# What is the empirical formula of magnesium chloride if 0.96 g of magnesium combines with 2.84 g of chlorine?

Oct 25, 2015

$M g C {l}_{2}$

#### Explanation:

Given that the weight of Mg is 0.96g and Cl is 2.84g and completely disregarding the possible chemical reactions, we need to convert each weight into the corresponding number of moles by multiplying the weight with its respective atomic weight.

$\text{Mg: " 0.96 cancel "g" xx "1 mol"/(24.305 cancel"g") = "0.395 mol}$

$\text{Cl: " 2.84 cancel "g" xx "1 mol"/(35.453 cancel "g") = "0.801 mol}$

Dividing the larger amount of mole with the smaller one (0.801 mol > 0.395 mol),

$\text{Mg: " "0.395"/"0.395} = 1$

$\text{Cl: " "0.801"/"0.395} = 2$

Therefore, empirical formula is $M g C {l}_{2}$