What is the empirical formula of a compound that contains 3.609 g calcium and 6.384 g chlorine by mass?

For a compound that contains 3.609g Ca; we shall calculate the number of moles as follows 
No of moles = mass/molar mass = 3.609/40 = 0.090 
Also for chlorine we have 6.384/35.4 = 0.19 
So we divide by the smallest ratio (0.090)  
I. e 0.090/0.090 and 0.19/0.090 this gives a ratio of 1 : 2. Hence the formula is Cacl2