Question

What is the empirical formula of a species that is `7.19%` by mass with respect to phosphorus, and `92.81%` by mass with respect to bromine? What is the molecular formula is the molecular mass is `431*g*mol^-1`?

Answer 1

I think you have quoted a question from this [site.](http://bergen.edu/Portals/0/Docs/Student%20Services/Tutor%20Resources/Chemistry/CHM%20100%20%20140%20-%20Empirical%20and%20Molecular%20Formulas%20%282%29.pdf)

Explanation:

And this gives (i) microanalysis of `P(7.19%)`, and `Br(92.81%)`. You are lucky that I bothered to search for the question, as I am no computer buff, and these data should have been included with the question.......

For these data we determine the empirical formula......by assuming an `100*g` mass of compound.

`"Moles of phosphorus"` `=` `(7.19*g)/(31*g*mol^-1)=0.232*mol`.

`"Moles of bromine"` `=` `(92.81*g)/(79.9*g*mol^-1)=1.161*mol`.

In each case I divided the elemental mass thru by the molar mass of the ELEMENT.

We divide thru by the lowest molar quantity (that of `P`), to get an empirical formula of `P:(0.232)/(0.232)=1`and `Br:(1.161)/(0.232)=5`.. to give an empirical formula, the simplest whole number ratio of constituent elements in a species, of `PBr_5`.

Now, we know that.........

`{"empirical formula"}xxn="molecular formula"`, and from (ii) the molecular mass, we determine that.....

`{31.00+5xx79.9}*g*mol^-1xxn=431*g*mol^-1`

Clearly, `n=1`, and so.........

`"empirical formula" = "molecular formula" = PBr_5`