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Old 07-02-2006, 01:27 PM   #1 (permalink)
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Chemistry equations... HELP ME/ GRADES ARRIVED

Okay, I'm in General Chemistry Lab I. Meant for pre-med types. I'm taking the lab but haven't actually taken the lecture yet (that's in the fall). So I'm doing my lab report and I'm having trouble figuring out some terms here...


Lab procedure/point:
We mixed acetic acid with zinc metal and iodine crystals to prove that quantities are conserved throughout the reaction to make zinc iodide. All different amounts of each to prove that every time, the ratio of zinc to iodine that actually reacted remains the same.

So on to the post lab questions:

It says to calculate moles of zinc reacted and moles of iodine reacted from one "good" trial. That seems simple enough. (I think 1 mole = atomic weight in grams.) But then it says to "Separately divide moles of zinc and moles of iodine by smallest mole value". Frankly, this sounds like they forgot to mention a detail that would make this clearer... I have no idea where to find the "smallest mole value". Smallest amongst the trials? Which one, zinc, iodine, or zinc iodide?

The numbers I'm using: Total zinc that reacted: 0.56g. Total iodine that reacted: 2.00g. Total zinc iodide produced: 2.64g. (Yes, I know, it's not precise.)

From here, by the by, I'm supposed to calculate the empirical formula.... yeah. Right.

So pretty please, help me. If you don't want to type it all out, respond or pm me asking for my cell number, and I will be your best friend forever and ever, more or less.

Here's hoping! THANKS!
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Old 07-02-2006, 01:54 PM   #2 (permalink)
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You have the definition of mole wrong, it is actually 6.022*10^23 units like atoms or molecules. So you need to find out how much 1 mole (or mol) of Zn, I and ZnI weighs in grams first, then you convert your weights to moles by dividing measured weight with mole weight. Then I think you'll pick the smalles value of the three and divide all the mole values with that, so you can see the proportions more clearly. One (the smallest) will be 1 and the rest will be bigger than 1.

I hope this helps.
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Old 07-02-2006, 01:58 PM   #3 (permalink)
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Oh, crapola, really???
Okay, I'm going to go check that out. You have my thanks in superfluous amounts! :*:*

Okay, hang on: so 1 mole of zinc = 65.38 * 6.022*10^23 ?
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Old 07-02-2006, 02:35 PM   #4 (permalink)
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jess

ok, you were essentially right in your first post, for the purposes of this problem. pip is right in that there are 6.023*10^23 molecules in a mole, but what you're looking for is that molecular weight, in grams / mole.

So, for Zn = 65.409 g/mol, I2 = 253.81 g/mol, ZnI2 = 319.22 g/mol

All you need to do is multiply/divide your measured masses by the atomic weights to get the moles of each one. Then, assume that the equation looks something like x*Zn + y*I2 => zZnI2.

The moles that you calculated in the first step are essentially those values of x,y, and z....only, you need to normalize them to make it sort of a pretty equation. So, if you divide them all by the smallest value, then your smallest number will now equal 1.000000000. The other values will be normalized against this coefficient.

Hope that helps
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Old 07-02-2006, 02:43 PM   #5 (permalink)
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So. If the measured amount of zinc (reacted) was 0.86g, the mole of that would be 0.86g/65.409g? or is 65.409/0.86? Or neither?
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Old 07-02-2006, 02:45 PM   #6 (permalink)
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Quote:
Originally Posted by JustJess
Oh, crapola, really???
Okay, I'm going to go check that out. You have my thanks in superfluous amounts! :*:*

Okay, hang on: so 1 mole of zinc = 65.38 * 6.022*10^23 ?
One mole of Zn is 6.022*10^23 atoms of Zn and it weighs 65.38 grams.

Quote:
Originally Posted by JustJess
So. If the measured amount of zinc (reacted) was 0.86g, the mole of that would be 0.86g/65.409g? or is 65.409/0.86? Or neither?
And this is where unit algebrae is your friend. You have g (measured weight)and g/mol (atomic weight) and you want mol. So you take the g and divide it with g/mol to get mol. x/(x/y)=y

Last edited by Pip; 07-02-2006 at 02:48 PM.. Reason: Automerged Doublepost
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Old 07-02-2006, 02:50 PM   #7 (permalink)
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jess

in this case, don't think...just follow your units.

