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Converting with moles (chem20)

  1. Nov 18, 2005 #1
    A chemist is studying one of the five oxides of nitrogen: N2O, NO, N2O3, NO2, or N2O5. He learns that 250mL of the gas has a mass of 0.335g. Which oxide is he working with?
    To get an answer, I would first calculate the mL of each substance, but how? I only know that one mol=22.4mL at STP, but I don't think these are at STP. Plz help. :confused:
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  3. Nov 19, 2005 #2


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    It seems you're a few decimal places off on molar volume..
    The molar volume of any gas at STP approximately 22.4L (not mL). Did you calculate the mass of a mole, for each of your oxides?

    Try with assumption that it is at STP. Using ideal gas law (PV = nRT), P=1 Atm, R=0.08206 (L Atm)/(mol degK) , T=273degK
    V=?? volume (L). The volume you are given in the problem 250mL (how many liters is that?)
    The only variable left, that you can now solve for is n (number of moles), n = PV/RT
    You will find it n<1 and therefore a fraction of a mole.
    If you already calculated the mass of a mole for each of your oxides, multiply each by this fraction and compare to the mass you were given.. You will find one of them agrees.

    (take care when applying the Universal Gas Constant (R). It will be a different value depending on the units you are choosing)
    Look at the table for R in the following reference to see what I mean)
    Last edited: Nov 19, 2005
  4. Nov 19, 2005 #3


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    Haha.. I just found a short cut.. You know that 1 mole = 22.4L at STP. Find out how many moles (n) are equal to 250mL. (don't forget to change to L). You will again find that n<1 and is the same fraction (to 5 decimal places) you would get using the Ideal Gas Law. (you might want to do it both ways as a double check)

    Then do the same as I described before, multiply the molecular masses of each of your oxides by this fraction and compare to the mass you were given. One of them will agree.
    Last edited: Nov 19, 2005
  5. Nov 19, 2005 #4
    the first way you mentioned was way to deep, I havn'tgotten to that yet, but I assumed it was at STP as you said, and I went 0.335g x 22.4L / molar mass andthat gave me the volume to compare with 250mL. I havn't learnt about Ideal gas Law yet, so it's kind of difficult for me.but thanks for your help.
  6. Nov 19, 2005 #5


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    Don't overcomplicate things. 1 mole at STP is 22.4 L and you know 0.25 L has mass 0.335 g. Calculate mass of 22.4 and assume it is a mass of one mole:

  7. Nov 19, 2005 #6


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    Nice job!!! As you found out, there's more than one way to solve the same problem.. Your method will work fine.. If you noticed that molar mass is a variable in your expression and the other 3 are known, you could algebraically manipulate your expression to solve for the variable (n = molar mass of oxide)

    (0.335g x 22.4L)/n = 0.25L
    so n = (0.335 x 22.4)/0.25L

    (which is the same expression Borek gave, assuming you calculate Borek's expression in a linear fashion)
    Solving for n, you will get the unknown molar mass. But of course, you still need to calculate the molecular masses of each of your oxides, to see which it is.

    Didn't mean to toss in there something you haven't covered yet. But one day when you do learn how to apply "Ideal Gas Law", you'll find that is how they derived 22.4L as the volume of 1 mole of any gas at STP.
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