Question

1. A 0.875g sample of aluminum was reacted totally with chlorine to produce 4.325g of a compound of the two elements. what is the empirical formula of the compound?

2. If the Magnesium nitride were not converted to magnesium oxide, how would the experimentally determined molar ratio MG:O have been altered?

Answer 1

1. Mass of Aluminium reacted,m1 = 0.875g molar mass of Aluminium = 26.98g/mol molar mass of Chlorine = 35.45g/mol Mass of product formed,m = 4.325g

Moles of Al = mass of Al/Molar mass of Al = 0.875g/(26.98g/mol) Moles of Al = 0.0324 mol   

Increase in weight = Mass of product formed - . Mass of Aluminium reacted = 4.325g - 0.875g Increase in weight =  3.45g so, 3.45g of Cl is combining with Al Moles of Cl reacting = mass of Cl / Molar mass of Cl = 3.45g/(35.45g/mol) Moles of Cl reacting = 0.09732 mol So, Al:Cl = 0.0324:0.09732 , dividing each number by the smallest number i.e 0.0324    Al:Cl = 1: 3

The empirical formula of the compound is AlCl₃

2) . Since air is mixture of various gases and not pure oxygen, some of the magnesium may react with the nitrogen in the air. This reaction will create unwanted magnesium nitride along with the magnesium oxide. Magnesium nitride is lighter than magnesium oxide due to the smaller atomic mass of nitrogen compared to oxygen, and if not removed it would alter the data and the ratio of oxygen to magnesium will appear to be high.This is because we are assuming that magnesium oxide is the only product but we have MgO and magnesium nitride both as product.