Register to reply

Help with Theoretical Yield

by blicker
Tags: chemistry, theoretical yield
Share this thread:
blicker
#1
Feb24-11, 12:31 AM
P: 19
1. The problem statement, all variables and given/known data
Here's the measurements from my expirement.
(reaction 1)
0.548 g Cu (s)
4.0 mL HNO3
100 mL DI water

(reaction 2)
30 mL 3.0 M NAOH

(reaction 3)
nothing was added.

(reaction 4)
200 mL of DI water
15 mL of 6.0 M HCl

(reaction 5)
2.095 g Zn
10 mL 6M HCl
5 mL methanol
5 mL DI water
40.767 g mass of evaporating dish
41.238 g mass of evaporating dish with solid


2. Relevant equations
Here are the reactions:
reaction 1: theoretical mass of copper (II) nitrate formed
Cu+4HNO3-->Cu(NO3)2+2NO2+2H2O

reactoin 2: theoretical mass of copper (II) hydroxide formed
Cu(NO3)2+2NaOH-->Cu(OH)2+2NaNO3

reaction 3: theoretical mass of copper (II) oxide formed
Cu(OH)2-->CuO+H2O

reaction 4: theoretical mass of copper (II) sulfate formed
CuO+H2SO4-->CuSO4+H2O

reaction 5: theoretical mass of copper formed
Zn+CuSO4-->ZnSO4+Cu

3. The attempt at a solution
I have no clue where to even start. my teacher said some of the measurements from the lab were needed in the calculations, but i dont know which measurements to use for each reaction.
Phys.Org News Partner Science news on Phys.org
An interesting glimpse into how future state-of-the-art electronics might work
Tissue regeneration using anti-inflammatory nanomolecules
C2D2 fighting corrosion
Borek
#2
Feb24-11, 04:05 AM
Admin
Borek's Avatar
P: 23,536
First of all - what are you expected to calculate?
blicker
#3
Feb24-11, 01:39 PM
P: 19
what I need to calculate (some of the measurements from the lab are used in the calculations):

the theorecital mass of copper (II) nitrate formed in reaction 1

the theoretical mass of copper (II) hydroxide formed in reaction 2

the theoretical mass of copper (II) oxide formed in reaction 3

the theoretical mass of copper formed

and the % yield of the overall reaction

im not asking you to do all those problems for me. but, i have no idea where to start. i need somebody to help show me how to do the first one, then i could do the rest on my own.

Borek
#4
Feb24-11, 01:49 PM
Admin
Borek's Avatar
P: 23,536
Help with Theoretical Yield

Quote Quote by blicker View Post
the theorecital mass of copper (II) nitrate formed in reaction 1

the theoretical mass of copper (II) hydroxide formed in reaction 2

the theoretical mass of copper (II) oxide formed in reaction 3
These can be calculated, no problem - but are you really asked to do it, or do you just guess you should do it?

Do you know what limiting reagent is?

the theoretical mass of copper formed
Think about mass conservation. You started with some amount of copper, did some tricks with it - but you have not added nor removed copper. You end with copper again. Assuming everything went perfectly well - is there any reason for the amount of copper to change?
blicker
#5
Feb24-11, 02:00 PM
P: 19
all of the reaction questions are apart of my assignment, so i have to do them all.

i think the limiting reactant is the reactant that limits the amount of product.
Borek
#6
Feb24-11, 02:24 PM
Admin
Borek's Avatar
P: 23,536
Quote Quote by blicker View Post
i think the limiting reactant is the reactant that limits the amount of product.
Good. Do you know how to check which reagent is limiting? If not, check out

http://www.chembuddy.com/?left=balan...iting-reagents

Start with the first reaction - what is a limiting reagent, copper, or nitric acid?

Hm, I see you don't have concentration of nitric acid. If so, assume there was an excess of nitric acid.
blicker
#7
Feb24-11, 04:19 PM
P: 19
i tried working it out. is this right?

