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Week  – 10/18 – 10/22

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10/18 – Monday – B Day – 2, 3b/4 Lab 

Main focus –                                                                                                                                                         
                                                  

     a) To perform Lab 5 and Lab 6 in determining the percent by mass or the empirical formula (in                       hydrates).

     b) To complete the error analysis of the hydrate labs

     c) Complete Mole / Particular concept presentation

Period 2: 

1. HW review – connected the problem concepts with stoichiometry of compounds and stoichiometry of chemical reactions

                        empiri&molec ditto hydrate combination KEY p.pdf
                        View Download

                        Stiochiometry 1 – balance yield key.pdf
                        View Download  

2. Mole concept lesson – diagram, conversions

3. Why we need a mole notes:  H + Cl –> HCl 
    We can use a scale to count!                                                                                                                                                                                 

4. Small movie / STM 

5. Complete the piezo electric discussion – quartz watch                     
                                                                                                             

Period 3/4 

1. Same as above plus

2. Complete Lab 5 and 6, conclusion questions.

                                                                                                                     

Mole Concept Diagram Lesson:
 
A mole is an abbreviated NUMBER that helps chemistry quantify the actual number of particles (atoms, ions, electrons, molecules, etc.) needed to DO chemistry. 
 
We need an abbreviated number because these microscopic particles ARE SO SMALL that we need a TREMENDOUS amount of them to have a significant amount of matter that we can measure with in the microscopic world.  Think how many atoms a mole of Al atoms stacked on each other would be?  (Lab 4!) It was over a 100 trillion meters even though each Al atom has 0.286 nm diameter!!  Light travels at about 10000 trillion meters per second so this distance would extend way beyond our solar system!
 
Please remember that the mole is a “how many quantity” that helps us actually Do Chemistry!!! From the mole concept we are able to:
 
1. Determine the atomic mass of elements (This is how we determined how elements  differed initially!!!).
2: Determine chemical formulas of compounds.
3: Determine the chemical equations of chemical reaction.
4: Predicting quantities like how much of product will we produce given a certain amount of reactants in a chemical reaction.
5:  Predicting quantities like how much of a reactant is needed to make certain amount of product.
6: Determining the quantities (V, P, T, n)  of a gas under changing conditions.
7:  Solve Electrolytic Problems (Amperage, how much metal is deposited on cathode, etc.)
8: Determine a concentration (Molarity = mole / Liter of solution) of a solution of ions etc.
9: Finding a quantity that is directly related to the mole.

The key is here is that we are able to use this How many Number not by counting particles which would be impossible but by USING A SCALE!   Because we decided to make the Relative Atomic Mass the equivalent in grams that 1 MOLE of every element would measure we can use a GRAM Balance or Scale to measure mass AND THEM convert to moles or vice versa.  
 
*OUR WINDOW INTO THE WORLD OF HOW MANY Particle IS THROUGH MASS 
   or volume if a gas.
 
Using moles to answer how many is called Stoichiometry!!!
 
Stoichiometry works because of the Law of Conservation of Mass.
Because of Antoine Lavoisier and the Modern Atomic Theory by John Dalton, atoms are neither created nor destroyed in chemical reactions and are just rearranged into different ratios of elements in compounds.  These ratios are a How Many Value!
 
Stiochiometry also works because of the Law of definite proportions.
Because of Joseph Proust and the Modern Atomic Theory, each compound has its own unique arrangement of atoms called a chemical formula which results in each unique compound having a unique percent by mass of each type of atom (element) in its molecule.  This is due to compound having a unique ratio of atoms in their chemical formulas.  Chemical formulas are a How many Value!
 
Since chemical reactions are a result of atoms, molecules, and ions colliding as individual particles then to calculate the outcomes of reactions we must consider the moles (THE HOW MANY VALUE) of the chemical species.

 

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10/15 – Friday Homework: –  Lab 5 & 6 we will complete in class Monday!

1.  Please complete the empiri&molec ditto hydrate combination.pdf worksheet
                                                                                                                                                                                                                              empiri&molec ditto hydrate combination.pdf
2. Review with the key below:                                                                                                                                                   
empiri&molec ditto hydrate combination KEY p.pdf
View Download
 
3: Please view Lecture 1.3 with the worksheet below.
     I will walk you through all questions except number 4.                                                                                                                                                          
4. Complete Worksheet Stiochiometry 1 – balance yield.pdf (number 4) and review with key below:
 
Stiochiometry 1 – balance yield.pdf
View Download
 

3: AP Lecture 1.3 : Stoichiometry and percent yield

 End of week 7!

