Carbocylic acid and Ethers

Carboxylic Acids are formed by the functional group:



-this is also the simplest Carboxylic Acids called Methanoic acid
-use standard rules but change the parent chain ending to "-oic acid"


ETHERS

-an Ether contains an oxygen group connected to 2 alkyl (carbon) chains



-name the smaller alkyl group first, then the second alkyl group followed by "ether"

Alicyclics and aromatics

We learned about alicyclics and aromatics

Carbon chains can form 2 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: "cyclo" is added in front of the parent chain
There are 3 different ways to draw organic compounds:

• Complete structural diagram
• Condensed structural diagram
• Line Diagram

Numbering can start anywhere and go clockwise or counterclockwise on the loop but side chain numbers MUST be the lowest possible!


CONDENSED STRUCTURAL DIAGRAMS

A condensed structural diagram always shows carbons and hydrogens in it
loops can also be side chains
same rules apply but the side chain is given a cyclo- prefix




NAMING COMPOUNDS


You see 2 lines connecting CH2 and C, so you know this organic compound forms a double bond and you must use the ending -ene
You count the longest carbon chain to be 5 so we use -pent
-pent + -ene = pentene and this bond occurs at 1 so: 1 pentene
we see a loop on #3 of the carbon chain so we must use to prefix -CYCLO and there are 3 of the CH's on it so we use propyl and carbon and hydrogen on #2
#2: 2 methyl
#3: 3 cyclopropyl
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
Careful analysis shows that all 6 C-C bonds are identical and really represent 1.5 bond
This is due to electron resonance for electrons are free to move all around the ring


AROMATIC NOMENCLATURE:

A benzene molecule is given a special diagram to show its unique bond structure


A benzene can be a parent chain or a side chain
As a side chain it is given the name phenyl

Draw line diagrams
a) 1, 4 diethyl 2 methyl Benzene
b) 1,3,5 triethyl 2,4,6 tripropyl benzene

Organic Chem + Alkanes

Organic Chemistry is the study of CARBON COMPOUNDS
- Carbon forms multiple covalent bonds


Carbon Compounds can form chains, rings 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


CH4 (Methane)



CH3CH3 (Ethane)


- Saturated compounds have no double or triple bonds
- Compounds with only single bonds are called Alkanes and always end in "-ane"




NOMENCLATURE


There are 3 categories of organic compounds
1) Straight chains
2) Cyclic chains
3) Aromatics



STRAIGHT CHAINS


Rules for naming straight chain compounds:
1) Circle the longest continuous chain and name the base chain
- meth, eth, prop, etc.
2) Number the base chin 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
5) List side chains alphabetically


Name the Alkanes!! :

Ketones and Aldehydes

KETONES & ALDEHYDES

A ketone is a hydrocarbon chain with a double bonded oxygen that is NOT on either end
An Aldehyde is a compound that has a double bonded oxygen at the END of the a chain.
Follow standard rules and add -one to the parent chain for Ketones
Follow standard rules and change the parent name ending to -al in Aldehydes

Propanone is a ketone. Propanal is an aldehyde. Their endings and where the oxygens are bonded

Here are some examples:

Draw the structural diagrams for the following Ketones:


a) 2, 4 dimethyl 3 pentanone
b) 2 chloro 4 methyl 3 hexanone
c) 1,2,2 trichloro 4,4 difloro 3 butanone





Name the following compound:



2 ethyl 4 floro 2 methyl 3 hexanone


ALDEHYDES


An aldehyde is a compound that has a double bonded oxygen at the END of a chain So Oxygen atoms will be found on the ends of these chains, and will have an ending of -al


Name the compound:








1) Count 4 carbons = Butane
2) Add ending "-al" to Butane = Butanal


Answer: Butanal

The simplest aldehyde is methanal otherwise known as FORMALDEHYDE:







Draw a structural diagram for
2,3 dibromo 4 propyl pentanal

Alkynes and Alkenes

In Chem class, we learned about double & triple bonds, trans & cis butene

Carbon can form double & triple bonds with carbon atoms
When multiple bonds form, fewer hydrogens are attached to the carbon atom
Naming rules are almost the same as Alkanes (single bonds) but instead we use 2 different endings
For DOUBLE BONDS (Alkenes) it ends with -ene (i.e Butene)
For TRIPLE BONDS (Alkynes) it ends with -yne (i.e Heptyne)
*The position of the double/triple bonds always has the lowest number and is put infront of the parent chain

