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Combustion analysis of 0.343 g of a compound containing C, H, and O produces 0.753 g of CO2 and 0.411 g of H2O. Mass spectral analysis shows that the compound has a molar mass around 120 g mol-1. What is the compound's:
Empirical Formula:-----
Molecular Formula:-----

  • Chemistry - ,

    The problem doesn't tell you how much oxygen so you must calculate that.
    Convert 0.753 g CO2 to grams C.
    Convert 0.411 g H2O to grams H.
    Add those two and subtract from 0.343 to find grams oxygen.

    Now convert each of those grams to moles.
    g C/12 = moles C
    g H/1 = moles H
    g O/16 = moles O.

    Find the ratio in small whole numbers to each other. An easy way to do that is to divide the smallest number by itself which assures you of getting 1.0000 for that one, then divide all the other numbers by that same small number. Round to the nearest whole number to obtain the empirical formula. Determine the mass of the molecule as if the empirical formula is the molecular formula. The problem gives additional information that the molecular formula is approximately 120; therefore, 120/your empirical formula mass = some number which you should round to the nearest whole number (which I will call x). Then the compounds molecular formula will be (empirical formula)x.
    To check your values, I obtained C3H8O for the empirical formula, x of 2 to make the molecular formula C6H16O2. Post your work if you get stuck.

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