Question

Determine the empirical formula of the compound formed when 1.2g of magnesium reacts with 3.55g of chlorine.Take the molar mass of magnesium and chlorine to be 24g and 35.5g respectively.

Answer 1

Explanation:

For Mg, (1.2 g Mg/24 g Mg) = 0.05 mol Mg.

For Cl, (3.55 g Cl/35.5 g Cl) = 0.1 mol Cl

So the ratio now is

Mg:Cl = 0.05 : 0.1 = 1:2

I got the 1:2 ratio by dividing both by the smallest number, which is 0.05 mol. Therefore, the empirical for formula of the substance is $MgCl_2$