Thursday, December 2

Today we started off class by reviewing the 3.4 Reading Sheet. After looking over that, Mr. Henderson passed out calculators and told us how to use them more efficiently. There is a button (sto) that can store a numbers in the calculator. For example, if you wanted to store a molar mass because you would be using it in the problem a lot, you would plug that number under sto 1. So when you used that number, you would not have to type it in; you would just have to the recall 1 button. He also said that we should either be comfortable with his calculators or bring your own to use. If you bring your own calculator it cannot be an inspire.

We then went over what would the format of the test would like like.

-14 multiple choice questions

-3 pages of short answers (stoichiometry problems)

-1 question on the lab

To study for these types of problems there are review questions on the website and moodle with answers. A great way to study would be to use your packet. Everyone should have almost a completely filled out packet with correct answers. You could rework a problem that is difficult for you and check your answer with your packet. To prepare for the test in class, we completed review questions 17-19 which simulated what the questions about the lab would look like.

Furthermore, Mr. Henderson lectured us about percent yield in detail. To find the percent yield of a substance you need the theoretical yield (mass of the expected product) and the actual yield (mass of the measured product). You discover the theoretical yield by using the normal stoichiometry conversion factors, and you discover the actual yield by massing what your results are in the lab. The equation for the percent yield is:

Percent yield = Actual Yield * 100

Theoretical Yield

Lastly, we attempted problems involving this percent yield concept on a worksheet Mr. Henderson passed out. The toughest problem was number 5; the answer to it was 709.75 grams. Today was a very productive class period as we got in needed review time for the test. Finally, I would like to wish everyone good luck!

Friday November 19th

Today, Mr. H began the class by going over Alex's blog and showing the variety of pictures that were from Wednesday's lab. He then refreshed our memories about the chemical reactions we had done the day before in our packets. (These types of conversions are going to be on tonight's webassign). Mr. H then proceeded to tell us that he had taken each lab group's silver precipitate, and put them in an oven to dry. He mentioned that he would take them out at the end of class so we could find the mass of each beaker that contained pieces of silver.

Mr. H then instructed the class to open our unit 5 Stoichiometry packets to page 6. There, we worked on the second half of the page; it was titled: Decomposition StoiCHEoMISTRY. The class worked on problems 12-15, and then Mr. H showed us the answers so we could correct our mistakes. He went on to answer any questions that students had about these problems. The answers to problems 12-15 are as followed:

12) 2H2O=> 2H2 + O2

13) 364.8 grams O2- this was a 2 step problem ( converting moles of H2O to grams of O2 )

14) 35.67 grams H2- this was a 3 step problem (converting grams of H2O to grams of H2)

15) 1.22 moles O2- this was another 2 step problem (converting grams of H2O to moles of O2)

In order to solve #13, this is how you would show your work:

         22.8 mol H2O * 1 mol O2 * 32g O2= 364.8g O2
                                  2 mol H2O   1 mol O2

Next, Mr. H gave a lecture on "Limiting Reactants." The notes he wrote on the board are as followed:

Limiting reactant- one reactant that produces the least amount of a product

   2 Al(s)             +          3CuCl2              ==>            3Cu +    2AlCl3

0.010 moles                  0.030 mol    =============> 0.03 mol Cu                           
not enough                    too much-precipitate
dissolves                      (excess reactant)
(limiting reactant)==========================> 0.015 molCu
                                                                                                                                                                  

0.030 mol                    0.030 mol ==============> 0.030 mol Cu
too much                      not enough
precipitate                    dissolves
(excess reactant)          (limiting reactant)
|
|
|======================================> 0.045 mol Cu

The reason why you choose the number that is less than the other one is because after you conduct 1 experiment, you'll run out of one of the elements and you won't have enough  of the second element to create another product.

Next, Mr. H introduced us to the next level of stoicheometry and used the notes he gave us in class to solve problem #12 on page 4. In this problem, you have to convert the number of grams of each element to the moles of Ti3N4. This is how you would solve this problem:

162g Ti * 1 mol Ti * 1 mol Ti3N4 = 1.13 mol Ti3N4  <== correct answer
                47.88g Ti   3 mol Ti

79 g N2* 1 mol N2 * 1 mol Ti3N4 = 1.41 mol Ti3N4
               28.01g N2   2 mol N2 

1.13 mol Ti3N4 is the correct answer, because as stated in the notes: you choose the number that is less than the other one because after you conduct 1 experiment, you'll run out of one of the elements and you won't have enough  of the second element to create another product.

