Question

An oxybromate compound, KBrOx, where x is unknown, is analyzed and found to contain 47.85 % Br. What is the value of x?

Answer 1

According to the given informations,  Assume 100 g of this compound

mass Br = 47.85 g

moles Br = 47.85 g/ 79.904 g/mol=0.598

Moles K = moles Br Both are same in the formula

Mass= number of moles * molar mass

mass K = 0.598 mol x 39.0963 g/mol=23.41 g

mass O = 100 - ( 23.41 + 47.85)=28.74 g

number of mole s= amount in g / molar mass

moles O = 28.74 g/ 15.999 g/mol=1.78

molar rati:

K= 0.598/0.598=1

Br = 0.598/0.598=1

O=1.78 / 0.598 =2.98 ≈3

x = 3

the formula is KBrO3