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Problem about the law of multiple proportions (includes solution)

  1. May 5, 2013 #1

    s3a

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    1. The problem statement, all variables and given/known data
    Problem:
    Three common gaseous compounds of nitrogen an doxygen of different elementary composition are known, (A) laughing gas containing 63.65% nitrogen, (B) a colorless gas containing 46.68% nitrogen, and (C) a brown, toxic gas containing 30.45% nitrogen. Show how these data illustrate the law of multiple proportions.

    Solution:
    According to the law of multiple proportions, the relative amounts of an element combining with some fixed amount of a second element in a series of compounds are in ratios of small whole numbers. Since percent means “parts per hundred” we can assume 100 g. Then, on the basis of 100 g of each compound, we tabulate beow the mass of N, the mass of O (obtained b the difference from 100), and the mass of N per gram of O.

    Compound A Compound B Compound C
    g of N 63.65 46.68 30.45
    g of O 36.35 53.32 69.55
    (g of N)/(g of O) 1.7510 0.8755 0.4378

    The relative amounts are not affected if all three amounts are set up in the form of a ratio and then divided by the smallest of the relative amounts.

    1.7510: 0.8755 : 0.4378 = 1.7510/0.4378 : 0.8755/0.4378 : 0.4378/0.4378 = 4.000 : 2.000 : 1.000

    The relative amounts are indeed the ratios of small whole numbers – 4.000 : 2.000 : 1.000 – within the precision of the analyses.

    The law of multiple proportions was an important contribution to the credibility of Dalton's atomic theory. It was discovered before relative atomic masses were well known (note that ##A_r## values were not involved in the calculation above.) However, it follows logically that all atoms of the same element have the same mass (which is unchangeable) and that compounds contain elements in the relative proportions of simple whole numbers.

    2. Relevant equations
    1) The use of percentages.
    2) The use of ratios.


    3. The attempt at a solution
    Towards the end of the solution, there is a mention of “##A_r##” values. What are these?

    Also, according to the law of multiple proportions “the relative amounts of an element combining with some FIXED AMOUNT OF A SECOND ELEMENT in a series of compounds are in ratios of small whole numbers.”

    The emphasized part from above holds because we fix the ratio of oxygen to be one gram (=the denominator of the fraction), right?

    If more information is needed, just ask. :)

    Any input would be appreciated!
     
  2. jcsd
  3. May 5, 2013 #2

    Borek

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    Staff: Mentor

    http://en.wikipedia.org/wiki/Relative_atomic_mass

    Doesn't matter if it is 1 gram, it can be anything. What matters is that you use the same amount each time, be it 1, 2 or 234323.789234.
     
  4. May 5, 2013 #3

    s3a

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    Edit:
    I made a few edits so, please read the post on this website rather than through the email update.

    I thought that each respective sample needed to have the same amount of mass for that “second element” however, if I'm now correct, what that part of the law meant (which I had misinterpreted) was that if you were to look at the amount of the “first element” PER A CHOSEN FIXED AMOUNT of the “second element” (which can be a different amount found from each of the compounds such that each compound can have a different amount of the “second element” from one another as well as be different than the chosen amount to compare the “first element” to).

    Is the relative atomic mass measured in amu?

    P.S.
    I'm sleepy at the time of thewriting of this reply but, I tried to write it well nonetheless. If it's too badly written, tell me that it's badly written and, I'll rewrite it when I feel more rested/alert.
     
  5. May 6, 2013 #4

    Borek

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    Staff: Mentor

    You've lost me somewhere in the middle.

    If I understand the situation correctly, you were right from the very beginning - I just pointed to the fact fixed amount doesn't have to be 1 g (even if it is a nice and easy amount to work with).

    Have you read at least the very first phrase of the wiki article?
     
  6. May 6, 2013 #5

    s3a

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    I do understand that the fixed amount can be any fixed amount and not just 1 g. And, I think I was wrong the first time and that I may have explained it poorly.

    I'll rephrase it just in case I am now right about being wrong the first time.:
    The amount of the “second element” does not need to be fixed in each compound (in this case compounds A, B and C). As an example to support this (what puzzled me and made me to start this forum thread), look at the compounds and how they each have a different amount of Oxygen. What IS fixed is the denominator (which is 1) in the following ratios: 1.7510, 0.8755, 0.4378. So, basically, there is 1.7510 g N PER ONE GRAM OF OXYGEN in compound A, 0.8755 g N PER ONE GRAM OF OXYGEN in compound B and 0.4378 g N PER ONE GRAM OF OXYGEN in compound C. So it's about choosing a fixed reference point for the calculations rather than actually having the same fixed amount of Oxygen in each compound (in this case compounds A, B and C).

    I feel this explanation of my logic above should be better than that from yesterday but, if you don't understand it perfectly, tell me and I will try to improve it again.

    Perhaps I was right in the beginning and wrong now but, I still wanted to give it a second shot at explaining this since yesterday's explanation was probably poorly written.

    Lol, I have read it but, reading it again with a bit of a fresher mind, it's basically a ratio so the units cancel out and is dimensionless as stated in the first sentence and has the same magnitude as the mass measurements in amu except that the amu unit is not carried along, right? :)
     
  7. May 8, 2013 #6

    s3a

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    Please do tell me if I still don't make sense.
     
  8. May 8, 2013 #7

    Borek

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    Staff: Mentor

    You are right - you can rescale your results, you don't have to do experiment using exactly the same mass each time. It is ratio that counts.
     
  9. May 8, 2013 #8

    s3a

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    Alright, thank you very much. :)
     
  10. May 8, 2013 #9

    s3a

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    I just have one last quick question.: Could I have done this problem with (g of O)/(g of N) ratios instead of (g of N)/(g of O) ratios?
     
  11. May 8, 2013 #10

    Borek

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    Staff: Mentor

    Simplest way to check: try.
     
  12. May 8, 2013 #11

    s3a

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    I already checked it, numerically, before replying to you. :)

    So, it works, numerically, but, I just wanted to make sure that it's not just a coincidence. Why did the solution choose to fix the mass of Oxygen rather than that of Nitrogen? Was it just a random choice?
     
  13. May 8, 2013 #12

    Borek

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    Staff: Mentor

    It is not a coincidence (law of multiple proportions doesn't say which element is the "main" one, so it has to work both ways), yes it was a random choice.
     
  14. May 9, 2013 #13

    s3a

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    Alright, thanks again. :)
     
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