Particle / Mole Theory Presentation:

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3b/4 Lab – Lab 5 – Percent mass of water in a Hydrate

LAB 5 – copper sulfate pentahydrate lab.pdf
View Download

Lets look at an ionic crystal that has no water called annhydrate.  
 
Lets look at copper (II) sulphate : CuSO4  :
 

 This image was build by X-ray imaging of salt.

Now it is more complex than the crystal above for NaCl because we have a polyatomic ion in the crystal.

The Yellow is the Sulfur attached to 4 oxygens (red).  The brown color (i think its brown) is the copper.

Notice regular repeating pattern.
The sticks represent the bonds or the attractions between the ions.

Its hard but if look inside the crystal there is one Cu per sulfate ion.
Lets look at the hydrate of the same salt : copper (II) sulphate pentahydrate  :
 
  Notice the Dot between anhydrate and water.  This dot means “WITH” and not to multiple.  Thus there are exactly 5 water molecules for every 1 Cu+2 and  1 SO4-2 in the crystal.
 

 You will notice that water (it has 2 white hydrogen atoms) molecules are situated inside the crystals at particular regions in the crystal in exact ratios.  This is a fixed ratio (stoichiometric ratio).

It is hard but you can see the 5 water molecules per 1 copper ion and 1 sulfate ion.

What makes it hard is that the crystal repeats in all directions.

The water can be removed from the salt by heating it.
                                                 CuSO4 • 5H20   ——->    CuSO4  (s)     +      5H20  (g)                                                                                                                               Heat                                         
                                                                                                                                                                                                                 Today we will be able to do the following:           
         a) Determine. the mass of your annhydrate (dried hydrate)
        b) Determine the mass of the water (THAT was in your crystal)
        c)  calculate the experimental percent by mass of water:
 
            % by mass =  Mass of the part (mass of water missing)         x     100  
                                    Mass of the whole (mass of original hydrate)
 
       d)  calculate the Theoretical percent by mass of water using the chemical formula of the hydrate:
                 
                                                                                    CuSO4 • 5H20
 
                    Given the formula above, determine the percent by mass of water of the hydrate:
 
                    Assume you have 1 mole of CuSO4 • 5H20:
 
                                Thus in 1 mole of the hydrate you have 5 moles of water:
 
                                For every one (CuSO4 • 5H20) you have 5 moles of water.
 
                                                                            1               :               5
                                               Ratios of how many!!! This is why we need a mole theory!                    
                                             % by mass =  Mass of the part (mass of 5 moles of water)         x     100  
                                                   Mass of the whole (mass of 1 mole of hydrate)
 
*Mass of 1 mole of any compound (with a unique chemical formula) is the sum of the atomic masses of all atoms in 
                                                                      the formula or molecule. You need the periodic table!    
 
___________________________________________________________________________________ 
Lab 6 – Determination of the Empirical formula of a Hydrate

               *Determine the chemical formula of a hydrate:  

LAB 6 – Empirical Formula analysis of Hydrate.pdf

Because of the conservation of mass by Antoine Levassuer, the mass of  water that is released from the crystal (hydrate) is the same mass that was is in the crystal.  THus the moles of water lost is the water that are in the crystal (in the formula ) .  

 
Formulas of compounds are fixed! (Thanks Joe Proust – Law of definite proportions)
 
Once we determine the mass of water lost, we at the same time determine the mass of remaining part of the salt called the annhydrate.  Convert grams of both parts of the hydrate to moles  and then get a ratio of how many (moles) by dividing by the lowest number of moles.  That ratio is what we will use for the fixed formula! 
 

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10/18 – Monday Homework: –  Lab 5 & 6 are due!

1.  Lab 5 & 6 are due for period 4. 
 
2.  Precipitation Analysis:  Now that we can determine chemical formulas from heating (Lab 5) or from combustion analysis (Analytical Chemistry I – determining chemical formulas.pdf) we can also do so by the number of dry grams of precipitate, as long as we use stoichiometry!
 
a)Watch Lecture 1.1  Below and and follow along with me with question one as we calculate the Molarity of the individual ions in the solution using stiochiometry.
 
b) Watch Lecture 1.2 and follow along with me to complete questions 4, and 5 as we use precipitation and stoichiometry to determine the chemical formulas of the salts (ionic compounds).
 
c) Complete question 5 on your own and review with the key below. This question puts it all together!
                                                                                                                                                                                                                    Analytical Chemistry II – Mol,Dilution, precipitation analysis.pdf
View Download 
 
Analytical Chemistry II KEY NEW – Mol,Dilution, precipitation analysis.pdf
View Download 

                                                                                                                                                                                                                                                                                                                                                                                                                                                 

a) Lecture 1.1 : Molarity of ions

b) Lecture 1.2 : Preciptation analysis

End of Monday..

