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Offline T

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Finding the element
« on: July 22, 2015, 05:30:20 AM »
Hello,

The question is:
The element A burns in oxygen to produce B, which can be catalytically oxidised further (using V2O5/K2O) to C. B reacts with water to produce a weak acid D, while C reacts with water to produce a strong acid E. In 2001, 165 tons of E were produced worldwide (more than any other chemical). Element A reacts with chlorine gas to form a toxic golden-yellow liquid F. F has two structural isomers. F can be further chlorinated to form a cherry red liquid G, which boils at 59oC, with molecular formula ACl2. Both F and G react with water to form a mixture of products, including B, D and E. The above information is summarised in the following diagram (which I have attached)

A 0.29 g sample of the element was exhaustively oxidised and the product (compound C) absorbed in water and titrated with 1.00 mol L–1 sodium hydroxide. The volume of hydroxide required was 18.0 mL. Use this information to deduce the identity of A.

So firstly I found how much sodium hydroxide in mol was used.
1 M × 0.018 = 0.018 mol of NaOH is used.

From a site they said that the most common strong acids are:
HCl, H2SO4, HI, HNO3, HBr, HClO4. So this means E is one of these molecules.

I already know the answer since the cherry red liquid was in the olympiad exam the year before, but without that, I would not be able to deduce what A is.

Other than these things I can not think of anything else, could someone please give me a hint on where to start?

Thanks

Offline mjc123

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Re: Finding the element
« Reply #1 on: July 22, 2015, 06:34:39 AM »
Quote
165 tons of E were produced worldwide (more than any other chemical).
Are you sure you read this right - 165 tons?
But this gives you a big hint right away. What is the most produced chemical in the world? - a fact every chemist should know.
A forms two oxides, each of which dissolves in water to give an acid. It also forms two chlorides, one ACl2, the other presumably of empirical formula ACl - but it can't be molecular ACl since it has two isomers. A2Cl2?
How many elements can you think of that satisfy the above?
If your guess is right, how many moles of acid E react with 0.018 mol NaOH (assuming the molecular formula of E contains one A atom)? Is this consistent with the formulation of E?

Offline T

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Re: Finding the element
« Reply #2 on: July 22, 2015, 07:03:39 AM »
I see thanks for the hints.

If your guess is right, how many moles of acid E react with 0.018 mol NaOH (assuming the molecular formula of E contains one A atom)? Is this consistent with the formulation of E?

So using this hint, (as the other 2 hints requires good common chemistry knowledge which I don't have) I would have just gone over my list of strong acids and trial and error to see which acid fits. Would you think this is the best way? Or is there a better way?

Thanks again

Offline Dan

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Re: Finding the element
« Reply #3 on: July 22, 2015, 07:10:24 AM »

A 0.29 g sample of the element was exhaustively oxidised and the product (compound C) absorbed in water and titrated with 1.00 mol L–1 sodium hydroxide. The volume of hydroxide required was 18.0 mL. Use this information to deduce the identity of A.

So firstly I found how much sodium hydroxide in mol was used.
1 M × 0.018 = 0.018 mol of NaOH is used.

From a site they said that the most common strong acids are:
HCl, H2SO4, HI, HNO3, HBr, HClO4. So this means E is one of these molecules.

Compare the parts I highlighted in red - not all of your compounds are consistent with the oxidation/hydration sequence.
My research: Google Scholar and Researchgate

Offline Arkcon

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Re: Finding the element
« Reply #4 on: July 22, 2015, 07:14:40 AM »
Well, if you know all the strong acids, that's a way to "cheat" at this problem.  Since I know which strong acid is produced most worldwide -- on the order of millions of tons -- that makes the problem trivial.  However, next time you won't have hints like this, so I would start attacking this problem's reactions.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline mjc123

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Re: Finding the element
« Reply #5 on: July 22, 2015, 08:53:00 AM »
Quote
So using this hint, (as the other 2 hints requires good common chemistry knowledge which I don't have) I would have just gone over my list of strong acids and trial and error to see which acid fits. Would you think this is the best way? Or is there a better way?
That's the point, really. The "better way" is to acquire "good common chemistry knowledge" - there's no way to cut corners on that. There is only a limited extent to which you can work out answers from first principles based only on knowledge of some general concepts (e.g. equilibria, Le Chatelier's priciple, ideal gas laws..), without a reasonable knowledge of factual chemistry. If you are working through a textbook, the problems with each chapter will test the knowledge learned from that chapter (along with what went before), but exam and olympiad questions will assume a certain amount of chemical knowledge. You would be expected to know, for example, the basic chemistry of common elements like sulfur; that sulfuric acid is the biggest industrial chemical; that ammonium chloride is a stable solid that doesn't decompose at room T and P; that carboxylic acids react with amines to give amides.

Another poster http://www.chemicalforums.com/index.php?topic=80837.0 has discovered that "knowing the concepts" isn't enough without the maths; likewise it's not enough without the factual chemistry.

