Question

Q1. In a compound of magnesium and nitrogen, 18g of magnesium had combined with 7g of nitrogen.

a. Find the number of gram atoms of magnesium and nitrogen in the compound.

b. What is the simple whole number ratio of these atoms?

c. what is the empirical formula of the compound?

Q2. 1.723g of a sample of aluminium oxide contains 0.912g of aluminium. Determine the empirical formula of the compound.

Answer 1

a) Number of moles of magnesium atom = 18/24 = 0.75

Number of moles of nitrogen = 7/14 = 0.5

Therefore, the number of grams atoms of magnesium in the compound = 0.75

The number of grams atoms of nitrogen in the compound = 0.5

b) Ratio of moles of magnesium and nitrogen atoms = 0.75 / 0.5

= 1.5 :1

= 3:2

Therefore, simple whole number ration of atoms of magnesium and nitrogen is 3:2

c) Since the ratio of magnesium and nitrogen atoms is 3:2, the empirical formula of the compound is Mg₃N₂.

2. Mass of aluminium oxide = 1.723 g

Mass of aluminium in the oxide = 0.912 g

Mass of oxygen in the compound = 1.732 - 0.912 = 0.82 g

Moles of aluminium = 0.912 / 27 = 0.034

Moles of oxygen = 0.82 /16 = 0.05

Ratio of moles of aluminium and oxygen = 0.033 : 0.5

= 1:1.5

= 2:3

Thus empirical formula of the compound = Al₂O₃