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Topic: Unknown Element X  (Read 12450 times)

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Offline darknietzsche

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Unknown Element X
« on: November 08, 2007, 06:31:54 PM »
I have a problem for my Chemistry homework for my class, and I have been having some difficulties. It involves an unknown element X in a compound reacting with water and I have to determine the unknown element X using the information given.

Q: A compound with the formula XOCl2 reacts violently with water, yielding HCl and the diprotic acid H2XO3. When .350g of XOCl2 was added to 50.0 ml of water and the resultant solution was titrated, 96.1 ml of NaOH was required to react with all the acid.

Find the atomic mass and identity of element X?

This is what I know:

XOCl2 + 2H20 ---> 2HCl + H2XO3

and

H2XO3 + 2NaOH ---> 2H2O +Na2XO3

If someone can just show the math to get to the number of moles of
XOCl2 which will allow you to get the molar mass of the compound. I may have messed up in the math somewhere, but when I did the calculations I got the mass of X to be negative and couldn't find my mistake.

Offline Borek

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Re: Unknown Element X
« Reply #1 on: November 08, 2007, 06:43:56 PM »
Show your caculations so that someone can locate problem. Right now it is impossible to check/do anything, as NaOH concentration is not given.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline ARGOS++

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Re: Unknown Element X
« Reply #2 on: November 08, 2007, 06:57:03 PM »
Dear DarkNietsche,

You may forgot, that you will also titrate the generated HCl and that means that :
      four mol NaOH is equivalent to only one mol of your H2XO3.

Now from the titration you should be able to calculate how much mol of Cl2XO will weight 0.350g.
I think that the last step is not too difficult for you.
(I assume that concentration of NaOH is equal 1N.)

 
Good Luck!
                    ARGOS++
« Last Edit: November 08, 2007, 07:15:46 PM by ARGOS++ »

Offline darknietzsche

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Re: Unknown Element X
« Reply #3 on: November 08, 2007, 08:54:01 PM »
Sorry, the molarity of NaOH is .1225M.

Offline darknietzsche

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Re: Unknown Element X
« Reply #4 on: November 08, 2007, 09:09:10 PM »
Okay, I have redid the math using the fact that four mol of NaOH = to one mol of H2XO3 and got a closer answer.

Num. of mols of NaOH= .0961L x .1225M= 1.177 x 10-2 mols

1.177 x 10-2 mols of NaOH X 1 mol of H2XO3/ 4 mols of NaOH= 2.94 x 10-3mols

2.94 x 10-3mols H2XO3 x 1 mol of XOCl2/ 1 mol of H2XO3= 2.94 x 10-3 mols of XOCl2

molar mass= .350g/2.94 x 10-3= 118.92 g/mol

118.92g- 35.45g x 2- 16.0g= Mass of element X or 32.02 g which is close to the mass of sulfur

This is my new math, and at least now I have an element. If this is incorrect, please correct the error.

Offline agrobert

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Re: Unknown Element X
« Reply #5 on: November 08, 2007, 09:37:19 PM »
Look up thionyl chloride, do your reactions make sense for this highly reactive compound?  I think you answered your questions.  Your weight may be off due to rounded atomic masses.
In the realm of scientific observation, luck is only granted to those who are prepared. -Louis Pasteur

Offline darknietzsche

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Re: Unknown Element X
« Reply #6 on: November 08, 2007, 09:57:23 PM »
Yeah, your right. Thanks.

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