Posted by Josh on Thursday, August 23, 2007 at 9:29pm.
Yes, we can help. You posted this earlier and Bob Pursley told you that 176 g sounded high to him and for you to show your work and we would find the error. So show your work how you obtained the 176.24 g theoretical yield.
Ok. For the theoretical yield I added up the atomic masses for CuSO4 + 4 NH3. Cu (12.01) + S(32.07) + O4(64.00) + (NH3(17.04) * 4 = 68.16) = 176.24. Am I finding the theoretical yield the wrong way?
NO NO NO read what I posted below please
First, that's not how to determine the theoretical yield.
Second, Cu is 63.5 something and not 12.01. You have used the atomic mass of C and not Cu.
Let me help you on this...
To find the limiting reactant first of all.
-You have to first find the moles you have of each reactant.
- then go and find out how much of each reactant do you need to consume the other reactant
- then compare it to how much moles you do have of each and the one that needs more to consume than you have is the limiting reactant
- use the limiting reactant to go and solve for how much is produced.
- convert moles to grams produced
I'll start with this and PLEASE post your work as I'm NOT going to do all the work =)
Ok, I will. Also, could someone tell me how to find the theoretical yield, since I'm apparently doing it wrong for some reason.
I just gave you instructions for finding the theoretical yield Josh.
Uh, christina, you just told me how to find the limiting reactant, not the theoretical yield.
-You have to first find the moles you have of each reactant.
- then go and find out how much of each reactant do you need to consume the other reactant
- then compare it to how much moles you do have of each and the one that needs more to consume than you have is the limiting reactant
theoretical yield is:
- use the limiting reactant to go and solve for how much is produced. (theoretical yield)
- convert moles to grams produced
Post work based on steps and if you have problems with the steps just post what you get up to and I'll analyze. =)
Ok, well, I have 1 mole of CuSO4 and 4 moles of NH3, correct?
No, of course not. You just calculated that you have 0.0626 mol CuSO4 and 1.76 mols NH3. Let's review what you and Christina have done.
Convert 10.0 g CuSO4 to mols.
mols CuSO4 = 10/159.6 = 0.0626 CuSO4,
Convert 30.0 g NH3 to mols.
mols NH3 = 30/17 = 1.76 mols NH3.
Determine limiting reagent.
IF we used ALL of the 0.0626 mol CuSO4 we would need how much NH3? We would need 0.0626 mols CuSO4 x (4 mols NH3/1 mol CuSO4) = 0.250 mol NH3. Do we have that much NH3? Yes, we have 1.76 mols.
IF we used ALL of the 1.76 mols NH3, how much CuSO4 would we need? We would need 1.76 mols NH3 x (1 mol CuSO4/4 mols NH3) = 0.440 mols CuSO4. Do we have that much CuSO4? No, we have ONLY 0.0626 mols CuSO4; therefore, CuSO4 is the limiting reagent and some of the NH3 will be used and leave some of it unreacted. This completes the determination of the limiting reagent.
Next, what is the theoretical yield? That is, how much product should we obtain using 0.0626 mols CuSO4? So we need to convert 0.0626 mols CuSO4 to mols Cu(NH3)4SO4. That is done in a similar fashion to the way we converted mols CuSO4 to mols NH3 and mols NH3 to mols CuSO4; i.e., using the coefficients in the balanced equation.
mols Cu(NH3)4SO4 produced = mols CuSO4 x (1 mol Cu(NH3)4SO4/1 mol CuSO4) = 0.0626 x 1/1 = 0.0626 mols Cu(NH3)4SO4 formed.
Now convert that to grams.
Remember you used mols = g/molar mass to convert grams to mols. We just reverse that to convert mols to grams. grams Cu(NH3)4SO4 = mols Cu(NH3)4SO4 x molar mass Cu(NH3)4SO4.
g Cu(NH3)4SO4 = 0.0626 x 227.7 = 14.25 g. SO, we will have 14.25 g of the product formed. This is the theoretical yield.
That last part of the question asks for percent yield if the actual yield was 12.5 grams.
%yield = (actual yield/theoretical yield) x 100 = ??. Just plug in the 12.5 and 14.25 and 100 and you have it.
I'm not looking at the problem at the moment. Did they ask how much of the NH3 remained unreacted. If so, you will need to calculate how much of the NH3 was used and subtract that from the 30 g with which you started.
g NH3 used = 0.0626 g CuSO4 x (4 mols NH3/1 mol CuSO4) = 0.250 g NH3 (actually you did this when you determined the limiting reagent).
So we started with 30.0 g NH3. We used 0.250 g and we have left 30.0 - 0.250 = ??
I hope this helps clear things up by putting all of it down at one time. Be sure and check my arithmetic and look out for typos. I didn't look up any of the molar masses since I have many of these memorized; therefore, you need to check them and go through the calculations to make sure I used them correctly.
I'm not looking at the problem at the moment. Did they ask how much of the NH3 remained unreacted. If so, you will need to calculate how much of the NH3 was used and subtract that from the 30 g with which you started.
g NH3 used = 0.0626 g CuSO4 x (4 mols NH3/1 mol CuSO4) = 0.250 g NH3 (actually you did this when you determined the limiting reagent).
