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?

Answer 1

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`