Functional Groups

Ether
- Contain an oxygen atom in the middle of two carbon chains
- Name by lowest number of carbons to highest
- functional group is "O" and ends in "ether"

Aldehydes
- An aldehyde is a compound that has a double bonded oxygen at the end of a chain
- The simplest aldehyde is methanal (also called formaldehyde)
- ends in -al

Carboxylic Acids
- Carboxylic acids are formed by the function group O=C-OH
- Use standard rules but change the parent chain ending to -oic acid
- The simplest carboxylic acid is methanoic acid

Esters
- Carbon chain separated by an oxygen
- Has double bonded oxygen
- ends in - oate
- Naming starts with carbon chain without double bonded oxygen

Nitro
- Standard naming rules
- Nitro in front of parent chain
- functional group : NO2

Amine
- Function group : N
- Separates carbon chains
- numbering starts from N & goes in alphabetical order
- ends in Amine
- bonded to either hydrogen or carbon
- Primary amines = 1 carbon chain
- Secondary amines = 2 carbon chains
- Tertiary amines = 3 carbon chains

Amides
- Functional group : O=C-N
- ends in -amide
- NH2 starts with double bonded oxygen

- JanCarlo Paysan

Aromatics

- Benzene (C6H6) is a cyclic hydrocarbon with unique bonds between the carbon atoms
- Structurally it can be drawn with alternating double bonds
- Careful analysis shows that all 6 C-C bonds are identical and really represent a 1.5 bond
- This is due to electron resistance
- electrons are free to move all around the ring

Aromatic Nomenclature
- A Benzene molecule is given a special diagram to show it's unique bond structure
- Benzene can be a parent chain or a side chain
- As a side chain it is given the name phenyl

- JanCarlo Paysan

Organic Chemistry

Nomenclature
- Organic chemistry is the study of Carbon compounds
   - Carbon forms multiple covalent bonds
- Carbon compounds can form chains, rins, or branches
   - there are less than 100 000 non-organic compounds
   - organic compounds number more than 17 000 000
- The simplest organic compounds are made of carbon and hydrogen
- Saturated compounds have no double or triple bonds
- Compounds with only single bonds are called alkanes and always end in -ane
- Isomer-two or more compounds with the same empirical formulas
- There are three categories of organic compounds
   - Straight Chains
      - rules for naming chain compounds
         - Circle the longest continuous chain and name this as the base chain -meth ,eth ,prop...
         - Number the base chain so side chains have the lowest possible numbers
         - Give each side chain using the -yl ending
         - Give each side chain the appropriate number
           - if there is more than one identical side chain numbers are slightly different
         - List side chains alphabetically
   - Cyclic Chains
   - Aromatics

- JanCarlo Paysan

Alicyclics

cyclopentane

-Are consists of 3 or more carbons
-Are single carbon bonds formed into a loop
-Prefix to name these kind of bonds are cyclo- (ex:cyclobutane)
-Side chains may also be alicyclic (ex: cyclobutyl)



-Paul Dinh

Halides

-Halides consists of F, Cl, Br, I.
-Are always side chains connected to carbon atoms
- When part of naming, Flourine would be Floro, Chlorine would be Chloro, Bromine would be Bromo, and Iodine would be Iodo. (ex: 1 floro 1 chloro 1 bromo 1 iodo methane)

-Paul Dinh

Alcohols and Ketones

Alcohols
-An alcohol is a part of the hydroxyl group
-Has an -OH on the end of the parent chain
- Follows the same nomenclature rules as other groups
-End in -ol (ex: ethanol)

Ketones
-A ketone is a part of the carbonyl group (carbon to oxygen)
-Has a double bond -O as a side chain from the parent group but never on the ends.
-Ends in -none (ex: butanone)

-Paul Dinh

Alkene(dbl bond) and Alkynes (triple bond)

Double bond (alkenes)

- Carbon can form double and triple bonds with other carbon atoms.
- When multiple bonds form fewer hydrogens are attached to the Carbon atom
  - The position of the double/ triple bonds will always have the lowest number
- double bonds alkenes end in -ene
- triple bonds end in -yne

