Answer: Calculate the empirical formula as follows:
Element | % Composition | Atomic Mass | Mole Ratio | Simple Ratio |
Carbon (C) | 38.71 | 12 | 38×7112=3.22 | 3.223.22=1 |
Hydrogen (H) | 9.67 | 1 | 9.679.67=1 | 9.67/3.22 = 3 |
Oxygen (O) | 100-38.71+9.67 =51.62 | 16 | 51×6216=3.22 | 3.223.22=1 |
C=1, H=3, O =1 = CH3O.
As a result, the compound’s empirical formula is CH3O.