A peek inside the everyday happenings of our classroom. This is an interactive learning environment for students and parents in my Honors Chemistry 173 class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.
In class on Thursday December 20th and Friday December 21st we did the Properties of Nuclear Radiation Lab. In this lab we were trying to figure out which type of radiation (alpha, beta, or gamma) would travel the furthest and which ones are easiest to shield. We did so by placing beta and gamma saucers in a Nuclear Scaler.
* We did not have access to alpha saucers, but Mr. Lieberman provided the results.
Beta and Gamma Saucers
Nuclear Scaler
There were 6 shelves in the Nuclear Scalers and the saucers were tested twice on each shelf.
Then the tests were repeated on the 2nd shelve from the top with different shields in place.
The shields could have been anything. Some examples were: Thick Lead, Thick Plastic, Thin Lead, Thin Plastic, and Paper.
Make sure everyone makes 2 graphs one for the different shelves and one for the different shields each with a line for the 3 types of radiation.
This Lab will be due when we return from break. Also, Everyone should start to study for finals.
Hey guys! So today in class we completed the Metal Reactivity Lab which had some pretty cool reactions in it. We started off the class, however, with watching some videos about Metal reactions and how the place on the periodic table corresponds with how strong a metal reacts with water (H2O). Both videos showed some pretty cool explosions and in case you want to watch it again, the episode of Braniacs is below even though everyone knows the explosions were fake :/
After the videos, we proceeded on with the lab. We began by taking a well plate and filling three wells with H2O, three wells with phenolphthalein, and three wells with HCl. Next, we took three pieces of magnesium ribbon, three pieces of aluminum ribbon, and three pieces of calcium metal and put one piece of each in their own wells in each of the liquids. We observed the reactions that took place and recorded them in a data table that will go into our lab books. It is a pretty straight forward lab and hopefully you understand the main idea that as you move down a column on the periodic table, elements become reactive in contrast to when you move across a period and the metals become less reactive. The images below show what happened in the the experiment. There are no picture of the elements in phenolphthalein because no reactions occurred.
Magnesium in HCl
Calcium in HCl
Aluminum in HCl
Aluminum in H20
Calcium in H2O
Calcium in H2O
Magnesium in H2O
I hope you guys have a good weekend and I would just like to remind you that the Alternate Universe Periodic table is due Monday along with this lab. The test is Tuesday and the Webassigns are due that days as well!
Next scribe will be Chris!! Don't worry it's not that bad :P
Yesterday was a great Monday in Chemistry! We started off class with some demos about elements patterns and properties. Mr. Lieberman brought us over to the flame hood and burned small samples of lithium nitrate (LiNO3), sodium chloride (NaCl), potassium nitrate (KNO3), copper (Cu), and strontium (Sr). Along with admiring the pretty colors that the flames gave off, we also observed that elements in the same group or family tend to have similar properties. We discussed how this might be a real-life application in investigating a crime scene with chemical residue, testing elemental properties, or making fireworks.
Here are the videos:
Using the information from this demo, we then played electron configuration battleship. Everyone was given a sheet with two identical periodic tables on it, but these tables were the extended version with the bottom two rows of the lanthanides and actinides placed in their original spot. Instead of using numbers to locate each element, we used the endings of their electron configurations. We observed some key patterns during this.
Here's a clear picture of the general patterns we found:
Here's one of every element individually labelled with their ending electron configuration:
(PLEASE NOTE: Helium (He) in the top right is different in that it ends with 1s rather than 1p.)
Towards the end of class, Mr. Lieberman demonstrated how we can use these configurations to abbreviate the entire electron configuration. Since, as good chemists, we only care about the electron that will react, we can use the previous noble gas to speed up the process. For example, if I wanted the electron configuration for uranium (U), I would write [Rn]7s2-5f3 rather than starting all the way from 1s1. If you're confused, make sure to come in for some extra help!
We didn't stamp the homework today, but the rest of the electron configuration packet it due for tomorrow! Also, keep in mind that tomorrow we have a sports assembly during period 2, so if you are missing that for marching band or a sport, make sure to check Moodle for the work you missed!
The next scribe will be... Rachel S.