Next, Mr. H showed us 7th grade Stoicheometry. This was on pages 7 and 8. This page included practicing finding limiting ingredients and the number of cakes produced. For each problem, we had to look off of the original equation, which was: 6 cups of flour + 4 cups of sugar + 2 tbsp baking powder + 3 cups of milk ==> 2 cakes, and find by how much each ingredient was being multiplied. By doing so, you could find the limiting reactant ( the lowest number) and then multiply that number by the original number of cakes produced from the given equation. The answers for page 7 are as followed:

a) Limiting ingredient: milk               # of cakes produced: 1

b) Limiting ingredient: baking powder            # of cakes produced: 3

c) Limiting ingredient: flour               # of cakes produced: 10

d) Limiting ingredient: baking powder                          # of cakes produced: 100

e) Limiting ingredient: baking powder           # of cakes produced: 100

On page 8, was more practice that involved finding the limiting reactant and the # of products produced. We only did #3 on this page. The answers for # 3 are as followed:

Molecules before reaction             Type and # of molecules after reaction                           Limiting Reactant
3 H2 + O2============>       2 H2O
                                                      1 H2                                                                                   O2

6 H2 + 3O2===========>       6H20                                                                                  none

18 H2 + 4 O2==========>      8 H2O
                                                     10 H2                                                                               O2

Mr. H ended the class by getting each lab groups' beaker from the oven, and then had 1 member of each lab group go and mass the beaker.The students that missed class that day, should ask one of their group members for that information. **Note, there is a QUIZ  on Monday, so be sure to study.

Thursday, November 18

This morning, we walked into class and were notified that there was a lab today. The lab is called ST1 Copper and Silver Nitrate Lab. The purpose is to use careful measurements to determine the mole ratio of copper(II) and silver in the single replacement reaction of silver nitrate with copper. We were instructed to take out our lab notebooks and copy down this purpose onto a clean right-side page. On the left side, Mr. H told us to write the chemical equation for the reaction, for it would be helpful later. It is:

Cu (s) + 2AgNO3 (aq) --> 2 Ag (s) + Cu(NO3)2 (aq)

We all then received a piece of paper describing the lab's procedure. 2 people from the lab group were to carry out the first procedure while the remaining 2 were to carry out the second procedure. The procedures are as followed:

1) Prepare the Silver Nitrate for Reaction:

• Mass an empty 50 mL beaker. Record.
• Add ~1.4-1.6 g of AgNO3.
• Mass beaker plus AgNO3. Record.
• Add 30 mL of water.
• Stir to dissolve solid; rinse stirring rod so as to avoid loss of the AgNO3.

2) Prepare Copper for Reaction and React:

• Obtain ~25 cm of copper wire.
  
• Mass wire. Record.
  
• Loosely coil the wire so as to rest of a wood splint as shown. It should be long enough to hang on the wood splint and dangle in the solution. Yet it should not touch the bottom of the beaker.
  
• Add 3 drops of nitric acid to the solution; avoid contacting the copper wire with the HNO3. Do not stir.

After we did what was required, we were instructed to leave our notebooks at our lab benches and come back to our seats. This time was set aside for the reaction to occur. When we were back at our desks, Mr. H told us to take out our packets and turn to page 6 for

E V E N M O R E S t o i C H E o M I S T R Y!!!!

Mr. H told us to do number 7; write the balanced equation for the synthesis of magnesium oxide from its two elements - magnesium and oxygen. The answer is as follows:

2Mg (s) + O2 (g) = 2 MgO (s)

Then we were told to do number 8; calculate the molar mass of each of the three reactants and products. The answer is as follows:

Mg: 24.305 g/mol O2: 32.0 g/mol MgO: 40.3 g/mol

After, as a class, we went through the next three problems and identified how many steps would be needed (using the Mole Island method, of course). Number 9 was a one step equation. Number 10 was a two step equation. And number 11 was a three step equation.

The answer to number 9 (Determine the number of moles of magnesium oxide produced by the reaction of 3.25 moles of magnesium.) is 3.25 molecules of MgO, simply found by using Mole Island to convert from moles of magnesium to moles of magnesium oxide.

The answer to number 10 (Determine the mass of magnesium oxide produced by the reaction of 5.22 moles of magnesium.) is 210.39 g MgO, again, simply found by using Mole Island to convert from moles of magnesium to moles of magnesium oxide to grams of magnesium oxide.

The answer to number 11 (Determine the mass of oxygen gas which will react with 65.2 grams of magnesium.) is 42. 92 g O2, aaaaaaand again, simply found by using Mole Island to convert from grams of magnesium to moles of magnesium to moles of oxygen to grams of oxygen.

Before we knew it, it was time to check back at our lab. Our whole lab group was to do the third and fourth procedures listed on the sheet of paper:

3) Retrieve Silver Product from Copper Wire:

• Obtain a clean, dry 100 mL beaker. Using a permanent marker, label it with your period and your names.
• Mass the empty beaker. Record.
• Carefully lift the copper wire (with silver attached) from the beaker.
• Hold wire over and into 100 mL beaker and rinse with a forceful stream of DI water. The goal is to knock the silver off the copper and into the beaker.
• Continue rinsing until all the silver crystals are removed.
• Set labeled beaker aside to dry. Once dry (the next class period), mass the beaker with the silver. Record.

4) Rinse and Mass Unreacted Copper:

• Rinse the remains of the copper wire in an acetone bath.
• Being careful not to break or crumble the copper, thoroughly dry it by dabbing it with a paper towel.
• Mass the remaining copper wire. Record.

We did just that. The untouched product (silver still connected in the original beaker) should look like this:The removal of the silver should look like this:

Mr. H then told us to clean up and reminded us that we would be finishing the lab tomorrow.

The homework is the Stoich WebAssign 2.