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#?

1 Answer
Jul 23, 2017

Answer:

I think you have quoted a question from this site.

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#