Calculating the Mass of a Precipitate

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  • Thread starter jackthehat
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  • #1
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Homework Statement:

To calculate the mass of a precipitate where one of the solutes has been previously prepared

Relevant Equations:

3MG 2+ + 2PO4 3- ---> Mg3(PO4)2 (Net Ionic equation)
MgCl2 + Li3PO4 ---> Mg3(PO4)2 + LiCl (Precipitation reaction equation)
Hi there,
I have a problem from an exam which I passed recently, however this was one problem I got wrong and I want to understand how to solve the thing. Now I shall detail the problem below. This is my subsequent attempt to solve it after the exam (so I dont know if this attempt is correct or not) and would appreciate any input reassuring me that this, my last attempt, is correct or if wrong where I am going wrong with this type of problem.

Problem
An aqueous solution containing excess magnesium chloride is mixed with the saturated solution of lithium phosphate prepared in problem #3. In problem #4, you wrote the net ionic equation for the reaction that results. How many grams of the new solid precipitates?


My solution

Note - The molarity of the prepared lithium phosphate mentioned above from a previous problem was 0.163 moles/litre (I know this is correct from the exam feedback). Also the net ionic equation talked about in the problem is listed in the equations section above, but I repeat here ..
3MG 2+ + 2PO4 3- ---> Mg3(PO4)2
So from the reaction equation - MgCl2 + Li3PO4 ---> Mg3(PO4)2 + LiCl
we can see that .. 1 mole of LI3PO4 produces 1 mole of Mg3(PO4)2 (which is the precipitate) .. so the molar ratio is 1/1 = 1
Now the previously prepared saturated solution of Li3PO4 has a calculated molarity of 0.163 M so as the molar ratio of this to our saturate is 1:1
then the molarity of Mg3(PO4)2 is also 0.163 M.
So we now convert from moles to mass (grams) of the precipitate by ..
Calculating it's molar mass (g/mol) and multiplying this by it's actual molarity (mol) giving the mass present of precipitate in solution after the reaction.
Molar mass of Mg3(PO4)2 = ((3 x 24.305) + (2 x 30.974) + (8 x 15.999)) g/mol = 269.855 g/mol
Multiplying this now by molarity we get ... (269.855) g/mol x (0.163) mol = 43.986 = 44.0 g (to 3 significant figures)

Is this correct ? If not where have I gone wrong ? Can anybody help ?
Regards,
Jack
 

Answers and Replies

  • #2
Borek
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Now the previously prepared saturated solution of Li3PO4 has a calculated molarity of 0.163 M so as the molar ratio of this to our saturate is 1:1 then the molarity of Mg3(PO4)2 is also 0.163 M.
No. Stoichiometry deals with amounts, not concentrations.

What was the volume of the saturated Li3PO4 solution? How many moles of phosphate did it contain?
 
  • #3
31
4
Hi Borek,
To answer your questions
1. The volume of the saturated Li3PO4 solution was 200mL.
2.The moles of phosphate contained in the solution was .. 0.0544 mol/L.
One other thing .. I am not quite sure about your statement that Stoichiometry deals with amounts, not concentrations ... I have used it before to calculate molar concentrations of ions in solution ???
Regards,
Jack
 
  • #4
Borek
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1. The volume of the saturated Li3PO4 solution was 200mL.
2.The moles of phosphate contained in the solution was .. 0.0544 mol/L.
No, moles/L is a concentration, I asked about amount. Besides, if the molarity (that you listed earlier) of the Li3PO4 was 0.163, how come it is now 0.0544? It can't be both at the same time. You are confusing something, no idea what.

200 mL of 0.163 M solution contains how many moles of solute?

I am not quite sure about your statement that Stoichiometry deals with amounts, not concentrations ... I have used it before to calculate molar concentrations of ions in solution ???
Knowing volume you can always convert concentration to the amount of substance and vice versa. That you can calculate one knowing other doesn't mean they are the same thing.

If all substances are dissolved in the same solution and the volume is constant throughout the process, it often cancels out in calculations and you can take a shortcut and do part of the calculations just following concentrations. It doesn't mean it was "concentration" that reacted. Plus, you can take shortcuts once you understand why and when the work. From what you wrote so far looks like you don't.

