### SUMMER INSTITUTE – Module 7 – Stoichiometry 2

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Module 7 – Stoichiometry 2 –

5 hours – due August 13th!

Link to the Module 7 worksheets:

Notes for this section is Test 2 study guide that we will use during the year: link to test 2 study guide

* Connections

So we have seen the importance of the mole concept in shaping chemistry and its effect on gas laws and in building empirical (lowest ratio of atoms) and molecular formulas (actual ratios of  atoms in molecules) from activity 6 in the last module.  We were able to to write chemical formulas because we had the ratio of moles (how many) for each element in the compound.  We determined this from percent by mass data that was collected.

In question 1 and 2 of the worksheet of the last module we used percent by mass data from unknown compounds.  That mass data was converted into moles and ratios (divide by smallest moles) gave us whole number ratios!! Another piece of evidence that atoms existed!!  Whole number ratios equals whole number of atoms!! The question is how do we get percent by mass data.  Do not forget that a percent is a part / whole x 100.

The whole will always be the Total Mass of some quantity. What quantity could we use?  Hmmm?? Yes you guessed it, 1 mole.  The mass of one mole is called the formula mass/molecular mass. We need that value to convert from mass to moles like we did in the above questions so that we convert a how heavy number (mass) into a how many number (moles).  This should makes sense to you because a how many number is what a chemical formula is.  There are  2 sodium (Na) atoms for every 1 sulfur (S) atom for every 3 oxygen (O) atoms.

We obtain percent by mass data through analytical chemistry experiments,  which we will do several in AP Chemistry laboratory activities.  In regents chemistry, we would typically accomplish this with the hydrate lab.

Activity 1:  SKILL : Nomenclature of salts

Read the notes on how to name and write chemical formulas for ionic compounds:

Unlike the math problems above which use the percent by mass of individual elements, we will determine the percent by mass of a compound (water) in an overall compound (Hydrate).   A hydrate is a salt, which we will learn is a crystal of positive and negative ions that has water molecules positioned in the crystal in regular repeating way.  This means that there is a fixed ratio of water in the crystal of positive and negative ions. We call this fixed ratio a stoichiometric ratio.

Lest look at a simple crystal of an ionic compound (salt).  I have a model of this on my desk.  Notice in the graphic they use element symbols when they should use charges next to the elements symbols.
Ionic compounds are positive ions that attract negative ions. THEY DO NOT MAKE MOLECULES!
They make CRYSTALs that repeat. We use EMPIRICAL formulas for these salts to show the lowest ratio of the ions in the crystal.  For example we have Na+ and Cl ions attracting themselves in the crystal formation below:

Ions are formed when atoms gain electrons (become negative) and atoms lose electrons (become positive). We write the formula of this compound to be NaCl (1 Na to 1 Cl) even though there are more than 1 Na and Cl ions in the crystal.  We use an empirical formula for all ionic compounds because they all make crystals!! If we did not we would have to write the exact number for every different crystal shape?  Thats walnuts!!

Notice that a +1 and -1 cancel out to zero. All chemical compounds are electrically neutral.

Now lets look at another salt (ionic compound) but this time we will not just use monoatomic ions like Na+ and Cl but instead use a polyatomic ion.  This is a cluster of atoms the collectively have a charge. I will explain the how a little later in the course.

NaNO3 example: Notice this is another example of a regular repeating crystal of ions like the example for NaCl above EXCEPT that the +1 will be from a the singular ion (fuscia = Na+) and the negative ion will result from a group of atoms bonded (polyatomic ion) called a nitrate ion  (NO3 -1 = 1 nitrogen and 3 oxygens bonded). Notice that 1 positive ion = 1 negative ion. Both NaCl and NaNO3 are examples of SALTS which are crystals of ions at room temperature.

In regents chemistry there is a table that I call , “what the heck is that Table” which is Table E that lists all of the polyatomic ions. Notice that these are clusters of atoms that have their own name: For example in NaCl we call that  sodium chloride   – (IDE if just single atom) If we have NaNO3 we call that sodium nitrate  – (nitrate is the name of  NO3-1) Notice that Na is the formula is really Na+1and it attracts 1  NO3– 1 cluster or polyatomic ion so that a +1 and  -1 = 0 What about CuSO4 ? What must be the charge of Cu (copper) if the formula is written this way?Answer in the bottom of the page.
Remember that formulas of ionic compounds are just the lowest ratios in the crystal and this ratio depends on the charge of the all the ions to be zero.

