What is the empirical formula of a sample that is 52.1% Carbon, 13.1% Hydrogen and 34.7% Oxygen?

Updated on August 6, 2022

2 Answers

  • C2H6O

  • Let’s do the calculation so that you will know how to do these problems in future:

    Divide % of each element by atomic mass:

    C = 52.1 / 12.011 = 4.3376

    H = 13.1*1.008 = 12.996

    O = 34.7/15.99 = 2.170

    Divide each by the smallest:

    C = 4.3376/2.170 = 1.998

    H = 12.996/2.170 = 5.9889

    O = 2.170/2.170 = 1.00

    Empirical formula = C2H6O

    This would be ethanol = CH3CH2OH