What is the empirical formula of a compound that contains 15.77% aluminum, 28.11% sulfur and 56.12% oxygen

Al₂(SO₄)₃

Explanation:

We are given percentage composition of elements in a compound;

Aluminium is 15.77%Sulfur is 28.11 %Oxygen is 56.12%

We are required to calculate the empirical formula of the compound.

Assuming the mass of the compound is 100 g then the masses of the elements is;

Aluminium = 15.77 g

Sulfur = 28.11

Oxygen = 56.12

We can determine the number of moles of each;

Moles of Aluminium = 15.77 g ÷ 26.98 g/mol

                                 = 0.585 moles

Moles of sulfur = 28.11 g ÷ 32.07 g/mol

                         = 0.877 moles

Moles of Oxygen = 56.12 g ÷ 16.0 g/mol

                            = 3.5075 moles

But, the empirical formula is the simplest whole number ratio of elements in a compound.Therefore; we need to get the ratio of moles of the above elements;

Aluminium : Sulfur : Oxygen

0.585 mol  :  0.877 mol : 3.5075 mol

0.585/0.585 : 0.877/0.585 : 3.5075/0.585

    1 : 1.5 : 6

But, we need whole number ratios, therefore;

= (1 : 1.5 : 6 ) × 2

= 2 : 3 : 12

Therefore; the formula of the compound is Al₂S₃O₁₂

The compound is written as Al₂(SO₄)₃

Thus, the empirical formula of the compound is Al₂(SO₄)₃

It’s c it stays the same. and the atomic number goes up by 1. a neutron has changed into a proton plus an electron. the electron is emitted as a beta particle.

Ci think not