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

Answer 1

C2H6O

answer 2

Let’s do the math so you know how to solve these problems in the future:
Divide the % of each element by the 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
That would be ethanol = CH3CH2OH

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