For your introduction you may leave your equations unbalanced and you can assume the mole ratio of al to the alum is 1 to 1.
Post all questions on the blog.
Thursday, October 16, 2008
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This blog is for RMHS students enrolled in the IB Chemistry course. The purpose of this blog is to facilitate communication between the teacher and IB Chem students, as well as between students in the course. All postings should be relevant to the IB Chem course, and must be school appropriate. If you have any questions, please contact Ms. Marostica. Thank you for visiting.
75 comments:
Ms. Marostica,
On the equation for Potassium Alum
AlK(S04)2 x 12H20 Is the water part of the potassium alum or is it seperate?
I had a question... you say in the intro we can leave it unbalanced. Does that mean not in processed data?
The water is part of the alum. That's why is a hydrate, might want to look that up. Sorry, shouldve said processed data not intro. Unbalanced equations can go in proc data.
Is independent variable everything we poured into the solution?
What was the mass of the filter paper again?
I have a uncertainty with aluminum because we originally wieghed it should I have any other uncertainties?
Ms. Marostica,
At lunch yesterday didn't you say that the only way you can get the theoretical yeild was from a molar ratio of the Al in the first part and the Al in the alum?
I think it was just the alum in the end, not aluminum, but I'm pretty sure that is the way to get the theoretical yield.
The balance has an uncertainty of +/- 0.1 right? I am...uncertain of this uncertainty because I remember the last lab's balances' uncertainty to be +/- 0.001
how can we get the theoretical yield without balancing the equation? Because without the balanced equation how are we supposed to know the mole ratios and thus the limiting reagent and yield etc........
and is the filter paper mass .2 or 1.2?
anyone found good websites?
and any good ideas on a title?
haha. the questions of a procrastinator
Well when you do balance it, it doesnt change what matters. And uh...I had 1.2
I believe that since Al is the limiting reagent, all of it will be used up. Therefore the theoretical yield will have the same amount of moles as the Al.
.....I am confused, this blog is made so that we can help each other.....but...didn't you say we can't get help from other students?
lol well..... we're not 'helping'. we're throwing thoughts out into the universe and they're coming back through diff people haha. lets go with that ;)
and sweet. thanx.
and would indep other reactants and dependant be mass of crystals?
what is the sig fig?
I did all my sig figs to 3 places because that was my least acurate measure.
but we are given 1 gram of Al. not 1.00. so wouldn't that make it only one sig fig?
o and if someone could help me with the hypothesis that would be sweet =]
I ignore that im counting things such as what I measured my self. Idk if thats right, but it seems ok to me
can someone explain to me exactly what the percent yield and theoreticla yield is?
Michelle theoretical yield is the thing i showed u in class but idk wat %yield is
percent yield = (actual. yield / theor. yield) times 100
i cant find a website as to why waiting overnight is better for crystal formation. help?
Percent yield is the % of how much product you got compared to how much you are suppose to get
For example: If i am suppose to buy 10 pounds of clay at a store, and I end up buying 6 pounds, then my percent yield of clay is 60%.
The 10 pounds of clay represents the theoretical yield(how much product you should theoretically get in a reaction)
muchos gracias. and same question as tia, website for why cooling overnight is better?
Ok I can not find the formula for the first reaction to save my life, and im starting to panic, help?(Basicly, Potassium Hydroxide+Aluminum=?)
Check out this site, its where i got my info on crystal stuffs: http://www.newton.dep.anl.gov/askasci/chem03/chem03139.htm
1st reaction is Al(s) + K+(aq) + OH-(aq) → [Al(OH)4]- + H2 (g) + K+(aq)
I talked to patrick and he told me that we wait overnight so the product is done reacting and can form into a solid. and chris, what i got down for the equation:
H2O + Al + KOH ---> [Al(OH)4] + KOH
ah crap i totally messed up on the equation. follow katie's equation.
Errmmm... would it be cheating if we stumbled upon the balanced equation online?...
I'll cite it I promise!!!
(Will give URL if its not considered lab write-up blasphemy. Hope that's ΔH<0 (cool*).)
*... D=
thankyou!!! so am i the only one who hasnt started the processed data yet lol?
um and 'in a problem very similar to the ones we are doing on this lab'.... what would we need percent unceratinty for?
