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March 27, 2017

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An unknown compound contains carbon, hydrogen, and oxygen. When burned, a 2.000 gram sample of this compound releases 3.451 g of carbon dioxide and 1.059 grams of water. What is the empirical formula of this compound?

Thank you very much.

  • Chemistry - ,

    Convert g CO2 to g C and g H2O to g H. Add grams C and grams H and subtract from total mass to find grams oxygen.

    Convert g C, g H, g O to moles.
    moles = grams/molar mass

    Then find the ratio of the elements to each other. I worked the Na2S2O3 problem for you earlier; this one is done the same way to find the empirical formula. Post your work if you get stuck.

  • Chemistry rework problem - ,

    Thank you Dr. Bob222

    An unknown compound contains carbon, hydrogen, and oxygen. When burned, a 2.000 gram sample of this compound releases 3.451 g of carbon dioxide and 1.059 grams of water. What is the empirical formula of this compound?

    Here is the problem I worked out:
    C=.94118 gC
    H=.1177 gH
    O=2.00-3.451-1.059 = .392

    if i divide everything by .1177, i get

    c=7.996
    H=1.00
    O=3.3305

    Now, I am stuck on how to get the ratios and final answer

  • Chemistry - ,

    First, you made an error before you get to that part.
    grams O = 2.00-grams C - grams H. You subtracted grams CO2 and grams H2O. Your grams C (and moles) is ok as is the grams H (and moles H) but the oxygen must be redone.

  • Chemistry - ,

    Okay, so my O=.82348

    then I divide everything by .1177, and i get

    c=7.996
    H=1.00
    O=6.99

    This looks like a ratio of 8:1:7

    C8H107??

  • Chemistry - ,

    No, it looks like 8:1:7 to me but I don't think that is right.
    I have C = 0.9421g (close enough to your answer) and H = 0.1176g (close enough to your answer) so O is
    2.000 - 0.9421 - 0.1176 = 0.9403 g O (the slight differences in our C and H won't account for this difference in your value and mine. I've checked mine; perhaps you should check yours.
    Then
    0.9421/12 = 0.0785 moles C
    0.1176/1 = 0.1176 moles H
    0.9403/16 = 0.05877 moles O

    0.05877/0.05877 = 1 for O
    0.1176/0.05877 = 2 for H
    0.0785/0.05877 = 1.33 for C and these aren't whole numbers. You can quickly see that multiplying all by 2 won't work because 1.33 becomes 2.66; however, multiplying by 3 should get it.
    O = 1*3 = 3
    H = 2*3 = 6
    C = 1.33*3 = 3.99 which rounds to 4.00
    I would write C4H6O3 for the empirical formula.
    Check my work.

  • Thank you DrBob222 - ,

    Thank you - you are an awesome teacher!!!!!

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