Double Bonds -ene



1. First determine the longest carbon chain. The longest Carbon chain here is 6, therefore we use the stem name -HEX


2. Next, you see a double bond represented by 2 lines (circled in red). Then you know our ending has to end with -ene. So it is 1 Hexene


3. Since you've found your parent chain (1 Hexene), determine the side chains. On #3 and #4 you see a 2 carbon group therefore it is 3,4 diethyl


4. So your answer is 3,4 diethyl hexene


Triple Bond: -yne

MULTIPLE DOUBLE BONDS:

More than 1 double bond can exist in a molecule
Use the same multipliers inside the parent chain



TRANS & CIS BUTENE

If 2 adjacent carbons are bonded by a DOUBLE BOND and have side chains on them 2 possible compounds are possible
CIS and TRANS:








Can you name this compound?









Answer: 4 ethyl, 4,5 dimethyl1 heptene

Functional Groups

Functional groups are organic compounds that can contain elements other than Carbon or Hydrogen. These organic compounds contain elements such as Oxygen (alcohols), group 7 elements (halides), double bonded oxygens (ketones/aldehydes), a combination of a double bonded oxygen and an OH group (carboxylic acids) and Nitrogens (Amines)

ALCOHOLS
HALIDES

We will learn about nine functional groups!!!

Alcohols
Halides
Aldehydes
Ketones
Carboxylic Acids
Ethers
Amines
Amides
Esters

ALCOHOLS:

An alcohol is a hydrocarbon with a -OH bonded
Its ending is -OL (i.e Methanol)
Same naming rules apply
If a compound has more than one -OH group number both and add -diol, -triol, -tetraol, etc
Examples:




After observing the structural diagram try to find some clues to start. From this diagram you see there is an OH, therefore you know this is an alcohol and has to end with -OL
Count the longest carbon chain. The longest carbon chain is 8. Therefore, it is 1 Octanol
Name the side chains from 1 Octanol. 3 ethyl, 4 propyl
The compound is 3 ethyl, 4 propyl 1 octanol
Name the following Alcohols:




On the left: You see 2 OH groups and a carbon chain of 2 (eth-). Since this is an alcohol with 2 OH groups, we name this Ethanediol
On the right: We know that this is a benzene molecule. But it is not exactly a benzene molecule because it has an OH group attached to it. This is called a Phenol.


HALIDES:



Group 7 elements (F, Cl, Br, I) can bond to a hydrocarbon chain
Naming follows standard rules with halides using
floro-, chloro-, bromo-, and iodo-





Left: a carbon chain of 2. (Ethane). The name of this compound is: 1,1,2,2 tetrabromo ethane
On the right: You see a carbon chain of 3, and one is double bonded (prop, -ene). The name of this compound is: 2,3 dibromo 3 chloro 1 Propene

 Name the following compounds:

Answers:
(left) 1,3 dibromo 2,4 dichloro cyclobutane
(middle) 2 bromo, 1 ethyl, 3 floro benzene
(right) 3,3 diphenyl 1 propanol

Intermolecular Bonds:

Types of Bonds:

1. Intramolecular bonds = within a molecule (*think intramural)
    - Ionic + Covalent

2. Intermolecular bonds = between molecules (*think international)
    - The stronger the intermolecular bonds the higher the BP or MP
    - 2 Types: Vander Waals bonds & Hydrogen bonds

Vander Waals Bonds:

• Based on electron distribution
• 2 Categories: 

1. Dipole - Dipole bonds 

• if a molec. is Polar, the + end of one molec will be attracted to the - end of another molec. 

     2.  London Dispersion Forces (LDF)

• present in all molecs
• creates the weakest bonds 
• if a substance is non-polar Dipole - Dipole forces don't exist 
• electrons are free to move around & will randomly be grouped on one side of the molec. 
• Creates a temporary dipole and can cause a weak bond to form
• the more e- in the molec. the stronger the LDF will be

EX//

- NH3 (10e) VS. C2H8 (18e)
  polar                  non-polar

- NH3 has the stronger bond because of it's Dipole-Dipole bond.  C2H8 is non-polar thus it's bond is LDF making it's bond weaker than NH3.

