Respuesta :
Answer: The empirical formula for the given compound is [tex]BrF_3[/tex]
Explanation:
We are given:
Percentage of Br = 58.37 %
Percentage of F = 41.63 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of Br = 58.37 g
Mass of F = 41.63 g
To formulate the empirical formula, we need to follow some steps:
- Step 1: Converting the given masses into moles.
Moles of Bromine =[tex]\frac{\text{Given mass of Bromine}}{\text{Molar mass of Bromine}}=\frac{58.37g}{79.90g/mole}=0.73moles[/tex]
Moles of Fluorine = [tex]\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{41.63g}{19g/mole}=2.19moles[/tex]
- Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.73 moles.
For Bromine = [tex]\frac{0.73}{0.73}=1[/tex]
For Fluorine = [tex]\frac{2.19}{0.73}=3[/tex]
- Step 3: Taking the mole ratio as their subscripts.
The ratio of Br : F = 1 : 3
Hence, the empirical formula for the given compound is [tex]BrF_3[/tex]