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10/19 – Tuesday – A Day – 2/3a Lab, 4  

Main focus –                                                                                                                                                         
                                                 

     a) To complete Hydrate Labs and the error analysis of the hydrate labs

b) To begin the conservation of mass presentation – balancing chemical reactions,                                           Stoichiometry  of  chemical reactions using the mole ratios that we get from balancing

Stoichiometry works because of the Law of Conservation of Mass.
Because of Antoine Lavoisier and the Modern Atomic Theory by John Dalton, atoms are neither created nor destroyed in chemical reactions and are just rearranged into different ratios of elements in compounds.  These ratios are a How Many Value!
 
Stiochiometry also works because of the Law of definite proportions.
Because of Joseph Proust and the Modern Atomic Theory, each compound has its own unique arrangement of atoms called a chemical formula which results in each unique compound having a unique percent by mass of each type of atom (element) in its molecule.  This is due to compound having a unique ratio of atoms in their chemical formulas.  Chemical formulas are a How many Value!

Period 2/3: 

1. HW review – Stoichiometry of ions, precipitation analysis

 Analytical Chemistry II KEY NEW – Mol,Dilution, precipitation analysis.pdf
 View Download  

2. Why we need a mole notes:  H + Cl –> HCl 
    We can use a scale to count!                                                                                                                                                                               
3. Conservation of Mass —-> Balancing provides the mole ratios needed to predict outcomes in chemical reactions. But these ratios of how many also provide chemistry the amounts needed to make the most efficient reactions = GOOD CAKE!

                                                                                                                                                                                                                                A.  Flaming Flask Demo:

Cyclohexane combusts more completely when poured out because the ratio of cyclohexane to oxygen is 1 : 9.           We get those ratios when we balance.   
 
1 C6H12  (l)  +    9 O2 (g)    —–>   6 CO2 (g)   +   6 H2O (g)    +   Energy
 
That means 9 O2 molecules per 1 C6H12  is needed to maximize the reaction.  The cyclohexane was not getting the right ratio in the flask but achieved the right ratio when it was poured out.
         
B. Hydrogen Balloon Demo: 
 
2 H2 (g)  +  O2 (g)  —>   2 H20  +   Energy 
                                                                                                             

This means  2 H2 molecules per 1 O2  is needed to maximize the reaction and thus maximize the energy released. If this ratios in the reaction are not  2 to 1 then the reaction will be less efficient with some unreacted reactants and a lower amount of energy released.  Sometime the reaction will not even run at all if the ratios are poor.

Period 4 

1. Same as above except will complete the LGI Hydrogen demo tomorrow                                                                                                                    

 

Conservation of Mass Presentation:

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10/19 – Tuesday Homework: – 

1.  Lab 5 & 6 are due for both classes. Lab 6 just has calculations and no conclusion but Lab 5 does have a conclusion.  Do not just list errors. Completely explain why your experimental outcome was lower or higher than the theoretical.  
 

2. Please complete the take-home quiz ( Analytic Chemistry Quiz 1 )ON the worksheet that was given out today                               or posted below:

   
           Analytical quiz 1 – Hydrate, precipitation, and Stioch.pdf
           View Download
 
3. Complete the Form below that will allow to place your answers into the form below.  The form will be on auto-grade to grade your responses.  You may submit as many as 3 times as you would like BEFORE 9:00 pm.  After 9:00pm you will be given 1 more response to submit.  The form after 9:00 pm will not be auto-grade.  I will send out the final grade with a hand written key at 10:00 pm.

                                                                                                                                                                                                                    

Analytic Chemistry Quiz 1 Form:

End Of Tuesday!

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10/20 – Wednesday – B Day – 2, 3b/4 Lab  

Main focus –                                                                                                                                                         
                                                

     b) To begin the conservation of mass presentation – balancing chemical reactions,                                           Stoichiometry  of  chemical reactions using the mole ratios that we get from balancing

Period 2: 

1. HW review – Form Review

           Analytical quiz 1 – Hydrate, precipitation, and Stoich KEY.pdf
           View Download

2. Why we need a mole notes:  H + Cl –> HCl 
    We can use a scale to count!                                                                                                                                                                               
3. Conservation of Mass —-> Balancing provides the mole ratios needed to predict outcomes in chemical reactions. But these ratios of how many also provide chemistry the amounts needed to make the most efficient reactions = GOOD CAKE!                      