You are obviously enthusiastic about learning chemistry, which is great, and I don't want to discourage you, but I think you're trying to run before you can walk. I would suggest leaving the olympiad questions until you have learned some more chemistry. We've all been there - we all knew nothing before we learned anything - but you'll never advance beyond a certain point without the sheer work of learning the facts. Hopefully you will find this interesting in itself, and will be able to relate it to your conceptual knowledge; e.g. don't just learn the chemistry of each element, but try to identify periodic trends, relate them to atomic structure etc. Organic chemistry is well organised in this way, the raw facts of millions of reactions being arranged into nucleophilic substitutions, eliminations, additions etc.

Offline T

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Re: Finding the element
« Reply #6 on: July 23, 2015, 02:03:34 AM »
Compare the parts I highlighted in red - not all of your compounds are consistent with the oxidation/hydration sequence.

So are you saying HCl, HI, HNO3, and HBr don't "match". Is this because there isn't 4 oxygens attached to it? If something is exhaustively oxidised with oxygen, can it turn oxygen into O + 2e- :rarrow: O2- instead of bonding with it?

However, next time you won't have hints like this, so I would start attacking this problem's reactions.

What do you mean by attacking this problem's reactions? The only thing I got from the reaction was A is non-metal as it forms a liquid with chlorine. A could form AO or AO2 or ACl or ACl2 so I can't really get the oxidation numbers of A.

You are obviously enthusiastic about learning chemistry, which is great, and I don't want to discourage you, but I think you're trying to run before you can walk. I would suggest leaving the olympiad questions until you have learned some more chemistry.

I completely understand what you are saying. I am in year 10 and have had chemistry for 6 weeks in February, and I was learning basic chem eg. properties of metals, properties of ionic compounds. My chem teacher told me to try for the olympiad and I have been studying myself since then. I really want to do well this year because it may be my only chance to do an olympiad. My olympiad exam is in two weeks and I doubt I will do as well as I hope but I will try my best.

It would help me so much if you could give me some more suggestions on learning factual chemistry in addition to the "identify periodic trends, relate them to atomic structure etc." I have been searching facts every chemistry student should know, do you think that is a good way?

Thank you everyone and especially mjc123 in this thread for telling me about the importance of basic chemistry knowledge.

Offline Dan

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Re: Finding the element
« Reply #7 on: July 23, 2015, 04:09:10 AM »
Compare the parts I highlighted in red - not all of your compounds are consistent with the oxidation/hydration sequence.

So are you saying HCl, HI, HNO3, and HBr don't "match". Is this because there isn't 4 oxygens attached to it? If something is exhaustively oxidised with oxygen, can it turn oxygen into O + 2e- :rarrow: O2- instead of bonding with it?

Let's take HI as an example. The acid C (correction, E) is produced by exhaustive oxidation of an element.

What is the oxidation state of I in HI?
What is the oxidation state of I in elemental I2?
Can oxidation of I2 produce HI?

You also suggested perchloric acid as an option for C (correction, E). Does this fit with the reactions A :rarrow: F :rarrow: G?

This question is not so crazy if you consider the information generally.

ACl2 is liquid - so A must be a non metal, and probably not a very heavy one.
A can exist in multiple oxidation states, some of which are even (you are told +2). An even oxidation state strongly suggests an even group number. Now we have narrowed it down  to only a handful of elements.
We know the highest oxide is strongly acidic in water, this limits the options even further, and at this point you can write out the formulae of the possible oxoacids E, look at the possible stoichiometries of the reaction with NaOH and solve for the atomic weight of A.

Alternatively, with very little chemical knowledge, you can probably assume D (correction, E) is either mono-, di- or triprotic. Write the 3 general equations for the reactions with NaOH, and solve for the atomic weight in each case and pick the one that makes sense.

e.g.

If A (correction, E) is monoprotic:

AOxH + NaOH :rarrow: NaAOx

18 mmol NaOH means: 18 mmol AOxH means: 18 mmol A (which weighs 290 mg) means: atomic weight of A is 16 g/mol.

Is that a sensible answer? If not, move on and calculate for diprotic... etc. until you reach a sensible answer.

Edit: Lettering mistakes corrected
« Last Edit: July 23, 2015, 08:15:25 AM by Dan »
My research: Google Scholar and Researchgate

Offline T

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Re: Finding the element
« Reply #8 on: July 23, 2015, 05:16:25 AM »
I see thanks Dan.

One last question, you said that the highest oxidation state is strongly acidic in water. So you are saying the oxidation state of A in C is the highest oxidations state element A can achieve. Is this because to oxidise A to C you need to use a catalyst so therefore it would be the highest oxidation state of A?

Thanks again

Offline Dan

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Re: Finding the element
« Reply #9 on: July 23, 2015, 06:06:23 AM »
One last question, you said that the highest oxidation state is strongly acidic in water. So you are saying the oxidation state of A in C is the highest oxidations state element A can achieve. Is this because to oxidise A to C you need to use a catalyst so therefore it would be the highest oxidation state of A?

We know it i the highest oxidation state because it says "exhaustively oxidised"; exhaustively = as much as possible.
My research: Google Scholar and Researchgate

Offline T

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Re: Finding the element
« Reply #10 on: July 24, 2015, 03:30:44 AM »
I see thanks Dan.

Why will sulfur + chloride turn into disulfur dichloride then sulfur dichloride and not directly to sulfur dichloride?
« Last Edit: July 24, 2015, 04:02:24 AM by T »

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