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Topic: Titanium compound dissolving - calculation.  (Read 9717 times)

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Offline Levi

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Titanium compound dissolving - calculation.
« on: February 21, 2009, 03:22:28 AM »
Hi! I'm new here :)

I tried to search for the question but nothing showed up. I'm trying to answer this question, but my main problem is getting the first equations. I've tried doing it, but i am pretty sure i'm doing it the wrong way. So it would be greatly appreciated if you could help me, thanks in advance!


"A 3.40g sample of a titanium compound dissolves in water to produce titanium ions and chloride ions. All the chloride ions are precipitated by the addition of excess silver nitrate soultion. After filtration and drying the mass of silver chloride was 9.47g.

(a) Calculate the number of moles of chloride ions in the silver chloride and hence the mass of chloride in the sample.

(b) Calculate the mass of titanium ions in the sample.

(c) Determine the empiracle formula of the compound."


Thank you very much!

-Levi-

Offline Arkcon

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Re: Titanium compound dissolving - calculation.
« Reply #1 on: February 21, 2009, 04:47:49 AM »
OK, can you start with part A?  Can you show us how you'd do that conversion?
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Levi

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Re: Titanium compound dissolving - calculation.
« Reply #2 on: February 21, 2009, 04:57:33 AM »
I don't know how to do the conversion. That's why i posted the question. I'm just really confused. Thanks for the reply though!

Offline Arkcon

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Re: Titanium compound dissolving - calculation.
« Reply #3 on: February 21, 2009, 05:05:10 AM »
I don't know how to do it. That's why i posted the question.

OK, but you said ...

Quote
I'm trying to answer this question, but my main problem is getting the first equations. I've tried doing it, but i am pretty sure i'm doing it the wrong way.

If you showed us your work so far, we could offer pointers.

Quote
I tried to search for the question but nothing showed up

Yeah, that's not shocking.  These are general concepts, and they can be applied to any chemical.  Goggling for this reaction won't solve it for you.

Quote
After filtration and drying the mass of silver chloride was 9.47g.

(a) Calculate the number of moles of chloride ions in the silver chloride and hence the mass of chloride in the sample.

Can you write a balanced chemical equation?  You can't easily, because the titanium is an unknown, but you can put a x for titanium, and carry it through.  It will come in handy later.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Arkcon

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Re: Titanium compound dissolving - calculation.
« Reply #4 on: February 21, 2009, 05:11:04 AM »

Oh, look, someone is doing much the same thing, in this thread over here:
http://www.chemicalforums.com/index.php?topic=31392.0;topicseen
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Levi

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Re: Titanium compound dissolving - calculation.
« Reply #5 on: February 21, 2009, 05:19:16 AM »
Well i didn't want to show you what i did because it's pretty embarrising. The sheet that the questions are on is a miscellaneous sheet. So i don't know what procedure to carry out to solve it, because it's not on a specific topic. We get an assignment every week to complete. But do you know how to solve it?

Thanks for the help and replies.

Offline sjb

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Re: Titanium compound dissolving - calculation.
« Reply #6 on: February 21, 2009, 05:22:12 AM »
How do you calculate the number of moles of anything, be it silver chloride, methanol, or gold?

Offline Astrokel

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Re: Titanium compound dissolving - calculation.
« Reply #7 on: February 21, 2009, 05:22:44 AM »
No, don't feel embarrassed! It's alright.
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline Levi

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Re: Titanium compound dissolving - calculation.
« Reply #8 on: February 21, 2009, 05:39:35 AM »
How do you calculate the number of moles of anything, be it silver chloride, methanol, or gold?

Well if it's 3.40g of a Titanium compound, then:

n=m/M

n(TiX)=3.40/M(TiX)

But i don't know the other element which is apart of the titanium compound, so i call it "X"?

I'm not sure where i'm going here  ???

Offline sjb

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Re: Titanium compound dissolving - calculation.
« Reply #9 on: February 21, 2009, 05:45:53 AM »
See, you do know something!

However, in this case it's probably best to start from the other end of the data, which is what the part questions are all about.

How many moles of silver chloride are there?

Offline Levi

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Re: Titanium compound dissolving - calculation.
« Reply #10 on: February 21, 2009, 05:51:38 AM »
n(AgCl)= m/M                          M(AgCl)= (1xCl)+1(Ag)
                                                       = 35.45+107.87   
          = 9.47/ 143.32                         = 143.32 g mol-1
          =0.066 mol

Offline Levi

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Re: Titanium compound dissolving - calculation.
« Reply #11 on: February 21, 2009, 05:58:58 AM »
Cl-(aq) + AgNO3(aq) = AgCl(s) + NO3-(aq)

Is that right?

So:

n(Cl-)=1/1 x 0.066
        = 0.066 mol

Therefore the mass of Cl in the sample is:

n=m/M

m= nxM
  = 0.066 x 35.45
  = 2.3397 g

So there is 2.3397 g of Cl in the AgCl sample?

Offline Borek

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Re: Titanium compound dissolving - calculation.
« Reply #12 on: February 21, 2009, 06:37:20 AM »
Close, you must have copied digits from your calculator wrong. It should be 0.0687 moles of AgCl, everything calculated later is slightly off - but the procedure is correct.

Now, how much titanium in the sample?
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Offline Levi

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Re: Titanium compound dissolving - calculation.
« Reply #13 on: February 21, 2009, 07:06:11 AM »
Oh really? I thought i did it correct, hmm.

Well if there is 2.3397 g of Cl, then there is 3.40-2.3397= 1.0603 g of Ti?


Offline Borek

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Re: Titanium compound dissolving - calculation.
« Reply #14 on: February 21, 2009, 08:43:00 AM »
Hmm, now I can't reproduce my result and 0.06608 seems to be correct. Some finger slip probably :-\ (seems like I may have entered 9.84 instead of 9.74).

OK, you know mass of titanium and mass of chloride - how many moles of each?  What's their ration? What's the compound formula?
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