TYPES OF REACTIONS (1)

There are 6 generlized types of reactions: Synthesis, Decomposition, Single Replacement, Double Replacement, Combustion, and Neutralization.

A: Synthesis
A synthesis reaction is a reaction that combines two or more reactions to form one product.

• general formula: A+B→C

B: Decomposition
A decomposition reaction is a reaction that breakes down one reactant into two or more products.

• general formula: A→B+C

C: Single Replacement
A single replacement reaction is one where an element replaces an ion in an ionic compound.  Metal elements replace positive ions (cations) and non-metal elements replace negative ions (anions).

• general formula: (A=metal) A+BC→AC+B
• general formula: (A=non-metal) A+BC→BA+C

Predicting Single Replacement Reactions
-Some metals are mpre reactive than other metals, and similarly some non-metals are more reactive than other non-metals.

-Using the "Activity Series" -An element higher up on the series replaces the ion below it on the table.

ACTIVITY SERIES:

METALS                                                                                                         NON-METALS
                                                                      Decresing Activity
lithium                                                                         ↓                                            fluorine
potassium                                                                  ↓                                            chlorine
calcium                                                                       ↓                                            bromine
sodium                                                                       ↓                                             iodine
magnesium                                                                ↓                                   
aluminum                                                                    ↓
zinc                                                                              ↓
chromium                                                                    ↓
iron                                                                              ↓
nickel                                                                          ↓
tin                                                                                ↓
lead                                                                            ↓
hydrogen                                                                    ↓
copper                                                                        ↓
silver                                                                           ↓
mercury                                                                      ↓
platinum                                                                     ↓
gold                                                                            ↓


Question:
How to get copper sulphate (CuSO4) if copper is less reactive than hydrogen and cannot replace it out from a acid (cannot react with surfuric acid)?

 

Balancing Equations Jan 25, 2011

Reactant side           Product side
# of atoms        =      # of atoms


Steps of Balancing equations.


1. First Balance the atoms which only occur in one molecule on each side of equation.

2. Balance whole groups whenever possible, rather than considering the atoms of group separately.

3. Be systematic do not jump all over and equation balancing a bit here and a bit there. Balance atoms and groups as you encounter them.

4. Balance atoms which occur in elemental form last. By elemental atoms we mean that the atoms are not combined with any of a DIFFERENT kind.

Example: __Mg(OH)2+ __(NH4)3PO4 →  __Mg3(PO4)2 + __NH3 + __H2O

Step1: Balance Mg.
             
                 3  Mg(OH)2+ __(NH4)3PO4 →  1 Mg3(PO4)2 + __NH3 + __H2O

Step2: Balance PO4 .
                   
               3  Mg(OH)2+ 2 (NH4)3PO4 →  1 Mg3(PO4)2 + __NH3 + __H2O             
Step4: Balance N, H and O.

                3  Mg(OH)2+ 2 (NH4)3PO4 →  1 Mg3(PO4)2 +  6 NH3 + 6 H2O      

Molar Volume of A Gas at ATP

Gases expand and contract (change volume) with changes in temperature and pressure.

We have a standard condition to compare volume of gas called STP (Standard Temperature and Pressure)

STP: 1atmosphere of pressure and a temperature of 0°C or 273.15K.

At STP, 1 mole of gas occupies 22.4L

Thus we can create the conversion factors: 22.4L/mol & 1mol/22.4L

Ex. Calculate the volume occupied by 32Kg of oxygen at STP

32*10^3/16=2000mol

200*22.4=45000L

Dilluting Solutions to Prepare Workable Solution Jan 1st

-Chemicals are shipped around the world in their most concentrated forms.( solids, concentrated acids, etc)
If they were not, we should be shipping lots of water along with the chemicals, which is less cost effective.
-Thus, we need to be able to make solutions of any concentration.
-Mole of solute is constant. ( Difference is that there is more H2O in the less concentrated solution)

Moles solute before= Moles solute after
or in terms of Molarity & Volume in L
M1L1=M2L2
"1"= before      "2"= after

Ex. I have                                                  I need
  1.00L of 16.0M HCl                            0.400L of 2.00M HCl

    M1L1=M2L2

16.0×L1=2.00×0.400
L1= 0.05 L                  0.400-0.05=0.35L
Take 0.05of 16.0 M  HCl and add  0.35L water.

