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Week 7 – 10/11 – 10/14

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10/11 – Monday – Columbus Day –

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10/12 – Tuesday – B Day – 2, 3b/4 Lab

Main focus –

a) Test 2 – Voltaic/Electrolytic Cells/Cubic cell calculations/alloys – Review

b) To Continue with particle and mole theory – 2/3a,4

a) Test 1 – makeup  – Review.

b) Particle/mole continues  –                                                                                                                                                                                                                                                                                                           Modern Atomic Theory – John Dalton!

Particle / Mole Theory Presentation:

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10/12 – Tuesday Homework: –

1:  Complete the Mole Concept Lecture (if I did not finish in class)  This will probably be for Period 2 who only had 1 period today.  I will post what time you can start as the lecture below starts from the G(r) eeks!

Please Start the video below at the 20:00 mark to the end to finish what I did with the other class.

2:  Complete the worksheet below by using the lecture below on the Law of Combining volumes and use the Law of Combining volumes tutorial below.  Review with the key or lecture.

Law of Combining Volumes Diagrams.pdf

Law of Combining Volumes Diagrams key new.pdf

3:  Complete the Mole Concept lecture Form below: You will have 2 only 2 submissions.

Mole Concept Lecture

Law of Combining Volumed Tutorial :

Mole Concept Lecture Form:

End Of Tuesday!

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10/13 – Wednesday – A Day – 2/3a Lab, 4

Main focus –

a) Psat today!

I will miss the 2 period day for my 2/3 students 🙁 so do not worry I will review the mole concept ideas that you viewed in my lecture (last night) in class Thursday.

We will however begin the type of calculations and labs that began after the Solvay Conference in 1860 when Canazzaro linked the work of Avogadro and Gay-Lussac to describe the chemical formulas and chemical reactions in the boxes.  Chemistry began and everyone was trying to figure out the chemical formulas of compounds using what we now call Analytical Chemistry.

Law of Definite Proportions = the same compound will always have the same percentage of different elements in it.

This really meant that the same compounds have  the same chemical formulas!!

And the different compounds have different chemical formulas MEANING different ratios of atoms CREATE the different compounds!!!

So the Law of Definite Proportions was a HUGE leap forward in Particle Theory! Thank you Joseph Proust!

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10/13 – Wednesday Homework: –

1. Watch lecture 0 below. In the lecture I am modeling how to complete question 1 on the first side of the  Analytical Chemistry I – determining chemical formulas.pdf worksheet and then you will complete question 2 on the first side on your own. (We are using percent by mass to get chemical formulas!!!).                                                                                                                            When we use the mole concept!

1: AP Lecture : Calculating chemical formulas from percent by masses (that are obtained through labs).

End of Wednesday..

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10/14 – Thursday – B Day – 2, 3b/4 Lab

Main focus –

a) Complete the particle and mole concept in class for period 2 –

Dalton –> Gay Lussac –>Amedeo Avogadro (Canazarro)

b) Determining the percent by mass of compound (hydrate) experimentally.  Our  homework has been based on starting with % by masses to determine the chemical formula (fixed ratio of how many atoms).  We will perform the type of experiment that was done after the Solvay conference which was first step to determining the chemical formulas.

This is the beginning of Analytical Chemistry.

Law of Definite Proportions = the same compound will always have the same percentage of different elements in it.

This really meant that the same compounds have  the same chemical formulas!!

And the different compounds have different chemical formulas MEANING different ratios of atoms CREATE the different compounds!!!

So the Law of Definite Proportions was a HUGE leap forward in Particle Theory! Thank you Joseph Proust!

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

LAB 5 – copper sulfate pentahydrate lab.pdf

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!

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10/14 – Thursday Homework: –

1Please watch lecture 0a  ( I model question 3) and complete question 4 on

Analytical Chemistry I – determining chemical formulas.pdf worksheet. This the other side of the
worksheet that we worked on last night.   Tonight we will use analytical chemistry (discovering                formulas using particle and mole theory in Combustion Analysis.

Combustion (burning):
A redox reaction (imagine that) that uses O2 as the oxidizing agent on a fuel (normally a hydrocarbon)

C?H?   +    O —–>    CO2     +     H2O

Using the law of conservation of mass by Lavoisier and the idea that every unique compound has a unique formula (law of definite proportions by Proust) we can determine the chemical formula of hydrocarbon fuel if we have the mass of the CO2 and  H2O that is produced.

The key to solving these problems are percent by mass!

1: AP Lecture 0a : Calculating chemical formulas from Combustion analysis (that are obtained through labs).

End of Thursday..

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10/15 – Friday – A Day – 2/3a Lab, 4

Main focus –

a) Determining the percent by mass of compound (hydrate) experimentally.  Our  homework has been based on starting with % by masses to determine the chemical formula (fixed ratio of how many atoms) or we had to get the % by mass through combustion analysis.  We will perform the type of experiment that was done after the Solvay conference which was first step to determining the chemical formulas.

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

This is Analytical Chemistry.

1. Today we will either be performing Lab 5 & 6 – period 2/3

Lab 6 – period 4

2. While we collect data we complete the following worksheet:

empiri&molec ditto hydrate combination.pdf

empiri&molec ditto hydrate combination KEY p.pdf

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2/3a, 4 Lab – Lab 5 – Percent mass of water in a Hydrate

LAB 5 – copper sulfate pentahydrate lab.pdf

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!
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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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10/15 – 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

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

3: AP Lecture 1.3 : Stoichiometry and percent yield

End of week 7!