0.86gZn * (mole Zn) / (65.409 g Zn) = mole Zn. So, its the first one. Use the atomic weights like conversion factors.

edit : or, you could just do as pip said
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Old 07-02-2006, 02:50 PM   #8 (permalink)
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Okay, that part is starting to make sense. Got that, atoms vs. weight, okay. And the weight is based on how many protons/neutrons actually make up the atom of the element.

BUT: you said the mole of how much reacted would be how much I measured (saying 3.00grams of iodine) divided by it's molecular weight (126.9045 for iodine). BUT Iodine crystals are I2 - so is that 3.00/126.9045 or 3.00/253.809?
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Old 07-02-2006, 02:53 PM   #9 (permalink)
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jess: the second. The atomic masses given in the periodic table correspond to the mass of 1 mole of the monatomic substance. So, for I2 its 253.809. Same for H2, Cl2, etc.

edit: because i'm a nerd, and a pure monatomic thing can't really be a compound.
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Old 07-02-2006, 02:59 PM   #10 (permalink)
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Crap on a stick, I don't know what I'd do without you two. Seriously. I don't understand my lab instructor nearly this well!

Okay, got the concept of how to find out how much moles actually reacted. So from here, they seem to be asking for a ratio, according to Pip, so I'll try that.

Off to try to apply new knowledge. Genuine smooches to my knights. :* :*

(Feel free to check in on the thread later... I just might be on here a smidgeon. Maybe. I hope not.)
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Old 07-02-2006, 03:11 PM   #11 (permalink)
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I wouldn't be too sure that atomic weight of I2 = 2*atomic weight of I since some energy/mass is lost or gained in forming bonds. But since you are dealing with only two significant numbers here anyway it probably won't matter in this case. I'm off to sleep now, good luck Jess!

(And I've never been able to understand the lab instructors either.)
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Old 07-02-2006, 04:15 PM   #12 (permalink)
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Does this make any sense to you guys? I thought the formula for zinc iodide was ZnI2. The following are my calculations.

Moles of zinc reacted: 0.86g/65.38g/mol = 0.013154
Moles of iodine reacted: 3.00g/253.809g/mol = 0.011820
M Iodine = 0.011820 = 1
0.011820 0.011820

M Zinc = 0.013154 = 1.11286
0.011820 0.011820
Integer values: 1 and 1
Chemical formula for zinc iodide: ZnI
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Old 07-02-2006, 05:42 PM   #13 (permalink)
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Iodine mol weight = 126.9 g/mol
Zinc mol weight = 65.49 g/mol

.86g zinc / 65.49 = 13.1 mM
3.00 iodine / 126.9 = 23.6 mM

I think the amount of zinc might be slightly off, but since the reaction required approx 2 moles of iodine for every 1 mole of Zinc, the formula should
be:

Zn(2+) + 2 I(-) <-> ZnI2

mM should be mMol

Last edited by rofgilead; 07-02-2006 at 05:43 PM.. Reason: Automerged Doublepost
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Old 07-02-2006, 06:11 PM   #14 (permalink)
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Quote:
Originally Posted by JustJess
Does this make any sense to you guys? I thought the formula for zinc iodide was ZnI2. The following are my calculations.

Moles of zinc reacted: 0.86g/65.38g/mol = 0.013154
Moles of iodine reacted: 3.00g/253.809g/mol = 0.011820
M Iodine = 0.011820 = 1
0.011820 0.011820

M Zinc = 0.013154 = 1.11286
0.011820 0.011820
Integer values: 1 and 1
Chemical formula for zinc iodide: ZnI
jess,

you're pretty close. you've got the molar (stiochiometric) ratios of Zn to I2 correct... Those are 1:1. However, don't forget that solid Iodine is diatomic (ie. it's I2 not I), so your reaction will look something like:

Zn + I2 => ZnI2.

Did you also recover the solid zinc oxide? If so, you should be able to do a separate molar balance on it. Look back to the original masses you posted: Those come pretty close to a 1:1:1 ratio for Zn, I2, and ZnI2.

rofgilead - I think your reaction would work well if it was an aqueous reaction. I'm sort of surprised that it's not, but apparently the Zn metal and I2 stay in solid form, despite the presence of the acetic acid.

And finally, I found this linky that seems pretty similar to what you're doing, jess.

pip - yeah, i think within the sig figs available here, those numbes are pretty accurate. considering that the amount of energy is in Joules, and if we look at an approximation like e=mc^2, m=e/c^2....that's going to be a *really* small mass that would gained or lost in bond rupture/formation.
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Old 07-02-2006, 07:22 PM   #15 (permalink)
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No, pigglet, I love you AND your piggy style. It's true.

