Question

What is the empirical formula of a barium salt that contains `17.14*g` of the metal, and `8.86*g` of chlorine?

Answer 1

You mean the `"empirical formula"...........` and here it is `BaCl_2`.

Explanation:

You mean the `"empirical formula"...........`

The `"empirical formula"` is the simplest whole number defining constituent atoms in a species. And it is calculated by taking percentage mass of each element, and dividing thru by atomic mass, and then modifying these percentages to give simple whole numbers.

And thus `%Ba` `=` `(17.14*g)/(17.14*g+8.86*g)xx100%=65.9%`

And thus `%Cl` `=` `(8.86*g)/(17.14*g+8.86*g)xx100%=34.1%`

You will note of course that the percentages sum to `100%`. Why `"of course?"`

And then we assume `100*g` of compound and divide thru by the atomic masses of each constituent:

`"Moles of barium"` `=` `(65.9*g)/(137.3*g*mol^-1)=0.480*mol.`

`"Moles of chlorine"` `=` `(34.10*g)/(35.5*g*mol^-1)=0.960*mol.`

If we divide thru by the smallest molar quantity, CLEARLY, we get an empirical formula of `BaCl_2`...........