How would we write a formula that needs more than one polyatomic ion?  For instance how would we write a formula of ionic compound that has more than one ion?  We use subscripts which are small numbers that tell us the mole ratio of ions in the formula.

Example:  What would be the formula of aluminum chloride?

In your periodic table you look up aluminum and Chlorine you will find that in the upper right hand corner the charge that aluminum and aluminum like to become:  Al likes to become +3: (by losing 3 electrons) This charge tells us that Al likes to become Al+3 when it reacts (loses 3 electrons)Because the ending of the chemical name ended in – ide we know it was a single monatomic ion. Cl likes to become -1: (by gaining an electron)*It can become +1, +5, and +7 but if Al CAN ONLY BE +3 then Cl must be -1 The ions involved:  Al+3  and Cl-1  thus we need 1 Al+3 per 3 Cl–     We write this this way:   AlCl3   (subscript 3 means 3 Cl ions) Notice that 1 Aluminum ion (+3)  and 3 Chloride ions (-1 each) must equal zero.What is formula of sodium oxide?  Answer in the bottom of the activity.
Look at the empirical formula that we found in question 1 at the top of the page: Na2SO3
This formula of the salt was obtained through the calculation in the connections before activity 1.

Is this a possibility? Does this exist? Yes of course.  We know Na like to become +1 and that SO3
has a negative 2 charge from “what the heck is that table (Table E)”.

Two Na+1  and 1 SO3-2  =  zero overall charge.

How do we deal with multiple polyatomic ions?

For instance if we have the chemical name: aluminum nitrate

What would be the chemical formula?

Aluminum likes to become +3 (from periodic table) and Nitrate (from table E) is -1 thus

1Al+3  and  3NO3-1    to make  Al(NO3)3

Parentheses are needed to show there are 3 nitrate ions (3 of the cluster).  Without them the chemical formula is incorrect (and does not balance to zero).

How many moles of oxygen atoms are in 1 mole of aluminum nitrate?  answer at bottom of activity.  How do we handle multiple charges that the positive ion can have? For instance copper because it can be +1, and +2 it can make 2 compounds with the nitrate ion and so can nickel. Some metals can become more than one charge. It can be CuNO3 when Cu is +1 and it can beCu(NO3)2 when Cu is +2 Since they have different chemical formulas they will have different properties and thus need 2 different names. They both cannot be copper nitrate is there are 2 versions? The way to get round this is to use something called the stock system:If the positive ion (always the first written in a chemical formula) has multiple positive charges we write a Roman Numeral to identify the charge of the first element (so we know which one the formula is using). Thus CuNO3 is copper (I) nitrate and Cu(NO3)2 is copper (II) nitrate. THE ROMAN NUMERAL IS THE EXACT CHARGE OF THE 1st ELEMENT! Why didn’t we call Al(NO3)3 aluminum (III) Nitrate?Because it only had 1 positive choice thus there was no need!!!

What is the name of Ni2O3?  Answer at the bottom of activity 1

So we have the following rules for naming SALTS (Ionic compounds):

1. All ions must equal zero and they must be the lowest Ratio since they are ions in a crystal!

Mg+2 and O-2 will become MgO and never Mg2O2!
Its is the lowest ratio of ions in a crystal!!  Empirical Formula always!!!

2.  The positive ion will always go first.

NaCl not ClNa EVER!!!!

3.  If the last ion is polyatomic (table E) the name ends with the name of that polyatomic ion.

NaNO3 is sodium nitrate.

4.  If the last ion is monoatomic (binary compound) then the name ends with -IDE ending.

NaCl is sodium Chloride

5. If the first ion (positive) has multiple positive ions it can become than it needs a
Roman numeral to  represent the positive charge that the element is using.

CuO  is copper (II) oxide

6. You can have salts made up polyatomic ions (TABLE E):

NH4+   – ammonium ion,    NO3-1  – nitrate ion

NH4NO3   =  AMMONIUM NITRATE

7.  CHECK TABLE S FOR NAMES OF ELEMENTS FOR EACH SYMBOL

Assignment for Activity 1 below:

1. Complete page 1 and 3 ONLY of the packet below.

UIC naming 4 pages – ditto 2.pdf

Table S (element name and symbols) and Table E (what the heck is that!) are your friends here!

2. To make sure you are on  the right path I posted a partial written key for the assignment.

IUC KEY partial key page 1 and 3 p.pdf

3. If you would rather watch me model a few of these problems please do so below:

I will collect these worksheets in September with your labs that you have completed.

