The empirical formula of the compound is calculated as follows
first calculate the mass of oxygen= 12-(4.09 +3.71)= 5.02g
then calculate the moles of each element, moles = mass/ molar mass
moles of K = 4.09g/39 g/mol(molar mass of K) = 0.105 moles
moles of Cl = 3.71g/35.5 g/mol(molar mass of Cl) = 0.105 moles
moles of O = 5.02g/ 16g/mol(molar mass of O) = 0.314 moles
then calculate e mole ratio by dividing each mole by the smallest number of moles ( 0.105 moles)
K=0.105/0.105= 1
Cl=0.105 /0.105=1
O= 0.314/0.105=3
therefore the empirical formula = KClO3
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