June 21, 2021, 03:21:05 PM
Forum Rules: Read This Before Posting


Topic: Problem of the week - 12/03/2012  (Read 8886 times)

0 Members and 1 Guest are viewing this topic.

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 26578
  • Mole Snacks: +1724/-402
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Problem of the week - 12/03/2012
« on: March 12, 2012, 06:27:55 AM »
Again, borrowed question, with source to be given later.

Sample of the potassium chloride contaminated with some other salt is not losing its mass on roasting in air nor in chlorine. When 2.44 g of the mixture was treated with concentrated sulfuric acid, 0.896 L of HCl was produced (STP). Obtained clear solution was added to the excess concentrated solution of K2CO3, precipitate was filtered and roasted, yielding 0.400g of non-volatile residue.

What substance (and in what mass percentage) was present in the sample?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline Caustikola

  • Regular Member
  • ***
  • Posts: 61
  • Mole Snacks: +4/-9
  • Gender: Male
    • Chemistry:The Central Science
Re: Problem of the week - 12/03/2012
« Reply #1 on: March 12, 2012, 07:57:32 AM »
From my calculations and assumptions, i think AlCl3 is present; 22.5%
« Last Edit: March 12, 2012, 08:24:09 PM by Arkcon »
Nothing gets higher than where God dwells. If there is an upper limit to the universe,call it the abode of God

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 26578
  • Mole Snacks: +1724/-402
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Problem of the week - 12/03/2012
« Reply #2 on: March 12, 2012, 08:18:21 AM »
I am not going to comment on results till at least Wednesday or Thursday, unless someone will show convincing and correct calculations together with the result.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline Wastrel

  • Regular Member
  • ***
  • Posts: 74
  • Mole Snacks: +5/-4
Re: Problem of the week - 12/03/2012
« Reply #3 on: March 15, 2012, 09:18:29 PM »
How difficult is this problem?  Should we ignore weirdness in the question and take the problem at face value?

If I had a fairly pedestrian answer that was out by about 1% should I work it through again, massage the maths a little and take it with a pinch of isotope variation or should I be throwing that away, correcting the HCl for deviation from ideal gas law and go looking for something squaratey?

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 26578
  • Mole Snacks: +1724/-402
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Problem of the week - 12/03/2012
« Reply #4 on: March 16, 2012, 04:48:59 AM »
If you are within 1% chances are you are OK. For sure you don't need corrections for non-ideality.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 26578
  • Mole Snacks: +1724/-402
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Problem of the week - 12/03/2012
« Reply #5 on: March 17, 2012, 06:52:14 AM »
OK, it is Saturday already.

It is not AlCl3. Easy to check: if the 2.44g sample contains 22.5% of AlCl3 it means it contains 0.5490 g of AlCl3 and 1.891 g of KCl, or 4.117x10-3 moles of AlCl3 and 0.02537 moles of KCl. So amount of HCl produced would be 3*4.117x10-3+0.02537=0.03772 moles. 0.03772 moles times 22.4 L (we are at STP) gives 0.845 L, clearly less than it was collected in the experiment.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline Wastrel

  • Regular Member
  • ***
  • Posts: 74
  • Mole Snacks: +5/-4
Re: Problem of the week - 12/03/2012
« Reply #6 on: March 17, 2012, 03:20:25 PM »
I make a mixture of 1.495g KCl and 0.945g MgCl2 (about 38.7%) quite close.  Residue of MgO would be 0.400g and HCl produced would be about 0.894 L.

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 26578
  • Mole Snacks: +1724/-402
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Problem of the week - 12/03/2012
« Reply #7 on: March 17, 2012, 06:59:44 PM »
I make a mixture of 1.495g KCl and 0.945g MgCl2 (about 38.7%) quite close.  Residue of MgO would be 0.400g and HCl produced would be about 0.894 L.

And that's the correct answer (book I use lists 38.9% of MgCl2).

Question was from 11th International Chemistry Olympiad.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline Wastrel

  • Regular Member
  • ***
  • Posts: 74
  • Mole Snacks: +5/-4
Re: Problem of the week - 12/03/2012
« Reply #8 on: March 19, 2012, 09:22:02 PM »
This was strange question.  Concentrated sulphuric acid was added to the chloride and the resulting solution was clear, but by inference that is a solution of the sulphate in concentrated sulphuric acid rather than water.  We are told HCl was produced but it doesn't tell us it was the only gas, it's an assumption.

I have tidied my method for posterity.

Pure KCl would not give enough HCl gas so the adulterating salt must contain chloride.  So the mixture is (1-p)KCl + pMCl where p is the mole fraction of adulterant and M is just inert mass. 

The amount of HCl produced is 0.896/22.4 = 0.04 Moles which is the amount of chloride in the mixture.  So,

2.44 / ((1-p)*(39.1+35.5)+p*(M+35.5)) = 0.04
...
p = 13.6 /(39.1 - M)

The residue after roasting is almost certainly either a carbonate or an oxide.

For a carbonate M2CO3 the amount of mass we would expect is the number of moles of MCl multiplied by the molecular mass of the carbonate, which is,

0.04*p*(M + 30) = 0.4
...
p = 10/(M+30)

We can now solve for M.

13.6 /(39.1 - M) = 10/(M+30)

13.6 (M+30) = 10(39.1 - M)

13.6M+408 = 391-10M
3.6M = -17

Negative masses for atoms are not valid so carbonates don't work and I struck off lithium chloride.

Same procedure for oxide, M2O.
p = 10/(M+8)
and solve,
13.6 /(39.1 - M) = 10/(M+8)
...
23.6M = 282.2

When done to higher precision M=12.26..

M can be any number or fraction of an atom, so long as there is 1 chloride ion per M of other stuff.  Close solutions with oxide products were magnesium chloride and titanium tetrachloride but unless it forms a double salt with potassium chloride the latter wouldn't have survived the roasting in air/chlorine part.  Aluminium trichloride does do this I think incidentally but the mass's weigh off. :)

After picking magnesium, it's atomic mass per chloride then provides two slightly different values for p according to which measurement you use, the volume of HCl or the mass of oxide.  Maybe an effect due to rounding atomic masses.

Sponsored Links