MOLAR CONCENTRATION OR MOLARITY OF SOLUTIONS!!

Molar Concentration:
in a solution: smaller quantity-solute
                    larger quantity-solvent

-The concentration is the amount of solute dissolved in a certain volume of a solution.
-MOLAR CONCENTRATION or MOLARITY is the number of moles of solute in one liter of a solution.  Use "M" to denote the molar concentration and it has the unit of "mole/L"

ex. A 6.00M solution is more concentratied than 1.00M.

• Formulae:

              -Molarity: moles of solute(mol)/volume of solution(L)
               or more simply: M=mol/L

ex. 0.510moles of NaOH in 1.400L of solution.

         M=mol/L    0.510mole NaOH/1.400L=0.364M NaOH

ex. in 1.30L of a 0.75M Ca(OH)2
         0.75*1.30=0.975moles

         0.975*74.1g/1mole=72g Ca(OH)2

                              

CALCULATING THE EMPIRICAL FORMULA OF ORGANIC COMPOUNDS!!!

The empirical formula of an organic compound (any substance that contains carbon) can be found by:
-burning the compound (react with O2)
-collecting and weighing the products
-from the mass of the products, the moles of each element in the original organic compound can be calculated


Ex. 5g of sample→15.0g of CO2 & 8.18g of H2O
     
      CxHy+zO2→xCO2+y/2H2O
     
      this is the balanced chemical equation for the burning of CxHy
      
      the moles of C and H in CxHy=moles of C in CO2 and the moles of H in H2O
     
     15g*(1mole/44g)=0.341mole C      8.18g*(1mole/18.02g)=0.454mole water*2=0.908mole H
      
     0.908 moles of H and 0.341 moles of C must have been in the original substance
CxHy=C0.341H0.908
    
    Carbon: 0.341/0.341=1           Hydrogen=0.908/0.341=2.66*3=8
     
    Empirical Formula=C3H8
      
    *always check the masses

Empirical+Molecular Formula

Empirical formula definition: gives the lowest term ratio of atoms (or moles) in the formula

*All ionic compounds are empirical formulas

Reference from wiki:

http://en.wikipedia.org/wiki/Empirical_formula

Ex. C₄H₁₀ (butane) à molecular formula

àC₂H₅

Ex. consider that we have 10.87g of Fe and 4.66g of O, what is the empirical formula

#1 convert g à mole

Fe 10.87g×1mol/55.8g = 0.195mol

O  4.66g×1mol/16.0g = 0.291mol

#2 divide both by the smallest molar amount

Fe 1    O 1.5

#3 scale ratios to whole numbers

Fe 1×2=2

O 1.5×2=3

Therefore empirical formula = Fe₂O₃

Molecular formula: is a multiple of the empirical formula and shows the actual number of atoms that combine to form a molecule.

To calculate multiple:

      Molar mass of the compound

N=--------------------------------------------------------------------

  Molar mass of the empirical formula

Reference from wiki:

http://en.wikipedia.org/wiki/Molecular_formula

Ex. A molecule has an empirical formula of C₂H₅ and a molar mass of 58g/mol, what is the molecular formula?

MM C₂H₅=12×2+1×5=29g/mol

N=(58g/mol)/(29g/mol)=2

Mf=2(C₂H₅)=C4H10

Ex. A compound contains 7.44g C, 1.24g H, and 9.92g O. the molar mass of the compound is 180g. What is the molecular formula?

C 7.74×(1mol/12g)=0.62  à1

H 1.24×(1mol/29g)=1.24  à2

O 9.92×(1mol/16g)=0.62  à1

EF=CH₂O àMM=30g

N=180g/30g=6 C₆H₁₂O₆(Glucose)