Respuesta :
Answer:
The empirical formula of the compound is = [tex]KBrO_3[/tex]
The name of the compound is potassium bromate.
Explanation:
Mass of potassium = 4.628 g
Moles of potassium = [tex]\frac{4.628 g}{39 g/mol}=0.1187 mol[/tex]
Mass of bromine = 9.457 g
Moles of bromine = [tex]\frac{9.457 g}{80 g/mol}=0.1182 mol[/tex]
Mass of oxygen = 5.681 g
Moles of oxygen = [tex]\frac{5.681 g}{16 g/mol}0.3551 [/tex]
For empirical formula of the compound, divide the least number of moles from all the moles of elements present in the compound:
Potassium :
[tex]\frac{0.1187 mol}{0.1182 mol}=1.0[/tex]
Bromine;
[tex]\frac{0.1182 mol}{0.1182 mol}=1.0[/tex]
Oxygen ;
[tex]\frac{0.3551 mol}{0.1182 mol}=3.0[/tex]
The empirical formula of the compound is = [tex]KBrO_3[/tex]
The name of the compound is potassium bromate.
