the standard rules and change the parent chain ending to "-AL"
   *** BE CAREFUL WHEN NAMING ALDEHYDES AND ALCOHOLS.

EX:
draw the following:
1,2,3,4 tetraethyl 1 pentyl nonanal

2,4 dichloro 1,5 dimethyl hexanal

name the following:

 

(NR) ALKENES & ALKYNES (double&triple bonds)

- Carbon can form double and triple bonds with Carbon atoms.
- When multiple bonds form fewer hydrogens 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

THESE ARE EXAMPLES OF ALKENES AND ALKYNES!

2 Butane:

5 ethyl 2 8 dimethyldecane

1 Butane

EXAMPLES: TRY THESE:
1) Name:

2) Draw: 2,3, Dimethyl-3,4,4 tetra tetrapropyl decane.
3) Draw: 3, 4, 5 Heptyne

(TG): BONDING & ELECTRONEGATIVITY

BONDING AND ELECTRONEGATIVITY 

Types of Bonds:

• There are 3 types of bonds: Ionic, covalent, metallic

Electronegativity: 

• Electronegativity is how much an atoms wants to join electrons
• Fluorine = 4.0
• Chlorine = 3.0
• Cesium = 0.8
• Atoms with greater en. attract electrons more
• Polar covalent bonds form an unequal sharing of electrons
• Non polar covalent bonds form from equal sharing
• The type of bond formed can be predicted by looking at the difference in en. of the elements
• En is greater than 1.7 = ionic 
• En is less than 1.7 = polar covalent  bond
• En = 0? Non polar covalent bond

EXAMPLES!:

Determine the difference in electronegativity between the following elements: (use your en. chart)

Helium - Lithium = 1.22

Potassium - Sulphur = 1.76

Strontium - Iodine = 1.71

Cobalt - Bromine = 1.08 

Identify the positive and negative sides of the polar bonds. 

 

P - Cu

 

Negative Side - P - Cu - Positive Side

Si - Cr

Negative Side - Si - Cr - Positive Side

 

THESE ARE ALL 100% ORIGINALLY MADE QUESTIONS BROUGHT TO YOU BY: THE ONE AND ONLY TG.

(NR) April 1st, 2011: LAB!

Polar and Non-Polar Solvents Lab

on friday mr. Doktor the Almighty Chemist assigned the class a lab!

The objective of the was to determine if Glycerin is a polar or non-polar.

The materials we used for this lab were...
- test tubes, stoppers and rack
- scupula
- safety goggle and apron
- sodium chloride
- sucrose
- iodine crystals
- paint thinner
- Glycerin

The class observed that when a polar solution mixes with another polar solution AND a non polar solution mixes with another non polar solution these solutions DISSOLVE!
while having to mix a polar solution with a non polar solution (vice versa) the solutions will NOT dissolve!

overall this lab was really fun! we should do more labs... but with explosives!!! hahaha LOL jk jk
WOOOOOOT!!! BLOCK G CHEM IS BETTER THAN BLOCK E YEAAAH BUDDDDY

(TG): CHEM LAB

 DILUTING AND CREATING SOLUTIONS LAB!

Today in class we had a lab where we had to create a solution of Copper Chloride with a concentration of O.1 M and then compare to samples\ Mr.Doktor set up. The solution that matched the solution you create (in terms of colour/shade) is the winner!

Materials:

• Beaker
• Graduated Cylinders
• Stir Rod
• Test Tube Stand

Procedure:

1. Tke about 5g of Copper Chloride and place it into a plastic dish
2. Take your plastic dish and pour the copper chloride onto a piece of wax paper
3. Take the wax paper and place it on your weighing scale
4. Weigh and record the total mass of the copper chloride
5. Put 25 mL of water in a graduated cylinder
6. Take 0.336g from the recorded amount of copper chloride with your scoopula
7. Put 0,336g  of copper chloride into the 25mL of water
8. Slowly stir the solution with a stir rod
9. Compare with Mr.Doktors solution and choose the closest shade of blue that matches yours
10. Repeat the procedure 3 times for consistent results

NECESSARY MATH:

0.025L x (.1mol/L\) x (134.5g/mol) = o.336g

Dont have any copper chloride? Well you could always....

