2.70g of aluminium was heated in oxygen until there was no further gain oxide ash formed had a mass of 5.10g. deduce the empirical formula of the oxide o show all your working.

The empirical formula is Al₂O₃.

Step 1. Calculate the mass of oxygen

Your reaction is

Aluminium + oxygen ⟶ aluminium oxide

  2.70 g     +   x g      ⟶       5.10 g

According to the Law of Conservation of Mass, the total mass of the reactants must equal the total mass of the products. Thus,

2.70 g + x g ⟶ 5.10 g

x = 5.10 – 2.70 = 2.40

Step 2. Calculate the moles of each element

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

Moles of Al = 2.70 g Al × (1 mol Al/(26.98 g Al) = 0.1001 mol Al

Moles of O = 2.40 g O × (1 mol O/16.00 g O) = 0.1500 mol 0

Step 3. Calculate the molar ratio of the elements

Divide each number by the smaller number of moles

Al:O = 0.1001:0.1500 = 1:1.499

Step 4. Multiply each number by a factor that makes the ratio close to whole numbers

Multiply by 2. Then

Al:O = 2:2.998 ≈ 2:3

Step 5: Write the empirical formula

EF = Al₂O₃