Sunday, December 9, 2012
Quantum Numbers
Thursday and Friday in class we learned about quantum numbers and electron configurations! The three coordinates that come from Schrödinger's wave equations are the principal (n),angular (l), and magnetic (m) quantum numbers. These quantum numbers describe the size, shape, and orientation in space of the orbitals on an atom. The principal quantum number (n) describes the size of the orbital. Orbitals for which n = 2 are larger than those for which n = 1, for example. Because they have opposite electrical charges, electrons are attracted to the nucleus of the atom. Energy must therefore be absorbed to excite an electron from an orbital in which the electron is close to the nucleus (n = 1) into an orbital in which it is further from the nucleus (n = 2). The principal quantum number therefore indirectly describes the energy of an orbital. The angular quantum number (l) describes the shape of the orbital. Orbitals have shapes that are best described as spherical (l = 0), polar (l = 1), or cloverleaf (l = 2). They can even take on more complex shapes as the value of the angular quantum number becomes larger. There is only one way in which a sphere (l = 0) can be oriented in space. Orbitals that have polar (l = 1) or cloverleaf (l = 2) shapes, however, can point in different directions. 3) We therefore need a third quantum number, known as the magnetic quantum number (m), to describe the orientation in space of a particular orbital. (It is called the magnetic quantum number because the effect of different orientations of orbitals was first observed in the presence of a magnetic field.)
I apologize for everyone who needed to read the blog yesterday because I wasn't feeling well and forget to write it.
In class on Wednesday, we learned about the development of the Atomic Theory.
First of all, everyone must know that in the olden times, philosophers and scientists could just claim their theories and people would have to believe them. As times when on, scientific data and evidence started being used. Dalton was the first philosopher to use data to support his atomic theory.
John Dalton(Late 1700)
-The atom were tiny, indivisible, and indestructible particles
-Each atom had its own individualistic properties to determine which element it was
-Each element was different because of their different masses
At Dalton's time period, it was amazing that a scientist had used proof and data to back up their theory. Although it was not exactly the correct idea, he had made a great discovery for his times. Then, a scientist named J.J Thomson came and discovered more about the atom. Between the time of Dalton and Thomson, electricity and magnetism had been discovered.
J.J. Thomson (1897)
Through the cathode ray experiment (a vacuum tube with all air expelled from it), he created the plum-pudding model of the atom.
-There were positive charges throughout the atom.
-Electrons were floating around in the cloud of positive charge
-Particles have to balance each other out.
Here's a video explaining the cathode ray experiment.
After Thomson came Ernest Rutherford who conducted the gold foil experiment to also improve the atomic theory. Rutherford was a student of Thomson and wanted to prove his mentor's model, but then ended up discovering his own model.
Ernest Rutherford (1871-1937)
-Discovered the nucleus
-The nucleus had small dense positive charge
This is a video of how Rutherford found this model.
Through this experiment he found the nucleus, but do not confuse this with the fact that he discovered the proton and neutrons, because he did not.
Around the end of class, we learned a little bit about Waves and forms of energy. The speed of light equals wavelengths times frequency. (λν = c,λ=wavelengths, ν=frequency, c=speed of light)
Visible colors to the naked human eye are only a small portion on the electromagnetic spectrum.
-The shorter the wavelength, the higher the frequency which is more dangerous to us humans. (towards the gamma ray end)
-The longer the wavelength, the lower the frequency. Generally, these pass right through humans and are not dangerous. (towards the radio end)
-On the frequency bar, the colorful rainbow in the middle is the place where frequencies can be seen by humans.
In class, Mr. Lieberman did an awesome demonstration with glow in the dark paper and a purple laser pointer. Unfortunately, it was too dark to tape and I don't think anyone would've appreciated flash. Plus, it might have cancelled the effect of the demo. In any case, here's a similar situation of what Mr. Lieberman did.
He did this with a red, green, and purple laser pointer. Purple worked best, though, because purple is on the end with shorter wavelengths.
Hope this all made sense to everyone. Sorry about being late on this post again.
The next scribe will be Courtney Smith. Good luck!!! ^.^
Today in class we worked on the Molar Volume of a Gas Lab. The goal of this lab is to measure the molar volume of Hydrogen gas at STP.
