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?

Question

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?

Natalie Anton · Accepted Answer

1- #P_2Br#
2- #PBr_2# 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#

Axel Alvarado · Answer

undefined To determine the empirical formulas of the two compounds, we need to calculate the ratios of the elements present in each compound using the given percentages. Then, we find the simplest whole-number ratio between the elements to determine the empirical formula.

For the first compound:
1. Calculate the ratio of phosphorus to bromine: 
   - 43.67% P / 16.23% Br ≈ 2.69

2. The simplest whole-number ratio is approximately 3:1 (rounded to the nearest whole number).

For the second compound:
1. Calculate the ratio of phosphorus to bromine: 
   - 16.23% P / 43.67% Br ≈ 0.37

2. The simplest whole-number ratio is approximately 1:3 (rounded to the nearest whole number).

Therefore, the empirical formulas of the two compounds are:
1. Compound 1: P3Br
2. Compound 2: PBr3