What is the empirical formula for a compound that is found to contain 10.15 mg #P# and 34.85 mg #Cl#?

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Answer 1 · 2016-05-29T19:39:53

#PCl_3#

#%P# #=# #(10.15*mg)/((34.85+10.15)*mg)xx100%# #=# #22.6%#

#%Cl# #=# #(34.85*mg)/((34.85+10.15)*mg)xx100%# #=# #77.4%#

We use the percentages and assume #100# #g# of compound.

#P:# #(22.6*g)/(30.9737*g*mol^-1)# #=# #0.730*mol# #P#

#Cl:# #(77.4*g)/(35.45*g*mol^-1)# #=# #2.18*mol# #Cl#

We divide thru by the SMALLER molar quantity to give an empirical formula of:

#PCl_3#.