Question

How would you determine the empirical formula of a compound found to contain 63.50% silver, 8.25% nitrogen, and the remainder oxygen.?

Answer 1

The empirical formula is `"AgNO"_3"`.

Explanation:

Percentages of Ag, N, and O:
`"Ag":` `63.50%`
`"N":` `8.25%`
`"O":` `28.25%`

Since the percentages add up to 100%, we can assume a 100 g sample, and we can rewrite the percentages as mass in grams.

`"Ag":` `"63.50 g"`
`"N":` `"8.25 g"`
`"O":` `"28.25 g"`

We need to determine the number of moles of each element using each element's molar mass, which is the `color(red)"atomic weight (relative atomic mass) on the periodic table in grams/mole (g/mol)"`. The molar mass is the mass of one mole of the element.

`63.50cancel"g Ag"xx(color(red)(1"mol Ag"))/(color(red)(107.8682cancel"g Ag"))="0.5887 mol Ag"`

`8.25cancel("g N")xx(color(red)(1"mol N"))/color(red)(14.007cancel"g N")="0.589 mol N"`

`28.25cancel"g O"xx(color(red)(1"mol O"))/(color(red)(15.999cancel"g O"))="1.766 mol O"`

Determine the mole ratios by dividing the number of moles of each element by the least number of moles.

`"Ag":` `(0.5887"mol Ag")/(0.589"mol")=0.999~~1 "mol Ag"`

`"N":` `(0.589"mol N")/(0.589"mol")="1.00 mol N"`

`"O":` `(1.766"mol O")/(0.589"mol")="3.00 mol O"`

The empirical formula is `"AgNO"_3"`. This compound is silver nitrate.