What is the empirical formula for a compound that is 36.1% Ca and 63.9% CI?

1 Answer
Nov 23, 2016

Answer:

#"CaCl"_2#

Explanation:

As with all these problems, we assume a #100*"g"# of compound, and divide through by the ATOMIC masses of each component element:

#"Moles of calcium"# #=(36.1*"g")/(40.08*"g"*"mol"^-1)=0.901 ·"mol"#.

#"Moles of chlorine"# #=(63.9*"g")/(35.45*"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 #"CaCl"_2#.

For ionic materials, we do not speak of the molecular formula, and the empirical formula is the formula we quote for reference.