4.  Flaming Flask demo

5.  Hydrogen balloon analysis from yesterday. When did we make GOOD CAKE? 

A.  Flaming Flask Demo:

Cyclohexane combusts more completely when poured out because the ratio of cyclohexane to oxygen is 1 : 9.           We get those ratios when we balance.   
 
1 C6H12  (l)  +    9 O2 (g)    —–>   6 CO2 (g)   +   6 H2O (g)    +   Energy
 
That means 9 O2 molecules per 1 C6H12  is needed to maximize the reaction.  The cyclohexane was not getting the right ratio in the flask but achieved the right ratio when it was poured out.
         
B. Hydrogen Balloon Demo: 
 
2 H2 (g)  +  O2 (g)  —>   2 H20  +   Energy 
                                                                                                             

This means  2 H2 molecules per 1 O2  is needed to maximize the reaction and thus maximize the energy released. If this ratios in the reaction are not  2 to 1 then the reaction will be less efficient with some unreacted reactants and a lower amount of energy released.  Sometime the reaction will not even run at all if the ratios are poor.

Period 4 

1. Same as above except will complete the LGI Hydrogen demo. 

2. Complete the Conservation of Mass Presentation

3.  Lab 7 – Collect data                                                                                                                   

 

Conservation of Mass Presentation:

Period 2 : Hydrogen Balloon Explosion Comparison:

2 H2 (g)  +  O2 (g)  —>   2 H20  +   Energy 

Period 4 : Hydrogen Balloon Explosion Comparison:

2 H2 (g)  +  O2 (g)  —>   2 H20  +   Energy 

______________________

3b/4 Lab – Lab 7 – Synthesis of Copper Iodide (and determination of the empirical Formula of copper iodide)

Lab 7 – Synthesis of Copper Iodide.pdf
View Download

This Lab will be our first formal lab.  I am giving out a data worksheet to keep you organized in your data collection but it will not be handed in.  You will write a data section and calculation section in your formal lab write-up that will be in a google doc that will be shared to you in next couple of days.

 Cu (s)  +  I2 (g)  —>  Cu? I?  

 Objectives:  1. To determine the empirical formula of the salt (product).

                        2. To determine the percent yield of the salt.

 We will take notes on how to perform this lab.

Lab 7 : Synthesis of Copper Iodide

______________________

10/20 – Wednesday Homework: – 

1. Please View Lecture 1.5 below on Molar stoichiometry.   I will model how to complete the first side of the worksheet and then you will complete the backside on your own and review with the key.
                                                                                                                                   
Stiochiometry 3 – Molarity limiting reagent.pdf
View Download
 
Stiochiometry 3 – Molarity limiting reagent Key.pdf
View Download

Lecture 1.5 : Molarity Stiochiometry

End of Wednesday..

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10/21 – Thursday – A Day – 2/3a Lab, 4 

Main focus –                                                                                                                                                         
                                                

     a)  To complete the conservation of mass presentation – balancing chemical reactions,                                   Stoichiometry  of  chemical reactions using the mole ratios that we get from balancing –                         applications to Challenger and Hindenburg

    b)  To perform an “adding mass lab” – Lab 7

Period 2/3: 

1. Complete the Challenger and Hindenburg discussion.

2. Flaming Flask demo and review of the balloon demos from the video below.         

3. Lab 7 – Note-taking setup/perform –
                                                                                                                                                                       

 Period 4: 

1. Complete the Challenger and Hindenburg discussion.

2. Review of the balloon demos from the video of both classes.

3. Lab 7 – Note-taking setup –

                                                                                                                                                                                                                          A.  Flaming Flask Demo:

Cyclohexane combusts more completely when poured out because the ratio of cyclohexane to oxygen is 1 : 9.           We get those ratios when we balance.   
 
1 C6H12  (l)  +    9 O2 (g)    —–>   6 CO2 (g)   +   6 H2O (g)    +   Energy
 
That means 9 O2 molecules per 1 C6H12  is needed to maximize the reaction.  The cyclohexane was not getting the right ratio in the flask but achieved the right ratio when it was poured out.
         
B. Hydrogen Balloon Demo: 
 
2 H2 (g)  +  O2 (g)  —>   2 H20  +   Energy 
                                                                                                             

This means  2 H2 molecules per 1 O2  is needed to maximize the reaction and thus maximize the energy released. If this ratios in the reaction are not  2 to 1 then the reaction will be less efficient with some unreacted reactants and a lower amount of energy released.  Sometime the reaction will not even run at all if the ratios are poor.

