Calculating Ca2+ Ions in 0.05 Moles of Ca(NO3)2 | Homework Equations

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Discussion Overview

The discussion focuses on calculating the number of Ca2+ ions in a given amount of calcium nitrate, Ca(NO3)2, specifically in 0.05 moles. Participants explore the necessary calculations and stoichiometric relationships involved in determining the number of ions present in a solution.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant asks how to calculate the number of ions in 0.05 moles of Ca(NO3)2.
  • Another participant suggests using the formula (0.05/molar mass) * 6.022 x 1023.
  • A third participant agrees with the calculation but emphasizes the importance of including the stoichiometric factor of 1 mole of Ca2+ ions per mole of Ca(NO3)2.
  • A participant expresses a desire to understand how to approach similar questions and mentions a habitual calculation method.
  • A former TA advises keeping track of units and stoichiometric ratios during calculations.
  • One participant inquires about calculating the number of oxygen atoms in 2 moles of Ca(NO3)2 and presents two different approaches.
  • Another participant points out that the stoichiometric ratio for oxygen would differ from that of calcium ions.
  • A participant proposes a calculation for the number of oxygen atoms based on their understanding of the stoichiometry involved.
  • Concerns are raised about the clarity of the calculations, particularly regarding the source of certain numbers and the relevance of the mass of calcium nitrate.
  • Another participant questions the relevance of the mass of calcium nitrate in the context of the problem, which specifies moles rather than grams.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to calculating the number of ions and atoms, with some agreeing on the necessity of including stoichiometric factors while others question the calculations presented. The discussion remains unresolved regarding the best method for approaching these types of problems.

Contextual Notes

Participants highlight potential confusion regarding stoichiometric ratios and the distinction between moles and grams in calculations. There are also unresolved questions about how calculations would change with different quantities of the compound.

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Homework Statement




Homework Equations


How many Ca2+ ions are there in 0.05 moles of Ca(NO3)2?

The Attempt at a Solution


how do we calculate the number of ions?
 
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(0.05/molar mass) * 6.022*10e23 ?
 
Correct, although you should include one more factor:
Each mole of Ca(NO3)[/sub]2[/sub] contains 1 mole of Ca2+ ions.

Since it's a 1:1 ratio, it won't affect the numerical value of your answer...but you don't want to start the habit of omitting that factor, since it would be a different story if you had been asked how many oxygen or nitrogen atoms were present.
 
can you tell me how it would be different. or what are the steps to approach these kind of questions!
because i have that habit of doing (number of moles/molar mass)*6.022*10e23
 
When I was a TA, I told my students to always keep track of units during calculations. This includes keeping track of which ions, atoms, or molecules you're measuring amounts of. That way you can tell at a glance whether you've inadvertently left out any stoichiometric ratios or conversion factors.

0.05g Ca(NO_{3}){_2}*\frac{1 mole Ca(NO_{3}){_2}}{molar mass Ca(NO_{3}){_2}}*\frac{1 mole Ca^{2+}}{1 mole Ca(NO_{3}){_2}}*6.022*10^{23}\frac{atoms}{mole}

The units for molar mass are obviously grams/mole.
 
how would it be if we have been asked to find how many oxygen atoms were present in 2 moles.
is it: 2/(2*3*16)*6.022*10e23
or 2/16 * 6.022*10e23 ?
 
The stoichiometric ratio
\frac{1 mole Ca^{2+}}{1 mole Ca(NO_{3}){_2}}
would be replaced with
\frac{6 moles O}{1 mole Ca(NO_{3}){_2}}
 
so it would be:
(2*6/164.1)*6.022*10e23
and that's 4.403*10e22 ?
 
Where is the 2 coming from, and why aren't you using the mass of calcium nitrate?
 
  • #10
2 came from (NO3)2 , we have 2N
and i used the mass which is 164.1
 
  • #11
There are several problems with this.
1. You stated you were looking for oxygen atoms, not nitrogen atoms.
2. Suppose you had been given 0.30 moles of the salt instead of 0.05 moles. How would your calculations have been different?
 
  • #12
PhaseShifter said:
Where is the 2 coming from, and why aren't you using the mass of calcium nitrate?

Why do you need the mass of calcium nitrate? The question asks for 0.05 mol, not 0.05 g.:confused:
 

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