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

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.