Stoichiometry



Stoichiometry is all about scaling the ratios of chemical formulas and balanced reactions up and down. Most reactions aren't run with perfect whole number amounts of moles. The amounts can be any amount. But no matter what that amount is, there is a perfectly matching amount of all the other species listed in the reaction based on the main balanced chemical reaction.

There are two types of stoichiometry in chemistry. The first type is composition stoichiometry. That is the one where you are only using the ratios of atoms that makeup specific molecules or formula units. The second type is reaction stoichiometry where you are looking at how many moles of a compound will react with how many moles of another compound to make some sort of chemical product. In order for stoichiometry to work, you have to have the correct chemical formulas for each substance (the composition) and the correctly balanced chemical reaction (the reaction, duh).

Once those two things are in place, stoichiometry is really just a matter of scaling things up or scaling things down. Let’s try a couple of examples of stoichiometry where things are scaled way up from the normal amounts we see in the formulas that we use.


Composition Stoichiometry Example

Iron(III)oxide has the formula Fe3O4 and is known in the mining world as magnetite. It is one of the the primary compounds in the ore that is mined so that we can produce iron, and more importantly - steel. So here is the question: how many kilograms of iron can be produced from 1.25 metric tons of iron ore in the form of magnetite? (note, on a question such as this, you should assume all the iron in the ore is extracted)

Answer: The first thing you should notice is the question is completely in mass units so we can answer this question by simply staying in mass units of iron oxide or magnetite. We first get its molar mass by taking the atomic weight of iron times three and the atomic weight of oxygen times four and adding the results. . The answer we get is 231.533 g/mol. Of that, 167.55 is iron. This can easily be turned into percent iron to get 72.36% iron. Now we use that percentage on the 1.25 metric tons of ore to get 0.905 metric tons of iron which is equal to 905 kg of iron. Here's what it looks like all written out.


Reaction Stoichiometry (2 Levels)

As previously stated, reaction stoichiometry is really about all the mole-to-mole ratios that are represented in the balanced chemical equation. There are actually two distinct levels of ratios depicted in a balanced chemical reaction. These are schematically shown in the figure below. Level 1 is the mole to mole ratio that is given in the balanced equation. Level 2 requires more conversions because the masses are given instead of moles. This means you have a pre-step and post-step to get grams into moles and moles back into grams.

The three steps shown are simple conversion steps:

  1. Convert the given amount in grams into moles.
  2. Convert the moles of the given substance into moles of the desired substance (what is asked for).
  3. Convert the moles of the desired substance into grams of that substance.

\[16~\bcancel{\rm CH_4\;molecules} \times {\rm 4~H~atoms\over 1~\bcancel{CH_4~molecule}}=64~{\rm H\;atoms}\]


Level 1: Mole-to-Mole Conversion

This is the main level that we work with. This is the mole-to-mole level. Each coefficient in the balanced equation is the number of moles of that substance used in the overall chemical reaction. This is the main level of reaction stoichiometry. We count in moles and all of our ratios of moles come out in nice and neat whole numbers. Any one of the reactants can be converted into any one of the products (or other reactants for that matter). Every possible mole ratio is contained in the balanced chemical equation. Take a look at the following fully balanced combustion reaction of butane.

2 C4H10   +   13 O2 \(\rightarrow\)   8 CO2   +   10 H2O

This balanced reaction has a total of 6 unique ratios that we can use to find equivalent amounts of each of the four substances listed. We can use these ratios as our unit factor or conversion factor to solve problems where we convert one amount into the equivalent amount of another substance. Here are all six of the dimensional analysis ratios we have at our disposal:

\[{\rm 2~mol\;C_4H_{10}\over 13~mol\;O_2}\]
\[{\rm 2~mol\;C_4H_{10}\over 8~mol\;CO_2}\]
\[{\rm 2~mol\;C_4H_{10}\over 10~mol\;H_2O}\]
\[{\rm 13~mol\;O_2\over 8~mol\;CO_2}\]
\[{\rm 13~mol\;O_2\over 10~mol\;H_2O}\]
\[{\rm 8~mol\;CO_2\over 10~mol\;H_2O}\]

Plus, you can invert each of those ratios and have 6 more. That one balanced equation has ALL that in it. Here are two questions based on these ratios for this reaction.

QUESTION 1: How many moles of water will be formed when we combust 15 moles of butane? To answer this we will use the water-to-butane ratio of 10:2

\[\require{cancel} \newcommand\ccancel[2][black]{\color{#1}{\bcancel{\color{black}{#2}}}} {\rm \left({15~\ccancel[red]{mol\;C_4H_{10}}}\right)\left({10~mol\;H_2O\over 2~\ccancel[red]{mol\;C_4H_{10}}}\right) = 75~mol\;H_2O}\]


QUESTION 2: Butane is burned (combustion!) such that 36 moles of CO2 is collected. How many moles of butane was burned? Now we will work backwards and use the butane-to-CO2 ratio of 2:8

\[\require{cancel} \newcommand\ccancel[2][black]{\color{#1}{\bcancel{\color{black}{#2}}}} {\rm \left({36~\ccancel[red]{mol\;CO_2}}\right)\left({2~mol\;C_4H_{10}\over 8~\ccancel[red]{mol\;CO_2}}\right) = 9~mol\;C_4H_{10}}\]

I'd go on and on about reaction stoichiometry but you can read up on it from OpenStax. Here is the OpenStax link about Reaction Stoichiometry.


Reaction Stoichiometry Example

Let's now consider a chemical reaction - a combustion. Propane reacts with oxygen to give carbon dioxide and water. How many pounds of carbon dioxide will be produced from 50 pounds of propane? How many kilograms is it?

A friendly reminder Make sure you know how to write and balance a chemical reaction. That was pointed out in the previous section. You can always get another perspective on the whole idea by clicking on that OpenStax link over there to the left. They have numerous examples and even practice problems for you there. So let's assume you've got that down. Back to the problem...

Solution Step 1: Step 1 is to write out the correct formulas for the reactants and products. Pretty easy - just look them up on the internet if you don't already know them. Propane (C3H8), oxygen (O2), carbon dioxide (CO2), and water (H2O). Now put in the coefficients (those whole numbers in front of the formulas) on each side until the total number of atoms on each side of the arrow match. There are SO many places on the internet you can go to figure all this out and learn it. Just do it.

Solution Step 2: Step 2 is "doing the math" based on the balanced reaction that you made in step 1. Just to make it feel real, I've done this for this problem and my work is shown below. First, I wrote out the entire balanced equation with all the correct chemical formulas. Next, I use the molar masses of the reactants and products to calculate the answer. Note how I immediately just worked the problem in pounds first - that is, I used mass units of the propane and carbon dioxide. I converted to kg on the last step.





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