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Hard emperical formula question for assignment

  1. Apr 22, 2007 #1
    Hi all, I'm quite new to posting; just been reading around the forums when I came upon this question for my assignment. I'm really quite stuck and no idea how to relate the different values (such as number of moles of Barium Sulfate) found back to the unknown compound.

    An unknown compound contains sulfur, oxygen and chlorine. When mixed with water the compound yields a mixture of sulfuric and hydrochloric acids. A 0.5404g sample of the compound was dissolved in water and split into two equal aliquots. The first aliquot, when treated with excess barium nitrate solution, yielded 0.4671g of barium sulfate. The second aliquot was titrated with 0.250 mol L-1 NaOH solution, requiring 32.0mL for complete neutralisation. Calculate the emperical formula of the unknown compound.

    If any one can help it would be greatly appreciated
    Steven
     
  2. jcsd
  3. Apr 22, 2007 #2

    symbolipoint

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    "Two equal aloquots", meaning what? WHAT VOLUMES? You must mean each portion contained 0.2702 grams of sample.

    The barium nitrate test would give the amount of sulfate.
    Ba+2 + 2NO3- SO4-2 ----> BaSO4, be sure to balance this if it will help.

    The second test, the titration, will give you total acidity. SOME of the acidity comes from SO4-2, which you just found from the barium test.
     
  4. Apr 23, 2007 #3
    hmm, still having quite a bit of trouble.
    If someone could give a slightly more in-depth explanation as to how to even approach this question it would be appreciated.

    I have found that the number of moles of BaSO4 is = m/M = 0.2702/233.37
    I have also worked out the number of moles of NaOH = cV = 0.25 x 0.032

    But from there what can i do in relation to finding the masses of sulfur, oxygen and chlorine.

    Thanks
    Steven
    - cheers symbolipoint for your response
     
    Last edited: Apr 23, 2007
  5. Apr 23, 2007 #4

    symbolipoint

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    from Steven10137:
    [tex] \[
    molesSulfur = 0.4671gBaSO_4 \frac{{moleBaSO}}{{233.4gBaSO}}\frac{{1moleS}}{{1moleBaSO}}
    \] [/tex]
    [tex] \[
    moles\,Oxygen = 4 \times molesSulfur
    \] [/tex]

    You chief concern is how many moles of each element; the samples analyzed are the same size. The sulfuric acid comes directly from the sulfur, 1:1 mole ratio. Each sulfuric gives two equivalents, so sum of equivalents of acid in sample equals number of equivalents of NaOH used in titration:
    [tex] \[
    moles\,H_2 SO_4 = 0.06417gS\frac{{moleS}}{{32.06gS}}\frac{{1moleH_2 SO_4 }}{{1moleS}}
    \]
    [/tex]

    You fill in the steps. I came up with [tex] \[
    SO_4 Cl_2
    \] [/tex]
     
  6. Apr 23, 2007 #5

    symbolipoint

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    I did not do the neatest typsetting and explaining job; sum of hydrochloric and sulfuric acid equivalents equals number of sodium hydroxide equivalents used.
     
  7. Apr 23, 2007 #6

    ok now i am totally confused by your working ....
    are u trying to say that:
    [tex] \[
    n(S) = \frac{{n(BaSO4)}}{{M(BaSO4)}} x \frac{{1moleS}}{{1moleBaSO4}}
    \] [/tex]
    ????
    and then using a ratio to determine the number of moles of Sulfur, or are you multiplying each of the components against each other?

    can you please just give a slightly more in-depth reasoning as to what you have done and why?
    and why have you used BaSO instead of BaSO4, is there a reason for that i am not sure???
    The answer in the book says:
    [tex] \[
    SO_2 Cl_2
    \] [/tex]
    so im not really sure what to do at all.
     
    Last edited: Apr 23, 2007
  8. Apr 23, 2007 #7

    symbolipoint

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    Steven10137

    I struggled using the TexAide for typesetting, and my explanation obviously is less clear than you or I want. Also, possibly I made a mistake, most likely regarding the titration information, something which should ordinarily be very simple for me. Maybe someone else might rectify some of what I have shown, but so far nobody yet has done so. I will recheck the problem description again and attempt a cleaner explanation.

