Asubstance contains 36.1 percent calcium and 63.9 percent chlorine by weight. what is the empirical formula of thegiven compound?

Assume 100 g of the compound is present. This converts the percentages into mass. Then, calculate moles of each element: 

Ca ---> 36.1 g / 40.08 g/mol = 0.9007 
Cl ---> 63.9 g / 35.453 g/mol = 1.802 

Next, look for the lowest whole-number ratio. We do this by dividing by the smallest value: 

Ca ---> 0.9007 / 0.9007 = 1 
Cl ---> 1.802 / 0.9007 = 2 

The empirical formula is CaCl2 

A) the enthralling of formation for pure element in its standard state is always positivehope this : )