A 1.24-g sample of a hydrocarbon, when completely burned in an excess of O2(g), yields 4.04 g CO2 and 1.24 g H2O Draw a plausible structural formula for the hydrocarbon molecule.

I think I understand how to get it to an empirical formula but I don't know how to get it into a structural formula.

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  1. based on the empirical formula and its contents, you can form the structural formula. what do you think the empirical formula is?

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  2. Sorry, I thought empirical gave only the simplest structure. I thought I would need to know the molecular to get the correct number of carbons and hydrogens.

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  3. What empirical formula did you get?

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  4. The empirical formula and structural formula are not the same thing... The empirical formula just shows the lowest whole number ratio of elements in a compound. The structural formula is a 2-dimensional or 3-dimensional graphic formula showing a plausible arrangement of elements.

    For this problem ...
    1. Determine the % Elemental Composition of CO2 and H2O using formula weights.
    2. Determine grams Carbon and Oxygen in 4.04 gram yield of CO2
    ... grams Carbon = Wt%C x 4.04g CO2 = 27.27% of 4.04g = 1.102g C
    ... grams Oxygen = Wt%O x 4.04g CO2 = 72.73% of 4.04g = 2.938g O
    3. Determine grams Hydrogen and Oxygen in 1.24 gram yield of H2O
    ... grams Hydrogen = Wt%H x 1.24g H2O = 11% of 1.24g = 0.136g H
    ... grams Oxygen = Wt%O x 1.24g H2O = 89% of 1.24g = 1.104g O
    4. Determine mass of Hydrocarbon consumed...
    ... grams of HC = grams C + grams H =
    4. Determine grams of Hydrocarbon consumed = grams C + grams H
    = 1.102g + 0.136g = 1.238g HC consumed
    5. Determine Empirical Formula from gram mass values ...
    a. Convert to % per 100 wt
    ... %C per 100wt = (1.102/1.238)100% = 89.014%
    ... %H per 100wt = (0.136/1.238)100% = 10.986%
    b. Calculate the empirical ratio ...
    % => grams => moles => mole ratio => reduce (divide by smaller mole value) => empirical ratio => empirical formula
    %C = 89.014% => 89.014g => (89.014/12) = 7.418 mole C
    %H = 10.986% => 10.986g => (10.986/1) = 10.986 mole H
    C:H mole ratio = 7.418 mole C : 10.986 mole H
    Reduce moles by dividing by smaller mole value => empirical ratio ...
    => 7.418/7.418 : 10.986/7.418 => 1:1.5 => 2:3 empirical ratio...
    Empirical Formula => C2H3
    Most likely Molecular Formula => 2(C2H3) => C4H6
    2C4H6 + 11O2 => 8CO2 + 6H2O which corresponds to moles of product yield (4.04/44 = 0.092 mole CO2) and (1.24/18 = 0.069 mole H2O) given in problem. Such requires 0.1263 mole O2 and 0.023 mole of C4H6.

    Structural Formula (using 'Lewis Formula' graphics) ...
    H2C = CH - CH = CH2

    Hope this helps, Doc :-)

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  5. Here's the Lewis Structure Analysis for C4H6 ...
    Available valence electrons = 4C + 6H = 4(4) + 6(1) = 22
    Electrons in All Bonds = 4C + 6H = 4(8) + 6(2) = 44
    No of Cov Bonds = Bonded Electrons - Valence Electrons / 2 = (44-22/2) = 11 Covalent Bonds in molecule...
    The most likely structure conforming to the octet rule is ...
    H2C = CH - CH = CH2 (containing 11 covalent bonds)

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