Percent Yield

The theoretical yield is the amount of product that can be made in a complete reaction. The actual yield is the amount of product that is actually obtained in an experimental setting. For many reasons your actual yield will be lower than the theoretical yield, that is why we do a percent composition.

Here is a percent yield problem we did during class.

C12H22O11 -----> 11H2O + 12C

In class we ran this experiment and got a mass of 241.84 grams. From this value you must subtract the mass of the beaker, 155.20 grams, and the mass of the sulfuric acid, 59.06 grams. This leaves you with the mass of the carbon which is 27.58 grams.

Now to get the precent yield you use the formula at the top of the post.

Your final precent yield is 93.5%. I hope this will help you understand percent yields.

The next scribe will be... Chris A.

The Wild Wild World of Limiting Reactants

Today was a pretty big day in terms of learning the concept of limiting reactions, so if you weren't here today, sucks to be you. Hopefully this post will help you a lot though.

Today we kicked off class finishing our Copper and Silver Nitrate Lab. If you somehow lost your lab, here is a link. We went to go get the masses of the silver left in our beakers. (hopefully you didn't spill all of your silver in the sink). After massing our silver, we went back to our desks and we wentthrough how to do the post lab.

In the post lab, you have to find the mass and moles of copper wire that reacted. Then you have to find the mass and molesof the silver metal that was produced in the reaction. Using this data you have to find a mole ratio between the two elements. With this mole ratio, you have tocreate a balanced chemical equation for the reaction. We used this example in class....

If the mole ratio between silver and copper was 3 Ag : 1 Cu, then the balanced chemical equation for this reaction would be....

The whole point of this lab is to find the balanced equation. Oh and of course also to make silver for Mr.Lieberman to sell. So that was pretty much the first half of class.

The second half we looked at a single replacement reaction between Aluminum and Copper Chloride.

The Reactants, Copper Chloride on the left (we dissolved it in water before reacting it) & Aluminum foil

The reaction over time

What happened was the Aluminum bonded with the Chloride and that left the copper (the redish chunks at the bottom) alone by itself. As you can see the liquid went from blue to clear because the copper was taken out of the liquid.

The balanced equation for this reaction is...

To begin with, we had .66 grams of Al and .64 grams of .

A limiting reactant is the reactant that runs out first, or the reactant that creates the least product. The balanced chemical equation and the mass of the reactants are crucial to finding limiting reactants. There are two ways to find the limiting reactant in this case....

Method # 1 : This is the way where we find out how much is needed to react with the .66 grams of Al.

This shows that you need 4.9 grams of Copper Chloride to react with .66 grams of Aluminum. Now you maybe thinking, "OH MY GOD!!!! WE ONLY HAVE .64 GRAMS OF COPPER CHLORIDE!!!!!" Well, congratulations my friend, you have just found the limiting reactant. Since there isn't enough copper chloride to react with .66 grams of aluminum, the copper chloride will run out first, making the copper chloride the limiting reactant. However, this way is not that great of an idea if you need to find out how much of a product is produced. For this, you probably shoud use method #2.

Method #2 : In this method, you take each of the two reactants and find out how much of the product that reactant produces.

(.66 grams of aluminum produces 2.32g of copper)

now you do the same thing with the other reactant

(.64 grams of Copper Chloride produces .30 grams of copper)

Since Copper Chloride produces less copper than Aluminum does, Copper Chloride is the limiting reactant. Also, with this method, you figured out that .30 grams of copper was produced by this reaction.

The last thing we learned how to do was find how much of the excess reactant will be left. Since we know that .30 grams of copper will be produced, we want to find out how much aluminum is needed to make .30 grams of copper. Than we subtract that amount from the amount of aluminum originally present, and, HA-ZAA! you have your excess reactant. Heres the work...

.085 grams of aluminum is needed to make .30 grams of Copper, so then you do .66 -.085=.575 There will be .575 grams of excess aluminum leftover.

That pretty much sums up todays class period. There is a quiz tomorrow so you better know your stuff, hopefully this post helped you learn your stuff. Also today Liebs gave us a homework sheet at the beginning of class which is like always due on the test day. If you forgot to pick it up or were absent today, you can find it here. It is called 02_stoichPrac.

Thats is for me guys. The next scribe will be Senor Trevor B. Have fun, you have five days to do this, so it better be good.

Limiting Reactants

Friday, in Liebs' class, we did four major things.

First we finished our labs by decanting the water of our beakers, and massing our silver.  This is largely uneventful, but necessary to complete the lab. The result probably looked something like this:

Which, let's be honest, is a little gross.

