Last Section of Stoich_ Percent of Yields

October 22nd 2012

Hello guys basically what we did today- those of you who were absent today- is we took stoich quiz #5 and learned last step of STOICH. Since we were working with Stoichiometry 2 weeks from now, we know how to convert gram to mole, mole to gram, gram to gram, and finally PERCENT OF YIELDS!!!!! :) 


As you see above, percent of yield is really really really simple and easy. 

• Percent of Yield is a comparison of the amount actually obtained to the amount it was possible to make. 

• Actual yield is the amount of product actually obtained from a chemical reaction.(Must be experimentally determined, cannot be calculated) Usually the problem or lab date gives the actual yields. 

• Theoretical yield is the maximum amount of product that can be get in a chemical reaction. You must calculate with stoich step to get theoretical yields. 

Hmm. Lets solve one problem together. 

Boom. You have given formula and question. 
This question is asking us to solve theoretical formula and find percent of yield. 
First, you need back up your memory to determine which one is limiting reactant. 
Lets find the limiting reactant :

1. Assume  
   C₂H₅OH is the limiting reactant and calculate how much C₂H₅Cl would be formed.
2. Also assume that  PCl₃ is the limiting reactant too. 

Oh, sorry for huge equations.

• The reactant producing the smaller amount of C₂H₅Cl is the limiting reactant which is PCL3 
• Corresponding quantity of C₂H₅Cl  is the theoretical yield

Limiting reactant= PCl3 

Theoretical yield= 48.9g 

Now calculating percent of yield is piece of cake, which is : (Sorry, huge pic!)

We found Percent of Yield!!!

Homework:

Finish percent of yield worksheet

Ready for Wednesday's EXAM

Our Next scribe is (Plz make a drum sound) Eddy Rohhhhhhhh! Good luck ;) 

Have an awesome midnight homework time guys.. I have to do my rest of the homework ehh. ;(

Limiting & Excess Reactants

What we did
Today in class, we measured the mass of the silver in the beaker and recorded the amount. We also went over a new concept, limiting and excess reactants. It's pretty straightforward, and as long as you can do gram to gram stoichiometry, the concept is similar.
Limiting &Excess Reactants

 Basically, a limiting reactant is the reactant that runs out the quickest, meaning it is the reactant that reduces the total amount of products that can be produced. An excess reactant, on the other hand, is the reactant you have left over, or the one that produces the most amount of product. When comparing limiting and excess reactants, the limiting reactant is naturally the smaller of the two yields.

Conceptual Problems

First, to get the concept of the limiting reactant, it's important to cover a few analogies.

1. You have 18 cars bodies and 43 tires. How many cars could you build?

A. You can only build 10 cars. This is because for each car, you need 4 tires. Although there are enough car bodies to build 18 cars, there is a lack of tires, therefore, the amount of cars built is dramatically reduced.In this particular problem, the tires would be the limiting reactant, and by process of elimination, the car bodies the excess reactant. 

2. In a  typical lunch, there are two napkins given along with one sandwich and three fruits. There are 100 napkins, 60 sandwiches, and 21 fruits. What's the most number of complete lunches you can make?

A. 7 lunches. The fruits are the limiting reactant.

Sample Problems

A 50.6 g sample of Mg(OH)₂ is reacted with 45.0 g of HCl according to the reaction: What is the theoretical yeild of MgCl_2?

Mg(OH)₂ + 2 HCl --> MgCl₂ + 2 H₂O The first step to find out the limiting reactant is to convert each grams of reactant to grams of any product, it doesn't matter whether you choose MgCl₂ or H₂O to convert to. However, please read the next question, as it asks for MgCl2 (Don't do extra work!) First, Convert grams of Mg(OH)2 to grams of MgCl2...  50.6 g Mg(OH)2 x 1 mol  Mg(OH)2/58.3 g x 1 mol MgCl2 /1 mol Mg(OH)2 x 95.g MgCl2/1 mol MgCl2 = 82.6 g MgCl2

Then, convert gams of HCl to to grams of MgCl2 as well

45 g HCl x 1 mol HCl/36.5 g x 1 mol MgCl2/2 mol HCl x 95.2g MgCl2 /1 mol MgCl2 = 58.7 g MgCl2

Okay, so now that you've found out how many grams of MgCl2 you will get by reacting with the two reactants, now you can compare to find out the theoretical yield.