First, we measured out 10 mL of Hydrochloric Acid and poured it into the eudiometer.
Then, we filled up the rest of the eudiometer with water.
Next, we took a small strip of Magnesium and shaped it into a little ball and wrapped the Copper wire around it, which was placed in the stopper.
After, we put in the stopper and placed the eudiometer upside down in the stand, somewhat in a beaker half full of water.
The eudiometer's contents started to bubble, which is the evidence of the chemical reaction, and created the gas. The volume of the eudiometer minus the volume of the liquid left in the eudiometer equals the volume of the gas created by the chemical reaction.
In order to find the mass of the Magnesium strip, you have to use dimensional analysis to convert the length of the strip to grams using the conversion of 10 cm of Mg = 0.1407 grams.
The barometric pressure is 30.04 in Hg (inches Mercury), which you have to convert to millimeters.
Your water vapor pressure depends on whatever temperature you got.
Don't forget to use Dalton's Law of P(total) = P(H20) + P(H2), so that you can find the pressure of the Hydrogen gas to then find the molar volume. The total pressure is the barometric pressure, and the water pressure is your water vapor pressure.
Hey Everyone! Today the lesson we learned was on Gas Stoichiometry. In class we went over our
Ideal Gas Law WKSHT that was due today (answers posted on moodle). We then spent the rest of the period learning Gas stoich.
Stoich using PV = nRT
Mr Lieberman pulled out the mole cannon today and used it as our demo for today's lesson. Here's the video. (Don't know why you guys are so frightened?)
1. So in order to find the pressure of the gas in the tube you have to make and balance a chemical equation. So this is it for the mixture of the solid and gases in the tube.
CaC_2 (s) + 2H_2 O ------> C_2 H_2 (g) + Ca(OH)_2 (aq) (sorry, underscore means subscript)
2. Next, you find the moles of gas formed from the reaction. Just a reminder, CaC_2 is the limiting reactant while the C_2 H_2 is the excess reactant.
.37 g CaC_2 1 mole CaC_2 1 mole C_2 H_2
_____________ x ________________ x _________________ = .006 moles.
1 64g 1 mole CaC_2
3. Next, you have to find the pressure created by the gas. In order to do this, let's use PV=nRT!!
P=? use PV=nRT or P= nRT/V
V= .87L (use the volume of cylinder that Mr. Lieb gave us)
T= 25 degrees Celsius + 273 = 298K (Has to be in Kelvin) (25 degrees is room temperature)
Moles= .006 Moles (work from above)
R= .082 atm/moles K (constant for gas)
Now that we have everything we need, we set up the problem!!!
(.006) x (.0821) x (298K)
P= _____________________
.87L
So in conlcusion, the pressure of the gas in the tube would be 0.17!
You could also solve stoich by using the Molar volume formula with STP. A.K.A. Standard Temperature Pressure
At STP, 1 mole of any kind of gas = 22.4L (Molar Volume)
Sample Problem: Sodium azide is used in airbags to cause them to inflate on impact. What volume of nitrogen gas is formed from the decomposition of 1.00g of sodium oxide (NaN_3)? Assume STP conditions.
So first is the balanced chemical equation.
3 NaN_3 -------> 4N_2 + Na_3 N
So we know this much....
T=273K
P=1 atm.
Let's use the formula
1.00g NaN_3 1 mole NaN_3 4 moles N_2 22.4L
____________ x ______________ x ______________ x ____________ = 0.46
1 65g 3 moles NaN_3 1mole N_2
See, you didn't have to use PV=nRT when you know you can sue STP! Time saver!!!
So that's pretty much what we learned in class today!
HMWK: Study for Quiz Tomorrow (Good Luck on it!)
Gas Stoich #1 WKSHT
Ideal V.S. Combined WKSHT
The next scribe is Danielle Sindelar!!! (Sorry to other people who wanted to be scribe)
Hi everyone. On Friday we had a very interesting lesson on the different factors that affect gases and how they affect them. These factors were pressure, temperature, and volume. There were several demos and explanations for these that I will go over. I have pictures of the actual demos we did in class and not ones off the internet.