(NR) March 14, 2011: Acid-Base Reactions

• Strong Acids dissociate to produce H+ ions.
  - HCl, H2SO4, HClO4
• Strong Bases dissociate to produce OH- ions.
  - NaOH, Ba(OH)2, LiOH
• When the Strong Acids and Strong Bases mix they form H2O and an ionic salt
  (The total volume changes!)
  
  
  pH and pOH- pH is a measure of the hydrogen ions present in a solution     > pH = -log[H+]
  - pOH is a measure of the hydroxide ions present in a solution > pOH = -log[OH-]
  
  
  EXAMPLE!
  0.200L of 0.500 M HCl is added to 0.200L of 0.500M LiOH.
  What is the mass of water produced?
  
  
  LiOH + HCl  --> H2O  + LiCl
  
  
  .500mol/L x 0.200L x 1/1 x 18g/mol = 1.8g
  
  
  What is the mass of the salt produced?
  
  
  0.500mol/L x 0.200L x 1/1 x 42.4g/mol = 4.24g

(DA) Feb 28, 2011: Titrations

• A titration is an experimental technique used to determine the concentration of an unknown solution

Terms & Equipment

• Buret - contains the known solution. Used measure how much is added
• Stopcock - 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 end point of the titration
• Stock Solution - Known solution

Example

• Jon the chemist wants to determine the concentration of a Sodium Hydroxide sample so he does a titration with HCl. He gathers the following data. Determine [NaOH]

NaOH sample = 10.00 mL                      [HCl] = 0.75M

Trial                              1                             2                          3                        4
Final Reading (mL)      13.3                        26.0                     38.8                  13.4
Initial Reading (mL)       0.2                        13.3                     26.0                   0.60
Volume Used (mL)      13.1                        12.7                     12.8                  12.8
                                                                                                          Vaverage = 12.85 mL


0.75 mol x 0.01285 L x 1 = 0.0964 mol
      L                             1

0.0964 mol x 1     = 0.964 mol/L
                 0.010L

(DA) Feb 24, 2011: Solution Chemistry

Example

• 100mL of 0.250 M Iron (II) Chloride reacts with excess copper. How many grams of Iron are produced?
  
  FeCl2 + Cu -> CuCl2 + Fe
  
  0.100L x 0.250 mol x 1 x 55.8g = 1.40 g
                        L          1     1mol
  

• How many moles of Copper (II) Chloride are produced?
  
  0.100 L x 0.250 mol x 1 = 0.0250 mol
                         L          1
  
• Determine [CuCl2]
  
  0.0250 mol x 1 = 0.250 mol/L
                    0.100L

(DA) Feb 22, 2011: Solution Stoichiometry

Solutions

• Solutions are homogenous mixtures composed of a solute and a solvent
  - Solute is the chemical present in lesser amount (whatever is dissolved)
  - Solvent is the chemical present in greater amount (whatever does the dissolving)
• Chemicals dissolved dissolved in water are aqueous
  - NaCl(aq); H2SO4(aq)

Molarity

• Concentration can be expressed in many different ways
  - g/L, ml/L, % by volume, % by mass, mol/L
• The most common (In Chemistry 11 & 12) is mol/L which is also called Molarity
  - mol/L = M
  - [HCl] = concentration of HCl
  
  Molarity = Moles
                  Volume 

Examples:

• What is the concentration of Sodium Chloride solution made from 0.75 mol of NaCl dissolved in 250 mL of water
  
  0.75 mol = 3.0 mol/L                or 3.0 M
  0.250 L 

(NR) DILUTIONS - March 3rd, 2011

Diluting solutions

- 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.  n1 = n2

- because concentration is mol/L 

   We can write…

                  

        C=n/V          AND          n= CV          AND          V=n/C

So… 

C1V1 = C2V2

(initial) = (final)

EXAMPLES

A 12.0 mL of 38 M of hydrochloric acid is diluted to a final volume of 380 mL. What is the concentration of the new solution?

*REMEMBER TO CONVERT mL to L!

C1V1   =   C2V2

(38M)(0.012L)   =   C2(0.380L)

0.456mol   =   C2(0.380L)

1.2M   =   C2

Therefore new solution has a concentration of 1.2M

Mr. Doktor the ALMIGHTY chemist is given a solution of 3.8M NaOH. After diluting the solution to a final volume of 0.12L the concentration is 0.83M. What volume of NaOH was Mr. Doktor the ALMIGHTY chemist initially given?