The theoretical yield of MgCl2 is 58. 7 g. As stated before, it is the smaller of the two numbers you get. Naturally, the limiting reactant is the HCl and the excess reactant was the Mg(OH)2.

Another question might be "Find the amount of excess reactant". To find the excess reactant, all you do is calculate how much of the non-limiting reactant (MgCl2) reacts with the limiting reactant (HCl).

So first, you take the amount of limiting reactant given, and convert that to grams of the excess reactant to compare.
45 g HCl x 1 mol HCl/36.5 g x 1 mol Mg(OH)2/ 2 mol HCl x 58.3 g Mg(OH)2/1 mol Mg(OH)2 = 35.9 g Mg(OH)2
That is the amount of Mg(OH)2 reacted, so to find out the excess amount of the reactant, all you do is subtract 35.9 grams from the initial amount of Mg(OH)2, which was 50.6 g. So...the excess amount is 14.7g
Tips 
-always make sure to convert the given amount of a reactant to grams of any product at least. If you can do at least this gram to gram stoichiometry, the rest will come naturally. (Once you have yields of both reactants, just pick smaller of the two and that is limiting reactant.)
-Just remember the idea that limiting is the smaller number, and excess is the larger.
-Always read ahead to see which product's yield it asks for, so you don't do extra work
- don't assume that just because one reactant has fewer moles it is the limiting reactant (it is not true in many cases)
Reminders
-Formula of a nitrate lab due Monday
-Several limiting & excess reactants worksheets due Monday as well
- Don't forget about the one webassign
In case you still don't understand...
http://www.youtube.com/watch?v=Vaiz0zLesHk
http://www.youtube.com/watch?v=-biLP2acKqg
The first video is actual sample problems using chemical formulas.
The second video is more about the concept, using hamburgers, etc.
Both videos helped solidify my understanding of this concept.

Another stoichiometry problem in case mine didn't make sense. This is again just converting grams to grams, and the smaller amount is again the limting reactant.
The next scribe will be Jaehee

Formula of a Nitrate Lab

Hello, today we worked on the Formula of a Nitrate Lab.

In this lab, we observed the reaction of copper metal and silver nitrate solution and tomorrow, we will use the masses in order to find and write the balanced chemical formula for this reaction.  Then we will be able to find out if a 2+ or 3+ ion of copper was used.

First we mixed 1.5 g of silver nitrate in 25 mL of water until all of the silver nitrate dissolved.  Then we cut a 25 cm piece of copper wire.  We loosely wound this piece of copper around a piece of would and placed it in the beaker with the silver nitrate solution so that the end of the copper was not touching the  bottom of the beaker.  

We let this sit for 15 minutes.

When we came back, the copper wire was covered with silver, and the water solution had become a bright, light blue liquid.  

After observing this, we took the removed the copper wire and sprayed with water over a massed beaker in order to clean all the silver off it.  We then put the clean wire in an acetone rinse beaker and then took the copper wire's mass.

 

Next we decanted the water from the beaker with the silver into a waste flask.

Tonight, we are allowing the rest of the water to evaporate from the beaker with the silver.  Tomorrow, we will take the mass of the silver, and we will be able to finish the lab.  The next scribe will be Christine.

Stoichiometry

Today in class, we learned the basics of stoichiometry.
We started out class with an experiment Mr. Lieberman did for the class.
He had a beaker of water and put some calcium carbide in it to produce acetylene gas.

After the reaction in the picture above, acetylene gas was produced.

The equation for this experiment was: CaC2 + H2O = C2H2 + CaO

Another experiment was done by taking a plastic tube and acetylene gas and oxygen was put in. After Mr. Lieberman pushed a button on the tube, it produced a large pop sound as carbon dioxide and water was produced.

This reaction is shown by: C2H2 + 5/202 = 2CO2 + H20

This part of the experiment didn't always work every time and give the pop sound.

This experiment was done to show that you need the correct mole to mole ratio in the substances among an equation so that the experiment can be successful. (Like the paint can experiment we did before!)

The Basics of Stoichiometry:

1. Foundation of stoichiometry-molar ratio or A.K.A. mole:mole ratio

2. molar ration is (Moles wanted/moles given) from balanced equation

3. To solve problems, think of it similarly to dimensional analysis

Homework: mole-mole worksheet and quiz on stoichiometry tomorrow

The next scribe will be...Autumn K.