Hydrogen Bonding:

• if hydrogen bonded to certain elements (F, O, & N) the bond is highly polar 
• this forms a very strong intermolecular bond.  

EX//

- H2O (10e) VS. CH4 (10e)
  Polar                 Polar

- They are both Polar and both have the same amount of e- but H2O wins for the highest boiling point due to its Hydrogen bond.

A very helpful Video... :)

 

Polar Molecules

What is polarity?

In chemistry polarity refers to a separation of electric charge leading to a molecule or its chemical groups having an electric dipole or multipole moment. Polar molecules interact through dipole–dipole intermolecular forces and hydrogen bonds. Molecular polarity is dependent on the difference in electronegativity between atoms in a compound and the asymmetry of the compound's structure.

*Things to remember

• Polar molecules have an average charge seperation
• Unsymmetrical molecules are usually polar
• Molecular dipoles are the result of un = sharing of electrons in a molecule

Predicting Polarity

• If a molecule is symmetrical, the pull of e- is usually balanced 
• Molecules can be un-symmetrical in 2 ways

-- Diff atoms

-- Diff # of atoms

A molecule can possess polar bonds and still be nonpolar. If the polar bonds are evenly (or symmetrically) distributed, the bond dipoles cancel and do not create a molecular dipole. 

Here's a video that will explain things a bit more clearly....

Bonds

There are 3 main types of bonds in chemistry

1. Ionic (metal to non-metal)
Electrons are transferred from metal to non-metal

2. Covalent (non-metal to non-metal)
Electrons are shared between non-metals

3. Last but not least, Metallic (metal)
Holds pure metals together by electrostatic attraction

Electronegativity (en) is a measure of an atom's attraction for electrons in a bond

Atoms with greater electronegativity attract electrons

Polar Covalent bonds form from an unequal sharing of electrons, a polar covalent bond is a bond between two non-metals with different electronegativities. Unsymmetrical molecules are usually polar.
Non-Polar covalent bonds form from equal sharing of electrons, and are symmetrical molecules.
The type of bond formed can be predicted by looking at the difference in electronegativity (en) of the elements



Here are some examples!


Predict the type of bond formed.


1. H- O

2.20 - 3.44 = 1.24
This bond between Hydrogen and Oxygen is a Polar Covalent bond!


2. K-F


3.99 - 0.82 = 3.17
This bond between potassium and fluorine is an Ionic bond!

Now that we went through some examples together, let's see if you can do some on your own!

QUIZ TIME!!!! (:

Ion Concentration

Disssociation:

-Ionic Particles are made up of 2 parst:

• Cation: + charged particles
• Anion: - charged particles

-When ionic compounds are dissolved in Water the cation and anion separate from each other
-This process is called dissociation.
-When writing dissociation equ'ns the atoms and charges must be ballanced
-The dissociation equ'n for NaCl is Na (+) + Cl (-)

EX// 

What are the following compound's Dissociation equ'ns?

-BaSO4 --> Ba (2+) + SO4 (2-)

-Al2(SO4)3 --> 2Al (3+) + 3SO4 (2-)

-If the volume doesn't change then the concentration of the individual ions depends on the balanced coefficients in the dissociation equ'n.

EX//

Determine the Ion Concentrations...

- 0.250 M of the sol'n KOH:

KOH --> K (+) + OH (-) 

0.250 mol/L x 1/1 + 0.250 M of [K+] 
*because there is a 1:1 ratio for this compound, [OH-] has the same ion concentration as [K+]. 

- A sol'n made from dissolving 2.5g of Ba3(PO4)2 in 50mL of water:

Ba3(PO4)2 --> 3Ba (2+) + 2PO4 (3+)

2.5g x 1 mol/601.9g= 0.00415mol / 0.050L = 0.831 mol/L x 3/1 = 0.249 M of [Ba 2+]
   "                     "                       "                           "              x 2/1 = 0.166 M of [PO4 3+]

A helpful video... :)

Solution Stiochiometry

What is Solution Stiochiometry? : Solution stoichiometry deals with reactions in solutions.

