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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: cœur_vaillant on June 03, 2008, 01:36:50 AM

Title: hi, another question
Post by: cœur_vaillant on June 03, 2008, 01:36:50 AM
hey, im having a bit of trouble with this question, if someone could help would be much appreciate. merci!  ;D
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In a series of experiments, a chemist prepared three different compounds that contain only iodine and fluorine and determined the mass of each element in each compound

Compound        mass iodine (g)            mass fluorine (g)

1                         4.75                       3.56

2                         7.64                       3.43

3                         9.41                       9.86



(a) calculate the mass of fluorine per gram of iodine in each compound.
(b) how do the numbers in part (a) support the atomic theory?

Title: Re: hi, another question
Post by: tamim83 on June 03, 2008, 09:51:25 AM
This question is pretty straight foward.  Reread part a, it tells you what to do practically. 

For part b, you need to know the postulates of atomic theory (see http://en.wikipedia.org/wiki/John_Dalton (http://en.wikipedia.org/wiki/John_Dalton)).  Look closely at number 4. 

Good Luck.   ;)
Title: Re: hi, another question
Post by: cœur_vaillant on June 03, 2008, 07:38:24 PM
well i tried the question and didn't get it (hence why i came on here for help..i did not just post it on here without doing any work).  it may be straightforward for you, but i do not see it as such. so, please be understanding.
Title: Re: hi, another question
Post by: macman104 on June 03, 2008, 07:45:17 PM
Mass of fluorine PER gram of Iodine, the "per gram" indicates division:

mass of fluorine / mass of iodine is mass per gram of iodine.

For the second part tamim has already given you a link and direction on where your next step should be.
Title: Re: hi, another question
Post by: tamim83 on June 04, 2008, 09:35:47 AM
Quote
well i tried the question and didn't get it (hence why i came on here for help..i did not just post it on here without doing any work).  it may be straightforward for you, but i do not see it as such. so, please be understanding.

Sorry, I wasn't trying to be mean or anything.  I usually tell my own students to reread questions that I think are pretty straight forward.  Sometimes a very careful rereading helps a great deal. 
Title: Re: hi, another question
Post by: klumba on November 15, 2009, 07:00:48 PM
i've got the same problem. I found the numbers but I can't associate it with the atomic theory. It doesn't apply to the law of multiple proportions either. Or..to make it work, I should solve this problem by molar masses. So, I am really confused about b) part.
Title: Re: hi, another question
Post by: Borek on November 16, 2009, 07:46:05 AM
What are these numbers? (list values)
Title: Re: hi, another question
Post by: klumba on November 18, 2009, 05:22:40 PM
0.749 g
0.449 g
1.05 g
does this suppose to support the law of multiple proportions?
Title: Re: hi, another question
Post by: Borek on November 19, 2009, 03:46:31 AM
Try to find a number so that each number on the list is a multiply of that number.

For example, 0.4, 0.6 and 1.0 are all multiplies of 0.2.
Title: Re: hi, another question
Post by: klumba on November 19, 2009, 09:37:43 AM
it should be whole number..unless i find molar masses, there's no way I could get the whole number multiple, what a stupid question it is :)
Title: Re: hi, another question
Post by: Borek on November 19, 2009, 09:48:12 AM
unless i find molar masses, there's no way I could get the whole number multiple

They all ARE whole multiples of some number and you don't need molar masses to find it out.

Well, at least I did it without using molar masses, so I know it is possible for me  :P
Title: Re: hi, another question
Post by: klumba on November 19, 2009, 10:37:07 AM
well...here 0.5 is the possible multiple, but diving .75 by .5 we get 1.5, so...i'm still very confused..
Title: Re: hi, another question
Post by: Borek on November 19, 2009, 10:55:34 AM
45, 75 & 105 - what is the GCD?
Title: Re: hi, another question
Post by: klumba on November 19, 2009, 11:19:35 AM
5...
0.05 would be for my numbers..
but don't they have to be divided by the smallest number among them and by gcd?
Title: Re: hi, another question
Post by: Borek on November 19, 2009, 12:32:40 PM
No, it is not 5.
Title: Re: hi, another question
Post by: klumba on November 19, 2009, 08:52:18 PM
7?
Title: Re: hi, another question
Post by: Borek on November 20, 2009, 02:55:41 AM
Is 45 divisible by 7?

Be warned: I am not going to play the game in which you will list all possible integer numbers asking if they are correct. You have one last try, finding gcd is not a rocket science.
Title: Re: hi, another question
Post by: klumba on November 20, 2009, 05:44:30 PM
no,no...i meant 15...  i divided 105 by 15 and got 7...so typed the wrong number..
but still...don't these numbers should be divided by the smallest number among them and not by gcd? ???
Title: Re: hi, another question
Post by: Borek on November 20, 2009, 07:24:08 PM
no,no...i meant 15...  i divided 105 by 15 and got 7...so typed the wrong number..

OK, so now you have muyour multiple proportions.

Quote
but still...don't these numbers should be divided by the smallest number among them and not by gcd? ???

No. They have to be converted into set of lowest integer numbers that have the same ratios that the original ones. 0.75/0.45 doesn't yield an integer, 0.75/0.15 does.
Title: Re: hi, another question
Post by: klumba on November 20, 2009, 08:09:25 PM
so, there is no direct relationship between obtained massed? I have to explain it " because all of them a multiple to 0.15 and give whole numbers, it supports the law of multiple proportions"? I am sorry, but I reeeeeeeeeeally don't get it :(
Title: Re: hi, another question
Post by: Borek on November 21, 2009, 04:26:03 AM
It doesn't matter multiples of what they are. It is their ratios that count and ratio of 1.05:0.75:0.45 is identical to 7:5:3 - three small integers. And fact that these are small integers is exactly what law of multiple proportions says.
Title: Re: hi, another question
Post by: klumba on November 21, 2009, 10:57:35 AM
ooooooooooooh, i got it now! thank you!!! :D