Hello on Thursday we had a lab to help us identify types of reactions. We did not get a hand out for this lab so here are some items that I believe will help you out.

Lab name: Chemical Reactions Lab Stations
Lab Goal: To identify the type of chemical reactions for 7 lab stations

Conclusion:















Just as a reminder the type of reactions are:
Combustion: Hydrocarbon+O2---->CO2+H2O
Synthesis: A+B--->AB
Decomposition: AB--->A+B
Single Replacement: A+BC--->AC+B(if a is a metal) or A+BC---> BA+C(if a is a non-metal)
Double Replacement: AB+CD---> AD+BC Remember ionic compounds.

When predicting products remember charges.
Always balance the equation. When balancing only change the coefficients, never change the subscripts.

Lab and Mole Stoich ws Due Monday10/15

The next scribe will be.......Danielle S.

Chemical Reactions and more...

October 4, 2012

Hello everyone, today we learned about the 5 types of chemical reactions.  Those include: synthesis reactions, decomposition reactions, single replacement reactions,  and double replacement reactions.

A synthesis reaction is when two reactants form to make a product

so it would be like A (reactant) + B (reactant) --->  AB (product)

an example of this is  8 Fe + S₈ ---> 8  FeS     

<sub>This is because you have two reactants (Fe and S) when reacted become a product.  

Also, here is a video demonstrating a synthesis reaction, and someone screaming in the background:
http://www.youtube.com/watch?v=gHkyO7pMj0U&feature=plcp 

In addition, another reaction is decomposition reactions.
 A decomposition  it breaks down into its component elements or simpler compounds. The anion determines what the product will be.  
Here is another awesome video directed by me about decomposition:
 http://www.youtube.com/watch?v=Nr5EXzDq28g

For example, an example of a decomposition reaction is 
 2Cl2O5 → 2Cl2  + 5O2  

 It goes from from a compound (2Cl2O5) into smaller elements (2Cl2  + 5O2) and thus is a decomposition reaction.

3rdly, we get to single replacement reactions, which are when one element replaces another element in a compound.
for example; element + compound --> product + product 
remember that the biggest mistake is people forgetting the cation goes first.</sub> 

<sub>Here is a picture of Mr. Lieberman demonstrating a single replacement reaction: 

4thly, we went over double replacement reactions.  which are when a metal replaces another metal in a compound and a nonmetal does the same thing to another nonmetal.
an example of this is: HCl +NaOH --> NaCl +H2O because one metal replaces another and one nonmetal replaces another.

Finally, we get to combustion, which we didn't really have time to go over in class, and it less important than the 4 previous reactions.  Combustion reactions are when hydrocarbon reacts with oxygen gas, also called burning.  to make things burn, you need a fuel, oxygen, and a spark.  So any equation with just hydrocarbon and oxygen is a combustion equation.

Hw: Chemical reactions wksht, quiz on friday.  3.4 WA is up, so maybe get started on that sometime before 3 am the night before the test. 
. 
Scribe for next week:  Kevin L.</sub> 

 

Empirical and Molecular Formulas

October 4th, 2012 

Hi everyone, I forgot to post yesterday and we did not really do anything today except take a homework quiz and work on the Web Assign review, so I will just go over what we did yesterday. Yesterday we learned about...

EMPIRICAL AND MOLECULAR FORMULAS!!

-Empirical formulas are the simplified formulas, which give the whole number ratio of the atoms in a compound-(example: H2O2 becomes HO)
               -always in whole numbers
               -can be the same as molecular formula


-Molecular formulas represent the total number of atoms per element in a compound

This might help you understand the difference between the two.

How to find the Empirical and Molecular Formulas If you have the percent of each element in a compound, assume that it is 100g, so that the percent is also the number of grams Convert the mass of each element to moles using the molar mass from the periodic table.  Divide each mole value by the smallest number of moles calculated.  Round to the nearest whole number These numbers make the subscripts for each element in the empirical formula Here is an example problem:

So in this problem, the empirical formula would be C14H18N2O5

Finding the molecular formula is very simple once you have found the empirical formula...

