Question

If 25.16 g of chlorine react with 12.99 g of manganese metal, what is the empirical formula of the compound?

Answer 1

We use the given masses of the reactants to calculate the moles of Mn and Cl. Empirical formula represents the simplest mole ratio of atoms present in a compound.

Moles of Mn = $12.99 g Mn * \frac{1 mol Mn}{54.94 g Mn} = 0.236 mol Mn$

Moles of Cl = $25.16 g Cl_{2} *\frac{1 mol Cl_{2}}{70.91 g Cl_{2}} * \frac{2 mol Cl}{1 mol Cl_{2}}$ = 0.710 mol Cl

Simplest mole ratio: $Mn_{\frac{0.236}{0.236}}Cl_{\frac{0.710}{0.236}}$

So the empirical formula is $MnCl_{3}$