Calculating Molarity & Volume of Cu(NO3)2

  • Thread starter Thread starter gurpalc
  • Start date Start date
  • Tags Tags
    Molarity Volume
Click For Summary

Discussion Overview

The discussion revolves around calculating the molarity and volume of Cu(NO3)2 based on given quantities of copper and nitric acid. Participants explore the implications of limiting reactants and the challenges of determining final volume in a chemical reaction context.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a homework problem involving 0.50 g of copper and 5 mL of 10M nitric acid, seeking help with molarity and volume calculations.
  • Another participant notes that the calculations are difficult to read but suggests that the number of moles of copper nitrate appears correct.
  • There is a suggestion that if no other information is provided, the final volume could be assumed to be 5 mL.
  • A participant expresses confusion about how to find the final volume of Cu(NO3)2, indicating a need for further clarification.
  • One participant discusses the consumption of acid and the approximation of volume change during the reaction, highlighting the limitations of using density tables for more accurate calculations.
  • Concerns are raised about the significant uncertainty in the measurements, noting that the given values have a range that affects the calculated concentration significantly.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the final volume or the best approach to calculate molarity, with multiple viewpoints and uncertainties expressed throughout the discussion.

Contextual Notes

Participants note limitations related to the accuracy of mass and volume measurements, as well as the lack of density tables for the specific mixture involved.

gurpalc
Messages
16
Reaction score
0

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.
 

Attachments

  • chem pic.jpg
    chem pic.jpg
    39 KB · Views: 468
Physics news on Phys.org


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.
 


Borek said:
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.
 


Have you read what I wrote?
 


Yes but I don't get it.
 


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.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
2
Views
4K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K