Triple bond (alkynes)

-Paul Dinh

Intermolecular Bonds

Types of Bonds
- Intramolecular bonds
   - exist within a molecule
   - ionic, covalent
- Intermolecular bonds exist between molecules
   - The stronger the intermolecular bonds the higher the boiling point or melting point
   - Two types of Intermolecular : Van der Waals Bonds and Hydrogen bonds

Van der Waals Bonds
- Based on electron distribution
- Two categories
1) Dipole - Dipole bonds
    - If a molecule is polar the positive end of one molecule will be attracted to the negative end of another      molecule
2) London Dispersion Forces (LDF)
    - LDF is present in all molecules
    - creates weakest bonds
    - if a substance is non-polar dipole-dipole forces don't exist
    - Electrons are free to move around and will randomly be grouped on one side of the molecule
    - This creates a temporary 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
- This forms a very strong intermolecular bond

Effects of Bonds
- Melting and boiling points are a measure of intermolecular bond strength

- JanCarlo Paysan

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 electrons is usually balanced
- Molecules can be unsymmetrical in two ways
   - different atoms
   - different number of atoms

- JanCarlo Paysan

Bonding

Bonds & Electronegativity

Types of Bonds
- There are 3 main types of bonds:

1) Ionic (metal - non metal)
     - electrons are transferred from metal to non metal
2) Covalent (non metal - non metal)
     - electrons are shared between non metals
3) Metallic (metal)
     - holds pure metals together by electrostatic attraction

Electro Negativity
- Electro negativity (en) is a measure of an atom's attraction for electrons in a bond
- Atoms with a greater en attract electrons more
- Polar Covalent bonds form from an unequal sharing of electrons
- Non-Polar Covalent bonds form from equal sharing

Bonds
- The type of bond formed can be predicted by looking at the difference in electronegativity of the elements
    - en > 1.7 = ionic bond
    - en < 1.7 = polar covalent bond
    - en = 0 = non polar covalent bond

- Jan Carlo Paysan

Ion Concentration

Dissociation

-Ionic compounds are made up of 2 parts:
 →Cation: positively charged particle
 →Anion: negatively charged particle
-When ionic compounds are dissolved in water the cation & anion separate from eachother
-This process is called dissociation
-When writing dissociation equations, the atoms & charges must balance.
The dissociation of Sodium Chloride is:
      NaCl(s) → Na+(aq) = Cl-(aq)
*have to write charges*

-Paulo Santillan

Dilutions

Diluting Solutions

-When 2 solutions are mixed the concentration changes
-Dilution is the process of the decreasing the concentration by adding a solvent (usually water)
-The amount of solute does not change

n1 = n2
C1V2 = C2V2

-Because concentration is mol/L we can write:
C = n/v
    &
n = CV → mol/L x L = mol
   so
C1V1 = C2V2

-EXAMPLE
-Determine the concentration when 100ml of 0.10M of HCL is diluted to a final volume of 400ml.
V2= 400mL
V1= 100mL
C1= 0.10M
C2= ?

C1V1 = C2V2
C1V1 / V2 = C2
(100)(0.1) / (400) = C2
C2 = 0.025M

-Paulo Santillan

Titrations

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



Terms and Equipment

-Buret: contains the known solution. Used to measure how much is added
-Stopcock: Valve used to control the flow of the solution from the Buret
-Pipet: Used to accurately measure the volume of the unknown solution
-Erlenmeyer Flask: Container for unknown solution
-Indicator: Used to identify the end point of the titration
-Stock Solution: Known solution

-Paulo Santillan

Solution Stoichiometry

Solutions
- Solutions are homogenous mixtures composed of a solute and a solvent
   - solute is the chemical present in the lesser amount (whatever is dissolved)
   - solvent is the chemical present in the greater amount (whatever is dissolving)
- Chemicals dissolved in water are aqueous
    NaCl(s) --H20--> NaCl(aq) : to show something dissolved by water

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

- JanCarlo Paysan

Limiting Reactants

- 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

ex. 2.5 mol of CuSO4 reacts with 2.5 mol of NaNO3. Determine the L.R. and calculate how many moles of Na2SO4 are formed

CuSO4 + 2NaNO3 -> Na2SO4 + Cu(NO3)2
2.5 mol * 2/1 = 5 mol of NaNO3 needed
2.5 mol * 1/2 = 1.25 mol of CuSO4 needed
- NaNO3 is the L.R.

