Question

A `0.967*g` mass of an oxide of phosphorus contains `0.422*g` of phosphorus. What is its empirical formula?

Answer 1

We gots `P_2O_5`.....

Explanation:

We calculate the empirical formula....

`"Moles of phosphorus"=(0.422*g)/(30.9737*g*mol^-1)=0.01361*mol`

`"Moles of oxygen"=(0.967*g-0.422*g)/(15.999*g*mol^-1)=0.0341*mol`

We divide thru by the SMALLEST MOLAR quantity to give an empirical formula of `P_((0.01361)/(0.01361))O_((0.0341)/(0.01361))`

`-=PO_(2.50)`....

Now the empirical formula is by definition the simplest WHOLE number ratio, so we double the trial formula to give.....

`P_2O_5`

The actual molecule is `P_4O_10` but we would need further data to assess this.