Question

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.

Answer 1

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