question img

I am new to chemistry as well as StackExchange network, so any advice is appreciated.

I tried to answer it myself, but I am sure it is not completely correct and there are certainly alternative solutions. There are also points I would like to know which I listed at the end of my solution.

Here's what I came up with: a) We assume $S+O_2 \rightarrow SO_2$. The equation is balanced and $1 \space mol \space S≏ 1 \space mol \space O_2 ≏ 1 \space mol \space SO_2$ thus, $0.1 \space mol$ $O_2$ is needed for $0.1 \space mol$ $S$ to react.

b) We assume $S+O_2 \rightarrow SO_3$. Balancing the equation gives $2S+3O_2 \rightarrow 2SO_3$ and $2 \space mol \space S≏ 3 \space mol \space O_2 ≏ 2 \space mol \space SO_2$ thus, $0.15 \space mols$ $O_2$ is needed for $0.1 \space mol$ $S$ to react.

c) Equipment given is an analytical balance, which have an accuracy of $0.0001$ to $0.00001$ grams. Since we know the molar mass of Oxygen (gas) is $31.999$ g and that of Sulfur is $32.065$ grams, we can weigh them on the balance to use correct number of moles. (For practical purposes, we can approximate both of these quantities as $32$ grams)

d) At room temperature, Sulfur Dioxide is a colorless gas meanwhile Sulfur Trioxide is a crystalline solid. Their physical properties can be used to identify them. I am unsure about the instruments.

e) I would use $1 \space mol \space S$ and $1 \space mol \space O_2$. If the assumption a) is true, then the reaction will result in one mole of Sulfur Dioxide. If no solid is left on the reaction vessel then assumption a) is supported. If the assumption b) is true, then $O_2$ will be the limiting reagent and products of the reaction will be $\frac{2}{3} \space mol$ Sulfur Trioxide and $\frac{1}{3} \space mol$ leftover Sulfur. Sulfur is also a solid in room temperature and if these two solids weigh 64 grams (that is, the weight of reactants) then assumption b) is supported. The last case is when the product is a mixture of $SO_2$ and $SO_3$, let's call it assumption c). Balanced equation then becomes: $4S+5O_2 \rightarrow 2SO_2+2SO_3$. Since Oxygen is still the limiting reagent, we expect products of the reaction to be $\frac{2}{5} \space mol \space SO_2$, $\frac{2}{5} \space mol \space SO_3$ and $\frac{1}{5} \space mol \space S$.

Points I would like to know:

  • What are the ways to distinguish between Sulfur Dioxide gas and Oxygen gas besides the impractical (downright foolish) idea of inhaling the resulting gas and checking for poisoning symptoms?
  • What could I have used the furnace for?
  • If I carry out the experiment, will I get Sulfur Dioxide or can initial conditions like temperature affect the results?
  • $\begingroup$ You cannot use $1$ mole $\ce{O2}$, because $1$ mol of any gas has a volume of about $25$ liters at room temperature and pressure. No laboratory vessel has a volume greater than $1$ liter. The furnace is here to heat the sulfur. Sulfur does not burn at room temperature. It has to be heated to about $200$°C to start burning. The amount of oxygen will be easier determined by knowing the volume of the vessel, and deriving the number of moles by $\pu{n = pV/RT}$ $\endgroup$
    – Maurice
    May 26, 2022 at 18:51

1 Answer 1


Sulfur is burned by the Megatonne to make sulfuric acid. All the conditions have been worked out and can be found in the literature or if you live in Florida, USA by a trip to the Bone Valley phosphate mines where they [used to] make sulfuric acid to convert phosphate rock into phosphoric acid and gypsum [with yellow cake as a side product].

This is a hellish chemical system for a newcomer to chemistry to even consider. It involves consecutive reactions, the handling of gases, and has always been a chemical engineering nightmare. Sulfur is easily burned to give SO2 but further oxidation to SO3 is difficult and requires a complicated reaction or catalysis. What I am trying to tell you is that any reaction sequences or analysis techniques that might be dreamed up would probably have no use in reality and would be useless in learning any chemistry. It took a lot of chemists a lot of work to work it out. One other comment: Equations such as you propose for SO2 +SO3 are not allowed. They are two separate equations and if there is a reaction between the two the stoichiometry is dependent on the various amounts; you don't get one SO2 for every SO3.

If you must tackle this! the approach should be to convert the sulfur to SO2 by burning in excess O2[not air, N2 might cause complications, more research to do]. [If SO3 were to be formed there would have to be excess of O2 so no need to use less or even stoichiometric amounts, just burn all the sulfur]. Then research the various methods of converting SO2 to SO3. Now you might need an analysis technique. I would recommend Raman spectroscopy on the gas phase. This will easily analyse for SO2, SO3, and O2and probably even for residual S8. BTW the analogy between the oxidation of carbon to CO and CO2 and Sulfur to SO2 and SO3 is at best a very weak one. The energetics are very different.


Your Answer

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

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