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

The empirical formula of the compound is
`C_(0.504)HO_(1.008)`.

Step-by-step 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 = (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\,(g)/(mol)` and
`m_(H) = 0.2\,g`, then the scale factor is:

`r = (1.008)/(0.2)`

`r = 5.04`

The molar masses of carbon (
`M_(C)`) and oxygen (
`M_(O)`) are
`12.011\,(g)/(mol)` and
`15.999\,(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) = (r\cdot m_(C))/(M_(C))` (2)

`n_(C) = ((5.04)\cdot (1.2\,g))/(12.011\,(g)/(mol) )`

`n_(C) = 0.504\,moles`

Oxygen

`n_(O) = (r\cdot m_(O))/(M_(O))` (3)

`n_(O) = ((5.04)\cdot (3.2\,g))/(15.999\,(g)/(mol) )`

`n_(O) = 1.008\,moles`

Hence, the empirical formula of the compound is
`C_(0.504)HO_(1.008)`.