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Aluminum Sulfide Formula (Aluminum Sulfide Chemical Formula)

What is the aluminum sulfide formula, and how is it derived from the charges on aluminum and sulfide ions?

Subject: General Chemistry Chapter: Chemical Compounds Topic: Naming Binary Compounds of Metals and Nonmetals Answer included
aluminum sulfide formula Al2S3 ionic formula binary ionic compound charge balance aluminum ion sulfide ion oxidation states
Accepted answer Answer included

Aluminum sulfide formula

The compound described by the keyword aluminum sulfide formula is a binary ionic compound made from aluminum (a metal) and sulfide (a nonmetal anion). The correct formula is found by making the total ionic charge equal to zero using the smallest whole-number ratio of ions.

Target condition (neutrality): the total positive charge must equal the total negative charge, so the sum of charges is \(0\).

Final result: \( \mathrm{Al_2S_3} \).

Step 1: Identify the ions and their charges

Aluminum commonly forms a \(+3\) cation because it is in Group 13 and tends to lose three electrons: \( \mathrm{Al^{3+}} \). Sulfur forms the sulfide anion with a \(−2\) charge (Group 16 gains two electrons): \( \mathrm{S^{2-}} \).

Ion Charge Meaning
Aluminum ion \( \mathrm{Al^{3+}} \) One aluminum ion contributes \(+3\) charge.
Sulfide ion \( \mathrm{S^{2-}} \) One sulfide ion contributes \(−2\) charge.

Step 2: Balance charges using the least common multiple

The charge magnitudes are \(3\) and \(2\). The least common multiple is \(6\). Make \(+6\) total from aluminum ions and \(−6\) total from sulfide ions:

  • \(2 \times (+3) = +6\) → use 2 aluminum ions.
  • \(3 \times (−2) = −6\) → use 3 sulfide ions.
Ion Charge per ion Number of ions Total charge
\( \mathrm{Al^{3+}} \) \(+3\) \(2\) \(2 \times (+3) = +6\)
\( \mathrm{S^{2-}} \) \(−2\) \(3\) \(3 \times (−2) = −6\)
Sum \(+6 + (−6) = 0\)

Step 3: Write the formula unit

The cation is written first, then the anion. The subscripts come from the balanced ion counts: \[ \mathrm{Al_2S_3} \]

A compact way to show the combination of ions into a neutral formula unit is: \[ 2\,\mathrm{Al^{3+}} + 3\,\mathrm{S^{2-}} \rightarrow \mathrm{Al_2S_3} \]

Criss-Cross Method for Aluminum Sulfide Formula A premium diagram showing how the +3 charge of aluminum and -2 charge of sulfur cross over to become subscripts in the Al2S3 formula. Al 3+ Cation S 2− Anion Al2S3 CRISS-CROSS METHOD
The charges \(+3\) and \(−2\) are balanced by the smallest whole numbers that produce equal total magnitudes (\(6\)), giving 2 aluminum ions and 3 sulfide ions, hence \( \mathrm{Al_2S_3} \).

Common mistakes and quick checks

  • Mistake: writing \( \mathrm{Al_3S_2} \) by placing charges directly as subscripts without balancing.
    Fix: enforce neutrality: \(2 \times 3 = 3 \times 2 = 6\) so the correct subscripts are 2 and 3.
  • Mistake: reducing incorrectly or leaving charges in the final formula.
    Fix: the final ionic formula contains only element symbols and subscripts; the net charge must be \(0\).
  • Quick check: compute total charge: \(2(+3) + 3(−2) = 0\). If the sum is not zero, the formula is not correct.

Final answer

The correct aluminum sulfide formula is: \[ \mathrm{Al_2S_3} \]

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