Post-Lab question #6 requires some outside knowledge.

It asks for the current market price is silver, which according to the most recent statistic on Wikipedia is on average $24.38 per troy ounce of silver.

The post lab gives the conversion 1 Troy ounce= 31.0 grams, so convert your mass to troy ounces, then multiply the times $24.38.

Next Liebs answered questions on the Super Mole Fun worksheet. He has the first three problems worked out here, just scroll down to the super mole fun pdf. (Public Data File)

After that we took our quiz, which covers what Tom talks about in his lovely post below. It was mostly converting grams to moles to grams again, in order to determine how much of a product would be produced, or reactant was need.

Next, were our notes on limiting reactants.

For some reason my computer hates me, and refuses to put the embeded slideshow of the notes in but the link may be found here starting on Slide 8.

A limiting reactant is the reactant that produces the least amount of product.  There are two ways to do this.  The first is to find to choose one of the actual amounts of  reactant used and determine how much of the other reactant it would need.  Then compare the number you found to the actual number used, and if it's less than what was used, then that is the limiting reactant. However, Liebs doesn't reccomend this method, instead he suggests the second one: determine in two equations how much of the product each reactant produces, and compare.  This method is better because it is clearer, and most labs/worksheets will also ask you how much of the product each reactant produces anyways.

Liebs ended class by tell him two things had made his day: one, was someone couldn't figure out how to turn off a solar-powered calculator. The second was this beauty right here....



"baby monkey, baby monkey..."

Next scribe is..... Amar! Congrats bud.

We are making silver!!!

Today we started off class by going over question #3 on the SuperMole Fun worksheet. The point of this problem is to find the amount of bacteria tissue (C5H7O2N) produced. In the problem you are given the kg of waste to start (10000 kg). You need to convert this into NH4+ ions so you multiply by 3.o kg NH4+ / 100 kg waste. Since the problem says the waste water contains 3% NH4+ ions, we can assume 100 kg. After converting to NH4+, multiply by 1000g / 1 kg so the equation is converted to grams. Next, you find the molar mass of NH4+, which is 18.04. Multiply the equation by 1 mol NH4+ / 18.04 g NH4+ to convert it into moles. After this you use the conversion factor 1 mol C5H7O2N / 55 mol NH4+ to convert the equation into moles of C5H7O2N. Next, you find the molar mass of the bacterial tissue which is 113 g and use the conversion factor of 113 g / 1 mol C5H7O2N to convert the equation into grams. The grams of bacterial tissue produced should be 34000 g. Once you have found this, multiply it by 0.95 (95%) because in the problem it says that only 95% of ammonium (NH4+) ions are consumed by the bacteria. If you had some trouble following this, the answers and work are posted here. After going over the worksheet, we started our Copper and Silver Nitrate Lab by going over the pre lab and procedures. This lab is going to take multiple days, so we haven't finished it yet, but it is due on tuesday of next week. Also, our class could probably make a huge profit by saving all our silver from the lab, just throwing that out there.

Everyone better watch the Bears own the Dolphins tonight. And the next scribe is...

Faith S.

A New Hope for Stoichiometry (Chemistry Honors Vs. Chemphys)

Today we continued our Intro to Stoichiometric Calculations. We had a very heated discussion about how Mr. Lieberman did not post the answers to the Intro To Stoich 1 worksheet until it was past some of our bedtimes (claims were made of that time being before 6:45 pm.... :s). Anyway after we got over that hump, we dove into a practice/example problem that Mr. Lieberman did on the board. It went a little like this:
Reaction #1: CaC2 + 2H2O --> Ca(OH)2 + C2H2
Then we plugged C2H2 into a combustion reaction: (It is balanced..DUH) 2C2H2 + 5O2 --> 4CO2 + 2H2O
Mr. Lieberman then posed to the question of: If .60 grams of CaC2 reacts, how many grams of H2O will form? ( Use the balanced combustion equation above)


The converstions look like this:

In the end you end up with .17 moles of H2O. Soon after this we witnessed the astonishing demo of Evil KinMOLble! This unlucky or lucky, (I'm not sure what was going through his head..... probably AAAAAAAAAAAAAAAHHHHHHHHHHHHH!!!!!!!!!!!!!) It was the projectilation of Mr. Mole from Mr. Lieberman's Mole Cannon using the combustion. The combustion was caused by the ignition of Acetylene from the reaction of H2O and CaC2(I think not sure on this). After that first demo, we took our short little quiz on Stoichiometry. Lastly, as a conclusion to our class period, Mr. Lieberman armed the launcher once again and fired it into the ChemPhys classroom next door. Causing complete academic Armageddon in the science wing. (ok I exhaggerated a little). Check out the Video in the post underneith this.

