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_2BrP2Br
2- PBr_2PBr2

Explanation:

Let us first discuss the first compound P_xBr_yPxBry.
1- We will assume 100g100g of this compound, therefore,
m_P=43.67gmP=43.67g and m_(Br)=56.33gmBr=56.33g

2- Find the number of mole of both elements:
n_P=(43.67cancel(g))/(30.97cancel(g)/(mol))=1.410 molnP=43.67g30.97gmol=1.410mol

n_(Br)=(56.33cancel(g))/(79.90cancel(g)/(mol))=0.7050 molnBr=56.33g79.90gmol=0.7050mol

3- Find the molar ratio by dividing by the smallest number of mole:
P:(1.410cancel(mol))/(0.7050cancel(mol))=2P:1.410mol0.7050mol=2

Br:(0.7050cancel(mol))/(0.7050cancel(mol))=1Br:0.7050mol0.7050mol=1

Therefore, the empirical formula is: P_2BrP2Br

Let us discuss the second compound P_zBr_wPzBrw.
1- We will assume 100g100g of this compound, therefore,
m_P=16.23gmP=16.23g and m_(Br)=83.77gmBr=83.77g

2- Find the number of mole of both elements:
n_P=(16.23cancel(g))/(30.97cancel(g)/(mol))=0.5241 molnP=16.23g30.97gmol=0.5241mol

n_(Br)=(83.77cancel(g))/(79.90cancel(g)/(mol))=1.048 molnBr=83.77g79.90gmol=1.048mol

3- Find the molar ratio by dividing by the smallest number of mole:
P:(0.5241cancel(mol))/(0.5241cancel(mol))=1P:0.5241mol0.5241mol=1

Br:(1.048cancel(mol))/(0.5241cancel(mol))=1.999 ~=2Br:1.048mol0.5241mol=1.9992

Therefore, the empirical formula is: PBr_2PBr2