Question

What is the ratio of Al ions to S ions in a neutral compound?

Answer 1

The ratio of Al ions to S ions in a neutral compound like aluminum sulfide is 2:3, ensuring electrical neutrality of the compound with the formula Al2S3.

Step-by-step explanation:

The ratio of Al ions to S ions in a neutral compound is determined by the requirement that the compound must be electrically neutral. This means the total positive charge from the cations (Al³⁺) must equal the total negative charge from the anions (S²⁻). In the case of aluminum sulfide, the ratio is 2:3 because two aluminum cations (each with a charge of +3) will balance with three sulfide anions (each with a charge of -2), resulting in the formula Al2S3, to ensure overall electric neutrality [(2 × +3) + (3 × -2) = 0].

When calculating the formula mass of the compound, it may be simplified as Al2S3O12, assuming the sulfide is in the form of sulfate ions (SO4²⁻). However, without the presence of oxygen, the straightforward form Al2S3 represents the correct stoichiometry for neutral aluminum sulfide.

Answer 2

Aluminum ions are usually +3 while sulfide ions are usually -2 which means that a neutral compound of aluminum sulfide would be Al₂S₃. 3×2=6 while -2×3=-6 and 6+-6=0.

I hope this helps. Let me know if anything is unclear.