    Also, I really did mean to put a subscript "4" at the end of what should have been Barium Sulfate (see, difficulty in typesetting, a mistake which I should have not made).

    symbolipoint
     
  9. Apr 23, 2007 #8
    cheers symbolipoint once again your help has been great, i just wish i culd understand the problem ...
     
  10. Apr 24, 2007 #9
    The thing here is that the 0.4671 g of baso4 were formed. Youve found the number of moles. Therefore, youve found the number of moles of SO4(-2) in the solution. This gives you the number of moles of H2So4. Twice the number of moles gives you the equivalents.

    For the second part, youre given the number of moles of NaOH required to neutralise the second solution. This is equal to the number of moles of HCl and H2SO4 present in the solution---(1). From this, subtract the number of moles of H2SO4 calculated above to get the moles of HCl.

    Now, the number of moles you get(from 1), multiply by two to get the total number of moles (H2So4+HCl) is equal to 0.5404grams of the substance. This is because the original solution was divided in two.

    Here is the interesting part, you have the weight of the two individual acids, and you know their formulae. So you know the weight of Hydrogen/oxygen/chlorine in both compounds.

    1)Divide the weight of each compound with its molecular weight
    2)The three values you get, divide each one with the lowest value.
    3)The values you get, round them off to the nearest integer.

    The integers you get are your ratios. So if the ratio of carbon to the lowest one is 3, hydrogen is 1 and chlorine is 3, then C3 H1 Cl2 is the empirical formula. Hit me back if you wanna clarify something, this post wasnt clear enough to get it on the first try, Im afraid.
     
  11. Apr 24, 2007 #10
    ok i have a better understanding of this now ... sort of

    1st test:

    [tex]
    S_xO_yCl_z+BaNO_3~\xrightarrow~BaSO_4+~?
    [/tex]

    [tex]
    n(BaSO_4)=\frac{m}{M}=\frac{0.2702}{233.37}=1.1578 \times 10^{-2}~mol
    [/tex]

    [tex]
    n(SO_4^{2-}) = n(BaSO_4)=1.1578 \times 10^{-2}~mol
    [/tex]

    [tex]
    n(H_2SO_4)=1.1578\times10^{-3}~mol
    [/tex]

    ^^ am i correct in saying this; you stated that: "Twice the number of moles gives you the equivalents" im not quite sure, are you saying that the number of moles of H2SO4 is 2x or what??

    2nd test:

    [tex]
    n(NaOH)=cV=0.25 \times 0.032=8.00\times10^{-3}~mol
    [/tex]

    [tex]
    n(NaOH)=n(HCl)+n(H_2SO_4)
    [/tex]

    [tex]
    n(HCl)=n(NaOH)-n(H_2SO_4)
    [/tex]

    [tex]
    n(HCl)=8.00\times10^{-3}-1.1578\times10^{-3}=6.8422\times10^{-3}~mol~HCl
    [/tex]

    [tex]
    Total~number~of~moles~n(HCl)+n(H_2SO_4)=2\times~(6.8422\times10^{-3})=1.3684\times10^{-2}~mol
    [/tex]

    Just to clarify; the emperical formula we are trying to find is that of Sulfur, Oxygen and Chlorine. The answer is infact [tex] SO_2Cl_2 [/tex]

    I am aware of how to find the emperical formula of the substance, I just need to be able to get the masses of S, O and Cl.

    [tex]
    n(H_2SO_4)=1.1578\times10^{-3}~mol
    [/tex]

    thanks for your help it has been great
     
    Last edited: Apr 24, 2007
  12. Apr 26, 2007 #11
    Yeah, you're right. There is a difference between the number of moles and the number of equivalents. In this case, since H2SO4 is dibasic, the n-factor is 2, so the number of equivalents is twice the number of moles. This is not really required, although its an alternative method. Don't worry about it.
    This is absolutely correct. Now, you see, you have the weights of HCl and H2SO4, and from those, you can find out the weights of H/Cl/S/O present in the compound. Since the compound is solvated, you would need to remove water from the formula you get (till all the hydrogens are removed), to get your original compound.
     
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