[futbolchica5]

So this is my first pre-lab assignment and I understood and solved all problems except this one. I'm sure I'm making a simple mistake. The professor put a lot of negligent information in the problems surrounding this one so here is the question and the work I did. Can someone tell me what I did wrong or how to solve it correctly? Any help would be much appreciated! 

If a KClO₃/KCl mixture weighing 1.54 g loses 0.340 g upon heating, what is the % KClO₃ in the mixture?

KCl0₃ = K + Cl + 0₃
                           = 39.10 g/mol+ 35.45 g/mol + (16 g/mol x 3)
                           = 122.55 g/mol

KCl = K+Cl
      = 39.10 g/mol + 35.45 g/mol
      = 74.55 g/mol


Percentage KClO₃ = 122.55 g/mol/197.1g/mol (Total of mixture)

                                             = .621765 or 62.2% (3 sigfigs)

Am I correct in thinking that the information about the change in mass and the heating of the mixture are negligent?

[Jorriss]

I don't see how what you did would give percentages of the present substances.
 

You need to pay attention to the mass of the mixture and the heating. Based on how you did it, how do you know there is ~1.4 grams of KClO3 and ~ .1 grams of KCl, which would totally skew the results in favor of KClO3.


I think you should go back to the drawing board.


Is this for general chemistry? Analytical?

[futbolchica5]

I've searched through the sample problems in class, but it's never mentioned what heating does. I did find on the forum a similar (but different too) problem where the heating did matter. It shows that the heating (and difference because of it) shows how many grams of O₃ was in the mixture.

I'm trying to run the figures now with that information... It's difficult!

[Jorriss]

I've searched through the sample problems in class, but it's never mentioned what heating does. I did find on the forum a similar (but different too) problem where the heating did matter. It shows that the heating (and difference because of it) shows how many grams of O₃ was in the mixture.

I'm trying to run the figures now with that information... It's difficult!

I'd look in that direction more. Can you link it to me?

[futbolchica5]

This is for a general chemistry class.

Here is the link to the other forum question that is very similar... I'm working with it now:
http://www.chemicalforums.com/index.php?topic=14543.0

[Jorriss]

That seems to be the exact same problem with different numbers.

Basically, heating the sample removed .340 grams of oxygen. You can figure out the moles of oxygen removed and since oxygen was only present in the potassium chlorate, you can figure out the moles of potassium chlorate.

[futbolchica5]

I found the correct answer... but only because I used the proportions of their answer to find mine (I was on my last attempt on my HW online!). I still don't understand the mechanics of the problem solving though. And... although I have a 100%, I'm ashamed to say I still don't get it 100%!

3g O₃/ 7.66g KClO₃ = .340 g O₃/ x g KClO₃   thus x=.8681 g of KClO₃


Thank you for all of your *delete me* I did start to get it a lot more. If anyone can write out how to do the problem easily I will checkup on the forum tomorrow so I can do well on the test.  Thanks again!

[Borek]

I still don't understand the mechanics of the problem solving though.

Jorris gave you very good hints:

Basically, heating the sample removed .340 grams of oxygen. You can figure out the moles of oxygen removed and since oxygen was only present in the potassium chlorate, you can figure out the moles of potassium chlorate.

[DrCMS]

If you did not know what happens when you heat potassium chlorate why did you not look it up rather than ignore it?

http://en.wikipedia.org/wiki/Potassium_chlorate

2KClO3(s) + heat → 3O2(g) + 2KCl(s)