I am afraid if you don't see the problem here your understanding on the stoichiometry and reactions is incomplete. Very incomplete, as this is a very basic concept :(
 
  • #5
mjc123
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Also note that you are going wrong because your precipitation reaction equation is not balanced. Doesn't your net ionic equation give you a clue?
 
  • #6
31
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Also note that you are going wrong because your precipitation reaction equation is not balanced. Doesn't your net ionic equation give you a clue?
Hi again Borek,
I am afraid I have given you an inadequate and erroneous description of the problem I seek to understand. If I give you details of the previous problem refered to in this problem and how I calculated the molarity of the Lithium Phosphate solution being used in this precipitation reaction maybe the situation will become clearer.
Below are the details of problem 3 (the prior problem referred to) and they are as follows ...

Problem 3 contains Question 13 to 20.
Again, two saturated aqueous solutions are prepared at 25 ºC.

One is made by dissolving lithium carbonate (Ksp = 8.15 x 10−4) in 100.0 mL of water until excess solid is present, while the other is prepared by dissolving lithium phosphate (Ksp = 2.37 x 10−4) in 200.0 mL of water until excess solid is present. Which solution has a higher lithium cation concentration?


Now I wont go into too much detail of how I calculated the Molarity of the Lithium Phosphate suffice to say I got this problem correct .. however I now realise that the figure of 0.163 M is actually not the molarity of Lithium Phosphate but the molarity of the [Li 1+] ion in solution... aaggh !
I calculated this from the Ksp equation substituting the variable 'x' for the molarity of the solution and then assigning the ion concentrations appropriate values of 'x' .. that is ..
Ksp = 2.37 x 10-4 = [3 x Li 1+] 3 [PO4 3-],
where Li+ = 3x and PO4 = x .. solving for 'x' we get .. x = 0.0544 M (Molarity of solute) and
[Li 1+] = 3 x 0.0544 = 1.63 M
(this is where the discrepency between 0.0544 M and 1.63 M comes from).
Actually I see now that I have made this mistake in my explanation to you I was using 0.163 M (which is molarity of Li 1+ ion in solution) whereas I should have used 0.0544 M (the molarity of Li3PO4) .. very careless of me .. I apologise for leading you astray here.
However this mistake apart .. is the rest of my methodology correct ? .. that is ..
Once I have the correct molarity of the Lithium Phosphate that is being mixed with the Magnesium Chloride and since the molar ratio of the two compounds .. the solute and the precipitate .. is 1 : 1 .. all I have to do is assign this molarity to the precipitate and then multiply this molarity by the formula (molar) mass of the precipitate to give the mass of precipitate formed ?
This time (with the corrected molarity of 0.0544 m) would give ..
0.0544 mol x 269.855 g/mol = 1`4.68 g or 14.7 g (to 3 significant figures) ???
am I on the right track here or am I still floundering in the dark ?
I again would appreciate your input because I am still very unsure .
Regards,
Jack
 
  • #7
mjc123
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No, your method is not correct.
First, you completely ignore, though you have quoted it, my statement that your equation is unbalanced. The ratio of the two compounds is not 1:1.
Second, mass is not molar mass times molarity. You must clearly understand the difference between amount of substance, measured in moles, and concentration or molarity, measured in moles/litre. These are different things, and if the difference is not clear to you, you need to go back and revise them, otherwise you will be constantly "floundering in the dark".
 
  • #8
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Hi mcj123,
Yes I see why you could see that from my last reply on this problem .. 2 points I would address .. in my haste to get all of the information down I completely missed that I had not being using the correct balanced equation which incidently I had written down correctly when doing the original problem but for some reason had transcribed incorrectly in my original post .. and so have been using them here incorrectly from the start ..
of course the correct equation should have been .. 3 MGCl2 + 2 Li3PO4 - ---> Mg3(PO4)2 + 6 LiCl
showing in fact that the molar ration between the precipitate and the phosphate solute is in fact 1 : 2 (and not 1 : 1) ... I can't understand how I missed that so consistently when I had the correct equation in the first place. Secondly I have also been very careless with my use of terms and concepts here as well .. I have been using the term Molarity when I really mean't and should have used Moles of a substance. I do that all of the time .. and I will need to be more self-disciplined in future when talking about scientific terms and not getting the names transposed. However i do take your point that I have then equated the Molarity of the Lithium Phosphate solute (which I calculated in an earlier problem) with the Molarity of the subsequent precipitate in this separate reaction .. and this has no particular validity in this case (I was mixing up the Molarity and Moles terms again and do this not because i dont know the difference but because of my bad habit of using the term Molarity haphazardly when sometimes I mean Moles .. then of course they get further mixed up in the working and then lose track of what I am describing) ..
My question now is if you cannot relate the Lithium Phosphate to the precipitate Magnesium Phosphate by a common Molarity (which i now see is not a valid relationship) .. then how do you relate the two substances quantitatively .. is it through the relation of moles to masses .. but then how do you get from Molarity of Lithium Phosphate to Moles of Lithium Phoshphate .. has it anything to do with the equation ..
Moles = Molarity x Volume ? .. hence you get Moles of the phospate solute ..
and from there and with the mole ratio you can find the moles of precipitate present .. and so convert this to mass .. is that the way to go ? I'd appreciate your input if you have the time to reply.
Regards
Jack
 
  • #9
Borek
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Assuming precipitation went to completion, if you know know number of moles of phosphate ions you can easily calculate number of moles of precipitate, that's what the stoichiometry is about.
 
  • #10
31
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Assuming precipitation went to completion, if you know know number of moles of phosphate ions you can easily calculate number of moles of precipitate, that's what the stoichiometry is about.
Hi again Borek,
I see principle behind the advice you have given in your last message .. however I am now a bit confused whether you mean I need to find the number of moles of lithium phosphate solute .. then link this via stoichiometric mole ratios to the precipitate in the full reaction equation .. or is it that i just have to find the number of moles of the phosphate ions alone and from there link it to the precipitate ... if it is the latter, can I use the net ionic equation of the reaction alone to calculate mole ratios as I normally would in the full reaction equation .. this is what i have usually done before in these types of problems.
Regards.
Jack
 
  • #11
Borek
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I am now a bit confused whether you mean I need to find the number of moles of lithium phosphate solute .. then link this via stoichiometric mole ratios to the precipitate in the full reaction equation .. or is it that i just have to find the number of moles of the phosphate ions alone and from there link it to the precipitate
Try both, see what you get.
 
  • #12
31
4
Try both, see what you get.
Hi Borek,
Well I tried both as you suggested and I get the same result forboth sets of calculations.
A quick description of method using Lithium Phosphate molarity ..
First with Li3PO4 .. I had calculated the Molarity (moles/L) of this solute as 0.0544 M In a solution of 200mL.
To convert this to the number of moles present I used equation .. Moles = Molarity x Volume
So .. Moles of Li3PO4 = 0.0544 mol/L x 0.2 L = 0.01088 mol
Now from reaction equation .. 3MgCL2 + 2LI3PO4 --> Mg3(PO4)2 + 6LiCl
we can see mole ratio of precipitate to Phosphate solute is 1 : 2
So to convert the number of moles of solute to moles of precipitate we multiply 0.01088 m x 1/2
So number of moles of precipitate present is 0.00544 mol. We now convert this to mass of precipitate by
multiplying the number of moles of precipitate by it's formula mass (262.855 g/mol) giving .. 1.4299 g
So .. Mass of precipitate = 1.43 g (to 3 sign. figures)
Please tell me I am on the right path with this calculation .. because I am on the verge of giving up ..
Regards,
Jack
 
  • #13
Borek
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Well I tried both as you suggested and I get the same result forboth sets of calculations.
You think it is accidental, or just reflects the fact number of moles of the only important reagent is the same in each case?

Mass of precipitate = 1.43 g (to 3 sign. figures)
And that's how it should be done from the very beginning.
 
  • #14
31
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You think it is accidental, or just reflects the fact number of moles of the only important reagent is the same in each case?



And that's how it should be done from the very beginning.
Hi Borek,
Thank you very much for your help .. I have to say at one point i never thought I'd get there but I suppose the old adage is apt .." If at first you don't succeed, try, try and try again."
Regards,
Jack
 
  • #15
31
4
No, your method is not correct.
First, you completely ignore, though you have quoted it, my statement that your equation is unbalanced. The ratio of the two compounds is not 1:1.
Second, mass is not molar mass times molarity. You must clearly understand the difference between amount of substance, measured in moles, and concentration or molarity, measured in moles/litre. These are different things, and if the difference is not clear to you, you need to go back and revise them, otherwise you will be constantly "floundering in the dark".
Hi mjc123,
Thanks a lot for taking the time to help me with my problem.
Regards,
Jack
 

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