# 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!?

##### 2 Answers

Because the equation was not properly balanced.

#### Explanation:

You're dealing with the combustion of *glycine*,

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

#1/2"N"_(2(g))#

makes sense because it describes **half** of one **nitrogen molecule**, **one atom of nitrogen**.

#1/2 xx "2 atoms of N" = "1 atom of N" color(white)(a)color(green)(sqrt())#

On the other hand, something like

#5/2"H"_ 2"O"_((l))#

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

#5/2 xx "2 atoms of H" = "5 atoms of H"color(white)(a)color(green)(sqrt())#

#5/2 xx "1 atom of O" = "2.5 atoms of O" color(white)(a)color(red)(xx)#

You **cannot** have **not** possible in ordinary chemical reactions.

The same goes for

#9/4 xx "2 atoms of O" = 9/2color(white)(a) "atoms of O"color(white)(a)color(red)(xx)#

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

#"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(g))#

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"NH"_ 2"CH"_ 2"COOH"_ ((s)) + 9"O"_ (2(g)) -> 8"CO"_ (2(g)) + 10"H"_ 2"O"_ ((l)) + 2"N"_ (2(g))#

If you want to have a **correct** balanced chemical equation and still use *fractional coefficients*, you can divide all the chemical species by

#2"NH"_ 2"CH"_ 2"COOH"_ ((s)) + 9/2"O"_ (2(g)) -> 4"CO"_ (2(g)) + 5"H"_ 2"O"_ ((l)) + "N"_ (2(g))#

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

#9/2 xx "2 atoms of O" = "9 atoms of O" color(white)(a)color(green)(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.

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**, **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

So, using the **enthalpy of combustion** for glycine (i.e. the enthalpy reaction for a combustion) of about

#DeltaH_"c"^@ = -"973.49 kJ"#

But, only because the reaction involves **equivalent** of glycine.

**WHAT IF I WANT NICER NUMBERS?**

If you feel less confident with fractions, just **scale the reaction** by

#color(blue)(\mathbf(1))"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(g)#

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

#color(green)(\mathbf(4))"NH"_2"CH"_2"COOH"(s) + 9"O"_2(g) -> 8"CO"_2(s) + 10"H"_2"O"(l) + 2"N"_2(g)#

#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.*