Question

Emperical formula for 0.0200 mole of al and 0.0600 mole of cl

Answer 1

The Al to Cl ratio is $\frac{0.0200}{0.060} = \frac{1}{3}$. Meaning that there is 1 Al to every 3 Cl, and so the empirical formula is $AlCl_{3}$.

Answer 2

Answer:

Empirical formula for the given elements = $AlCl_3$

Explanation:

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

Moles of aluminum = 0.0200 mol

Moles of chlorine = 0.0600 mol

For empirical formula , divide the smallest number of moles from each moles of element.

Aluminium = $\frac{0.0200 mol}{0.0200 mol} =1$

Calcium = $\frac{0.600 mol}{0.0200 mol} =3$

Empirical formula for the given elements = $AlCl_3$