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Molecular formula of S in benzene

  1. Oct 14, 2011 #1
    The molal freezing point depression constant of benzene is 5.07 K kg/mol. A .45% solution of monoclinic sulfur in benzene freezes .088K below the freezing point of pure benzene. Find the molecular formula of sulfur in benzene.

    Can someone tell me if the answer is S2 ?

    I guess I'm not 100% sure about what exactly I'm looking for.

    What I did was say that if there was 100g of solution, then
    100g *(.45)*(1/32.1)=1.402 moles of sulfur
    100g *(.55)*(1/78.1)=0.704 moles of benzene

    then divided them which gave about 2 moles of sulfur per benzene.

    But 1) I never used the molality from the freezing point depression which makes me think I'm definitely wrong and 2) I'm just not sure if it makes sense to find moles of sulfur per moles of benzene.

    any insight? This was on my quiz today and I won't rest until I know how wrong I was! lol.
  2. jcsd
  3. Oct 14, 2011 #2


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  4. Oct 14, 2011 #3
    Well, I know that the molality is moles of solute per kg of solvent.

    So for this problem, the molality of sulfur in benzene is .088/5.07= 0.017357

    But I'm not sure what I can do with this to get the molecular formula of sulfur.
  5. Oct 14, 2011 #4


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    Use this information to calculate molar mass of solute.
  6. Oct 14, 2011 #5
    I am getting something obviously very wrong.

    if the molality is .017 mol s /kg benz then...

    (.017 mol S / 1000 g B)*(.55 g B / g total) * (1 g total / .45 g S) = .0000208 mol S / g S

    so the molar mass would be the recipcrocal or 48128 g S / mol S. Which is just ridiculously high.

    So i obviously still don't know what i am doing. I would assume, if this were right, that i would take this molar mass and divide it by the molecular weight of S.

    But with this number that would give me S 1468 , which just can't be right.

    What am I doing wrong?
  7. Oct 14, 2011 #6


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    I must confess I never use dimensional analysis for these things and I am not even sure if it makes any sense here. This means I can't help you showing what you did wrong. However, I can help you solve the problem using slightly different approach.

    How many moles of solute is present in 1000g of solution? What is mass of this solute? Can you combine these two numbers to calculate molar mass?
  8. Oct 14, 2011 #7
    Well, there should be

    (.017 mol S / 1000 g B) * (.55 g B / g sol) = .00935 mol S / 1000 g solution

    But without knowing how much solution there is, I can only say that there is

    (.00935 mol S / 1000 g sol )*(32.1 g S / 1 mol S)= .3001 g S / 1000 g sol = .0003 g S / g sol

    But this doesn't help me see anything differently

    to get back to molar mass I would have to take .0003 g S / g sol and divide by 32.1 (basically undoing what I had done in the previous step) and then dividing by .45.

    this gives me the same number I got before.

    so lost
  9. Oct 14, 2011 #8


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    I told you to assume 1000g of solution. Use molality definition - rearrange it to find number of moles (iow solve it for number of moles). 1000g of solution contains 99.55% of solvent (100%-solute).

    Actually you can assume any other amount of solution (like 100g, 500g, 3.1415g and so on), final result will be identical. You can even assume msolution and solve the problem using symbols - in the end msolution will cancel out.
  10. Oct 14, 2011 #9
    OH...now i see what my mistake was.

    I didn't realize it was .45 PERCENT. I was wondering where you got 99.55%. I was basiclaly doing my calculations for a 45 percent solute solution and 55 percent solvent.

    So I end up with a molar mass of 265.9 which is 8.28 times larger than the molar mass of sulfur.

    so I guess the molecular formula should be S8 ?

    ::sigh:: which means I was definitely wrong on my quiz today.
  11. Oct 14, 2011 #10


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  12. Oct 14, 2011 #11
    You are definitely in my class, and i can't believe we are the only two nerds checking for this on a friday. lol. if it makes u feel any better, i got that question wrong too. I calculated the mb, im hoping for partial credit. :-(
  13. Oct 14, 2011 #12
    hahha i remember you! I definitely was a little grumpy after this quiz...especially now that I know how to solve it and can see how easy it was! I don't like when he throws random surprises like that on the quizzes. The second I looked at the question I think my brain shut off because I panicked since it wasn't something I remembered solving in class or homework. Maybe if he would have given more time I would have figured it out, or not. But these last few quizzes have definitely not gone so hot for me!
  14. Oct 14, 2011 #13
    lol. kudos for solving it tho! I was still scratching my head, if he gives us a 're-take' we will be ready. ha ha. and yes these last couple of quizzes have been awkward... lol. but i gained a lil confidence after the re-take he gave on wednesday (he should do more of those) lol. by the way was the first diagram a non-ideal azeotrope at low boiling?
  15. Oct 14, 2011 #14
  16. Oct 14, 2011 #15
    I don't remember, did it ask about an azeotrope? All I remember was that A boiled at a lower temperature than B.
  17. Oct 15, 2011 #16
    i don't remember what the question was exactly. It didn't say azeotrope though. If it said at two different boiling points, then you might be right. From what I remember it said "Draw a Liquid-vapor diagram that boils at lower temperature." Which is why i drew azeotrope. If he did include A & B then it would be the ideal solution diagram you described.
  18. Oct 15, 2011 #17
    This is what 'I' remember the questions being.

    For the first diagram it said. Draw a Temperature vs. Composition diagram at low boiling temperature. Draw 2 theoretical plates.

    For the second it said. Draw a Temperature vs. Composition diagram of two mixtures A & B and something about the eutectic.

    For the second graph I'm sure he mentioned two different components. for the first graph im not sure he mentioned any components.
  19. Oct 15, 2011 #18


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    This is off topic. You can discuss it using PMs.
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