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

1 Answer
Sep 11, 2017

Answer:

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.