- JanCarlo Paysan

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

Evergy & Percent Yield

-Enthalpy is the energy stored in chemical bonds
-Symbol of Enthalpy is H
      - units of joules (J)
- Change in Enthalpy is ΔH
- In exothermic rxns enthalpy decreases
-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)
There are related by the equation: ΔH = mC ΔT

ex: calculate the heat required to warm a cup of 400g of water (C= 4.18J/g °C) from 20.0 °C to 50.0 °C.
 ΔH=mCΔT
 ΔH= (400)(4.18)(50.0-20.0)
 ΔH=50160 J = 50.2 kJ


-Paul Dinh
 

Other Conversions : Volume & Heat

-Volume at STP can be found using this conversion factor 22.4 L/mol
-Heat can be included as a separate terms in chemical reactions (This is called Enthalpy)
-Rxns that release heat are exothermic
-Rxns that absorb heat are endothermic
-Both can be used in stoichiometry

ex: If 5.0 g of Potassium chlorate decompose according to the reaction below, what volume of Oxygen gas (at STP) is produced?

2KClO3 > 2KCL + 3O2
5.0 g x 1mol/122.6 g x 3/2 x 22.4 L/ 1 mol = 1.4 L of Oxygen


ex2: When Zinc reacts with hydrochloric acid exactly 1.00 L of hydrogen gas is produced (at STP). What mass of Zinc was reacted?


Zn +2HCl > ZnCl2 + H2
1.00 L x 1mol/22.4 x 1/1 x 65.4 g / 1 mol = 2.92 g of Zinc


- Paul Dinh

Mass to Mass Conversions

mass to mass problems are additional conversions.

Grams of A                                                                      Grams of B
            |                                                                                     |
Moles of A -  >>Multiplied boy the stoichiometric ratio >>> Moles of B

ex: lead (IV) nitrate reacts with 5.0 g of Pottasium iodide. How many grams of lead (IV) nitrate are required for a complete reaction?

Pb (NO3)4  + 4KI   > 4KNO3  = PbI4
5.0g x (1 mole over 166 g/mol) x 1/4  x (455 over 1 mole) = 3.4 g

ex2: how many grams of O2  are produced ffrom the decomposition of 3.og of Pottasium chlorate?


2KClO3 >> @K + Cl2 + 3O2


3.0 g x (1 mole over 122.6 g  x 3/2 x (32 over 1 mol) = 1.2 g


-Paul Dinh

Mole to Mass & Mass to Mole

ex How many grams of Bauxite (Al2O3) are required to produce 3.5mol of pure aluminum?

Step 1 - Write a balanced equation (2Al2O3 - > 4Al + 3O2)

Step 2 - Change moles of aluminum to moles of bauxite using "what you need over what you have" (2/4)

Step 3 - Multiply the moles of bauxite by the molar mass of bauxite (1.75mol * 102g/mol)

=1.8*10^2g of bauxite

How many moles of Lead(II)Nitrate are consumed when 4.5g of sodium sulphide completely react?

Step 1 - Write a balanced equation (Pb(NO3)2 + Na2S -> PbS + 2NaNO3)

Step 2 - Change grams of sodium sulphide to moles using molar mass ( 4.5/78.1)

Step 3 - Change moles of sodium sulphide to moles of Lead(II)Nitrate using "what you need over what you have" (1/1)

= 0.058mol of Lead(II)Nitrate


- JanCarlo Paysan

Mole to Mole Conversions

- Coefficients in balanced equations tell us the number of moles reacted or produced
- They can also be used as conversion factors
ex 3X + Y -> 2Z    ratios :  x to y 3:1, x to z 3:2, y to z 1:2