Types of Solution Reactions

   - Precipitation

   - Acid-base

   - Oxidation-reduction

   - Titrations

Example:

100 mL of 0.250 M Iron (II)chloride reacts with excess copper. How many grams of Iron are produced?

FeCl2 + Cu----> Fe + CuCl2

0.250 mol    x 0.100 L x 1     x 55.8 g  = 0.0250g 

           L                          1         mol

Example:

A beaker contains 100 mL of 1.5 M Hcl. Excess Zinc added to the beaker. Determine how many litres of hydrogen gas should be produced.

2HCl + Zn ---> H2 + ZnCl2

1.5 mol    x 0.100L x 1   x    22.4 L  = 1.68 L

        L                        2         mol     = 1.7 L

Here is a step-by- step video that shows you how to solve solution stoichiometry problems

Just in case if you didnt understand the previous video...

Solutions and Molarity

Solutions are homogeneous mixtures composed of a solute & a solvent
Solute: is the chemical present in lesser amount
(whatever is dissolved)
Solvent: the chemical present is the greater amount


Molarity


Concentration can be expressed in many ways. g/L, g/mL,mol/L, etc..

• But the most common is Molarity mol/L= M
• [HCL] concentration of HCl
• [NaCl] concentration of NaCl

Heres a video that teaches you how to calculate molarity problems :)

Molarity= moles    

                Volume

Example:

What is the concentration of a Sodium chloride solution made from 0.75 mol of NaCl dissolved in 250 mL of water?

0.75 mol x     1      =   30 mol/L

Example:

What is the concentration of the solution when 2.75g of Sodium Carbonate are dissolved in 50 mL of water?

2.75g x     1   mol    = 0.0259

                 106 g      = 0.0260

              

Titrations

A Titration is a common laboratory experimental technique used to determine the concentration of an unknown solution. Because volume measurements play a key role in titration, it is also known as volumetric analysis.


Needed for a titration lab:


1. Buret
2. Stopcock
3. Pipet (Glass tube)

4. Erlenmeyer Flask
5. Indicators- Used to identify the end point of filtration
6. Stock solution - Known Solution


Mike the explorer completed a titration of 0.330 M NaOH with 15.00 mL samples of HI of unknown concentration. The data he gathered is below. Determine the concentration of HI.




Subtract the final reading from the initial reading to get the Volume used.


Next we add up 11.2 mL + 10.9 mL + 11.4 mL = divide by 3 = 11.16666667 mL
Doing this gets us the average, but notice how we don't count the volume that was far off.


Next we change 11.16 mL into L

11.16 x 1L/1000 mL
= .0111L

Then, we take our concentration and change that into Moles of HI

0.330 M x .011 L = .00363 x 1 = 0.00363 mol
L 1


Finally, we take 0.00363 mol
and divide it by the 15.00 mL, but remember to convert into L!!



15.00 mL x 1 L/1000 mL
= .015 L


0.00363 mol/ 0.015 L
= .246 M

TItration Quiz!!

Dilutions

Dilution is the process of decreasing the concentration by adding a solvent (usually water)
the amount of solute does not change (n1=n2), but the concentration does change!

Because concentration is mol/L we can write:

C= n/C, n= CV, and V= n/C
n=  number of moles, v= volume in litres


So we can use:

C1V1= C2V2
C1V1 is the intial concentration and volume
C2,V2 is the final concentration and volume


Here are some examples to follow by:


Example 1) A 30.0 mL of 11.9 M of Hydrochloric Acid is diluted to a final volume of 30 mL. What is the concentration of the new solution?


C1V1= C2V2


(11.9mol/L)(.03L)= C2 (0.3L)
.357 mol = C2 (0.3L)
C2= 1.19 M


Example 2) 100 mL of 0.330 M Li2CO3 is added to 600 mL 0.125 M Li2Co3. Determine the [Li2Co3]


0.10 L x (0.330 mol/L) = 0.033 mol/L

0.60 L x (0.125 mol/L) = 0.075 mol/L

0.033 mol + 0.075 mol = 0.108 mol

Take your volumes (100 mL + 600 mL) Add them together then convert it into Liters.


