Answer:
The empirical formula of the compound is [tex]C_{0.504}HO_{1.008}[/tex].
Explanation:
We need to determine the empirical formula in its simplest form, where hydrogen ([tex]H[/tex]) is scaled up to a mole, since it has the molar mass, and both carbon ([tex]C[/tex]) and oxygen ([tex]O[/tex]) are also scaled up in the same magnitude. The empirical formula is of the form:
[tex]C_{x}HO_{y}[/tex]
Where [tex]x[/tex], [tex]y[/tex] are the number of moles of the carbon and oxygen, respectively.
The scale factor ([tex]r[/tex]), no unit, is calculated by the following formula:
[tex]r = \frac{M_{H}}{m_{H}}[/tex] (1)
Where:
[tex]m_{H}[/tex] - Mass of hydrogen, in grams.
[tex]M_{H}[/tex] - Molar mass of hydrogen, in grams per mole.
If we know that [tex]M_{H} = 1.008\,\frac{g}{mol}[/tex] and [tex]m_{H} = 0.2\,g[/tex], then the scale factor is:
[tex]r = \frac{1.008}{0.2}[/tex]
[tex]r = 5.04[/tex]
The molar masses of carbon ([tex]M_{C}[/tex]) and oxygen ([tex]M_{O}[/tex]) are [tex]12.011\,\frac{g}{mol}[/tex] and [tex]15.999\,\frac{g}{mol}[/tex], then, the respective numbers of moles are: ([tex]r = 5.04[/tex], [tex]m_{C} = 1.2\,g[/tex], [tex]m_{O} = 3.2\,g[/tex])
Carbon
[tex]n_{C} = \frac{r\cdot m_{C}}{M_{C}}[/tex] (2)
[tex]n_{C} = \frac{(5.04)\cdot (1.2\,g)}{12.011\,\frac{g}{mol} }[/tex]
[tex]n_{C} = 0.504\,moles[/tex]
Oxygen
[tex]n_{O} = \frac{r\cdot m_{O}}{M_{O}}[/tex] (3)
[tex]n_{O} = \frac{(5.04)\cdot (3.2\,g)}{15.999\,\frac{g}{mol} }[/tex]
[tex]n_{O} = 1.008\,moles[/tex]
Hence, the empirical formula of the compound is [tex]C_{0.504}HO_{1.008}[/tex].
