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 $\ce{S8}$?
Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.
Sign up to join this communityI 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 $\ce{S8}$?
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 $\ce{SO2 + 2 H2S}$).
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.