C1V1   =   C2V2

(3.8M)V1   =   (0.83M)(0.12L)

(3.8M)V1   =   0.0996mol

V1   =   0.026L

(TG) Limiting Reactants! FUNFUNFUN.

LIMITING REACTANTS.

Before we dive into the wonderful world of Chemistry I believe it is important to first understand exactly what a limiting reactant is. With limiting reactants it is best, and most important to understand exactly what are you doing, as opposed to just memorizing a process. 

A limiting reactant, is one of the reactants in a chemical reaction that gets used up in the rxn first, and "limits" the rest of the reaction. A great comparison to this process in chemistry appears in our cafeteria: On pancake day one meal consists of:

1. 2 pancakes 
2. 1 sausage

Every single breakfast has to have these components. Now if we have 30 sausages and 2 pancakes how many meals can we make? Only one! Then pancakes are the limiting reactants because they limit/prevent the making of any further breakfasts. The same is true in chemical reactions.

Now on to some examples:


Identify the limiting reactant when 1.22 g of 0(2) reacts with 1.05 t of H(2) to produce H(2)0

2H(2) + 0(2) = 2H(2)O

1.22g x mol/ 32g x 2/1 x 2 = .15g

Now because we only need .15 g of H and we have 1.05 g of it. We know that Hydrogen is NOT the limiting reactant leaving Oxygen as our only option :)!

(TG) Heat and the Stoichiometric Method

* I know that these two topics were taught several weeks apart, but I want to keep the blog updated, and 1 post is simpler than 2.

ENERGY (in kJ's)

• Often times in chemistry you will get a chemical reaction that is either: EXOTHERMIC, or ENDOTHERMIC. These words may seem complicated but really they just mean, releasing heat (exo) and absorbing heat (endo)
• Whenever you see these in a formula (ie. NaBr + I = NaI + Br + 234kJ) all you have to do is treat the kJ's like another element.
• Since energy will only appear on one side of the equation you dont even have to balance it
• Whenever using energy pretend as if there is the coefficient of "1" infront of it

Now lets look at an example:

How much energy is released when 2.0 mol of I is used up?

2NaBr + I(2) = 2NaI + Br(2) + 345kJ

2.0 x 1/2 = 1 x 345 = 345kJ

(NR) Feb 8, 2011 : ENERGY AND PERCENT YIELD

• Enthalpy is the energy stored in chemical bonds
• Symbol of Enthalpy is H
  - units of Joules (J)
• Change in Enthalpy is delta H
• In Exothermic reactions enthalpy decreases
• In Endothermic reactions enthalpy increases
  
       CALORIMETRY
  • To experimentally determine the heat released we need to know THREE things:
    
    1. TEMPERATURE CHANGE (delta T)
    2. Mass (m)
    3. Specific heat capacity (C)
       Cw= 4.19 J/g degrees C
     These are related bythe equation:
          delta H = mCdeltaT            
                     = mC(Tf-Ti)
    
    
    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
    
    

(NR) Jan 31, 2011: LAB!

today we did a lab, "Testing the Stoichimetric Method"

PROBLEM: Does Stoichimetric accurately predict the mass of products produced in chemical reactions?

PROCEDURE:

1. Carefully measure about 3.00g of Copper (II) sulphate.
2. Crush the Copper (II) sulphate into a fine powder using a mortar and pestle.
3. Dissolve the Copper (II) sulphate in 50mL of water. 
4. Carefully measure 2.00s of Strontium nitrate and dissolve it in 50mL of water
5. Slowly pour the two solutions together.
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 an Erlenmeyer flask, place the filter paper in funnel. Slowly pour the mixture into the funnel.
10. Pour the filtrate into the waste collection bottle.
11. Place the filter paper in the drying oven and record the mass when its dry.

