The study of chemistry is concerned with understanding how and why chemicals react with one another. A critical tool in building this knowledge is called stoichiometry. The word originates in Greek roots for elements and measurement. This is appropriate because stoichiometry is a mathematical approach to understanding the proportions in which elements (and compounds) react in balanced chemical equations. These proportions are based on the wholenumber ratio of molecules (or formula units) which react with one another. For example, consider the combustion of butane in oxygen to make carbon dioxide and water:
In this balanced chemical equation two molecules of butane (C_{4}H_{10}) are shown to require 13 molecules of oxygen (O_{2}) to react to make carbon dioxide (CO_{2}) and water (H_{2}O). Based on this observation it’s straightforward to conclude that 4 molecules of butane would require 26 molecules of oxygen to react. And 8 molecules of butane require 52 molecules of oxygen and so on. This mathematical relationship continues to hold no matter how many molecules of butane you start with. Consider the case when 6.022 × 10^{23} molecules of butane react with oxygen. In that case,
13 molecules O_{2} 6.022 × 10^{23} molecules C_{4}H_{10} ×  = 3.914 × 10^{24} molecules O_{2} 2 molecules C_{4}H_{10}
You may recall that 6.022 × 10^{23} is the number of items in one mole (1 mol). By this reasoning it is just as true to say that two moles of butane react with thirteen moles of oxygen as to say that two molecules of butane reacts with thirteen molecules of oxygen. So the equation states that two moles of butane react with thirteen moles of oxygen to make eight moles of carbon dioxide and ten moles of water.
In fact, expressing the mathematical proportions of compounds and elements in chemical equations in terms of moles is far more useful to chemists than expressing them in terms of molecules. A mole of a substance is a measurable amount of it: one mole of butane has a mass of 58.122 g. Remember, molar mass is calculated by adding up all the atomic mass in the chemical formula: C_{4}H_{10} has a molar mass of 4 × 12.011 g/mol + 10 × 1.00794 g/mol = 58.122 g/mol. Single molecules of butane are far too small to measure or work with. One molecule has a mass of 9.65 × 10^{23} grams.
Mathematical proportions from chemical equations are expressed as molar ratios. A molar ratio is the relative number of formula units of two chemicals in a balanced chemical equation. The full set of possible molar ratios from the chemical equation for the burning of butane is as follows:


