1
$\begingroup$

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 $\ce{S8}$?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ 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, 2018 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, 2018 at 0:34
  • $\begingroup$ See this: chemistry.stackexchange.com/questions/8438/… $\endgroup$
    – Nilay Ghosh
    Feb 22, 2018 at 3:30

2 Answers 2

Reset to default
1
$\begingroup$

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

Improve this answer
$\endgroup$
0
$\begingroup$

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

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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