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    a compound contains C 62.08%, H 10.34% and O 27.58% by mass. find its emperical formula given that its relative molecular mass is 58
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    (Original post by itsastudentyo)
    a compound contains C 62.08%, H 10.34% and O 27.58% by mass. find its emperical formula given that its relative molecular mass is 58
    You should post a question like this in chemistry section in future.

    62.08/12 : 10.34/1 : 27.58/16

    5.17 : 10.34 : 1.72

    3 : 6 : 1

    Empirical formula = C3H6O
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    (Original post by TeachChemistry)
    You should post a question like this in chemistry section in future.

    62.08/12 : 10.34/1 : 27.58/16

    5.17 : 10.34 : 1.72

    3 : 6 : 1

    Empirical formula = C3H6O
    thanks i already got that answer but i thought it was wrong becauce i thought u were to divide somwthing with the molecular mass '58' are you not meant to do anytging with that ?
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    (Original post by itsastudentyo)
    thanks i already got that answer but i thought it was wrong becauce i thought u were to divide somwthing with the molecular mass '58' are you not meant to do anytging with that ?
    No. You would only need that to confirm a molecular formula.
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    I always did these with a formula i workedout on the toilet, take the molecular mass, multiply it by the percentage/100, then divide by the element with that percentage.

    Eg. Carbon; (58 x 0.6208)/12 = 3

    Repeat for other elements

    This helps because it gives you the exact number of atoms of each element in the compound
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    (Original post by Spudgunhimself)
    I always did these with a formula i workedout on the toilet, take the molecular mass, multiply it by the percentage/100, then divide by the element with that percentage.

    Eg. Carbon; (58 x 0.6208)/12 = 3

    Repeat for other elements

    This helps because it gives you the exact number of atoms of each element in the compound
    But that assumes you know the molecular mass which is not always the case.
 
 
 
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