0.108 mol/ 0.7L
= 0.154 M




Example 3) If 250 mL of water are added to 50.0 mL of 1.00 M BaCo3, what will the final concentration be?

(1.00 mol/L) (.05 L) = C2 (.25L)
0.05 mol = C2(0.25 L)
C2 = 0.2 M

Try some Dilution questions!!

Other Conversions:

·         Volume @ STP can be found using the conversion factor 22.4L/1mol

Ex.  If 6.0g of KCLO3 decompose according to the reaction below, what volume of O2 (@ STP) is produced?

1.      Balanced Eq’n: 2KClO3 à 2KCL + 3O2

2.      6.0g x 1mol/122.6g x 3/2 x 22.4L/1mol

= 1.6L

·         Heat can be included as a separate term in chemical reactions (enthalpy)…

·         Things to keep in mind when solving these eq’ns are that rxns that release heat are exothermic and rxns that absorb heat are endothermic. 

Ex.  Find the amount of heat released when 2.0mol of H2 are consumed accordingly to the reaction…

1.      Balanced Eq’n:  N2 +3H2 à 2NH3 + 46.2 kJ

2.      2.0mol x 46.2/3

=77kJ

Ex.  To find the amount of heat released when 675g of ammonia for according to the reaction below…

1.      Balanced Eq’n:  N2 + 3H2 à 2NH3 + 46.2kJ

2.      675g x 1mol/17g x 46.2/2

= 46.2kJ

Mole to Mass & Mass to Mole Conversions

Moles to Mass conversions…

·         Give you an amount of Moles and ask you to determine the mass.

·         Converting Moles – Mass only requires one additional step. 

Ex.  When silver reacts with 3.45 moles of zinc phosphate what mass of silver phosphate would be produced?

1.      Balanced Eq’n:  6Ag + Zn3(PO4)2 à 3Zn + 2Ag3PO4

2.      3.45mol x 2/1 x 418.7g/1mol

= 2889.03 ~ 2.89 x 10^3g

Ex.  You are given 4.00g of methane and are expected to burn it in order to initiate a chemical reaction.  How many moles of water vapour would be produced? 

1.      Balanced Eq’n:  CH4 + 2O2 à 2H2O + CO2

2.      4.00g x 1mol/16g x 2/1

= 0.500 mol

Mass-to-Mass conversions…

·         Involve one additional step to the “Moles to Mass” conversions

Ex.  When aluminum reacts with Iron (III) oxide to produce solid iron and aluminum oxide, what is the mass of aluminum oxide produced along with 19.55g of iron?

1.      Balanced Eq’n:  2Al + Fe2O3 à 2Fe + Al2O3

2.      19.55 x 1mol/55.8g x ½ x 102g/1mol

= 17.87g

·         To achieve the mass of aluminum oxide as shown above, we simply take its amu and multiply it by the final amount of moles. 

Ex.  Sodium iodide reacts with lead (II0 nitrate to produce sodium nitrate and lead (II) iodide.  What mass of lead (II) nitrate will be required to produce 150g of sodium nitrate?

1.      Balanced Eq’n:  2NaI + Pb(NO3)2 à NaNO3 + PbI2

2.      150g x 1mol/85g x ½ x 331.2/1mol

= 292g

Limiting Reactants

In Chemical reactions, usually one chemical gets used up first before the other. The chemical used up first in a chemical reaction is called the limiting reactant.

Once it is used up, the reaction stops! Limiting reactants determines the quantity of products formed.

To find the Limiting Reactant, assume one reactant is used up. Determine how much of this reaction is required!

Remember the very important equation for this:

Actual / Theoretical x 100% = % yield

Try some limiting reactant problems!

Energy and Percent Yield

• Enthalpy is the energy stored in chemical bonds
• The Symbol of Enthalpy is H, Units of Joules (J)
• Change in Enthalpy is ∆H
• In exothermic rnxs Enthalpy decreases
• in edothermic rnxs Enthalpy increases

Here are the definitions of exothermic and endothermic reactions, in case you forget:

 

EXOTHERMIC:

 

The term exothermic describes a process or reaction that releases energy from the system, usually in the form of heat, but also in the form of light, electricity or sound.

 

 

 

ENDOTHERMIC:

 

The word endothermic describes a process or reaction in which the system absorbs energy from the surroundings in the form of heat.

