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Empirical Formula

  1. Dec 11, 2007 #1
    [SOLVED] Empirical Formula

    Hi, I have a example question in my textbook that I don't fully understand. If someone could please help me out I would really appreciate it!! Thanks!


    In this example, 15.5 g of a compound containing carbon, hydrogen and oxygen is completely burned in air; 22.0 g of CO2 and 13.5 g of water vapour are produced. Find the empirical formula.

    Moles of CO2 = 22.0 g / 44.0 g / mol = 0.500 mol

    Moles of H2O = 13.5 g / 18.0 g / mol = 0.750 mol

    moles of C atoms = 0.500 mol and moles of H atoms = 0.750 mol x 2 = 1.50 mol

    We must now find the number of of moles of oxygen atoms.
    (I don't understand how you can get mols of C atoms and H atoms, but not mols of O atoms at this point.)

    Mass of carbon = 0.500 mol x 12.0 g / mol = 6.00 g
    Mass of hydrogen = 1.50 mol x 1.00 g / mol = 1.50 g
    mass of oxygen = 15.5g - (6.00 + 1.50) g = 8.0 g

    moles of oxygen atoms = 8.0 g / 16.0 g / mol = 0.50 mol

    Hence mole ratio is
    C = 0.500 mol
    H = 1.50 mol
    O = 0.50 mol

    Multiply ratios by 1.5 to get whole numbers
    C = 1 mol
    H = 3 mol
    O = 1 mol

    Therefore, the empirical formula = CH3O
  2. jcsd
  3. Dec 12, 2007 #2
    To answer your question, first they are computing the masses of carbon and hydrogen in the original compound. You cannot say directly how much O is involved because oxygen is also a reactant (visualize the combustion equation), therefore, since you don't know how much O2 is used in the combustion, you cannot say how much O is in the original compound without deducing it from the other masses.

    Basically, the O that is in the products came from the original compound AND the O2 used in the combustion, so you cannot tell how much came from the original compound versus the oxygen used in combustion.
  4. Dec 12, 2007 #3
    Thanks dwintz02 :)

    So, because the equation is specifically a combustion equation, you have no way of knowing the O atoms.

    Okay, so I have another problem relating to this...

    A compound containing titanium (Ti) and chlorine is analyzed by converting all the titanium into 1.20 g of TiO2 and all the chlorine into 6.45 g of AgCl. What is the simplest formula for the original compund?

    In the problem if I compute the masses of the reactants TiO2 and AgCl the same way above, will I then be able to directly know how much of ALL atoms (Ti, O2, Ag, and Cl) are involved since this is NOT a combustion equation?
  5. Dec 12, 2007 #4
    why don't use just convert the masses to moles so you can tell how many moles of each you will have? Or you could even do percent composition since you know that all of the Ti is in the 1.2 gram of TiO2
    Last edited: Dec 12, 2007
  6. Sep 27, 2011 #5
    Re: [SOLVED] Empirical Formula

    Hey ace123,
    Can you tell me that whether we can/should conserve the moles of Cl2 as well in the last question by mystix??
  7. Sep 27, 2011 #6


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    Staff: Mentor

    Re: [SOLVED] Empirical Formula

    You do realize that this thread is 4 years dead?

    All chlorine was converted into AgCl, this type of analysis is based on mass conservation principle.
  8. Sep 27, 2011 #7
    Re: [SOLVED] Empirical Formula

    kk. Sorry, I din't realize that earlier. But I still want the answer. Can you help me with this, Borek??
  9. Sep 27, 2011 #8


    User Avatar

    Staff: Mentor

    Re: [SOLVED] Empirical Formula

    I already gave you an answer, didn't I?
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