Aidan Bradley

Finding percentage of reagent in a mixture

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So I've been stuck on this problem for about three days or so. It involves precipitation and some reaction, but the reaction is not given:

"A 1.0 g mixture of magnesium chloride and sodium nitrate is dissolved in water. Then an excess of silver nitrate solution is added and a precipitate is collected and dried. The mass of the precipitate is 1.43 g. What is the percentage of magnesium chloride in the original mixture?"

What I have tried so far is to find out the reaction first, then work backwards to find out how much magnesium chloride reacted with the silver nitrate, but none of the answers I have found match the multiple choice answers listed. I really need to figure this out because I am so lost and confused.

-Thanks

Aidan B.

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2 hours ago, Aidan Bradley said:

What I have tried so far is to find out the reaction first

You are right this is the key to it.

So what do you think the reaction is or reactions are?

What do you know about the solubility of the four possible salts involved?

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16 hours ago, studiot said:

You are right this is the key to it.

So what do you think the reaction is or reactions are?

What do you know about the solubility of the four possible salts involved?

Well, from what  I can tell, all of them are soluble except when MgCl2 mixes with AgNO3 and NaNO3 to make AgCl and NaCl and Mg(NO3)2, to which AgCl is always insoluble. However, Mg(NO3)2 is also insoluble in many non-polar solvents. I assume they mean water here for the sake of High School AP Chem. So in this case AgCl is our precipitate. Now shall I divide the mass of the precipitate by the molar mass of AgCl? I think that would set me down the path of moles so I could convert into the original MgCl2. But Im second guessing myself because I'm not sure I have the reaction right.

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1 minute ago, Aidan Bradley said:

However, Mg(NO3)2 is also insoluble in many non-polar solvents. I assume they mean water here for the sake of High School AP Chem.

Unless they carefully say otherwise, you can assume they are working with aqueous solutions.

 

3 minutes ago, Aidan Bradley said:

MgCl2 mixes with AgNO3 and NaNO3 to make AgCl and NaCl

Why would it form NaCl?

All the chloride ions fall out of solution as AgCl.

Otherwise, you seem to be on the right track.

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7 minutes ago, John Cuthber said:

Why would it form NaCl?

All the chloride ions fall out of solution as AgCl.

That makes sense since the Ag+ ions are strong and would force the reaction towards AgCl rather than NaCl

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5 hours ago, John Cuthber said:

Unless they carefully say otherwise, you can assume they are working with aqueous solutions.

 

23 hours ago, Aidan Bradley said:

"A 1.0 g mixture of magnesium chloride and sodium nitrate is dissolved in water......

An additional hint in the OP.

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2 hours ago, Aidan Bradley said:

I have no idea what OP means...

Original Post (first post) in this case but it can also mean Original Poster if that's the context.

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Posted (edited)
On 4/19/2019 at 7:25 PM, Aidan Bradley said:

Then an excess of silver nitrate solution

 

What does the excess mean?

 

John Cuthber has already told you that the precipitate is all silver chloride.

Can you see why?

If not does anything stand out below?

Did you look up solubilities as I suggested?

In solution there are (before precipitation)

Cations : Mg2+, Na+ and Ag+

Anions : Cl- , NO3-

Possible salts and solubilities

MgCl2 very soluble
NaCl very soluble
AgCl low solubility

 

Mg(No3)2 very soluble
NaNO3  very soluble
AgNO3  soluble

 

So your 1.43 grams of precipitate is made up of what?

 

Can you think of the next two steps?

 

Edited by studiot

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6 hours ago, studiot said:

 

What does the excess mean?

 

John Cuthber has already told you that the precipitate is all silver chloride.

Can you see why?

If not does anything stand out below?

Did you look up solubilities as I suggested?

In solution there are (before precipitation)

Cations : Mg2+, Na+ and Ag+

Anions : Cl- , NO3-

Possible salts and solubilities

MgCl2 very soluble
NaCl very soluble
AgCl low solubility

 

Mg(No3)2 very soluble
NaNO3  very soluble
AgNO3  soluble

 

So your 1.43 grams of precipitate is made up of what?

 

Can you think of the next two steps?

 

Ahh I see, and to explain, my only confusion in the beginning was the reaction. I must divide the final precipitate weight by the molar mass of AgCl which will give me both moles of Ag+ and Cl-. From there I can divide the moles of Cl- by 2 since MgCl2 requires 2 Cl- ions to form. So this would then give us the total moles of MgCl2, from which we can divide the weight of the moles by the weight of the original sample, which is 1.0 g. I think this is right, and I will post back whether it works or not.

Aha, so it worked. My answer came out to .475 * 100 --> 47.5% of the original sample. Thank you all for the help, I understand this much better now!

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Some much better and more satisfying to have done (most of) it yourself.  +1

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