Question

In an experiment to determine the empirical formula of magnesium oxide a student obtained the following data:

Mass of crucible and lid: 29.1234 g

Mass of crucible and lid + magnesium: 29.9345 g

Mass of crucible and lid + magnesium oxide: 30.7450 g

Obtain the empirical formula of magnesium oxide using the above data

Answer 1

Ans. Step 1: Mass of Mg metal =

(Mass of crucible & lid + Mg metal) - Mass of crucible & lid

                                    = 29.9345 g – 29.1234 g

                                    = 0.8111

# Moles of Mg metal = Mass / Molar mass

                                    = 0.8111 g / (24.305 g/ mol)

                                    = 0.033372 mol

Step 2: Mass of O-atoms reacted with Mg metal=

(Mass of crucible & lid + Mg-oxide) - (Mass of crucible & lid + Mg metal)

                                    = 30.7450 g - 29.9345 g

                                    = 0.8105 g

# Moles of O-atom = 0.8111 g / (15.9994 g/ mol) = 0.050658 mol

# Step 3: Molar ratio of O-atom and Mg-metal = Moles of O / Moles of Mg

                                    = 0.050658 mol / 0.033372 mol

                                    = 1.5 : 1

                                    = 3 : 2            (smallest, whole number ration)

That is there is 3 O-atom for every 2 Mg-atom.

Empirical formula represents a molecule as the smallest whole number ratio of moles of individual elements in the compound.

Therefore, empirical formula of Magnesium oxide = Mg₂O₃

Note: Mg₂O₃ does not exists naturally.