Calorimetry: To experimentally determine the heat released we need to know 3 things:

1. Temperature Change (∆T)
2. Mass (m)
3. Specific heat capacity (C)

These are related by the equation:

 

∆H = mC∆T

Example:

Calculate the heat required to warm a cup of 400 g of water (C= 41.8J/g°C) from 20.0°C to 50.0°C.

 

∆H = mC∆T

 

∆H =(400g)(4.18J/g°C)(50.0°C-20.0°C)

 

=50160 J

 

 

 

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?

 

********THIS EQUATION IS VERY IMPORTANT*********

 

Actual / Theoretical x 100% = % yield

 

                      or

 

Actual amount of product / Expected amount of product x 100% = % yield

 

 

Example: A student makes a single displacement reaction that produces 2.755 grams of copper. He determines that 3.150 grams of copper should have been produced. Find the student's percentage yield.

 

 

                                      actual amount of product

percentage yield = ------------------------------------------- x 100

                                    expected amount of product

 

                                   2.755g

percentage yield = --------------- x 100

                                   3.150g

 

percentage yield = 87.46 %

 

 

 

 

 

 Try these percent yield questions!

 

 

Mass to Mass

Today we learned about Mass to Mass problems!

Example:

How many grams of O2 are produced from the decomposition of 3.0 g of Potassium Chlorate?


This question gives us grams, but we must change it to grams of another element. First we must write a balanced Chemical equation:

2KClO3 ----> 2K + Cl2 + 3O2


Than we take what we are given, which is 3.0 grams. Change 3.0 grams into moles, than find what you need over what you have, than with the moles you are given, find grams!


3.0 g x 1 mol/122.6 g = 0.024 mol x 3 O2 mol/ 2 KClO3 = 0.036 mol x 32.0 g/1mol
= 1.2 g

Let's try one more!

Example:

Determine the mass of lithium hydroxide produced when 0.38 g of lithium nitride reacts with water according to the following equation:

Li3N +3H2O ---> NH3+ 3LiOH


0.38 g x 1 mol/34.7 g = 0.011 mol x 23.95 g LiOH/1mol = 0.033 g x 23.9 g/ 1 mol
= 0.78 mol


Try some mass to mass problems, too!

mOle to mole conversions

What is a Mole Ratio?  A mole ration is a ration between the amounts in moles of any two compounds found in a chemical reaction. Mole ratios are used as conversion factors between products and reactants in many chemistry problems.

Key things to remember when doing Mole to Mole conversions:
1. Write the chemical equation
2. Balance the chemical equation
3. Understand what the question is asking
4. Now place what you need over what you have. It may not be easy to grasp so here's an example..
If 0.15 mol of methane are consumed in a combustion reaction. How many moles of CO2 are produced?
CH4 + 02 ---> CO2 + H2O

(If you forgot..refer to the steps, for solving mole to mole conversions)


the answer is....
CH4  + 202---> CO2 + 2H2O

Want more problems??

How many moles of bauxite (Aluminum oxide) are required to produce 1.8 mol of pure Aluminum?


Al2O3 ----> Al + O2           




Answer:
2AlO3------> 4Al+3O2






When 1.5 mol of Copper react with Fe (II) chloride. How many moles of Iron should be produced?
2Cu + FeCl2------> 2CuCl+Fe


Answer: 
nothing is changed

Mole to Mole conversions

What is a Mole Ratio?  A mole ration is a ration between the amounts in moles of any two compounds found in a chemical reaction. Mole ratios are used as conversion factors betweenproducts and reactants in many chemistry problems.

Key things to remember when doing Mole to Mole conversions:
1. Write the chemical equation
2. Balance the chemical equation
3. Understand what the question is asking
4. Now place what you need over what you have. It may not be easy to grasp so here's an example..

If 0.15 mol of methane are consumed in a combustion reaction. How many moles of CO2 are produced?
CH4 + 02 ---> CO2 + H2O

(If you forgot..refer to the steps, for solving mole to mole conversions)


the answer is....
CH4  + 202---> CO2 + 2H2O

Stoichiometry (Quantitative Chemistry)

Stoiciometry is a branch of Chemistry that deals with the quantitative anaysis of chemical reactions

• It is a generalization of mole conversions to chemical reactions
• understanding the 6 types of chemical reactions is the foundation of stoichiometry

There are 6 types of reactions we use in stoichiometry, they are...

