-2
$\begingroup$

I learnt that sulfur dioxide and hydrogen sulphide react to form sulfur and water. The reaction given was: $\ce{SO2 + 2 H2S -> 2 H2O + 3S}$

So why is the product sulfur written as $\ce{S}$ and not as $\ce{S8}$?

$\endgroup$
2
  • 1
    $\begingroup$ Welcome to the site. Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers. $\endgroup$
    – Buttonwood
    Commented Sep 30, 2021 at 11:57
  • 1
    $\begingroup$ It also forms polythionic acid . Used to conduct stress corrosion tests on sensitized stainless steels . Sometimes called " Wachenroter ?" or "Samans" solution. $\endgroup$ Commented Sep 30, 2021 at 14:44

2 Answers 2

4
$\begingroup$

Because you don't really need to identify the sulfur as eight-atom molecules. The empirical formula $\ce{S}$ is sufficient for rendering chemical reactions and avoids encumbering the reaction with rather large coefficients on the other species ($\ce{8 SO2 + 16 H2S}$ vs $\ce{SO2 + 2 H2S}$).

$\endgroup$
1
  • $\begingroup$ OP: An analogue is H3O+, as it can exist in different hydrated form. $\endgroup$
    – Alchimista
    Commented Oct 1, 2021 at 10:02
2
$\begingroup$

The focus of the reaction equation

$$\ce{SO2 + 2 H2S -> 2 H2O + 3S}$$

is to balance the number of atoms on the left hand side (starting materials) with those on the right hand side (the products). It eventually depends on the reaction conditions (e.g., pressure and temperature) if sulfur will form the famous $\ce{S8}$ cycles, or rings larger/smaller than that (cf. e.g., with wikipedia's entry about the allotropes of sulfur). For balancing the reaction (i.e., stoichiometry), this subsequent formation of these rings is not relevant.

Small note: chemists prefer the writing sulfur.

$\endgroup$
1
  • 2
    $\begingroup$ Pure modifications of sulfur are only rarely accessible. It's basically always a wild mixture. $\endgroup$
    – Karl
    Commented Sep 30, 2021 at 12:17

Not the answer you're looking for? Browse other questions tagged or ask your own question.