Next class we will check our filter paper and weigh it

(TG) Mass to Mass Conversions

MASS TO MASS CONVERSIONS: A Process

• Mass to mass problems involve one addition conversion
• This can best be described through the following diagram/process:     

 The Three Step Process:

START WITH GRAMS OF 'A'

(1 step)

CHANGE TO MOLES OF 'A'

(2 step)

CHANGE TO MOLES OF 'B'

(3 step)

END UP WITH GRAMS OF 'B'

EXAMPLES:


Lead (IV) Nitrate reacts with 5.0g of Potassium Iodide. How many grams of Lead (IV) Nitrate are required for a complete reaction? (this example includes the LONG way of the process)

Pb(NOthree)four + 4KI ----> PbIfour + 4KNOthree

5g of KI (multiplied by) 1 mol/166g = .0301 mol

.0301 (multiplied by) 1/4 = .007525

.007525 (multiplied by) 455.2g/mol =

3.4 GRAMS of LEAD (IV) NITRATE 

How many grams of Otwo are produced from the decomposition of 3.0g of Potassium Chlorate (This example includes the FAST way of the process)

2KClOthree ----> K + Cltwo + 3Otwo

3.0 grams (multiplied by) mol/122.6g (multiplied by) 3/2 (multiplied by) 32g/mol =

1.2g

ALWAYS REMEMBER TO COUNT YOUR SIG FIGS!!!

(TG) STOICHIOMETRY; quantitative chemistry

LEARNING ABOUT STOICHIOMETRY

• Stoichiometry is a branch of chemistry that deal with the quantitative analysis of chemical reactions
• It is a generalization of mole conversions to chemistry reactions
• To learn about stoichiometry we must have a good grasp on balancing equations  
• Balancing equations involves only simple math, and a periodic table

THE 6 TYPES OF EQUATIONS:

1. Synthesis
2. Decomposition (reverse synthesis)
3. Single Replacement
4. Double Replacement
5. Neutralization (a form of double replacement)
6. Combustion

When balancing equations you have to ensure there are the same number of atoms of each element on either on both sides of the equations. When doing this you can add coefficients to the beginning of an atom/molechule.

When balancing combustion equations a good trick to know is to first balance the Carbons then the Hydrogens and then finally oxygens.

REMEMBER: never change the subscripts when balancing equations.(only when finding the chemical formula)


EXAMPLES:

Al +Ftwo ------> AlFthree = 2Al +3Ftwo ------> 2AlFthree




HthreePOfour ------> Htwo + Pfour + Otwo =  4HthreePOfour ------> 6Htwo + Pfour + 8Otwo



 CHfour + Otwo ------> COtwo + HtwoO =  CHfour + 2Otwo ------> COtwo + 2HtwoO


**** Also remember that in some equations you may have to use fractions balance your equations. As well always reduce to lowest terms.






 

(NR) Jan 10th, 2011: Empirical Formula & Molecular Formula

- Empirical Formula are the simplest formula of a compound.

- Show only the simplest ratios, not the actual atoms.

• The empirical formula for Chlorine gas is Cl, not Cl2 ([*chlorine is diatomic] its reduced to its lowest factor)
• N2O4 becomes NO2 (reduced and divisible by 2)

- Molecular formulas give the actual number of atoms

- To determine the Empirical formula we need to know the ratio of each element.

When you're making your chart be sure to have the following for FULL marks!

1. ATOM
2. MASS
3. MOLAR MASS
4. MOLES
5. MOLE/SMALLEST MOLE
6. RATIO

*MUST BE MEMORIZED!

- The simplest ratio may be decimals. For certain decimals you need to multiply everything by a common number.

MOLECULAR FORMULAS

- The empirical formula to find the molecular formula you need the molar mass. 
*With this chart you should be given one information for each side!
** YOU MUST ALSO KNOW AND MEMORIZE THIS CHART FOR FULL MARKS!


_________________EMPIRICAL FORM.                   MOLECULAR FORM.______

FORMULA:

MOLAR MASS:

(DA) Jan. 7, 2011: Percent Composition

• The percentage by mass of an element in a compound is always the same
• To find the percent by mass, determine the mass of each element present in one mole.

EXAMPLE:

• Determine the percent by mass of Carbon in Octane (C8H18)

 8     =    30.8% Carbon             96     =    84% Hydrogen
26                                             114

• Determine the percent of mass of each element in Zinc Chloride ZnCl2   

65.4+71             Zn 65.4 = 47.9%
                              136.4      
                                                                            
                          Cl  71  = 52.1%
                              136.4             

*Keep to 3 decimal places!