The next scribe is Tom btw.

Stoichiometry

We started class today with an in-class blowup. With the help of the whole class believing and some intense movie-quality diving out of the way, we got the magical blowing up elves to make the magical paint can blowup.
Then we got down to business. We went over the notes of the first two slides in Stoich.

Stoichiometric calculations

View more presentations from gbsliebs2002.

To figure out the problem on the second slide, you need to know that the mole ratio = the moles wanted/the moles given. We also had a worksheet given to us along with the objectives and calendar. The next scribe is going to be Erica G.

Intro Stoich Answers

They are posted here

Honors chem Goes viral

So today we started with a demo that was pretty awesome. 

The reaction taking place was
Pb(No3) + KI

This is lead nitrate and potassium Iodide

So back to what happened. the molecular formula for this reaction was PbI2(s) + KNO3

This is then balanced to Pb(No3) + 2KI --------> PBI2 + 2KNO3

The NIE was Pb^+2 + 2I- -------> PbI2(s)

That was pretty much our day after that we worked on the worksheet that can be found here and the lab which can be found here

Things you might want to know - Hali = ion form of a halogen

and there is a review sheet up here

After that we had some tough decisions to make because when our video goes viral we need a name and a mascot. We decided on Chris as our mascot ( the shirts we make will have his face on the back) and we havent came up with a good name, but we will. Tomorrow in class will be a review for the test on unit 4. Bring questions about the book problem,labs, worksheets and anything else about the unit.

Our next scribe is Sarah. Have fun.

Reactions, Reactions, Reactions

Today, after a long weekend of waiting, was the day of the "Double Replacement Reactions and Solubility Lab." We started out class talking about the lab procedure. After briefly discussing the post lab (questions 1 and 2 are now combined into one question, just write the Molecular Equation on one line and the Net Ionic Equation on another), we headed to the lab stations.

Our group first put a dropper-full of each sodium _____ in each respective row, and then added the various nitrates. The finished tray looked something along the lines of this:

The combinations found to form precipitates for our group are as follows:

Sodium Carbonate + Aluminum Nitrate

Sodium Phosphate + Aluminum Nitrate

Sodium Phosphate + Barium Nitrate

Sodium Sulfate + Barium Nitrate

Sodium Carbonate + Copper(II)Nitrate

Sodium Hydroxide + Copper(II)Nitrate

Sodium Iodide + Copper(II)Nitrate

Sodium Phosphate + Copper(II)Nitrate

Sodium Carbonate + Iron(III)Nitrate

Sodium Hydroxide + Iron(III)Nitrate

Sodium Phosphate + Iron(III)Nitrate

Sodium Carbonate + Silver Nitrate

Sodium Chloride + Silver Nitrate

Sodium Hydroxide + Silver Nitrate

Sodium Iodide + Silver Nitrate

Sodium Phosphate + Silver Nitrate

Sodium Carbonate + Zinc Nitrate

Sodium Phosphate + ZInc Nitrate

Here is an example for one part of numbers 1 and 2 in the post-lab:

Molecular Equation: 3Na2CO3 +2Al(NO3)3 --> Al2(CO3)3+6NaNO3

Net Ionic Equation: 2Al^3+ +3 CO^3- -->Al2(CO3)3

You can tell which one is the solid based on the fact that sodium and nitrate salts are soluble in water...(the one that isn't NaNO3).

Homework for tonight:

The lab is due Friday

as are all the worksheets and book problems because...

the test is friday. Yay!

The only matter left to be settled is who will have the challenge of following this excellent scribe post. I believe his name starts with B and ends with -obby S.

Bobby S

Ionic Equations Galore!

Friday's class was mostly a notes and independent work day. Mr. Lieberman started off by continuing his lecture about molecular and ionic equations.

