Emperical formula for 0.0200 mole of al and 0.0600 mole of cl
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:
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 = \(AlCl_3\)