Question

A sample of a compound was found to contain `29.16%` nitrogen, `8.39%` hydrogen, `12.50%` carbon, and `49.95%` oxygen. What is its empirical formula?

Answer 1

The empirical formula of the compound is `"N"_2"H"_8"CO"_3` or `("NH"_4)_2"CO"_3"`.

Refer to the explanation for the process.

Explanation:

An empirical formula represents the simplest whole-number ratio of elements in a compound. The basic procedure is to first determine the moles of each element, and then divide the moles of each element by the least number of moles. This will give you the whole-number ratios for the empirical formula.

If you do not get whole-number ratios, then you must multiply them by a factor that will give whole-number ratios.

Since the percentages of the elements equal `100`, they can be directly converted to mass in grams. We can determine the moles of each element by dividing the given mass by its molar mass (atomic weight on the periodic table in grams/mole).

`"mol N":` `(29.16color(red)cancel(color(black)("g N")))/(14.007color(red)cancel(color(black)("g"))/"mol")="2.082 mol N"`

`"mol H":` `(8.39color(red)cancel(color(black)("g H")))/(1.008color(red)cancel(color(black)("g"))/"mol")="8.32 mol H"`

`"mol C":` `(12.50color(red)cancel(color(black)("g C")))/(12.011color(red)cancel(color(black)("g"))/"mol")=color(red)("1.041 mol C"`

`"mol O":` `(49.95color(red)cancel(color(black)("g O")))/(15.999color(red)cancel(color(black)("g"))/"mol")="3.122 mol O"`

Determine the mole ratios by dividing the moles of each element by the least number of moles, `color(red)(1.041`.

`"N":` `(2.082)/(1.041)="2"`

`"H":` `(8.32)/(1.041)="7.99` `~~8"`

`"C":` `(1.041)/(1.041)="1"`

`"O":` `(3.122)/(1.041)="2.999` `~~3`

Empirical Formula: `"N"_2"H"_8"CO"_3`, or `("NH"_4)_2"CO"_3"`