Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Finding Empirical Formula Question

  1. Feb 6, 2010 #1
    1. The problem statement, all variables and given/known data
    A 2.203g sample of an organic compound was extracted from a plant. When it was burned in oxygen, the hydrogen in the compound was converted to 1.32g of water and the carbon was oxidised to 3.23g of CO2. Find the empirical formula of the compound

    2. The attempt at a solution
    So I found the number of moles of H2O that were reacted using the n = m/Mr formula (so m(H2O) = 1.32g, Mr(H2O) = 18g/mol therefore n(H2O) = 0.0733mol).

    Also, I found then number of CO2 the same way, and came out nicely that n(CO2) = 0.0733mol

    Here was my first problem. The ratio of n(H2O) : n(CO2) = 0.0733 : 0.0733, but since there are two H atoms in H2O does that mean there is the equivalent of 0.1466mol of H2O (ie: 2 * n(H2O)) or does it mean there are 0.03665mol of H2O (through: 0.5 * n(H2O)). Which one is it and why?

    The second problem I had is this. The answer says that the empirical formula of the compound is: CH2O. What I don't see is how they found the number of O. Also, I'm confused as to how they knew there was O in the compound, does Organic mean it has O in it? Because I always thought that organic meant it had carbon, and that was the only prerequisite?

    Thanks for your help!
     
    Last edited: Feb 7, 2010
  2. jcsd
  3. Feb 8, 2010 #2

    chemisttree

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That analysis only calculates C and H. Anything left over must be due to O.

    1. Find the # grams of CO2 due only to carbon itself. (moles CO2/FW CO2)*(12grams C/ 1 mole C)

    2. Repeat #1 using H2O.

    3. Sum result of 1 & 2. Is it equal to 2.203 g? If not, what is missing?

    4. Calculate moles of O.

    5. Calculate empirical formula given results in 1, 2 & 4.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook