Question

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

Answer 1

`"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.