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-   -   Copper silver nitrate equation (https://www.askmehelpdesk.com/showthread.php?t=510914)

  • Sep 25, 2010, 11:23 PM
    mindyabbey
    copper silver nitrate equation
    I have completed the lab and understand the concepts but seem to be missing the forest for the trees. The only question I have not answered is the "purpose of the lab". Any thoughts?
  • Sep 25, 2010, 11:25 PM
    Unknown008

    What are your thoughts first?

    What would happen if you didn't have a laboratory?
    Where would you carry out experiments?
    How would you get the necessary reagents?
    What about the waste products?
    What if they are harmful to the environment?
  • Sep 26, 2010, 12:19 AM
    mindyabbey

    I understand the purpose of having labs in general, but for my lab report, I am unsure as what to put for my hypothesis as it seems too obvious. It seems like the lab is about theoretical yield and quantitive relationships, but I don't understand exactly how that helps us.
  • Sep 26, 2010, 12:33 AM
    Unknown008

    How much substance are you expected to give?
  • Sep 26, 2010, 01:08 AM
    mindyabbey

    This is our first lab so I am not clear on how much is "enough". Do you have any further suggestions other than what I previously wrote? I need to state the problem and why it is worth studying. Thank you!
  • Sep 26, 2010, 01:13 AM
    Unknown008

    Could you post all what you have put into your hypothesis? At least, the main points?

    I'll reply later on. I'm having lunch shortly.
  • Sep 26, 2010, 02:52 AM
    mindyabbey

    For my hypothesis, I put that I predict the Copper will replace the Silver in the reaction. We can include the percentage yield but it is optional. When calculating the percentage yield for silver, I got more silver than the theoretical yielrd. Could this be due to a lab error? And would you probably not include it?
  • Sep 26, 2010, 02:55 AM
    Unknown008

    Oh! That's what I was wondering. I was finding no connection with your title at all.

    Ok, first the equation.



    You can never get more than theorical yield. There must be some error somewhere. Could you describe me steps you took for the experiment?
  • Sep 26, 2010, 03:14 AM
    mindyabbey

    Yes, I'll include the main points:
    We poured silver nitrate crystals into a test tube with water, and then after coiling a copper wire, we placed the wire in the tube so that there was part of it sticking out. After about 15 minutes, we shook off the crystals and left the wire in to make sure the reaction was done. Then we washed as much of the silver crystals off the wire into a beaker as we could. Then poured the contents in the test tube and decanted. Then left the silver to dry and weighed it
  • Sep 26, 2010, 03:16 AM
    mindyabbey

    For sources of error I put that even after 15 min. the reaction may not have been completely finished, water might have not been enough to wash all the crystals off the copper wire, and while decanting, some of the silver crystals that were floating on top might have been thrown away.
  • Sep 26, 2010, 03:28 AM
    Unknown008

    Well, this is strange. Are you sure you did your calculations right? According to the procedures, the errors should give you a lower yield rather than a yield greater than 100%.

    Except if the bits of silver metal were not completely dry.

    How did you find the percentage yield?

    Did you take:

    (Mass of silver metal obtained / Mass of theorical silver metal to be obtained) x 100% ?

    May I have some of your values, especially:
    1. Mass of Silver nitrate used.
    2. Mass of Silver metal obtained.
  • Sep 26, 2010, 04:32 AM
    mindyabbey

    Yes, came up with 1.739 grams of silver nitrate used and 1.021 grams of silver formed. I think I might have calculated it wrong. What would be the correct way to find the theoretical yield? I tried following examples in my textbook but I was a little bit confused
  • Sep 26, 2010, 04:45 AM
    Unknown008

    Ok, theorical yield from 1.739 g of Silver nitrate:

    1.739 g = 1.739/(108 + 14+48) = 0.0102294 mol = 0.0102 mol

    Number of moles of Ag formed = 0.0102 x 1 = 0.01023 mol

    Mass of Ag formed = 0.01023 x 108 = 1.105 g

    Percentage yield =

    There, the calculations are good now :)
  • Sep 26, 2010, 05:02 AM
    mindyabbey

    Thank you! I understand it now. I appreciate the time you put in to tutor me :)

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