The elements phosphorus and bromine combine to form two binary compounds each containing 43.67% and 16.23% phosphorus, respectively. How would you determine the empirical formulas of these two compounds?

1 Answer
Nov 7, 2015

1- #P_2Br#
2- #PBr_2#

Explanation:

Let us first discuss the first compound #P_xBr_y#.
1- We will assume #100g# of this compound, therefore,
#m_P=43.67g# and #m_(Br)=56.33g#

2- Find the number of mole of both elements:
#n_P=(43.67cancel(g))/(30.97cancel(g)/(mol))=1.410 mol#

#n_(Br)=(56.33cancel(g))/(79.90cancel(g)/(mol))=0.7050 mol#

3- Find the molar ratio by dividing by the smallest number of mole:
#P:(1.410cancel(mol))/(0.7050cancel(mol))=2#

#Br:(0.7050cancel(mol))/(0.7050cancel(mol))=1#

Therefore, the empirical formula is: #P_2Br#

Let us discuss the second compound #P_zBr_w#.
1- We will assume #100g# of this compound, therefore,
#m_P=16.23g# and #m_(Br)=83.77g#

2- Find the number of mole of both elements:
#n_P=(16.23cancel(g))/(30.97cancel(g)/(mol))=0.5241 mol#

#n_(Br)=(83.77cancel(g))/(79.90cancel(g)/(mol))=1.048 mol#

3- Find the molar ratio by dividing by the smallest number of mole:
#P:(0.5241cancel(mol))/(0.5241cancel(mol))=1#

#Br:(1.048cancel(mol))/(0.5241cancel(mol))=1.999 ~=2#

Therefore, the empirical formula is: #PBr_2#