Help with Theoretical Yield

  1. 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

    reactoin 2: theoretical mass of copper (II) hydroxide formed

    reaction 3: theoretical mass of copper (II) oxide formed

    reaction 4: theoretical mass of copper (II) sulfate formed

    reaction 5: theoretical mass of copper formed

    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.
  2. jcsd
  3. Borek

    Staff: Mentor

    First of all - what are you expected to calculate?
  4. 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.
  5. Borek

    Staff: Mentor

    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?

    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?
  6. 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.
  7. Borek

    Staff: Mentor

    Good. Do you know how to check which reagent is limiting? If not, check out

    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.
  8. 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
  9. Borek

    Staff: Mentor

    Yes and no. In general you are on the right track.

    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.
  10. sorry i didnt mean to stick that g in there. what should i have done differently besides rounding?
    Last edited: Feb 24, 2011
  11. Borek

    Staff: Mentor

    Other than things I listed what you did looks OK. Try the next step.
  12. is there more i need to do for the first reaction? and would i do the second reaction the same way?
  13. Borek

    Staff: Mentor

    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.
  14. would i use the answer from the first reaction to calculate the second reaction?
  15. 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
  16. Borek

    Staff: Mentor


    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?
  17. 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?
  18. Borek

    Staff: Mentor

    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.
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