1. Synthesis (formation)
2. Decomposition
3. Single Replacement (SR)
4. Double Replacement (DR)
5. Neutralization
6. Combustion

SYNTHESIS

A synthesis reaction is when two or more simple compounds combine to form a more complicated one. The general form for this reaction is:

A + B ---> AB

DECOMPOSITON

A decomposition reaction is the opposite of a synthesis reaction - a complex molecule breaks down to make simpler ones. These general form for this reaction is:
AB ---> A + B

Example: 2 H2O ---> 2 H2 + O

 
SINGLE REPLACEMENT

This is when one element trades places with another element in a compound. The general form for this reaction is:

A + BC ---> AC + B (A is a metal)
OR
A +BC---> C+ BA (A is a non-metal)
Example: Mg + 2 H2O ---> Mg(OH)2 + H2


DOUBLE REPLACEMENT

This is when the anions and cations of two different molecules switch places, forming two entirely different compounds. These general form for this reaction is:
AB + CD ---> AD + CB
Example: Pb(NO3)2 + 2 KI ---> PbI2 + 2 KNO3

NEUTRALIZATION

Also known as Acid-base neutralization: This is a kind of double displacement reaction that takes place when an acid and base react. The H+ ion in the acid reacts with the OH- ion in the base, causing water to form. The product of this reaction is some ionic salt and water:
HA + BOH ---> H2O + BA

Example: HBr + NaOH ---> NaBr + H2O

COMBUSTION

Combustion reactions happen when oxygen combines with another compound to form water and carbon dioxide. These reactions are exothermic.

Example:  C10H8 + 12 O2 ---> 10 CO2 + 4 H2O

TEST YOUR KNOWLEDGE ON REACTIONS! (:

Empirical Formulas & Molecular Formulas

* Are the simplest formulas of compounds


Ex. Empirical Molecular
* C4H9 - C8H18
* NO2 - N2O4
* CH7O2 - C3H21O6


* They show only the simplest rations, not the actual atoms.
* Molecular formulas give the actual # of atoms…
* To determine an empirical formula we need to know the ratio of each element. This is why we use the table below; to gather all the information we may need to solve each problem.

Atoms	Mass	Molar Mass	Moles	Mol/ Smallest Mole	Ratio
C	8.4	12.0	0.7	2
H	2.1	1.0	2.1	6
O	5.6	16.0	0.35	1

Step # 1. Fill in the chart with what’s already given to you (the names and masses of the three atoms, plus their molar masses which you can find on the periodic table).

Step #2. Calculate the number of moles simply by following the conversion chart.
* 8.4g x 1mol / 12g = 0.7 mol
* 2.1g x 1mol / 1g = 2.1mol
* 5.6g x 1mol / 16g = 0.35mol

Step # 3. Find the smallest mole (0.35mol in this case) and divide it into itself and the rest of the mole #’s.


* 0.7 / 0.35 = 2
* 2.1 / 0.35 = 6
* 0.35 / 0.35 = 1
* If your answers are not “whole #’s you must multiply everything by a common #. For example:
* Say you end up with the numbers 1.0, 1.5, and 6.0… you would use the number 3 to change 1.5 into a whole # while still keeping the other two numbers whole as well. Thus you would end up with 2.0, 3.0, and 12.

= C2H6O

Atom	Mass	Molar     Mass	Moles	Mole/ Smallest mole	Ratio
Pd	42.56	106.4	0.3992	1
H	0.80	1.0	0.80	2

Step # 3.  You must be given the compound’s molar mass (in this case it’s 216.8 g/mol) in order to figure out its molecular formula.  Besides the molar mass, you also know from your previous calculations that the empirical formula is PdH2. You then calculate the empirical formula’s molar mass (by looking at the periodic table).  In this case it is 108.4.  You can then divide 216.8 by 108.4 to figure out what number to use for the molecular formula’s subscripts.  It is 2… thus you change Pd into Pd2 and H2 into H4.  And there you have your molecular formula!  


Informational and VERY helpful video: