Answer:
The empirical formula of the compound is
.
Explanation:
We need to determine the empirical formula in its simplest form, where hydrogen (
) is scaled up to a mole, since it has the molar mass, and both carbon (
) and oxygen (
) are also scaled up in the same magnitude. The empirical formula is of the form:

Where
,
are the number of moles of the carbon and oxygen, respectively.
The scale factor (
), no unit, is calculated by the following formula:
(1)
Where:
- Mass of hydrogen, in grams.
- Molar mass of hydrogen, in grams per mole.
If we know that
and
, then the scale factor is:


The molar masses of carbon (
) and oxygen (
) are
and
, then, the respective numbers of moles are: (
,
,
)
Carbon
(2)


Oxygen
(3)


Hence, the empirical formula of the compound is
.