The empirical formula is Al₂O₃.
<em>Step 1</em>. Calculate the <em>mass of oxygen</em>
Your reaction is
Aluminium + oxygen ⟶ aluminium oxide
2.70 g + x g ⟶ 5.10 g
According to the <em>Law of Conservation of Mass</em>, the total mass of the reactants must equal the total mass of the products. Thus,
2.70 g + <em>x</em> g ⟶ 5.10 g
<em>x</em> = 5.10 – 2.70 = 2.40
<em>Step 2</em>. Calculate the <em>moles of each element</em>
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.
<em>Moles of Al</em> = 2.70 g Al × (1 mol Al/(26.98 g Al) = 0.1001 mol Al
<em>Moles of O</em> = 2.40 g O × (1 mol O/16.00 g O) = 0.1500 mol 0
<em>Step 3</em>. Calculate the <em>molar ratio</em> of the elements
Divide each number by the smaller number of moles
Al:O = 0.1001:0.1500 = 1:1.499
<em>Step 4</em>. Multiply each number by a factor that makes the ratio close to whole numbers
<em>Multiply</em> by 2. Then
Al:O = 2:2.998 ≈ 2:3
<em>Step 5</em>: Write the <em>empirical formula</em>
EF = Al₂O₃
