A 12.82 g sample of a compound contains 4.09 g potassium (K), 3.71 g chlorine (Cl), and oxygen (O). Calculate the empirical formula.

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