# Why are there fractions in this chemical equation? NH_2CH_2COOH (s) + 9/4 O_2 -> 2 CO_2 (s) + 5/2 H_2O (l) + 1/2 N_2 (g) Thank you!?

May 8, 2016

Because the equation was not properly balanced.

#### Explanation:

You're dealing with the combustion of glycine, $\text{NH"_2"CH"_2"COOH}$, which produces carbon dioxide, ${\text{CO}}_{2}$, water, $\text{H"_2"O}$, and nitrogen gas, ${\text{N}}_{2}$.

Now, the idea with using fractional coefficients to balance chemical equations is that you must make sure that you don't end up with fractions of atoms.

For example, something like

$\frac{1}{2} {\text{N}}_{2 \left(g\right)}$

makes sense because it describes half of one nitrogen molecule, ${\text{N}}_{2}$, which is equivalent to one atom of nitrogen.

$\frac{1}{2} \times \text{2 atoms of N" = "1 atom of N} \textcolor{w h i t e}{a} \textcolor{g r e e n}{\sqrt{}}$

On the other hand, something like

$\frac{5}{2} {\text{H"_ 2"O}}_{\left(l\right)}$

does not make sense because it implies that you're dealing with

$\frac{5}{2} \times \text{2 atoms of H" = "5 atoms of H} \textcolor{w h i t e}{a} \textcolor{g r e e n}{\sqrt{}}$

$\frac{5}{2} \times \text{1 atom of O" = "2.5 atoms of O} \textcolor{w h i t e}{a} \textcolor{red}{\times}$

You cannot have $2.5$ atoms of oxygen because that would imply the splitting of atoms, which is not possible in ordinary chemical reactions.

The same goes for $\frac{9}{4} {\text{O}}_{2}$, which is equivalent to

$\frac{9}{4} \times \text{2 atoms of O" = 9/2color(white)(a) "atoms of O} \textcolor{w h i t e}{a} \textcolor{red}{\times}$

So, the best-case scenario here is that this version of the equation

${\text{NH"_ 2"CH"_ 2"COOH"_ ((s)) + 9/4"O"_ (2(g)) -> 2"CO"_ (2(g)) + 5/2"H"_ 2"O"_ ((l)) + 1/2"N}}_{2 \left(g\right)}$

represents a step in the process of balancing the equation that describes the combustion of glycine. To get the correct balanced chemical equation, you'd have to get rid of the fractional coefficients.

To do that, multiply all the chemical species by $4$. You will have

$4 {\text{NH"_ 2"CH"_ 2"COOH"_ ((s)) + 9"O"_ (2(g)) -> 8"CO"_ (2(g)) + 10"H"_ 2"O"_ ((l)) + 2"N}}_{2 \left(g\right)}$

If you want to have a correct balanced chemical equation and still use fractional coefficients, you can divide all the chemical species by $2$. This will get you

$2 {\text{NH"_ 2"CH"_ 2"COOH"_ ((s)) + 9/2"O"_ (2(g)) -> 4"CO"_ (2(g)) + 5"H"_ 2"O"_ ((l)) + "N}}_{2 \left(g\right)}$

This time, the fractional coefficient makes sense because it gets you

$\frac{9}{2} \times \text{2 atoms of O" = "9 atoms of O} \textcolor{w h i t e}{a} \textcolor{g r e e n}{\sqrt{}}$

So remember, you can use fractional coefficients to balance chemical equation, but make sure that the they make sense at the level of the atom.

Any fractional coefficient that gives you fractions of an atom is not used correctly.

May 9, 2016

Although Stefan has good reason to be concerned about the fractional stoichiometric coefficients, it is not unusual to see fractions in reactions that form a compound whose enthalpy of reaction is being referenced.

So, here's another interpretation.

ENTHALPY OF REACTION IS NORMALIZED ON A PER-MOL BASIS

Standard enthalpy of reaction, $\Delta {H}_{\text{rxn}}^{\circ}$, has units of $\text{kJ/mol}$, relative to the compound of interest being produced. In this case, the enthalpy of reaction is relative to $\text{1 mol}$ of glycine (it is 'normalized on a per-mol basis').

So, in some textbooks, you may see formation reactions that have fractional coefficients, and that is OK.

HOW DOES IT AFFECT CALCULATIONS?

The equation is still balanced, and you can still use it to correctly perform calculations (provided your math is sound).

You should just note that when using fractions, you cannot have a fraction of an atom, so maybe consider it in terms of mass. You can have half a gram. That's OK.

It is a bit convenient actually, because in this case, you do not have to consider whether or not you have to multiply the enthalpy of formation by the stoichiometric coefficient in front of glycine in the reaction---it's $1$!

So, using the enthalpy of combustion for glycine (i.e. the enthalpy reaction for a combustion) of about $\Delta {H}_{\text{rxn"^@ = DeltaH_"c"^@ = -"973.49 kJ/mol}}$, you can easily say this as well:

$\Delta {H}_{\text{c"^@ = -"973.49 kJ}}$

But, only because the reaction involves $\setminus m a t h b f \left(1\right)$ equivalent of glycine.

WHAT IF I WANT NICER NUMBERS?

If you feel less confident with fractions, just scale the reaction by $4$; you will get integer coefficients, but you have to use $4$ times $\Delta {H}_{\text{c}}^{\circ}$ as well when you convert to units of only $\text{kJ}$, so don't forget that.

$\textcolor{b l u e}{\setminus m a t h b f \left(1\right)} {\text{NH"_2"CH"_2"COOH"(s) + 9/4"O"_2(g) -> 2"CO"_2(s) + 5/2"H"_2"O"(l) + 1/2"N}}_{2} \left(g\right)$

color(blue)(DeltaH_"c"^@ = -"973.49 kJ/mol" -> \mathbf(-"973.49 kJ"))

$\textcolor{g r e e n}{\setminus m a t h b f \left(4\right)} {\text{NH"_2"CH"_2"COOH"(s) + 9"O"_2(g) -> 8"CO"_2(s) + 10"H"_2"O"(l) + 2"N}}_{2} \left(g\right)$

color(green)(DeltaH_"c"^@ = -"973.49 kJ/mol" -> \mathbf(-"3893.96 kJ"))

Mainly, if nothing else, at least pay attention to which product corresponds to the enthalpy of reaction, and make sure the reaction itself is balanced, no matter how weird it looks.