Boyle's Law The first three demonstrations were proving this law. The law states that as pressure decreases, volume increases or P1V1=P2V2 where P is pressure and V is volume.
Demo #1 Balloon in Vacuum Pump
As you can see, there is a tube hooked up to the container that sucks out all of the pressure around the balloon. When this happens, the volume of the balloon increases...
The same procedure is done with shaving cream and the result is the same.
Again, the pressure decreases and the volume increases...
This was also done with a marshmallow, but unfortunately i could not get a picture. The same exact thing happened with that.
Charles's Law
This law states that as temperature decreases, volume decreases, or volume is directly proportional to temperature. The formula for this is V1/T1=V2/T2 where V is volume and T is temperature.
The example of this was a demonstration that used liquid nitrogen, which is extremely cold, to lower the temperature of the air inside of the balloon resulting in a decrease in volume.
Here the balloon is being stuffed into the liquid nitrogen. It takes a while for the balloon to deflate and shrivel up. It was difficult to get a picture of the shriveled up balloon because once it was taken out of the liquid nitrogen its volume started to increase rapidly because of the rise in temperature.
Gay Lussac's Law This Law states that the pressure and temperature of a gas are directly related. The equation for this is P1/T1=P2/T2.
To demonstrate this, and instrument that measures pressure was put into the liquid nitrogen to see if the pressure would lower along with the temperature.
Before the it was put in, the device read 15 atm and after it read 4.25 atm, so the pressure did decrease by a lot.
The Combined Gas Law puts all three equations into one, which makes it much easier to remember.
The equation is P1V1/n1T1=P2V2/n2T2. P is pressure, V is volume, n is moles, and T is temperature. This equation is used in problems where there are two sets of variables, such as numbers 9, 11, and 12 in the homework.
Finally, there is the Ideal Gas Equation, in which the constants of the previous laws have been combined to form the gas constant R. This equation is PV=nRT where P is pressure, V is volume, n is moles, R is a constant for pressure (either 0.0821 atm L/ mol K or 8.314 kPa L/ mol K), and T is temperature. This is used in a problem where there is only one set of variables, which includes nearly all of the homework problems.
To end class on a good note, we had fun with liquid nitrogen and did activities such as watching Mr. Lieberman chuck frozen orange at the wall, pouring liquid nitrogen on the floor, and pouring it on Charles.
Homework: Ideal Gas Law worksheet and work on WA The next scribe is.....Kevin L. who sits next to me
Hello all, today in class discussed and demonstrated how gas and pressure are related. We did 3 demos, and although due to mitigating circumstances, neither video nor photos are available for you to view these demos, I think that my eloquent writing style better exemplifies these demos. Therefore, there were 3 demos on pressure, and we took some notes. 1. Can Crushing http://www.youtube.com/watch?v=QVayky_b-6U
In this demo, we poured a little boiling water to can, then we put a flame under the can to heat it, then we put it upside down into the a bowl of water, and a there was slight "bang" and the can crumpled. One might wonder why this would occur, it seems to defy the laws of chemistry. However, if one thinks this, then one does not really know the laws of chemistry at all in the first place, but only a select few do. As for the explanation, the can crumpled because the air pressure in the can is equal to the air pressure outside it. Thus there are no overall forces. However, when the water is boiled, lots of steam is produced, which then pushes air out the can, until the can is completely filled with steam. The steam has the same pressure as the air outside the can, so nothing occurs yet. But once the can is placed upside down in the bowl of water, the steam cools and turns back into a liquid, and it thusly does not have the same pressure as the air outside the can. The inside of the can becomes a vacuum, and with no pressure inside the can to fight the much higher air pressure outside the can, the air crushes the Coke can. I also lied, i will be showing video and photos, just not from class.
2. Stuck to the ground http://www.youtube.com/watch?v=zqyuaMqOEGA The second demo, Mr. Lieberman put a vacuum device clamped onto a table and then onto the floor. He then challenged anyone to try to lift up the device. Because of the intense air pressure, no one without the use of performance enhancing drugs would be able to lift it up because the vacuum sucks up all the air. When you lift up the clamps however, it is possible for you to lift it up, because the intense air pressure is let out.
