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chem

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A 0.01020 mg sample of an unknown compound was found to contain 0.0084 mg of C, and 0.0018 mg of H. It was also determined that the unknown compound had a molecular weight of 58.12 g/mol. What is the molecular formula for this compound?

  • chem -

    You really should convert mg to gams BUT that give you so many zeros before the decimal point that it becomes confusing. It's ok to leave them in mg if we do them all the same way.
    0.0084/12 = about 7E-4
    0.0018/1 = 1.8E-3
    Find the ratio of C and H. The easy way to do that is to divide both numbers by the smaller value.
    7E-4/7E-4 = 1.00
    1.8E-3/7E-4 = 2.6 and I would multiply both by 2 to give C2H5
    empirical mass = (2*12 + 5*1) = 29
    empirical mass x ?whole number = 58.12
    ? number = 58.12/29 = 2+ and round to a whole number.
    (C2H5)2 = C4H10 = molecular formula.

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