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

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