Yes, I went back and noted that I should have calculated the molar weight of reacting iodine with 126.xx instead of 253.xx, so the ratio ends up iodine to zinc, 2:1... which would make sense since it's supposed to be ZnI2. !!!!

We didn't get into zinc oxide at all - I assume that's because this is a basic sort of beginner's class. But I don't think we're doing too badly since I haven't taken any Chem since high school!

Seriously... I'm getting all mushy here. It's time for bed. Thanks so so so much for all of your help - you kept me sane!
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Old 07-02-2006, 08:26 PM   #16 (permalink)
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Quote:
Originally Posted by Pip
I wouldn't be too sure that atomic weight of I2 = 2*atomic weight of I since some energy/mass is lost or gained in forming bonds. But since you are dealing with only two significant numbers here anyway it probably won't matter in this case. I'm off to sleep now, good luck Jess!

(And I've never been able to understand the lab instructors either.)
The mass loss/gain for chemical reactions is completely negligible even when using a greater number of significant figures.
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Old 07-02-2006, 10:29 PM   #17 (permalink)
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Quote:
Originally Posted by JustJess
No, pigglet, I love you AND your piggy style. It's true.
Awwwww....shucks...

If you want to run this stuff by me later, pop me an im. I got the results using the molecular weight of I2 being 253.xx as well...either way, your original numbers in the earlier post work out to give the equation as being the equation given in the link, with some room for experimental error.

If you assume that the equation looks like

Zn + I2 => ZnI2

then you should get a 1:1:1 ratio, using the molecular weight of I2 = 253.xxx.

If you use the equation as being:

Zn + 2I(-) => ZnI2

Then you would get a ratio of 1:2:1, using the molecular weight of I = 126.xx

Does that make any sense?

Its a factor of 2, which shows up either in the equation, or in the molecular weights. Let me know if I can do anything else; I'm a nerd for hire
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Old 07-03-2006, 02:43 PM   #18 (permalink)
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This is where having the lecture along with the lab is VERY important. In fact, I've never heard of having the lab first, then the lecture. The lab is supposed to be a supplement to the lecture.
The one tip I can give you is for diatomic molecules (HOFBrINCl....pronounced hawf-breenkl). Hydrogen, Oxygen, Fluorine, Bromine, Iodine, Nitrogen, and Chlorine.
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Old 07-04-2006, 10:59 AM   #19 (permalink)
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Pfsh. Student laborations are 95% about gaining practical knowledge about laboration methods(*) and 2% about gaining a deeper understanding of the theory. Some of the first labs I did in college were on basic mechanics stuff, high school level really, but we had to set up the experiments ourselves while the lab assistants laughed their asses off/hid in their break room. We learned more from those labs than by any later ones.

(*) How to handle chemicals, how to use the equipment, how to determine what parameters to study, how to put out fires, how to explain away why the measured data didn't agree with the theory (equipment malfunction, contaminated samples and local magnetic fields).

The remaining 3%? Giant lasers.
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Old 07-04-2006, 11:14 AM   #20 (permalink)
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So, you guys feel like tutoring long distance come the fall?

Thanks again to the Knights of P! :*:*

And I have to agree with Pip, these labs are seriously junior level. We're doing POSTER PRESENTATIONS! The lab manual suggests friggin' construction paper to make it colorful. Seriously. It's just that it's been 11 years since I had any form of Chemistry, and the only reason I'm not totally screwed is because my Bio class covered lots of the basics of biochem.

I can't wait til we get to the giant lasers... Muahahahahaa!
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Old 07-20-2006, 06:24 AM   #21 (permalink)
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Hey guys, guess what? We got an A in Chem I lab!! Without having taken lecture or any other class in Chem in oh, 11 years!

I heart Pip and Pigglet a LOT. :*:* rofgilead is pretty keen too.

I'll see y'all on this thread in September, if you can bear to look at it...

Wheeee!!
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Old 07-20-2006, 06:42 PM   #22 (permalink)
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Congrats jess, but you're the one who got the A...we're just sort of an extended study group Come Sept., you know we'll be here - that's what we do
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Old 02-24-2007, 01:17 PM   #23 (permalink)
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Hey guys - guess what? I made it to Chem Lecture 2.... and I have more questions!

