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

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To calculate the number of Ca2+ ions in 0.05 moles of Ca(NO3)2, one must recognize that each mole of Ca(NO3)2 yields one mole of Ca2+ ions. The correct calculation involves multiplying the number of moles by Avogadro's number, 6.022 x 10^23, without omitting stoichiometric ratios. For example, to find the number of oxygen atoms in 2 moles of Ca(NO3)2, the calculation would include the factor for the six oxygen atoms present per formula unit. It is crucial to keep track of units and stoichiometric relationships to avoid errors in these types of calculations.
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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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