 

______________________

2/3a Lab/4 – Lab 7 – Synthesis of Copper Iodide (and determination of the empirical Formula of copper iodide)

Lab 7 – Synthesis of Copper Iodide.pdf
View Download

This Lab will be our first formal lab.  I am giving out a data worksheet to keep you organized in your data collection but it will not be handed in.  You will write a data section and calculation section in your formal lab write-up that will be in a google doc that will be shared to you in next couple of days.

 Cu (s)  +  I2 (g)  —>  Cu? I?  

 Objectives:  1. To determine the empirical formula of the salt (product).

                        2. To determine the percent yield of the salt.

 We will take notes on how to perform this lab.

Lab 7 : Synthesis of Copper Iodide

______________________

10/21 – Thursday Homework: – 

                                                                                                                                                                                                                                     1. Please the Flaming demo video below.  This demo is what the calculations in the preceding lecture are based.

What is the Limiting Reagent in this demo?  Why isn’t the flame coming out of the tube and burning my room into oblivion?

2. Please view the Limiting reagent Lecture that teaches the chart method of calculating the limiting reagent. Complete side one with me. How could this method made last night’s homework easier?                                                               

Limiting reagent intro demo new worksheet.pdf

1 : The Flaming Tube Demo – for the lecture below!

2 : Limiting Reagent Lecture –  Teaches the chart method!

End of Thursday..

 

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10/22 – Friday – B Day – 2, 3b/4 Lab

Main focus –                                                                                                                                                         
                                                

   a)  To introduce Limiting Reagent concept in class and review the “chart method”.

b)  To perform an “adding mass lab” – Lab 7 – Synthesis of copper iodide

Period 2: 

1.  We will review the concept of limiting reagent through the chart method, using a previous homework problem that Liam said was hard.

2. Flaming tube demo.

3. Lab 7 discussion of the write-up requirements.

                                                                                                                                                                       

 Period 3/4: 

1. Same as as number 1. and 2.

2. Lab 7 note-taking and perform experiment. 

 

Classwork today:

Please complete Stiochiometry 3 – ICE Table Molarity Salsa.pdf  which is really the second problem the homework Last night (Stiochiometry 3 – Molarity limiting reagent.pdf).  So please redo that limiting reagent problem using the ICE Table method. Don’t be mad that this way is simpler!!!                                                                                                                                         
                                            *Remember that moles will be calculated through Molarity (M * V(in Liters) = moles.                                                                      
                                                                     Molarity ( mole / Liter )   X   Volume (Liter)  = mole
 
Stiochiometry 3 – ICE Table Molarity Salsa.pdf
View Download
 
Stiochiometry 3 – ICE Table Molarity Salsa key BEST p.pdf
View Download

 

______________________

2, 3b/4 Lab – Lab 7 – Synthesis of Copper Iodide (and determination of the empirical Formula of copper iodide)

Lab 7 – Synthesis of Copper Iodide.pdf
View Download

This Lab will be our first formal lab.  I am giving out a data worksheet to keep you organized in your data collection but it will not be handed in.  You will write a data section and calculation section in your formal lab write-up that will be in a google doc that will be shared to you in next couple of days.

 Cu (s)  +  I2 (g)  —>  Cu? I?  

 Objectives:  1. To determine the empirical formula of the salt (product).

                        2. To determine the percent yield of the salt.

 We will take notes on how to perform this lab.

 

______________________

10/22 – Friday Homework: – 

                                                                                                                                                                                                                                    1. Complete the backside of the Limiting reagent intro demo new worksheet.pdf (questions 3 and 4 ONLY) that was homework (front side) from Thursday night using the video posted below.  You could try these 2 questions on your own first and then view the video if you are feeling adventurous!

Limiting reagent intro demo new worksheet.pdf with the video posted below.  Start the video at 16:26 as you are not held accountable to questions 1 and 2 yet. We will be using the concept of limiting reagent in different type of AP Multiple Choice Problems. 
 
Back side (questions 3 and 4) of the following worksheet:
Limiting reagent intro demo new worksheet.pdf

2. Complete the Short Form from a Tall Teacher below:  This form is based on a recent free response question. You have 3 submissions to this form. Use the ICE Chart!

3.  Complete the Lab 7 calculations on the handout I gave you. 

If you need help with Lab 7 I have posted a review video below the weekends form.

You should be able to calculate or determine the following:

a) Mass of the copper reacted

b) Mass of the iodine reacted

c) Mass of the product

d) empirical formula for the salt

e) Calculate the theoretical amount of product (salt) using the empirical formula (that you found).

f) Calculate a percent yield.

1 : Side 2 of Thursday nights homework –  Teaches the chart method!

End of Thursday..

 

 2: Limiting Reagent Form – A short form

 

3 : Lab 7 review –  This might help with your calculations!

End of week 8!