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

1 Answer
anor277 · Monzur R.
Nov 23, 2016

#"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.

Related questions
See all questions in Empirical and Molecular Formulas
Impact of this question
11735 views around the world
You can reuse this answer
Creative Commons License