A hydrocarbon Y on combustion gives 0.352g of carbon(iv)oxide and 0.18g of water. Calculate the empirical formula of the compound and If the relative molecular mass of Y is 58g, determine the molecular formula of Y.?

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  1. ...Y + O2 ==> CO2 + H2O
    ...................0.352 g + 0.18 g
    g C in Y is 0.352 x (12/44) = 0.096
    g H in Y is 0.18 x (2/18)) = 0.02

    mols C in Y = 0.096/12 = 0.008
    mols H in Y = 0.02/1 = 0.02
    Now find the ratio. The easy way is to divide both numbers by the smallest number; i.e.,
    C = 0.008/0.008 = 1
    H = 0.02/0.008 = 2.5
    Make whole numbers which will be C = 2 and H = 5 so the empirical formula is C2H5. The empirical formula will be (C2H5)x so x is a multiple of the empirical formula. The empirical formula is 29 so 29*what number = 58.
    The molecular formula is (C2H5)2 or C4H10. Is the molar mass 59?
    4*12 = 48 and 10*1 = 10
    48+10 = 58. Bingo.

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