The Way You SHOULD Be Solving Stoichiometry Problems, Part 1

If you are taking Chemistry for the first time, you may encounter a class of problems known as stoichiometry problems.  These are problems to calculate the amount of product generated (or reactant required) in a particular chemical reaction, given the quantity of some other product or reactant in the reaction.

Most Chemistry teachers teach students to solve these problems using something called dimensional analysis, which looks like a series of multiplied fractions:

And, truth be told, this is not a bad way to work through this type of problem.  What I have discovered, however, is that some of my tutoring clients find this method confusing in the beginning, as they were not properly shown the strategy for deciding which fractions to use and what order to put them in.  I am going to show you a method that will get you the same answer as if you had used dimensional analysis, but in a way which makes the strategy more understandable.

When solving a stoichiometry problem, there are only two actions I am allowed to take:

1. I can convert from moles of one substance to any other unit of the same substance (and vice versa) using a conversion factor.  The conversion factor I have to use depends on what units I am converting between:
   • If I am converting between moles and grams, the conversion factor is the molar mass.  Molar masses are always given in grams / mole.
   • If I am converting between moles and liters of gas at STP (standard temperature and pressure), the conversion factor is 22.4 liters / mole.
   • If I am converting between moles and liters of solution, the conversion factor is the molarity, which is given in moles (of solute) / liters (of solution).
2. I can convert from moles of one substance to moles of a different substance using the ratio of coefficients in the reaction equation.  Note that I can only perform this conversion if both substances are measured in moles.

I can systematize both of these allowed actions by putting them into a table like the one below.  Each substance is assigned a column, and each unit in the problem is assigned a row.  I always want to include a row for moles, but I only need the other rows if the associated units are actually mentioned in the problem.

So in the context of the table:

• Action #1 means that if I have a value in the MOLES row on the table, I can compute any other value in the same column using the associated conversion factor.  Likewise, if I have a value in any row other than the MOLES row, I can convert to the moles value in the same column.  Note that all the allowed transitions are to and from moles only.  These allowed conversions are shown as purple arrows on the table.
  • Because of the order in which I placed the rows on the table, I have to multiply by the conversion factor to compute a value in a higher row from a value in a lower row, and I have to divide by the conversion factor to compute a value in a lower row from a value in a higher row.  I can remember this as, “Multiply to go up; divide to go down.”
• Action #2 means that if I have a value anywhere in the MOLES row of the table, I can compute the values for other columns in the MOLES row only using the reaction equation coefficients. These conversions are shown as green arrows on the table.  Note that I cannot convert values across any row other than the MOLES row.

If a proposed conversion is not supported by an arrow, I can’t do it!

Let’s look at a particular example to illustrate the method:

Fe₂O₃ (s) + 2 Al (s) → 2 Fe (s) + Al₂O₃ (s)

The above equation refers to the thermite reaction, which is used for certain kinds of welding.  If 52.7 grams of iron (III) oxide reacts with an excess of aluminum, what mass of iron can be produced?

Note that the chemical equation is already balanced, but if it were not for some reason, I would have to balance it as my first step.  If you have not been introduced to balancing chemical equations, there are resources online for this, such as at OpenStax or Khan Academy.  Also, please note that the phrase “excess of aluminum” just means that there is more aluminum than needed for the reaction, such that the iron (III) oxide is the limiting reactant.

Now I will create my table.  Since the only units mentioned in the problem are grams, my table will need a grams row and a moles row (since I always have to include a row for moles) and no others.

Since the conversions are going to be between grams and moles, I know that the conversion factor I will need is the molar mass.  I am going to compute the molar mass for each substance and include it with the name of the substance in the column headers.  (If you don’t know how to compute the molar mass of a substance, you can read more about that here.)

• Fe₂O₃ :  2 × (55.85) + 3 × (16.00) = 159.70 g / mol
• Al : 26.98 g / mol
• Fe : 55.85 g / mol
• Al₂O₃ :  2 × (26.98) + 3 × (16.00) =  101.96 g / mol

I am also going to add the given mass of 52.7 grams of iron (III) oxide, and place a question mark ? for the mass of iron, since that is what I am trying to find:

I remember that I am not allowed to go across the table in the grams row, but I am allowed to convert the mass of Fe₂O₃ in grams to moles.  I am going down the column so I divide by the molar mass of Fe₂O₃:

(52.7 g Fe₂O₃) ÷ (159.70 g / mol) = 0.330 mol Fe₂O₃

So I can enter that value into the table in the Fe₂O₃ column and moles row.