The molecular equation is just another name for a balanced equation, which we learned how to do several classes ago. It's important that your molecular equation is written correctly before going any further, so make sure you pay attention that everything is correctly balanced. If not, the whole process afterward will be incorrect. Next, Mr. Lieberman introduced two new terms: complete ionic equations and net ionic equations. The complete ionic equation is the balanced equation with all the strong electrolytes written in as separate ions. So, for example, to make this molecular equation:

KCl + AgNO₃ ➝ KNO₃ + AgCl

into it's complete ionic equation, you would find out the electrical charge of each of the elements except for the precipitate. So, it would look like this:

K⁺ + Cl^- + Ag⁺ + ➝ AgCl + K⁺ +

Every element shows it's electrical charge except AgCl since that is the precipitate [solid] formed. Lastly, Mr. Lieberman talked about a net ionic equation, which only includes the ions that undergo a change. That means that the spectators [ions that do not undergo a change] are not included. So, looking at the complete ionic equation above, the net ionic equation would be:

Cl^- + Ag⁺ ➝ AgCl (s)

And that was basically all we did for the notes! After that, we had the rest of class to work on the Net Ionic Equation Worksheet. If you don't have it or have lost it, you can download it here and if you would like to check your answers, click here. Mr. Lieberman also mentioned that you should have done the Dissolving worksheet we got a couple days ago before attempting this one. The answers for that worksheet is here.

So that was Friday's class! Homework for Monday is the Chapter 4.2 Webassign and the Pre-Lab for the Double Replacement Reactions and Solubility Lab [to download, click here]. Also for the lab, make sure you create a data table! It's supposed to be a 6 x 8 grid with the rows and columns labeled with the anion solutions and cation solutions respectively.

Oh, and the scribe for Monday will be Declan G. Good luck! :]

Solutions and Electrolytes

    We began today's class by going over the answers to the Classifying Chemical Reactions post lab.  In case you missed it:


Reaction 1:  2Mg + O2 ---> 2MgO
                  Synthesis
Reaction 2:  2HCl + Mg ---> MgCl2 + H2
                  Single Replacement
Reaction 3:  (NH4)2CO3 ---> CO2 + 2NH3 + H20
                   Decomposition
Reaction 4:  CaCO3 + 2HCl ---> CaCl2 + H2O + CO2
                  Double Replacement
Reaction 5:  CuCl2 + Zn ---> ZnCl2 + Cu
                  Single Replacement
Reaction 6:  3CuCl2 + 2Na3PO4 ---> Cu3(PO4)2 + 6NaCl
                  Double Replacement
Reaction 7:  NaOH + HCl ---> H2O + NaCl
                  Double Replacement


    Next, Mr. Lieberman went over some new notes.  We talked about solutions and electrolytes.  A Solute is what is being dissolved in a substance (in this case we mainly focused on water).  A Solvent is what is doing the dissolving.  For example:  You have water and salt.  Water is the solvent and salt is the solute because it is being dissolved in the water.  A solvent can be anything from a liquid, solid, or gas.  Water dissociates ionic compounds into its ions (meaning that it takes them apart).  Example:  NaCl ---> Na+ + Cl-.  Another example we used is the rock salt you pour on your driveway and sidewalks in the winter.  The formula for it is CaCl2 but when it dissolves in the snow (water) you get Ca (charge of 2+) and 2Cl (charge is -).  


    If the compound happens to be covalent, the formula stays the same but changes whether it is a gas, solid, or liquid.  For example:  CO2 (g) ---> CO2 (aq).  

    Electrolytes are basically ionic compounds that carry a charge.  Strong electrolytes are a solution containing enough ions to carry a current efficiently (complete ionization).  A weak electrolyte is a compound that doesn't completely dissociate (small amount of ionization).  


    Electrolytes are essential to our lives.  Without them we would die.  Athletes drink gatorade because they want those electrolytes.  Mr. Lieberman also brought up your nervous system.  Neurotransmitters work as ions too.  Your nervous system passes electrical currents all the time.  


    Mr. Lieberman conducted an experiment to show the presence of electrolytes.  The goal was to light a lightbulb.  At first this was attempted by placing a beaker of salt under the lightbulb where the metal ends were.  The lightbulb did not light.  Then the same thing was done with sugar.  The bulb still did not light.  Water was used next but the electrolytes were to weak to light the bulb.  When adding a little salt to the water the electrolytes were strengthened and were able to light the bulb.  When even more salt was  added the electrolytes were even stronger and the light was brighter.  

Strong Electrolytes

No Electrolyte

Weak Electrolytes

    That sums up today's class.  We received a worksheet titled Dissolving that you should begin to complete.  The unit test has been postponed to next friday.  Hope everyone has a happy Thursday and a good night!!  

The next scribe will be..................... Nicole C.

Good Luck! :)

chemical reaction notes and chemical reaction lab

This post is a combination of what happened in chemistry today and yesterday!