3. The Vacuum http://www.youtube.com/watch?v=Ct4AqODwhI8
In the third demo and final demo, Mr. Lieberman put someone in a plastic garbage bag with a very tight air seal. With such a tight air seal, there was no air in the bag. with no air in the bag, Alex the demonstrator began to feel more compressed and with nothing to resist, the air pressure outside the bag compressed on to him. It was very interesting and one thinks we learned a lot because chemistry is better explained through demos than through lectures.
Throughout the class we took some interesting notes on pressure. We discussed some formulas and concepts on pressure in gases.
Formula for pressure:
We also learned some standard pressure units and values that include:
A. 1 standard atmosphere B.101.3 kPa C.760 mm Hg D.760 torr E.14.7 lbs/in2
Apparently you use these to convert from unit to unit. One may wonder why there is not just one unit for all science, but that is merely wishful thinking. Due to that, here is how you convert from pressure units. I. Converting between atmospheres and millimeters of mercury. One atm. equals 760.0 mm Hg, so there will be a multiplication or division based on the direction of the change.
Example #1: Convert 0.875 atm to mmHg.
Solution: multiply the atm value by 760.0 mmHg / atm.
Notice that the atm values - one in the numerator and one in the denominator - cancel, leaving mmHg. So like in stoich, u want to the units to cancel, and once the units cancel, the rest is fairly easy multiplication done on a calculator. It is important not to have a nervous breakdown about the complex-sounding units, for they merely sound and are written like that to inspire awe and wonder. There you go, and I hope you learned a lot about pressure in gases.
Homework: Do pressure worksheet and begin completing webassigns unless they are not up yet, in which case, do not complete the webassigns, because it is technologically feasible. I suppose you could make your own webassign and then complete that, but one doubts the likelihood of that. Thank you.
Next Scribe: Bram Hill who sits in the front row to the left.
Today in class we learned about the properties of gases and had a few gas demos, after going over and reviewing our tests.
The first gas demo was with iodine crystals. The crystals were heated up and turned into a pinkish/purple color of gas.. There is a myth that since you cannot see gases, they are not real, which this experiment obviously proves wrong.
The next demo was with Hexane. Hexane evaporates very easily into the air, and the gas acts as if it is a liquid. After a little shaking, we were able to pour it down a pipe into a flame, igniting the flame and the rest of the gas coming down the pipe.
The final demo was with baking powder, acetic acid, and a candle. Two beakers, one empty, and one filled with baking powder, are presented. A lit candle is held in both of the beakers, but nothing happens. But when you pour acetic acid into the baking powder and it reacts, if you put the lit candle in it will extinguish due to the carbon dioxide from the reaction causing oxygen, which the flame needs, to be pushed out. This reaction makes the gas act like water and can be poured into the empty beaker, which will extinguish a lit flame as well. The final part of the demo with the tube and the carbon dioxide did not work so well, but was meant so that gravity pulls carbon dioxide into an empty beaker where the candle can be extinguished.
Homework: Gases Chemthink and Behaviors of Gases worksheet.
Colligative Properties are properties of substances such
as vapor pressure lowering, boiling point elevation,melting point depression,
and osmotic pressure. The main ones we will focus on is boiling point elevation
and melting point depression. Also, these properties are only changed by the
number of solute particles, not by the identity of the solute.
Boiling Point Elevation
The boiling point elevation property is simply
this, ΔTb(subscript) = Kb(subscript) x
m. ΔTb is added to the normal boiling point to get the new boiling
point. Kb is the constant which will be given to you. m is the molality.
That is how you figure out the boiling point when adding one substance to
another.
Freezing
Point Depression
The freezing point depression property is, ΔTf(subscript) = Kf(subscript) x m. The symbols in the
equation meant the same thing as above. Again, the constant (Kf) will be given
to you in problems. Below is an experiment that we did in class. First, Club
Soda was chilled to about -8 degrees Celsius. This was done by adding
normal table salt to ice making the freezing point of the water go way down,
this enabled the water and ice to get very cold. Then when the Club
Soda was removed from the water the bottle was opened to release the carbon
dioxide and flash freeze the soda.
