Question

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.

Answer 1

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

Answer 2

The empirical formula of the oxide is Al₂O₃ when 2.70g of aluminium was heated in oxygen until there was no further gain oxide ash formed and had a mass of 5.10g.

What is an empirical formula?

A chemical formula showing the simplest ratio of elements in a compound rather than the total number of atoms in the molecule $CH_2O$ is the empirical formula for glucose.

First, calculate the mass of oxygen

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

Now, calculate the moles of each elements.

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 = $\frac{mass}{molar \;mass}$

Moles of Al =  $\frac{2.70 g Al}{26.98 g Al}$ = 0.1001 mol aluminium.

Moles of O = $\frac{2.40 g O}{16.00 g O}$ = 0.1500 mol oxygen.

Now, calculate the molar ratio of the elements.

Divide each number by the smaller number of moles

Aluminium : Oxygen = 0.1001:0.1500 = 1:1.499

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

Multiply by 2.

Al:O = 2:2.998 ≈ 2:3

The empirical formula will be:

Al₂O₃

Thus, the empirical formula of the oxide is Al₂O₃ when 2.70g of aluminium was heated in oxygen until there was no further gain oxide ash formed and had a mass of 5.10g.

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