(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!