- What you need over what you have (stoichiometric ratio)
ex

if 0.15mol of methane are consumed in a combustion reaction how many moles of CO2 are produced?
CH4 + 2O2 -> CO2 + 2H2O
0.15 * 1/1 = 0.15 mol of CO2


When 2.1*10^-2 mol of Aluminum Hydroxide react with Sulphuric acid how many moles of Aluminum sulphate are produced?
2Al(OH)3 + 3H2SO4 -> Al2(SO4)3 + 6H2O
2.1*10^-2 * 1/2 = 1.1*10^-2 moles of aluminum sulphate 


How many moles of bauxite (Aluminum Oxide) are required to produce 1.8mol of pure aluminum?
2Al2O3 -> 4Al + 3O2
1.8 * 2/4 = 0.90 mol of aluminum


- JanCarlo Paysan

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative analysis of chemical reactions
- it is a generalization of mole conversions to chemical reactions
- understanding the types of chemical reactions is the foundation of stoichiometry

6 types of reactions
1) synthesis
2) decomposition
3) single replacement
4) double replacement
5) neutralization
6) combustion

Synthesis
- A + B = AB
- usually elements -> compounds
ex 2Al + 3F2 -> 2AlF3
     4K + O2   -> 2K2O
     CoCl + 6H2O -> CoCl2 * 6H2O
     2SO2 + O2 -> 2SO3


Decomposition
- AB -> A + B
- Reverse of Synthesis
  - Always assume the compounds decompose into elements during decomposition unless told different
ex 4H3PO4 -> 6H2 + P4 + 8O2
     Mn(C2O4)2 - > Mn + 4C + 4O2
     2C12H22O11 - > 24C + 22H2 + 11O2


Single Replacement
- A + BC -> B + AC
ex Ca + 2KCl -> 2K + CaCl2
     3Mg + 2Al(NO3)3 -> 2Al + 3Mg(NO3)2


Double Replacement
- AB + CD -> AD + CB
ex MgCl2 + K2SO4 -> MgSO4 + KCl
     Mn(ClO4)4 + 2CaCO3 -> Mn(CO3)2 + 2Ca(ClO4)2


Neutralization
- Reaction between an acid and a base
ex H2SO4 + 2KOH -> 2H2O + K2SO4
     3Ca(OH)2 + 2H3PO4 -> 6H2O + Ca3(PO4)2


Combustion
- Reactions of something (usually hydrocarbon) with air
- Hydrocarbon combustion always produces CO2 and H2O
ex CH4 + 2O2 -> CO2 + 2H2O
     2C8H18 + 25O2 -> 16CO2 + 18H2O
     2C11H23OH + 33O2 -> 22CO2 + 24H2O

- JanCarlo Paysan
 

Molecular Formula

Molecular formulas give the actual number of atoms
- if you know the empirical formula, to find the molecular formula you need the molar mass

example
empirical formula - CH2O  molar mass - 60.0g/mol  Determine molecular formula
step 1 - find molar mass of CH2O (30.0g/mol)
step 2 - divide molar mass of molecular formula by molar mass of empirical (60.0/30.0 = 2)
step 3 - multiply all subscripts of the empirical formula by the answer (2)
molecular formula = C2H4O2


- JanCarlo Paysan

Empirical Formulas

Empirical formulas are the simplest formula of a compound
- examples
    - empirical    |   molecular
        C4H9      |     C8H18
        NO2        |     N2O4
        CH7O2   |     C3H21O6
           P2O5     |     P4O10
        C5H11   |     C10H22
        C2H6O  |     C6H18O3
        C5H12O|     C5H12O
        NO2       |     N2O4
        H2O       |      H2O
- show only the simplest ratios, not the actual atoms
   - the empirical formula for chlorine gas is Cl
   - dinitrogen tetraoxide is not equal to N2O4
- To determine the empiricaL formula we need to know the ratio of each element
- To determine the ratio fill in the table below