3 : Lecture 3.0 – Naming and Writing Formulas for Binary Salts

3 : Lecture 3.1 – Naming and Writing Formulas for Ternary (polyatomic) Salts

Charge of Copper :+2

Formula of sodium oxide: Na2O

Moles of oxygen: 9 moles of oxygen ( 3 nitrate ions – each one has 3 oxygens thus 3 x 3 = 9)

Chemical Name of  Ni2O3  :  nickel (III) oxide  (monoatomic ion = ide, nickel has more than 1
positive charge.  We need 2 Ni+3 to equal 3O-2 to equal zero charge!

Activity 2:  SKILL : Percent Water in a Hydrate.

Let’s look at an ionic crystal that has no water called an anhydrate.  All the formulas that we wrote and named in skill 1 were anhydrates:

Let’s 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 ion. Notice regular repeating pattern.The sticks represent the bonds or the attractions between the ions. If you look carefully inside the crystal there is one Cu per sulfate ion.

Lets look at the hydrate of the same salt : copper (II) sulphate pentahydrate  :

 CuSO4  ·  5 H2O 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.  Stiochiometric ratios 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.
Assignment for Activity 2 below:

1. View the animations and read notes above.
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2. Use the video (Diehl Lectures 1) below to start the worksheet. Complete the rest of worksheet and review with the key.

Hydrate intro 1213 regents.pdf

Hydrate intro 1213 regents key complete.pdf

3.  Please watch only from 24:50 and on from the following lecture (Diehl Lectures 2) and complete question 7 and 8 in the following worksheet.  Review with the key.

worksheet:
empiri&molec ditto hydrate combination 1314 .pdf

Key:
empiri&molec ditto hydrate combination regents 1314 COMPLETE key .pdf

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Activity 3:  SKILL: Hydrate assessments:

1. Please answer the following questions below from the Hydrate – regents 2015.pdf worksheet and write into the auto reply Form below:

Hydrate – regents 2015.pdf

2.  Please complete the RAT 38 Form – Institute below and fill in with auto reply form below it.

Readiness Assessment Test 38 – math 1 .pdf

Activity 4:  SKILL: Conservation of Mass – Stoichiometry of Chemical reactions an identifying reactions.

Connections: We have spent some time looking at the ratios of atoms in chemical formulas. In the percent of a hydrate problem we have introduced this concept.  How are we able to figure out the chemical formula of a hydrate if amount of atoms and type atoms are different in the reactant side and the product side. If that is the case then atoms are destroyed and created in chemical reactions AND THAT CANNOT BE THE CASE!  One of Daltons’s postulates in his Modern Atomic Theory is that in a chemical reaction atoms are just rearranged into new combinations from the starting compounds (reactants.)

MgCO3 · 5H2O  (s)   →    MgCO3 (s)    +    5 H2O (g)

hydrate                                 anhydrate             +          water

Notice that the 5 water molecules released as gas came from reactant that also had 5 water molecules.
Notice this is the case for every individual type of atom.

For example there are 8 oxygens atoms in the reactant side (left of arrow) and 8 oxygen atoms in the product side.  This is true for magnesium and carbon atoms as well.  Knowing that the same type of atoms and their moles must be the same on each side of the equation.  In math 3x = 9 because each side must be equivalent.
In Chemistry, both sides of a chemical equation must be equal with the same number and type of atoms. We use an arrow instead of an equal sign but concepts are similar.

This must be the case and for all of our Analytical Chemistry Experiments that we figure out chemical formulas we need chemical reactions to be balanced this way.  The power in chemistry comes from the mole concept and and the particles of atoms that rearrange in chemical reactions (balanced chemical equations).

1.  Please complete question 1 in the worksheet below and review with the key.

Balancing and identifying reactions 2014.pdf

Balancing and identifying reactions Key 2014.pdf

2.  Follow along with me during the rest of the lecture (Diehl Lectures 3) with the worksheet above. You can see all of the concepts that we have learned are continuing throughout this next set of skills.

2 : The Diehl Lectures 3 – Balancing and Reaction Identification

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3.  Please watch the the 2 demo below and read the notes for each. Notice the reaction only works when the correct stoichiometric ratio of oxygen is reached in the Pringles can.

