Question: what is the empirical formula for a compound that is 36.1 ca and 63.9 cl?
Answer: Moles of calcium =36.1⋅g40.08⋅g⋅mol−1=0.901⋅mol . Moles of chlorine =63.9⋅g35.45⋅g⋅mol−1=1.802⋅mol . We divide each molar quantity through by the smallest molar quantity (that of calcium) to give an empirical formula of CaCl2 .