# 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?

Nov 7, 2015

1- ${P}_{2} B r$
2- $P B {r}_{2}$

#### Explanation:

Let us first discuss the first compound ${P}_{x} B {r}_{y}$.
1- We will assume $100 g$ of this compound, therefore,
${m}_{P} = 43.67 g$ and ${m}_{B r} = 56.33 g$

2- Find the number of mole of both elements:
${n}_{P} = \frac{43.67 \cancel{g}}{30.97 \frac{\cancel{g}}{m o l}} = 1.410 m o l$

${n}_{B r} = \frac{56.33 \cancel{g}}{79.90 \frac{\cancel{g}}{m o l}} = 0.7050 m o l$

3- Find the molar ratio by dividing by the smallest number of mole:
$P : \frac{1.410 \cancel{m o l}}{0.7050 \cancel{m o l}} = 2$

$B r : \frac{0.7050 \cancel{m o l}}{0.7050 \cancel{m o l}} = 1$

Therefore, the empirical formula is: ${P}_{2} B r$

Let us discuss the second compound ${P}_{z} B {r}_{w}$.
1- We will assume $100 g$ of this compound, therefore,
${m}_{P} = 16.23 g$ and ${m}_{B r} = 83.77 g$

2- Find the number of mole of both elements:
${n}_{P} = \frac{16.23 \cancel{g}}{30.97 \frac{\cancel{g}}{m o l}} = 0.5241 m o l$

${n}_{B r} = \frac{83.77 \cancel{g}}{79.90 \frac{\cancel{g}}{m o l}} = 1.048 m o l$

3- Find the molar ratio by dividing by the smallest number of mole:
$P : \frac{0.5241 \cancel{m o l}}{0.5241 \cancel{m o l}} = 1$

$B r : \frac{1.048 \cancel{m o l}}{0.5241 \cancel{m o l}} = 1.999 \cong 2$

Therefore, the empirical formula is: $P B {r}_{2}$