any ratio can be expressed as a fraction: 2 mol C_{4}H_{10} 13 mol O_{2}  or  13 mol O_{2} 2 mol C_{4}H_{10} 
When the ratios are expressed as fractions (see above, at right) they can be used as conversion factors of a sort. If you remember your dimensional analysis this will be familiar. Molar ratios are really proportions that can be used to find out how many moles of any of the other chemicals correspond to a given amount of the first. For example, 2 mol of butane corresponds to 13 mol of oxygen; how many moles of oxygen correspond to 4 mol of butane? Twentysix:
13 mol O_{2} 4 mol C_{4}H_{10} ×  = 26 mol O_{2} 2 mol C_{4}H_{10}
To the wellinformed this means that to burn four moles of butane you need twentysix moles of oxygen gas. How about if you have a number like 3.19 moles of butane? It works even then:
8 mol CO_{2} 3.19 mol C_{4}H_{10} ×  = 12.8 mol CO_{2} 2 mol C_{4}H_{10}
In this calculation, as in all of these examples, you use the molar ratio to convert moles of one substance to moles of another substance. In the example immediately above we found that burning 3.19 mol C_{4}H_{10} produces 12.8 mol CO_{2}.
In principle none of this is any different from making a batch of cookies from scratch. All the ingredients are present in a particular proportion: if you add too much sugar or too little flour or too much vanilla or too few chocolate chips then the cookies don’t come out right. If you double the recipe (2 mol of butane instead of one in the chemical equation ‘recipe’) then you double all the other ingredients and you double the amount of your product. You could even multiply the recipe by a number like 3.19 and, as long as there were no eggs to divide, you could make all the ingredients come out in proportion.
Now let’s look at a few examples in the way they might be phrased in a set of chemistry problems or on a quiz.
How many moles of carbon dioxide result from the burning of 8.9 moles of butane?
8 mol CO_{2} 8.9 mol C_{4}H_{10} ×  = 35.6 mol CO_{2} (because 8.9 × 8 ÷ 2 = 35.6) 2 mol C_{4}H_{10}
If 42 mol of carbon dioxide were produced then how many moles of oxygen were used up in making it?
13 mol O_{2} 42 mol CO_{2} ×  = 68 mol O_{2} (because 42 × 13 ÷ 8 = 68.25) 8 mol CO_{2}
It should be clear why it is that chemists prefer not to try to measure out individual molecules in the lab. It’s just not possible. But why are moles any better? Moles are better because they can be measured in the lab. Every chemical has a characteristic called its molar mass. The molar mass of a chemical is the sum of the atomic masses of the atoms in its chemical formula. If you are reading this then you have probably already learned about molar mass and how to calculate it. Even so, here is an example of how to calculate molar mass:
Molar Mass of Fe_{2}O_{3} (iron(III) oxide, aka rust) Fe: 55.845 g/mol and O: 15.9994 g/mol 2 × 55.845 g/mol + 3 × 15.9994 g/mol = 159.69 g/mol 159.69 g The molar mass of Fe_{2}O_{3} is  1 mol
So one mole of iron(III) oxide has a mass of 159.69 g. Butane has a molar mass of 58.122 g/mol. It is far easier to measure a chemical by mass than by counting atoms and molecules. Moles enable a chemist to correctly understand the proportions of reactants and products in chemical reactions. These proportions are based on numbers of atoms and molecules and not on mass. Yet by measuring masses of chemicals and converting these masses mathematically to numbers of moles the chemist can use a chemical equation to determine how much of each chemical is required for, or produced by, a reaction.
How many grams of carbon dioxide result from the burning of 392 grams of butane?
A  1 mol 392 g C_{4}H_{10} ×  = 6.74 mol C_{4}H_{10} 58.122 g 
B  8 mol CO_{2} 6.74 mol C_{4}H_{10} ×  = 27.0 mol CO_{2} 2 mol C_{4}H_{10} 
C  44.009 g 27.0 mol CO_{2} ×  = 1.19 × 10^{3} g CO_{2} 1 mol 
1 mol 392 g C_{4}H_{10} ×  58.122 g 
8 mol CO_{2} ×  2 mol C_{4}H_{10} 
44.009 g ×  = 1.19 × 10^{3} g CO_{2} 1 mol 
In step A the mass of butane in grams is changed to the number of moles of butane using the molar mass of butane. In step B the moles of butane is changed to moles of carbon dioxide using the molar ratio from the balanced chemical equation. In step C the moles of carbon dioxide is changed to the mass of carbon dioxide in grams. All of these calculation steps can be combined into one calculation, as shown above. This simplifies matters and can mean fewer entries into a calculator, which can help to avoid errors due to putting the wrong number into your calculator. Just remember to ‘multiply by the top and divide by the bottom’ as your learned doing dimensional analysis.
The following exercises were written in the order given to help you to develop the skills necessary to master the topic of stoichiometry. You will need a calculator and a periodic table to complete them. You may also use your ions reference sheet.
Remember Dimensional Analysis  


Basic Stoichiometry5C + 2SO_{2} CS_{2} + 4COSolid carbon reacts with sulfur dioxide to make carbon disulfide and carbon monoxide.  


Basic Stoichiometry with Reactions in WordsWrite a balanced chemical equation for each problem. Then use stoichiometry to answer the question. (Hint: you are being given an opportunity to practice your nomenclature skills).  


Stoichiometry with Masses  
2KClO_{3}
2KCl +
3O_{2}
Potassium chlorate decomposes to form potassium chloride and oxygen gas. 
N_{2} +
3H_{2}
2NH_{3} Nitrogen and hydrogen gases react to form ammonia. 


More Stoichiometry with Masses  
Tin(II) fluoride and hydrogen gas can be produced by the reaction of metallic tin with hydrogen fluoride gas. Balanced Equation: 
Limestone (calcium carbonate) can be decomposed into carbon dioxide and calcium oxide by heating to high temperature. Balanced Equation: 

