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How to figure volume and molarity of product when you have g of react. & V/M of other

  • Thread starter gurpalc
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  • #1
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Homework Statement



My problem is in the attachment. I am trying to find the molarity and volume of Cu(NO3)2 while I only know Cu has .50 g and HNO3 is 10M and is 5mL.

Homework Equations





The Attempt at a Solution



I have done my calculations on the right. What I have circled is what I believe to be the moles of Cu(NO3)2. The copper is limiting I assume. I don't know where to go from there. Please help.
 

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Answers and Replies

  • #2
Borek
Mentor
28,401
2,800


Your calculations are barely readable. Number of moles of copper nitrate seems to be OK.

You need final volume to calculate concentration. Hard to help not seeing the question, but if you are not given any other information, assume 5 mL.
 
  • #3
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Your calculations are barely readable. Number of moles of copper nitrate seems to be OK.

You need final volume to calculate concentration. Hard to help not seeing the question, but if you are not given any other information, assume 5 mL.
How do I find final volume of Cu(NO3)2. That's what I'm stuck on. I don't know how to make the calculations.
 
  • #4
Borek
Mentor
28,401
2,800


Have you read what I wrote?
 
  • #5
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Yes but I don't get it.
 
  • #6
Borek
Mentor
28,401
2,800


Some acid was consumed and replaced with copper. Even if the volume of the liquid changed during the reaction, it didn't change by much. This is only an approximation, but there is no better one available.

On a very general level it would be possible to calculate more exact volume given density tables of nitric acid solutions and nitric acid/copper nitrate solutions. Two problems with this idea. First, I have never seen density table for a mixture of nitric acid and copper nitrate (and I have seen plenty of density tables). Second, you are limited by the accuracy of mass and volume, as they are given with one significant digit only - using very accurate method for inaccurate data won't give a better result.

Don't read if you are already confused:
0.5 g means anything between 0.45 and 0.55 g, 5 mL means anything between 4.5 mL and 5.5 mL. That in turn means amount of copper is between 7.08x10-3 and 8.66x10-3 moles, and concentration something between 7.08x10-3mol/0.0055L=1.29M and 8.66x10-3mol/0.0045L=1.92M - almost 50% difference. Error you are making assuming final volume of 5mL is most likely not larger than 10% - so much smaller than the one already intrinsic to the data.
 

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