What is the empirical formula of a compound composed of 43.64% P and 56.36% O by mass?

1 Answer
Mar 22, 2016

Answer:

#P_2O_5# is the empirical formula, the simplest WHOLE number ratio that defines constituent atoms in a species.

Explanation:

We assume that there are #100# #g# unknown, and work out the proportion of phosphorus and oxygen atoms.

In #100*g# of stuff there are #43.64*g# phosphorus and #56.36*g# oxygen.

We divide thru by the ATOMIC masses of each constituent:

#O:# #(56.36*g)/(16.00*g*mol^-1)# #=# #3.52*mol" oxygen"#

#P:# #(43.64*g)/(31.00*g*mol^-1)# #=# #1.41*mol" phosphorus"#.

We divide thru by the SMALLEST molar quantity, we want to get whole numbers:

#(3.52*mol)/(1.41*mol)# #=# #2.5" oxygen"#

#(1.41*mol)/(1.41*mol)# #=# #1" phosphorus"#

A formula of #PO_(2.5)# is no good to us at all. Why, because as we have specified, the empirical formula is the simplest #"WHOLE NUMBER RATIO"# defining constituent atoms in a species. So, we simply double the ration to give #P_2O_5# as required. If the molecular mass of this material is #283.89# #g*mol^-1#, what is the molecular formula of this material???