So we started with 30.0 g NH3. We used 0.250 g and we have left 30.0 - 0.250 = ??
What?? Dr.Bob
g NH3 used = 0.0626 g CuSO4 x (4 mols NH3/1 mol CuSO4) = 0.250 g NH3
You used g of CuSO4 to get moles??
But 0.0626 is the moles of CuSO4 found not the grams. Then from what I did before don't you have to go and convert using the ratio to find the moles used up by CuSO4, THEN go and use the molar mass of NH3 to convert the moles of NH3 to grams of NH3 used up ?? HOW did you get 0.250g ...I think I got:
______________________________________
for the grams of excess reagent
since you found that the limiting is
CuSO4 then what you have left is only NH3 right?
to find the amount left:
you have 30g of NH3 in the begining so how much did CuSO4 use up?
use the same equation to find out the limiting to find out how much was used up (you already found this)
0.2504mol NH3 used up (17.034g/1mol NH3)= 4.26g used
subtract the amount you have from the ammount used
30.0g NH3 have - 4.26g NH3 used = ____?
_______________________________________
Please check my thinking Dr.Bob =)
Yes, you are correct. I didn't convert 0.250 mols NH3 to grams and I should have done so. That's what I get for not looking at the problem I was trying to do it from memory and I forgot a step.
So 0.250 mol NH3 x 17 g/mol = 4.25 and 30-4.25 = g NH3 remaining unreacted. That's also why I always tell the student to check my work.
If your speaking of moles in the 1st step No you don't have 1mole...
starting with grams you find moles:
10gCuSO4 (1mol/159.62g)=
30gNH3 (1mol/17.034g)=
So I have 10/159.62 = 0.062 moles, and 30/17.034 = 1.761 moles?
yes but watch the significant figures
CuSO4 moles= 0.0626 mol
NH3 moles= 1.76 mol
And the limiting reactant is CuSO4?
not necessarily you have to go to the next step and find out how much do you need of each to consume the other.
use the moles that you found (minding the significant figures) using the equation to see how much the ratio is between the NH3 and CuSO4.
#moles CuSO4(#mol NH3/ #mol CuSO4)= moles NH3 needed to consume all CuSO4 you have
#moles NH3(#mol CuSO4/ #mol NH3)= moles CuSO4 needed to consume all NH3 you have
then compare that to the amount of each that you have and then you see which you do not have enough to consume the other of.
to clarify that
the moles of each multiplying
CuSO4 + 4 NH3 ----> Cu(NH3)4SO4
#moles CuSO4(4mol NH3/ 1mol CuSO4)= moles NH3 needed to consume all CuSO4 you have
(the moles # is the amomunt you found before I just wanted to clarify the moles in the ratio is from the equation and it's the same except you flip the ratio of course to find the moles of CuSO4 needed to consume all the NH3 you have)
So 0.0626(4 / 1) = 0.2504 mole NH3 and 1.76(1 / 4) = 0.44 mole of CuSO4?
And the limiting reactant is CuSO4 since I have enough NH3?
Yes exactly Josh..
now using the limiting amount of CuSO4 finding the ammount of product you can produce will be the theoretical yield.
so...
0.0626mol CuSO4 (1mol Cu(NH3)4SO4/1mol CuSO4)= ?
then taking that ...
mol Cu(NH3)4SO4 produced (1mol / molecular weight of Cu(NH3)4SO4)
molec weight = add up all the elements involved in the product
Yes...
so compared to what you have
0.0626mol of CuSO4
1.76 mol of NH3
If you want to consume all the
CuSO4 you'd need 0.250mol NH3
while if you want to consume all the
NH3 you'd need 0.440 mol of CuSO4
which one would you have enough to consume and which one do you not have enough to consume the other?
(limiting reagent?)
0.0626(1 / 1) = 0.0626, then 0.0626(1 / 227.78) = 0.00027 is the theoretical yield? (227.78 is the molec weight?)
molecular weight is a little off...
it's actually from what I calculated
227.756g
so the theoretical yield is
0.0626 mol Cu(NH3)4SO4 ( 227.756g /1mol Cu(NH3)4SO4)= ?
I'm just going to say I got 14.26g of Cu(NH3)4SO4
Josh, I have to seriously get to sleep so I'll just post this to help you out with the rest and I'll check tommorow morning to see what you post if anything =D
for the grams of excess reagent
since you found that the limiting is
CuSO4 then what you have left is only NH3 right?
to find the amount left:
you have 30g of NH3 in the begining so how much did CuSO4 use up?
use the same equation to find out the limiting to find out how much was used up (you already found this)
0.2504mol NH3 used up (17.034g/1mol NH3)= 4.26g used
subtract the amount you have from the ammount used
30.0g NH3 have - 4.26g NH3 used = ____?
_______________________________________
for the percent yield:
the equation for that is
percent yield = actual yield (given)/ theoretical yield
Hope this helps =)
Thanks christina, for all your help. Also, in case you check, 12.6 / 14.26 = 88%.
Yep I just went back to look since I forgot to say x100 for the percent yield but you figured it out yourself
Yes it's 88.4% in significant figures
Your welcome, I help when I can of course =)