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 * (1 mol Mn)/(54.94 g Mn) = 0.236 mol Mn`

Moles of Cl =
`25.16 g Cl_(2) *(1 mol Cl_(2))/(70.91 g Cl_(2)) * (2 mol Cl)/(1 mol Cl_(2))` = 0.710 mol Cl

Simplest mole ratio:
`Mn_{(0.236)/(0.236)}Cl_{(0.710)/(0.236)}`

So the empirical formula is
`MnCl_(3)`