0.548g Cu*(1 mol Cu/63.55g Cu)=0.009g mol Cu

0.009 mol Cu*(1 mol Cu(NO3)2/1 mol Cu)=0.009 mol Cu(NO3)2

0.009 mol Cu(NO3)2*(125.56g Cu(NO3)2/1 mol Cu(NO3)2=1.08g Cu(NO3)2
Borek
#8
Feb24-11, 04:32 PM
Admin
Borek's Avatar
P: 23,536
Quote Quote by blicker View Post
i tried working it out. is this right?
Yes and no. In general you are on the right track.

0.548g Cu*(1 mol Cu/63.55g Cu)=0.009g mol Cu
What is "g mol"?

0.009 seems to be rounded down way too far. You should use at least the same number of significant figures as you have in the copper mass.
blicker
#9
Feb24-11, 04:35 PM
P: 19
sorry i didnt mean to stick that g in there. what should i have done differently besides rounding?
Borek
#10
Feb24-11, 05:00 PM
Admin
Borek's Avatar
P: 23,536
Other than things I listed what you did looks OK. Try the next step.
blicker
#11
Feb24-11, 05:02 PM
P: 19
is there more i need to do for the first reaction? and would i do the second reaction the same way?
Borek
#12
Feb24-11, 05:08 PM
Admin
Borek's Avatar
P: 23,536
First reaction is OK. In general you may assume all other reactions can be done the same way, although - to be on the safe side - you should check each time if added reagents are in excess.
blicker
#13
Feb24-11, 05:17 PM
P: 19
would i use the answer from the first reaction to calculate the second reaction?
blicker
#14
Feb24-11, 05:47 PM
P: 19
i dont know if its right, but i used the answer from the first reaction to calculate the second reaction since all these reactions are connected.

1.08g Cu(NO3)2*(1 mol Cu(NO3)2/187.57g Cu(NO3)2)=0.00576 mol Cu(NO3)2

0.00576 mol Cu(NO3)2*(1 mol Cu(OH)2/1 mol Cu(NO3)2)=0.00576 mol Cu(OH)2

0.00576 mol Cu(OH)2*(97.566g Cu(OH)2/1 mol Cu(OH)2=0.562g Cu(OH)2
Borek
#15
Feb25-11, 02:47 AM
Admin
Borek's Avatar
P: 23,536
Quote Quote by blicker View Post
0.009 mol Cu(NO3)2*(125.56g Cu(NO3)2/1 mol Cu(NO3)2=1.08g Cu(NO3)2
[/quote]

Quote Quote by blicker View Post
1.08g Cu(NO3)2*(1 mol Cu(NO3)2/187.57g Cu(NO3)2)=0.00576 mol Cu(NO3)2
It can't be OK - what you wrote means 0.009 = 0.00576. I have not checked your math earlier, just the logic, obviously that wasn't enough.

What is molar mass of copper nitrate?

Other than that you are on the right track. Question: do you have to calculate number of moles of each substance each time? Or are they somehow dependent?
blicker
#16
Feb25-11, 05:05 PM
P: 19
yeah i realized my math was wrong on the first reaction. i fixed it. and my teacher said we are suppose to work in that 4.0 mL of HNO3. its suppose to be turned into mols. how would i do that?
Borek
#17
Feb25-11, 05:57 PM
Admin
Borek's Avatar
P: 23,536
You can't without concentration. Perhaps it was just a stock concentrated solution, that would mean around 68% w/w. You have to check the solution density.


Register to reply

Related Discussions
Theoretical yield Biology, Chemistry & Other Homework 10
Theoretical Yield, Molar Mass, and Percent Yield Biology, Chemistry & Other Homework 1
Calculating theoretical yield of resulting alum from water dissolving...help asap! Biology, Chemistry & Other Homework 4
Theoretical Yield in Multi-Step Synthesis Biology, Chemistry & Other Homework 1
Theoretical yield Biology, Chemistry & Other Homework 7