Question

A compound contains 1.2 g of carbon, 3.2 g of oxygen and 0.2g of hydrogen. Find the formula of the compound

Answer 1

Answer:

The empirical formula of the compound is $C_{0.504}HO_{1.008}$.

Explanation:

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

$C_{x}HO_{y}$

Where $x$, $y$ are the number of moles of the carbon and oxygen, respectively.

The scale factor ($r$), no unit, is calculated by the following formula:

$r = \frac{M_{H}}{m_{H}}$ (1)

Where:

$m_{H}$ - Mass of hydrogen, in grams.

$M_{H}$ - Molar mass of hydrogen, in grams per mole.

If we know that  $M_{H} = 1.008\,\frac{g}{mol}$ and $m_{H} = 0.2\,g$, then the scale factor is:

$r = \frac{1.008}{0.2}$

$r = 5.04$

The molar masses of carbon ($M_{C}$) and oxygen ($M_{O}$) are $12.011\,\frac{g}{mol}$ and $15.999\,\frac{g}{mol}$, then, the respective numbers of moles are: ($r = 5.04$, $m_{C} = 1.2\,g$, $m_{O} = 3.2\,g$)

Carbon

$n_{C} = \frac{r\cdot m_{C}}{M_{C}}$ (2)

$n_{C} = \frac{(5.04)\cdot (1.2\,g)}{12.011\,\frac{g}{mol} }$

$n_{C} = 0.504\,moles$

Oxygen

$n_{O} = \frac{r\cdot m_{O}}{M_{O}}$ (3)

$n_{O} = \frac{(5.04)\cdot (3.2\,g)}{15.999\,\frac{g}{mol} }$

$n_{O} = 1.008\,moles$

Hence, the empirical formula of the compound is $C_{0.504}HO_{1.008}$.