So on Tuesday we finished up the notes and started our classifying chemical reactions lab which we finished today.

There are five types of chemical reactions synthesis, decomposition, single replacement, double replacement, and combustion.

We finished talking about the last three:

3. Single replacement. This type of reaction occurs when there are three elements and one replaces another. Element + compoundà product + product. A + BCà AC + B (if A is a metal) or A + BCà BA + C (if A is a nonmetal).

4. Double replacement. This type of reaction occurs when a metal replaces a metal and a nonmetal and nonmetal. Compound + compound product + product. AB + CD à AD + CB.

5. Combustion. This type of reaction occurs when a hydrocarbon reacts with oxygen gas. In general CxHy + O2 à CO2 + H2O.

Then we did the classifying chemical reactions lab. For this lab you did a series of 7 reactions which each represented 4 out of the 5 reactions (everything but combustion).

So this is what you had to do and the observations me and my partner got:

Reaction #1: You had to heat a piece of magnesium metal and let it light and then hold it over an evaporating dish. Our observations for this were, obviously, that the magnesium light up had a bright glow.

Reaction #2: for this you add hydrochloric acid to a small test tube and then add strip of Magnesium metal. Then you almost immediately light a wood splint and place it in the mouth of a test tube. For this we saw it fizz and make a POP! sound.

Reaction #3: for this you put a piece of ammonium carbonate in a test tube and heat it for 30 seconds. Then you piece of wet litmus paper in the mouth of the test tube and afterwards test the gas with a wood flint. For this we saw that the stick immediately went out after we put it in and the litmus paper turned blue very fast.

Reaction #4: For this reaction we put calcium carbonate in a test tube and added hydrochloric acid to it. Afterwards we light a wood splint and put the burning splint halfway down the test tube. Our observations for this were that there was no temperature change and the splint went out.

Reaction #5: For reaction five you had to add copper (II) chloride solution into a small test tube and add 1 piece of mossy zinc to the test tube. We observed that the mossy zinc instantly turned black.

Reaction #6: For this reaction you had to add copper (II) chloride to a small test tube and add sodium phosphate to that. Our observation for this was that the mixture precipitated and clumped.

Reaction #7: For the final reaction you add sodium hydroxide solution and phenolphthalein into a test tube and mix the solution. Then you add hydrochloric acid until there is a permanent color change. So as you can guess in this reaction we observed a color change. It went from a clear liquid to a pink liquid.

That was all for the lab and class! Tonight’s homework is to finish this lab, it’s due tomorrow.

The next scribe is ……….. Stephanie K.

Balancing Equations and Chemical Reactions

Today we began our discussion on the balancing of equations. One of the most important things to remember as you balance a chemical equation is the Law of Conservation of Mass, or in other words, what you start with you have to end up with. When dealing with an equation, you may only change the coefficients, NOT the subscripts. Two things you can do when balancing an equation is first: to start simplifying the atoms that only appear once in the formula, and second: to always balance and oxygen last.

There are five types of chemical reactions. The two we began discussing in class today were:

1. Synthesis reactions. These types of reactions occur when two substances (generally elements) combine and form a compound. Reactant + Reactant --> 1 product or A + B --> AB. An example of a synthesis reaction is 2 H2 + O2 --> 2 H2O

2. Decomposition reactions. This kind of reaction occurs when a compound breaks up into the elements or in a few to simpler compounds. 1 reactant --> product + product or AB --> A + B. An example of this type of reaction is 2 H2O --> 2 H2 + O2

In order to further explain these two reactions, Mr. Lieberman showed us two demonstrations.
http://www.youtube.com/watch?v=8ncN7AdROrY

The first video demonstrates a synthesis reaction. Liebs used a balloon filled with hydrogen gas and a candle as the two reactants. When the open flame touched the candle, it reacted violently with the hydrogen inside the balloon, causing the balloon to pop (although it was more of an explosion) and the product of the reaction was water.

The second video shows an example of a decomposition reaction. A large graduated cylinder was filled with hydrogen peroxide and some dish soap. When Liebs added the catalyst, the hydrogen peroxide reacted and foamed over the top of the graduated cylinder. Liebs calls this demonstration "elephant toothpaste."
http://www.youtube.com/watch?v=f_ftXxCwvDw

So that pretty much sums up what we learned in class today. Sorry everybody, my computer won't let me upload/embed the videos, so I just posted the web addresses to each of the videos. Hopefully they work.
Also, the next scribe is................. Nirali P.