Atom | Mass | Molar Mass| Moles | Mole/Smallest Mole | Ratio

- The simplest ratio may be decimals. For certain decimals you need to multiply everything by a common number
           Decimal           | Multiplying Coefficient
    0.5                         |                2
    0.33 or 0.66           |                3
    0.25 or 0.75           |                4
    0.2, 0.4, 0.6, 0.8    |                5

- JanCarlo Paysan

Density & Moles

Density is a measurement of mass per volume
formula is -> d=m/v

examples
Water has a density of 1.0g/m. Mass of 11.5 mL of water? How many moles?
d=m/v  d * v = m  1.0 * 11.5 = 11.5g
11.5 * 1mol / 18 = 0.639mol of water


Unknown compound with molar mas of 65.0g/mol. If 0.25mol occupies a volume of 50mL, determine the compound density.
0.25mol * 65.0 / 1mol = 16.25g / 50mL = 0.33g/mol


The density of aluminum is 2.70g/mL. A solid piece of aluminum has a volume of 45.0mL. Find the number of atoms.
d = m/v   d*v = m  2.70 * 45.0 = 121.5g * 1mol / 27g * 6.02*10^23 / 1mol = 2.71*10^24


- JanCarlo Paysan

Moles ↔ Atoms / Molecules

[ MOLES ] → [ MOLECULES ] → [ ATOMS ]
       Avogadro's Number       Subscripts

Example:
• How many water molecules are there in 0.65 mol?
0.65mol x 6.02x10^23molec / 1 mol = 3.9 x 10^23molec

•How many hydrogen atoms are there? How many oxygen atoms?
3.9x10^23molec x 2 atoms / 1 molec = 7.8 x 10^23atoms }Hydrogen

3.9x10^23molec x 1 atoms / 1 molec = 3.9 x 10^23atoms } Oxygen

- Paulo Santillan 

Mole Conversions Chart

[DENSITY]
         ↑                                                                                                                    
         ↑                                                                                                             
         ↨                                                                                                           
         ↓                                                                                                               
         ↓   Molar Mass                 Molar Volume                                        
[MASS] ← ← ↔ → → [MOLES] ← ← ↔ → → [VOLUME (@STP)]
                                              ↑
                                              ↑
                                              ↨   Avogadro's Number
                                              ↓  
                                              ↓
                                   [MOLECULES]
                                              ↑
                                              ↑
                                              ↨   Subscripts            
                                              ↓
                                              ↓
                                       [ATOMS]

-Paulo Santillan

Mole Ratio Lab (November 30, 2011)

In this lab, we were to determine the ratio btwn. the moles of iron → moles of copper. (Copper II Chloride)

1) Weigh both nails and filter paper
2) Collect copper (II) chloride w/ scoopula
3) Place the 8g of weighed copper (II) chloride on top of filter paper
4) Add 8g of copper (II) chloride to beaker, then fill w/ 50mL of water
5) Stir the copper (II) chloride & water
6) Determine the mass of nails
7) Place the 2 nails into the solution for 10 mins
8) Place funnel in clamp
9) Fold filter paper into 4 divisions & put in funnel in the shape of a cone
10) Use the tongs to pick up the 2 nails from the beaker & scoop the copper (II) chloride off the nails into the solution
11)Put stirring rod in center of funnel & pour solution into funnel
12) After all the solution is in the beaker, pick the filter paper w/ the tongs
13) Place filter paper in oven for about 30 mins
14) Take paper & weigh on scale
15) Clean up everything

-Paulo Santillan

Molar Volume Lab (November 22, 2011)

In this lab we followed the steps to determine the molar volume of a gas (Butane = C4H10)

1) Fill the sink about 3/4 w/ water
2) Weigh your butane lighter on the scale
3) Place the graduated cylinder in water (base facing upwards)
4) Fill the whole graduated cylinder w/ water
5) Hold the butane lighter under the cylinder (water)
6) Click the butane lighter to release butane in cylinder
7) Make a measurement of the amount of butane filling the test tube
8) Dry the butane lighter in oven for about 10-15 min
9) Weigh the butane lighter again
10) Make calculations
11) Clean up

-Paulo Santillan