1. Find the molar mass of the empirical formula by using the periodic table

2. Divide the molar mass of the compound (which should be given to you) by the molar mass of the empirical formula 

3. Round the number from step 2 to the nearest whole number and multiply it by the subscripts for each atom in the empirical formula to get the molecular formula 

OK.....that's it. So here are the reminders....

Homework:

-study for unit 2 test 

-finish all web assigns 

THE NEXT SCRIBE IS.... ZACK 

Works cited: http://www.chem.tamu.edu/class/majors/tutorialnotefiles/empirical.htm

                   

Percent Composition/ The Bubble Gum Lab

What's up guys! I hope you all enjoyed today as we stepped back into 90s and learned what NOT to wear to school by looking at some pretty ridiculous outfits (my clothing was a fine example). But in all seriousness, the bubble gum lab. Haha just kidding it was pretty fun to chew gum and listen to music in class. :P

Okay. So we started off class today by reviewing Avogadro's number (6.02x10^23) and how to use it as a conversion factor. SEE PICTURE BELOW

This is a pretty handy dandy model to remember so make sure you remember this!! 

PERCENT COMPOSITION

Today, we learned how to find the percent composition of an element and it's super simple. First, you have to find the molar mass of an element. For example, CuBr2 would be 223.55 grams. Then, you take the mass of each individual element and divide that by the total mass. For example, Cu would be 63.55 divided by 223.55. You then take that number and multiply it by 100 to find the percentage. SO EASY! You do that with the rest of the elements and that's it! 

The Lab

The next thing we did, was the amazing bubble gum lab. The goal is to find the % sugar in a piece of bubble gum and write an empirical formula for it. (GxSy) REMEMBER G is a FICTIONAL symbol for Gum. Get it? G for Gum? And S is a FICTIONAL symbol for Sugar. S for Sugar? Yeah, I know it's super clever. But the mass of the GUM is 513 g/mol and mass for SUGAR is 342 g/mol.

Basically we took a piece of gum, weighed it, chewed it, and weighed it again. Super simple I know. However, we then need to calculate the mass of the sugar in the gum, the gum before the chewing, the gum after the chewing without sugar, the % composition of the sugar, and find how many moles of sugar and gum there are. We needed to use the data and measurements obtained from weighing the gum to find that. Finally, we have to calculate the ratio between the moles of sugar to moles of gum which will then be the empirical formula. AKA the answer.

Reminders

• Hydrate lab due tomorrow
• Bubble gum lab due Thursday
• % Composition worksheet due tomorrow
• Test on Friday!
• Do your WEBASIGNS!!!
• DRESS AS YOUR FAVORITE 90s CHARACTER SO THE GENIE COMES OUT!!

NEXT SCRIBE WILL BE: BRAM!!!!!

Formula of a Hydrate Lab

HAPPY MONDAY!

In class:

We did the Formula Hydrate Lab (hence the title)

*Extra: Be aware of the spirit days this week as it is possible we might get a visit from the "Homecoming Genie" (I have NO idea what that means--but just show your spirit!)

For HW:

This lab (in the carbonless paper notebook of course) is due on wednesday.

DON'T FORGET: Test on Friday--Remember to do those webassigns!

Before Reaction

During Reaction

After Reaction

Let's get started.  So the purpose of this lab was to determine the amount of moles of water that are present in the compound per mole of Copper (II) Sulfate.

EQUATION OF Copper (II) Sulfate: CuSO4 * nH20 we are finding n in the lab)

Basically, we took blue hydrated Copper (II) Sulfate crystals, put them in a test tube, and heated them with a Bunsen burner until they turned white (allowing the water to separate from the substance). Here is a video of the process of the change from hydrated to anhydrous salt:

But, we had to do some weighing too.

1. Finding the mass of the empty test tube
2. Finding the mass of the test tube with hydrated salt
3. Finding the mass of the test tube with anhydrous salt

AND some calculations after you collected TWO trials of the data:

• Calculating the mass of the hydrated salt 
• Calculating the mass and moles of water lost 
• Calculating the mass and moles of anhydrous salt
• calculating the ratio between moles of water and moles of anhydrous salt (this being your n from before)

Make sure to show all your work and answer all the questions at the bottom of the lab sheet!

**Don't forget--you should have used Avagadro's number at some point in your calculations.

Next Scribe will be: Rachel! :)