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I would like to ask a question about Sulfur in the reaction with Nitric Acid, $\ce{HNO3}$ to produce Sulfuric Acid, $\ce{H2SO4}$

I read that as Nitrogen and Sulfur are reduced and oxidised respectively, the reaction equation can be constructed as follows:

$\ce{S + 6HNO3 -> H2SO4 + 6NO2 + 2H2O}$

Now, when I read an article on wikipedia regarding sulfur, the introductory paragraph states the following:

Under normal conditions, sulfur atoms form cyclic octatomic molecules with a chemical formula $\ce{S8}$.

So why is sulfur represented in the equation as $\ce{S}$ and not $\ce{S8}$?

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    $\begingroup$ Because that doesn't make much of a difference. Sure, you may multiply everything by 8, if you'd like. $\endgroup$ – Ivan Neretin Feb 21 '18 at 21:18
  • $\begingroup$ Just call $\ce{S}$ an empirical formula. Empirical formulas work perfectly well in chemical reaction equations. $\endgroup$ – Oscar Lanzi Feb 22 '18 at 0:34
  • $\begingroup$ See this: chemistry.stackexchange.com/questions/8438/… $\endgroup$ – Nilay Ghosh Feb 22 '18 at 3:30
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The most stable allotropic form of sulfur exists in the formula of S8. In other words, sulfur occurs naturally in S8 formula. Multiply all the stoicheomtric coefficient by 8 (other than S) to obtain the balanced chemical equation

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Balance the equation by taking S8. Yes, it is true that naturally, sulfur does exist in S8 (crown sulfur) form.

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