Balancing Reactions

I am surprised at how far balancing equations can take you on the MCAT. Typically, it will allow you to get to a 50:50 and at times all the way to the correct answer. The nice thing about starting by balancing is that it is quick, easy, and applies across multiple content areas within chemistry. We will explore this skill in two different scenarios since they both have slightly different approaches.

Equations As Answers

When your answer choices are balanced chemical equations we will first start by balancing the charges. While not the typical order taught in general chemistry classes it is usually the best place to start since more often than not the elements will be balanced on the MCAT, while the charges aren’t. In this way, we can be more efficient and spend less time on these types of questions. Let’s take a look at the following question to see how this works.

During fermentation in yeast Pyruvate (CH₃C(=O)CO₂^– ) is fully reduced to Ethanol (CH₃CH₂OH) by NADH. What is the balanced reaction for this metabolic process?

1. CH₃C(=O)CO₂^– + 2 H⁺ + NADH → CH₃CH₂OH + CO₂ + NAD⁺
2. CH₃C(=O)CO₂^– + H⁺ + 2 NADH → CH₃CH₂OH + CO₂ + 2 NAD⁺
3. CH₃C(=O)CO₂^– + H⁺ + NADH → CH₃CH₂OH + CO₂ + NAD⁺
4. CH₃C(=O)CO₂^– + 2 H⁺ + 2 NADH → CH₃CH₂OH + CO₂ + 2 NAD⁺

Due to the conservation of charge whatever charge you start with on the reactant’s side has to equal the charge on the product’s side. To begin we will sum up the charge on the left and right sides of each answer choice making sure to account for the stoichiometric coefficients present. Let’s breakdown the first answer choice together charge wise then try the rest on your own and eliminate answers that don’t make sense.

[latexpage]
In the first answer choice, there is one pyruvate with a -1 charge and 2 hydrogens with a +1 charge each on the reactants side. While on the products side this a NAD with a +1 charge and no other charged molecules.

\[
Charge\;Total_{Reactants} = -1 + (2\times1) = +1
\]

Therefore we will go from a +1 charge on the reactants side to a +1 charge on the products side meaning that the charges are balanced and that this is a potentially correct answer.

Now go through the rest of the answer choices and eliminate any answers where the charges are unequal and answer this question below.

Needing Equations to Answer

In other scenarios we will need to generate a balanced reaction and use it to answer a question. This occurs anytime we need to calculate the product yield, determine the limiting reagent, or otherwise use stoichiometric coefficients (e.g calculating ∆G, ∆H, or ∆S). In these circumstances it is best to first focus on balancing the molecules present then reconciling any charges that remain.

To begin we want to pick out an atom that is only in one molecule on both sides and determine if that same number of atoms are present on each side. Here carbon and hydrogen are only in one molecule on each side. For carbon in CH₄ on the reactants side and CO₂ on the products side while hydrogen is present only in CH₄ and only in H₂O on the products side. This is in contrast with oxygen which shows up in both CO₂ and H₂O.

Therefore we would start by seeing if the carbons and hydrogens are matched on both sides. In this case, the carbons are with one showing up in the reactants and one in the products. However, there are four hydrogens on the reactants side and only two on the products side. We can correct this by looking back at our initial reaction and adding a stoichiometric coefficient to balance it out.

A stoichiometric coefficient defines how many moles of a molecule is needed or produced in a chemical reaction. They are always whole numbers so you shouldn’t ever end up with a fraction or decimal as a coffecient once the equation is balanced. If we have two mols of water we will have 4 mols of hydrogen which will balance out with the four mols we have on the reactants side. Additionally, the products gained an additional oxygen molecule that we need to account for when balancing.

Since all of the atoms that show up in only one molecule on either side (this was carbon and hydrogen) are now balanced we will move onto balancing the oxygen atoms. Here we can see that we only have two oxygen on the reactant side and four on the product side so we need to see if we can add in a coefficient to balance it out.

Since we have O2 on the reactants side we can add one more and in the process get the additional two oxygens required for both sides to have the exact same number of moles of each atom.