# Can we use fractional coefficients in stoichiometric equations?

## Is ${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow {H}_{2} O \left(l\right)$ correct?

Aug 16, 2017

So you ask about half-integral coefficients.....? Can you use them? I dunno, ask your chemistry professor. If he/she uses them on the whiteboard I would say yes....

#### Explanation:

So is ${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow {H}_{2} O \left(l\right)$ right?

Now I cannot have HALF a dioxygen molecule, but I certainly can have half a mole, i.e. a $16 \cdot g$ mass of dioxygen. And this combines with a mole of dihydrogen, i.e. a $2 \cdot g$ mass of ${H}_{2}$, to give a molar quantity of water, i.e. an $18 \cdot g$ mass of water.

The stoichiometric coefficients used in the given equation are thus no more than tools. They may be appropriate or inappropriate given a specific scenario. When we assess stoichiometric equivalence, I maintain that a half-integral coefficient can certainly be useful, and correct, and therefore right. And, for me at least, when we write out the alternative.....

$2 {H}_{2} \left(g\right) + {O}_{2} \left(g\right) \rightarrow 2 {H}_{2} O \left(l\right)$

....when we calculate equivalent masses, there is the possibility of ballsing something up. Do I divide by two, or do I multiply by two...to find mass equivalence? And still, after many years, it is NOT immediately obvious to me.....on the other hand...for...

${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow {H}_{2} O \left(l\right)$

..the mass equivalence is directly obvious to me.

We know that all chemical equations observe stoichiometry.

And the answer to your question is that by the criteria I have advanced.....${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow {H}_{2} O \left(l\right)$ IS RIGHT! They are a means to an end to represent stoichiometry, i.e. conservation of mass and charge. The equation is there to serve me, to help me in calculating stoichiometric equivalence. I am not here to serve some abstract notion of chemical equations.

Likewise.....$H C \equiv C H \left(g\right) + \frac{5}{2} {O}_{2} \left(g\right) \rightarrow 2 C {O}_{2} \left(g\right) + {H}_{2} O \left(l\right)$ is also correct, as is...

${C}_{2} {H}_{6} \left(g\right) + \frac{7}{2} {O}_{2} \left(g\right) \rightarrow 2 C {O}_{2} \left(g\right) + 3 {H}_{2} O \left(l\right)$ because the representation absolutely CONSERVES MASS and CHARGE. Happy? Not everyone will be.