Hello all, today we went over molarity and dilution. Using molarity is pretty simple if you follow some easy steps.
Primero, remember that molarity= moles solute/liters of solution
Por ejemplo, if you have an equation like : Dissolve 4.00 g of NiCl2 x 6H20 in enough water to make 150 mL of solution. Calculate the molarity, so first we do what we learned in stoichometry last unit, 4.00 g x 1 mol/ 237.7 g (which is the molar mass) and you will get0.0168 mol.
Entonces you simply follow the formula mentioned previously: 0.0168 mol (the moles solute you found) / 0.150 L (after you convert milliliters to liters) and your final answer is 0.112 M.
For using molarity, remember that moles= mass x volume so simply make any necessary conversions in any given equation, the do the calculations to find moles. once moles are found. Convert the moles to grams by multiplying the mols by the g/mol. that is ur final answer.
Ademas, 2 other ESSENTIAL concentration units is Molarity, m which is : m of solution= mol solute/ kg solvent and % of mass= grams solute/ grams solution. Basically, if you understand the formulas and know them relatively well, all you have to do is make sure your conversions are correct, and you put the numbers in the right places to be multiplied or divided,
Tambien, it is necessary to understand that diluting a solution is just adding more water and therefore the amount of solute stays the same. Henceforth nb =na and n= MV . MbVb=MaVa
Very important in dilution equations is to first find all the variable you know for sure. for example, if in a given equation M1, V1, and V2 are given for sure. all you have to do is simple algebra to solve for M2. once you get M2 you put all 4 variables back into the equation andipso facto you have your answer. In our level of chemistry you are almost always given 3 variables and thus you only have to solve for one. Don't get caught up over the different letters for variables, because that is just the chemistry gods trying to mess with your brain.
Finalmente, molarity and stoichiometry equations are literally the same things as mentioned previously. just find out all the variables you know for sure, solve for the variable you dont know. and make all the necessary conversions. If you did not receive a 61.6 % on a test like some people did, you should understand it pretty well. That is all for today, and I hoped i somewhat atoned for my good, but not spectactular showing yesterday.
Our class today began with a recap of yesterday's solubility lab. We were showed some tricks for equation balancing and determining solubility before we delved into today's main lesson.
Our main lesson today was the rules for the solubility of common compounds. We learned that certain compounds are soluble in water, unless containing certain ions that make them insoluble. For example, compounds with Chloride, Cl-, are soluble in water unless they contain the ions Ag+, Hg2+2, or Pb2+.
Inversely, some compounds have rules in which they are insoluble in water unless they contain certain ions that make them soluble. For example, sulfide compounds, S2-, are insoluble unless they contain ions of K+, Na+, or NH4+.
We had the rest of the class to work on the lab or the 1st Reaction Solubility worksheet.
There were some great costumes today! Mr. Leiberman had a stellar Captain America costume on!
We started off class by checking in our two homework assignments. If you missed class, need the worksheets, or the answers, make sure to refer back to our Moodle site for all those links!
We spent the second half of class on our Solubility Rules Lab. We mixed six different sodium solutions with eight different nitrate cations to see which reactions would create a precipitate. We looked for "cloudiness" to help determine if a precipitate formed. Just looking at the color change was NOT enough to distinguish a precipitate.
The anions were sodium...
carbonate
chloride
hydroxide
iodide
phosphate
sulfate
The cations were nitrates of...
aluminium
potassium
barium
calcium
copper (II)
iron (III)
silver
zinc
It was a quick and organized lab, so we spent the last 10-15 minutes of class working on the post-lab questions. Also, since some people did not know who Captain America is, we watched a trailer of Captain America: The First Avenger.
Remember, the lab is due Friday! Have a great November!
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 Yieldis 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 yieldis 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 :
Assume
C2H5OH is the limiting reactant and calculate how much C2H5Cl would be formed.
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. ;(
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, thelimiting reactantis 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 MgCl2?
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
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.
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.
Sunday, October 14, 2012
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.
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
This is because you have two reactants (Fe and S) when reacted become a product.
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.
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.
