The problem in my textbook goes like this:

Burning $0.68\ \mathrm{g}$ of an unknown substance, we obtained $1.28\ \mathrm{g}$ of sulfur oxide (IV) and $0.36\ \mathrm{g}$ of water. Find the chemical formula of the burned substance, providing that it was a complete combustion.

It's quite easy to see that $0.2\ \mathrm{mol}$ of both $\ce{SO2}$ and $\ce{H2O}$ have been produced, there are $0.04\ \mathrm{mol}$ of hydrogen in the water and $0.02\ \mathrm{mol}$ of sulfur in the oxide. Calculating the masses and summing them up, we get $0.68\ \mathrm{g}$, meaning the original substance contained only sulfur and hydrogen, and it's $\ce{H2S}$.

However, the solution volume offers this equation as part of the solution process:

$$\ce{S_{$x$}H_{$y$}O_{$z$} + \frac{x + y}{4} O2 -> $x$ SO2 + \frac{$y$}{$z$} H2O}$$

The volume is full of typos (apparently a plus sign missing here), but still there must be some explanation for the coefficients which I don't get. Why do the coefficients for $\ce{O2}$ and $\ce{H2O}$ equal $(x+y)/4$ and $y/z$? If there's zero oxygen in the first term, won't we get division by zero?

How do we come up with coefficients of this sort? Maybe these particular coefficients are erroneous? (the solutions volume is quite low in quality)

  • 4
    $\begingroup$ It is supposed to be $y/2$ not $y/z$. That arises from balancing hydrogen. Another typo - perhaps it was a scan and something went wrong during the character recognition. $\endgroup$ Apr 30, 2016 at 17:38
  • 1
    $\begingroup$ Please use \ce{...} in the maths environment for chemical formulae. Upside: It automatically does things such as subscript numbers in formulae: $\ce{H2O}$ automatically gives $\ce{H2O}$ $\endgroup$
    – Jan
    Apr 30, 2016 at 17:45

2 Answers 2


The equation is a little dubious, as I mentioned in the comments.

How do we come up with coefficients of this sort?

Let's go from scratch. We don't know the coefficients at this point so we'll just call them $a$, $b$, and $c$. Note that the coefficient of $\ce{S$_x$H$_y$O$_z$}$ is taken to be 1. It doesn't have to be 1; however, for the sake of simplicity you might as well take it to be 1. If you take it to be 2, then you just have to double all the other coefficients.

$$\ce{S$_x$H$_y$O$_z$ + $a\,$O2 -> $b\,$SO2 + $c\,$H2O}$$

In the reactants, we have $x$ sulfur, $y$ hydrogen, and $z + 2a$ oxygen. In the products, we have $b$ sulfur, $2c$ hydrogen, and $2b + c$ oxygen.

Our aim is to express $a$, $b$, and $c$ in terms of $x$, $y$, and $z$. Equating all three, we get

$$\begin{align} x &= b \\ y &= 2c \\ z + 2a &= 2b + c \end{align}$$

which immediately tells you that $b$ and $c$ are equal to $x$ and $y/2$ respectively. Substituting these into the last equation,

$$\begin{align} z + 2a &= 2x + \frac{y}{2} \\ a &= \frac{4x + y - 2z}{4} \end{align}$$

So the full equation is:

$$\ce{S$_x$H$_y$O$_z$ + $\frac{4x + y - 2z}{4}$O2 -> $x\,$SO2 + $\frac{y}{2}\,$H2O}$$

  • $\begingroup$ ...and I'd far prefer to write the last equation as $\ce{S$_x$H$_y$O$_z$ + $(x + \frac14 y - \frac12 z)$O2 -> $x\,$SO2 + $\frac12 y\,$H2O}$, to get rid of the confusing cancellation in the $\ce{O2}$ coefficient. $\endgroup$ Apr 30, 2016 at 19:37

They are very erroneous. No oxygen in the reactant would give you a division by zero as you state, so the last fraction is definitely wrong. Here’s how I would proceed:

  1. We know that there is only one source of sulphur which fully reacts to $\ce{SO2}$; so the amount of sulphur atoms must be equal to $\ce{SO2}$ hence the $x$ before $\ce{SO2}$ is correct.

  2. For every two hydrogens a water molecule is formed, but the amount of oxygen present in the original compound is irrelevant. Hence that fraction should be $\frac{y}{2}$, not $\frac{y}{z}$.

  3. That leaves us the oxygen. Well, it is easy to see that we need one oxygen atom for two hydrogens and two oxygen atoms for a sulphur atom. What about oxygens originally present in the sample? They need to be taken away (subtracted) from the oxygen that we add in, leaving:

    $$2x + \frac{y}{2} - z\ \text{oxygen atoms}\\ \frac{2x + \frac{y}{2} - z}{2}\ \text{oxygen molecules } (\ce{O2})\\ \frac{2x + \frac{y}{2} - z}{2}= \frac{2(2x + \frac{y}{2} - z)}{4} = \frac{4x + y - 2z}{4}$$

    Thankfully, this is in full agreement with Orthocresol’s result.

  • $\begingroup$ Thank you, Jan! It took me awhile to understand the "They need to be removed from the oxygen that we add in" bit, but once I got it, it all clicked together. $\endgroup$ Apr 30, 2016 at 18:19

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