What is the empirical formula for a compound that is 60.9% arsenic and 39.1% sulfur?

Question

6363 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-02T08:07:54

$As_2S_3$

How did we get this formula?

We assume $100.0$ $g$ of compound, and there are $60.9$ $g$ of arsenic, and $39.1$ $g$ of sulfur.

We divide each mass thru by the molar mass of each element:

$As$ $=$ $(60.9*g)/(74.92*g*mol^-1)$ $=$ $0.813$ $mol$ .

$S$ $=$ $(39.1*g)/(32.06*g*mol^-1)$ $=$ $1.22$ $mol$ .

Now, we divide thru by the LOWEST molar quantity, $0.813$ $mol$ :

We get $AsS_(1.500)$ . But we want the whole number ratio, so we get $As_2S_3$ as required.