An oxybromate componud, KBrOx, Where X is unknown , is analysed and found to contain 59.19% Br.?

moles Br = 59.19 g/79.904 g/mol=0.7408
moles K = 0.7408
mass K = 0.7408 mol x 39.1 g/mol= 28.96

59.19 + 28.96= 88.15
moles O = 100 - 88.15 / 16= 0.740

the ratio is 1 : 1 : 1
so X is 1

and KBrO is the compound