Im working on the bloody processed data >:( and its not making me a happy camper
Um ok, now on the fisrt reaction, did we ever measure the actual yeild? or do i just move on? if anybodys allowed to tell O.o
ditto. and i say that the balanced equation is a work of god. but i'm not the person grading it haha and she said we can use the unbalanced one. is it much different in terms of AL to alum mole ratio?
well.... in a theoretical lab we would have measured the mass of product plus the filter paper, and the filter paper. so actual yield in processed data would look like (prod+filter paper) - (filter paper) = actual yield.... theoretically of course ;)
Im off to bed but If u people get up at 5:00 am I should be online for helo im done with my lab except for the conclusion and im pretty sure I have percent uncertainty correct too! good night and good luck.
Griffin out.
stephen wanna hook me up with percent uncertainty tomorrow morning if you get the chance? thanx =)
ahhh, i cant think straight. lol
anyone wanna explain dependent and independent variable of this lab?
Well I'm just about to give up on this lab out of confusion. I don't see how i can figure out accurate data for three chemical equations without balancing the equations. I am getting different results than everyone I talked to so ...uh... I guess I'll just wing it. Wish me luck :(
You don't need to think about it too hard.
The mole to mole ratio of Al to potassium alum is 1 to 1.
anyone find another website about the temp change?
anyone want to explain to me how to figure out the percent error?
chris had a really good website... look his up
we worked on the equations during 6th period if you want them.
the equations for percent error?
wait you worked on balancing the equations? i though we didnt have to do that.....
aaaaahhhhh
Wow... I am stupid. I just realized I don't have to calculate the actual yield. Now I feel much better. Thanks for the help, I am just confused on why my result is different from a friend's on the theoretical yield.
no, we don't have to balance the equations. I just have the corrected versions we did with ms. marostica if you want to check them or anything.
geoff, what mol:mol ratio did you use?
if it's not 1:1, that might have thrown you off.
ummmm why isn't it 1:1? she said it was. aaahh.
Tia ignore all of this I was confused, it is 1:1
yes, it is 1:1.
i'm just saying that if that's not what you used, it would have thrown off your calculations,
how do you figure out the percent error?!?! im dying here...someone please help!
So what's the difference between letting the crystals sit out over night v. putting it in the ice bath?
Edward, to find the percent error, you do this:
(Uncertainty/Measurement) x 100.
are there three equations?!? anyone?
lauren look at the website chris posted earlier.....
http://www.newton.dep.anl.gov/askasci/chem03/chem03139.htm
no percent error is (experimental results-theoretical yield)/theoretical yield x 100
there are three or four equations
i got a negative percent error...and im sure thats wrong. my experimental result was smaller than my theoretical result...
My results were like that too
I was talking about % uncertainty not % yield. Sorry I guess I was confused. % yield is (Actual yield/theoretical yield) x 100%. I am not exactly sure what % error means though...
percent error is like when you take all the percent uncertainties and add them up and then divide by result. i think. i'm not sure. look at handout she gave us. do limmiting reagents need percent uncertainties?
and for sig figs.... do we need to do sig figs at each step? and wont that really be inaccurate if we do?
Tia did you ever come up with a good title?
If your printer isd broke is broken or down, you need to find a way to get me a hard copy. I will not be printing anyone's labs. This is YOUR responsibility. Figure out a way o get it to me.n print pprinter printer
If your printer is broken or down, you need to find a way to get me a hard copy. I will not be printing anyone's labs. This is YOUR responsibility. Figure out a way to get it to me.
Tia is right about the percent error thing. The handout it's on is called propagation of uncertainties.
Synthesis of potassium alum. how do you do limmiting reagent on first equation? its Al right?
before i hit the hay.... ms marostica, aren't you so proud of us for using this awesome blog?
Cheah! 72 comments on the LAB OF DOOM!
Though, was it really productive at all or was there just a lot of confusion going on and people asking the same question yet never being answered?
(Look at this timestamp... almost 5 am. Awesome. I get free sleep deprivation tonight.)
good morning chemistry procrastinators =] sooooo..... for those of u stuck on percent yield heres an equation to use:
actual yeild
_____________
theoretical yield
multiply the decimal u recieve for a percent yield.
srry multiply that decimal by 100 to get a percent yield
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