Now that I have Fe₂O₃ in moles, I can convert to moles of any other substance.  Since I am trying to get to the question mark, I know that I want to find the quantity of iron in moles.  To do this, I will look at the ratio of coefficients in the reaction equation:

Fe₂O₃ (s) + 2 Al (s) → 2 Fe (s) + Al₂O₃ (s)

This says that 2 iron atoms are created for each iron (III) oxide molecule that is consumed.  So the ratio of moles of Fe to moles of Fe₂O₃ is 2 : 1.  

x = 0.660 mol Fe

I can put this into my table:

Finally, I can convert moles of iron to grams of iron, which is what I wanted for my final answer.  I will use the molar mass of iron, and since I am going up the table, I will multiply by the molar mass:

(0.660 mol Fe) × (55.85 g / mol) = 36.9 grams Fe

Here is the same table without the arrows:

So the reaction can produce a maximum of 36.9 grams of iron.  

Let’s look at a reaction that involves units other than grams.

2 C₈H₁₈ (l) + 25 O₂ (g) → 16 CO₂ (g) + 18 H₂O (g)

The above equation represents the combustion of octane, one of the components of gasoline.  How many liters of CO₂ gas at STP is produced when 2660 grams of octane (roughly one gallon) is burned completely?

As before, I want first to verify that the equation is balanced, and happily it already is.  Since the problem mentions both grams of octane and liters of carbon dioxide gas, I will need THREE rows in the table: one for grams, one for liters of gas, and the obligatory one for moles.  I will also place my known mass of octane and a question mark for the liters of carbon dioxide gas.

I am starting from grams of C₈H₁₈, but again I can’t go to any other substance on the table without converting to moles first.  I need to go down the table, so I divide by the molar mass of C₈H₁₈:

(2660 g C₈H₁₈) ÷ (114.26 g /mol) = 23.3 mol C₈H₁₈

Now I am ready to convert from moles of C₈H₁₈ to moles of CO₂.  The reaction equation says that 16 molecules of CO₂ are produced from 2 molecules of C₈H₁₈, so the ratio is 8 : 1.

x = 186 mol CO₂

Finally, I can convert moles of CO₂ to liters of CO₂ at STP.  I know that the conversion factor is 22.4 liters / mole, and since I am going up the table, I know that I have to multiply:

(186 mol CO₂) × (22.4 L / mol) = 4180 liters CO₂

Again, here is the same table without the arrows:

So my one gallon of gasoline will generate over 4000 liters of carbon dioxide when I burn it in my vehicle’s combustion engine.

Let’s do one final example, this time with solutions.

Ca(OH)₂ (aq) + CO₂ (g) → CaCO₃ (s) + H₂O (l)

The above reaction can be used to precipitate calcium carbonate from calcium hydroxide solution.  How many grams of calcium carbonate could be produced from 1.35 liters of 0.100 M Ca(OH)₂ solution with excess carbon dioxide gas?

It looks like the rows for the table will be grams, liters of solution, and the always-required moles:

As usual, I will have to convert my given value to moles before doing anything else.  This time, I am starting with liters of solution, so I know that the relevant conversion factor is the molarity, which I am told is 0.100 M.  I am going up the table, so I know I have to multiply by this factor:

(1.35 L Ca(OH)₂ solution) × (0.100 M) = 0.135 mol Ca(OH)₂

I can now convert to moles of CaCO₃; this time the ratio is 1 : 1 so I don’t even have to multiply by anything:

And finally I can convert from moles of CaCO₃ to grams of CaCO₃.  I use the molar mass of CaCO₃, and I am going up the table so I multiply:

(0.135 mol CaCO₃) × (100.09 g / mol) = 13.5 g CaCO₃

And without the arrows:

So I will get a little more than 13 grams of calcium carbonate from the entire solution.

Most of my clients find this table approach helpful, but some of them have come back and told me that their teachers will not let them solve these problems using this method.  What to do in that case?  I can show you how to translate the method shown here into the dimensional analysis method that some teachers require.  We will look at that question in Part 2.

I hope you find my post helpful!  If you would like more help in Chemistry, I would love to work with you on 1:1 tutoring.  Please see my website at https://andrewjeungtutoring.com .