Rate of reaction stuff:
equation is H2 + I2 -> 2HI
If rate of disappearance of I2 is .0037mol L-1 s-1, what is the rate of formation of HI?

I think it's -1/2 * (-.0037), but the book says -2*(-.0037). ????? Am I crazy? Doesn't half as much HI get made as I2? Or is is that twice as much gets made? I'm confused as to which logic path to follow.
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Old 02-24-2007, 01:58 PM   #24 (permalink)
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Each time you split an I2 you make 2HI

think of it like this:

H-H + I-I =========> H I H I ========> H-I + H-I

Because you are taking two molecules of gas, and making two molecules of gas there is no change in volume or pressure. The point that is confusing you is that only HALF of the reacted gas is Iodine.

If you start with a 50/50 mis of Hydrogen and Iodine, and Iodone is consumed at 0.0037mol/L/s, then so is Hydrogen.

Therefore, unless you have invented a way of creating atoms from vacuum, you have generated HI molecules at 2 x 0.0037, which equals 0.0084mol/L/s.

I hope I've got that right, or they'll make me give my degrees back...
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Old 02-24-2007, 02:03 PM   #25 (permalink)
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Okay, I think that makes sense. At least, it's going to have to. Remind me not to wait on doing recitation homework, would you?
Thank you! I'll be back... it's going to be a long weekend.

Btw, I don't suppose you remember the difference between ln and e (in terms of log stuff)?
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Old 02-24-2007, 02:45 PM   #26 (permalink)
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jess, daniel has spot on. for the ln and e bit, they're inverse functions. so exp(ln(x)) = x, and ln(exp(x))=x. The wiki is pretty ok with this one. If you're still confused about the reaction stoichiometry stuff later, post back and we'll try to clear it up. hope the ln/exp stuff helps.
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Old 02-24-2007, 03:07 PM   #27 (permalink)
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Thanks, pigglet! (I was hoping you'd come back to this thread. )
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Old 02-24-2007, 07:30 PM   #28 (permalink)
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for you babe? anything :P

jess,

because i'm a nerd, let's go ahead and hit that stoich junk. Tell me if this helps:

write the equation as :

aH2 + bI2 => cHI, where a, b, and c are constants.

you're given that dI2/dt = -.0037 mol/L/s

you want d(HI)/dt.

so, you write d(HI)/dt = - d(I2)/dt * c/b

The idea is that for every b moles of I2, there will always be c moles of HI.

a=1
b=1
c=2.

Many times in reaction chemistry, we take the (-) sign out of the conversion equation I put up, and write

a=-1
b=-1
c=2

to reflect that H2 and I2 are being consumed. It also makes the math more standardized. But as long as you keep track that some stuff is being generated, and some other stuff is being consumed, then it all comes down to the same thing. Production of HI shouldn't be a negative number.
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Old 02-25-2007, 03:29 AM   #29 (permalink)
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Quote:
Originally Posted by pigglet
for you babe? anything :P

jess,

because i'm a nerd, let's go ahead and hit that stoich junk. Tell me if this helps:

write the equation as :

aH2 + bI2 => cHI, where a, b, and c are constants.

you're given that dI2/dt = -.0037 mol/L/s

you want d(HI)/dt.

so, you write d(HI)/dt = - d(I2)/dt * c/b

The idea is that for every b moles of I2, there will always be c moles of HI.

a=1
b=1
c=2.

Many times in reaction chemistry, we take the (-) sign out of the conversion equation I put up, and write

a=-1
b=-1
c=2

to reflect that H2 and I2 are being consumed. It also makes the math more standardized. But as long as you keep track that some stuff is being generated, and some other stuff is being consumed, then it all comes down to the same thing. Production of HI shouldn't be a negative number.
That's what I said, but with mathematics...

It's the maths that caused me to not enjoy this type of stuff when I was in University.

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Old 02-26-2007, 05:24 AM   #30 (permalink)
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Piggy - yes, that did help, actually. Enough so I could explain it to a classmate! Daniel - yeah, math is why I hated physics too. Loved the theories, hated the application of the theory.

Let no one ever disparage Chemistry Nerds in my presence, ever. Ever. I'm on to cramming a bit more info into my head today at work (I hope) for the exam this afternoon... I really have to practice this crap more than the weekend before the test, I think.

On the bright side, you are wonderful nerds! If you want info on anatomy or something along those lines, I can nerd out on that instead.
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