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Homework Help: Law of multiple proportions

  1. Sep 12, 2006 #1

    I guess I posted this in the wrong string first I didn't notic the homework helper until after I put it in the chem forum.

    I have a question... it is actually from my homework. I am trying to figure out what exactly this problem is wanting. I am thinking a proportion of Carbon and Oxygen.

    In addition to Carbon Monoxide (CO) and carbon dioxide (CO2), there is a third compound of carbon and oxygen called carbon suboxide. If a 2.500g sample of carbon suboxide contains 1.32g of C and 1.18g of O, show that the law of multiple proportions is followed.

    I have been working on this for a good portion of the day and I can't seem to figure out what it really wants. It also confuses me why they would put Carbon Monoxide and Carbon Dioxide with it unless it has something to do with the problem. Unless it is just extra information. first I was thinking of doing 1.32g/1.18g to get the ratio for carbon suboxide. I am not sure if that is what they are asking for though.

    The next problem I have to do is

    What is a possible formula for carbon suboxide (problem 2.35)?

    I am thinking they are wanting C3O2 but I guess I need to show I got that from the last problem.
  2. jcsd
  3. Sep 12, 2006 #2


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    CO and CO2 don't have much relevance for this problem, to solve the problem you need to be concerned with the stoichiometry of carbon suboxide.

    First, state the law of multiple proportions, it's a simple theory. Then show your approach for the solution to this problem.
  4. Sep 12, 2006 #3
    well the law of multiple proportions is Elements can combine in different ways to form different substances, whos mass ratios are small whole-number multiples of each other. there is an example in the book that shows there are twice as many atoms of O in NO2 than NO. They use the law of multiple proportions for showing this. The reason I am getting confused is because for all the examples in this book are showing the difference between different elements and with this problem it is just showing one. I have gotten as far as

    carbon suboxide: C:O mass ratio = 1.32g C/1.18g O = 1.12

    after that I am lost... I have had a long day after having Calc 2 and then trying to get this homework done and study for my chem quiz and chem lab safty quiz tomorrow. I have a feeling it is going to be a long night lol.
  5. Sep 13, 2006 #4


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    Well, it's similar to the example with NO and NO2.

    You are told the mass ratio of the suboxide - it's 1.12. Can you now calculate the mass ratios in CO and CO2 as well (using the atomic masses)? What numbers do you get?

    Now compare these 3 numbers.
  6. Sep 13, 2006 #5
    Ok here is what I got out of Recitation from my Chem TA

    1.12 = 28/25

    28 is similar to 25 so he is saying it is more or less 50/50


    CxOy… we know that the amu for C is about 12, and the amu for O is about 16

    12x = 16y

    12(3) = 16(2)

    36 is approximately 34

    so the outcome is C3O2

    Also another way to do it is

    1.32g * ((1 C atom)/(12.0107amu)) * (1amu/1.66x10^-24 g) = 6.6206x10^22

    1.18g * ((1 O atom)/(15.9994amu)) * (1amu/1.66x10^-24 g) = 4.4429x10^22

    take 6.6/4.4 = 3/2 ratio

    finally getting

  7. Sep 14, 2006 #6


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    Yuck! :yuck: That first part is completely wrong! The second bit is correct, but it is still quite badly done and itdoes not answer the question - it does not demonstrate the validity of the law of multiple proportions.

    I'll try and undo the havoc your TA has wreaked upon your class when I next get a longish break.
    Last edited: Sep 14, 2006
  8. Sep 14, 2006 #7
    yeah it was really hard to understand how that worked. Your help would be very much appreciated :u)
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