2 H2 (g) +  O2 (g)   –>   2 H2O (g)  +   Energy

*Remember that we originally obtained the chemical formula for water BEFORE the mole concept?  We used gases and their property that the same volume = same number of particles (Avogadro’s hypothesis) to obtain the formulas AND THEIR RATIOS!! Pringles Can Demo:

A Pringles Can is filled with hydrogen gas, H2 by a tube that carries the very light gas (lighter than air) from the reaction vessel.  There is a hole on top of the can (that is upside down)  and a hole on the bottom of the can.  I filled the can with hydrogen from the bottom and it filled the can upward because it is lighter than air.  The hydrogen flushed out the atmospheric air out of the container leaving only pure hydrogen. I disconnected the lower tube and lit the pure hydrogen on the top.  As the pure hydrogen burns Air with oxygen is pulled into the can from the bottom hole and an explosion only occurs when the ratio of hydrogen to oxygen reached 2 : 1.  Thus there is a delay when the hydrogen is lit UNTIL the correct ratios are reached. We get these ratios (or ingredients) when we balance a chemical equation!

Hydrogen Balloon Explosions Demo:

I fill 2 balloons. One with 100% hydrogen and one with 66.6% Hydrogen and 33.3% (2:1 Ratio) and compare the explosions.  Remember from reaction of hydrogen and oxygen above, that Energy is given off exothermically in the reaction. If more energy is given off then the reaction is more efficient due to the correct ratio of particles that is needed to react is present.  When ratios are not ideal then reactions are slower and sometimes do not react at all with a lot of waste.

Activity 5:  SKILL:  Identifying reactions and Balancing Reactions

1. Please follow along with me with the lectures below that complete the backside of the worksheet in activity 4.
The video is broken into 2 parts. Once you get the hang of the skills, please feel free to go ahead and then review with the key below.

Balancing and identifying reactions Key 2014.pdf

Part 1 Review of backside of Balancing and identifying reactions Key 2014.pdf

Part 2 Review of backside of Balancing and identifying reactions Key 2014.pdf

2. Complete the new worksheet below and  review with the key.

Balancing Chemical Reaction Concoction 2015.pdf

Balancing Chemical Reaction Concoction 2015 KEY.pdf

Activity 6: SKILL: Use Mole ratios of chemicals in a chemical reaction to predict the quantity of another chemical in the balanced chemical reaction.

*Connections – Now that we have used the mole concept to build chemical formulas (empirical), learned how to name chemical compounds, identify type of chemical reactions, predict products and balance, we are ready to use balanced chemical reactions to calculate with.   Think about what a balanced chemical formula does for us.  It gives us the mole ratios (Coefficients) that determine how much of one substance (element or compound) reacts with another to produce a certain amount of product(s).  It is very much like a recipe!  How much of the reactants do we need to make a certain amount of product. (How much initial ingredients is needed to make a certain size cake.)  Remember the tank sizes for liquid oxygen and liquid hydrogen.  Why were they in a 1 : 2 ratio?

Not only can we predict the products (in activity 4 & 5), we can also calculate how much of the product will be made given a certain amount of the reactants (initial ingredients).  We can also calculate the amount of reactants needed to make a certain amount of products (cake).  This is Stoichiometry! What makes it different from what we have done so far is that we are using information from one chemical to get information about another.
Stoichiometry is something we will use throughout the entire AP course.

1.  Please view the introductory lesson on Stoichiometry problems (Lecture 5.13R).  In the video I modeled the last problem in 3 separate steps but you can use the dimensional analysis (module 1) to do it using your units in one large step. The key below uses dimensional analysis.

2. Complete the first side, except question 4 and then complete questions 2 & 3 from the backside and review with key. Try to use dimensional analysis instead of solving for each step separately.

stoichiometry ditto 1 0607.pdf

stoichiometry ditto 1 KEY 0607.pdf

3.  Please view the lecture below (Stoichiometry Homework) that reviews question 6 on the from page of the worksheet below and review dimensional analysis in all the many different problems on the back page. Please start the lecture at 3:16 to begin how to do the #6 in the worksheet below.  Follow along with me and when I model a few problems in the back and try stopping the video to complete one and then see how you did.  This is a very important video to watch. I have included the key if you want go ahead and try the other problems that I did not review. They are all very similar!!

stoichiometry ditto 2a and 2 combined 0607.pdf

stoichiometry ditto 2 key 0607 p.pdf

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Module Test 7Reaction Identification/ balancing /Stoichiometry/ Assessment –

– available below

Once you are done with all activities please complete Module Test 7 – by using the file below and the form below:

Module Test 7.pdf

Remember none of this is graded but I am getting a number to see how well you understand the topics.  If you are enjoying this madness then you are at the right place.  If this is very painful, I am sorry but you need these skills for AP Chemistry!!!

You must have completed all assignments above first before you begin your module Test 7.

Module Test 7: ( Yes it will send you your submission graded automatically) – make sure your email is correct when submitting in the form.

END OF MODULE 7!