2.4 Reaction Stoichiometry

The concept of the mole is especially useful when considering chemical changes on the macroscopic scale - that is, involving quantities of materials that we can actually see and conveniently measure. This is because the coefficients in any balanced equation are relative so we can view those as corresponding to individual molecules of reactants and products, or to moles of them. Thus, as stated earlier and shown in the balanced equation, six molecules of CO₂ react with six molecules of water to yield one molecule of glucose and six molecules of O₂. Likewise six moles of CO₂ react with six moles of water to yield one mole of glucose and six moles of O₂. 

\[ \ce{6 CO2 + 6 H2O    ->     C6H12O6 + 6 O2}  \]

Relationships between the amounts of reactants and products of chemical reactions are analyzed through the application of stoichiometry, a term that loosely translates from the Greek as "measuring elements". To illustrate, we'll frame a typical stoichiometry problem in the context of carbon sequestration, a topic of considerable interest because of its relevance to the remediation of climate change. Let’s say we were interested in the amount of CO₂ that could be removed by plants through photosynthesis: how would we do it? 

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Question: If one 1.00 kg of glucose is produced from CO₂ and water according to the above equation, what mass of CO₂ will have been removed from the atmosphere? 

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Before we begin punching numbers into a calculator, let's consider what information is provided and what we are being asked to do. First, we are given a mass of glucose. Now, if the balanced equation provided us a mass ratio, we would simply need to use that ratio to directly calculate the mass of CO2 produced. For example, if the mass ratio of CO2 consumed to glucose produced was 6:1, the mass of CO2 used in the process would be 6.00 kg. But that is NOT what the balanced equation means. It's worth repeating: the balanced equation gives us the ration of molecules (and moles) between the compounds involved in a given chemical reaction. In other words, it gives us the number of moles CO₂ consumed per mole glucose produced, which is 6:1. This ratio, which is called the stoichiometric factor, is written as a fraction with moles of glucose in the denominator and moles of CO₂ in the numerator; this allows the units of the calculation to appropriately cancel out.  Once we have the moles of CO₂ produced, we can find the corresponding mass. The scheme below illustrates the logic.

 

 

In these sorts of calculations you always need a "starting point" that gives you the information you need to determine the number of moles of a substance involved in the reaction you are interested in. In this case that starting point was "1.00 kg glucose". From this we can get the moles of glucose from its molar mass. Then, from the balanced equation, we generate a stoichiometric factor, which allows us to determine the moles of CO₂ that are consumed to generate the amount of glucose given in the problem. The stoichiometric factor is simply the ratio of the coefficients that appear in the balanced equation corresponding to the compounds you are interested in, in this case CO2 and glucose. In this case the stoichiometric factor is:

\[ stoichiometric\ factor\ = \dfrac{6\ mol\ \ce{CO2}}{1\ mol\ \ce{C6H12O6}} \nonumber \]

It is a common error to "flip" coefficients, for example by writing the following instead:

\[ (incorrect)\ stoichiometric\ factor\ : \dfrac{1\ mol\ \ce{C6H12O6}}{6\ mol\ \ce{CO2}} \nonumber \]

You can easily avoid this mistake by making sure that the units cancel appropriately such that the only remaining unit corresponds to the value you are looking for, in this case the mass of CO₂. So, rather than writing a shortened version of the stoichiometric factor, such as \( \dfrac{6 mol}{1 mol} \) ,or, even worse, \( \dfrac{6}{1} \), writing the units out completely, such as " \( mol\ \ce{CO2} \)"  and "\(mol\ \ce{C6H12O6} \)" will allow you to see if they are cancelling the way they should.

The final step of the problem is to convert the moles of CO₂ to the corresponding mass using its molar mass. The entire solution written out in one expression is shown below. While it is not necessary to write things out in this way (some students prefer writing each step on its own, then using the solution of one step in the next one), it is helpful to get accustomed to this approach and will save you time in the long run.

\[ \text{mass}\  \ce{CO2} = (1.00\ \cancel{kg}\ \ce{C6H12O6})(\dfrac{1000\ \cancel{g}}{1\ \cancel{kg}})(\dfrac{1\ \cancel{mol\ \ce{C6H12O6}}}{180.16\   \cancel{g\  \ce{C6H12O6}}})(\dfrac{6\  \cancel{mol\ \ce{CO2}}}{1\ \cancel{mol\ \ce{C6H12O6}}}) (\dfrac{44.01\ g\ \ce{CO2}}{1\cancel{mol\ \ce{CO2}}}) = 1.47\ \times 10^3\ g\ \ce{CO2} = 1.47\ kg\ \ce{CO2} \nonumber \]

The scheme above will be similar for the vast majority of stoichiometry problems you may encounter, either in your study of chemistry or in laboratory work. These calculations are some of the most commonly done in the lab and in other contexts. They are essentially a form of bookkeeping where the quantities involved are not dollars and cents, but the quantities of elements and compounds involved in a reaction, and are a manifestation of the Law of Conservation of Matter. Such problems all follow a very clear logic that can be summarized graphically (Figure 2-7). The general pattern is as follows: you are given a mass of a particular product or reactant and, from that, you need to find the mass of another reactant or product (any permutation of reactants and/or products are possible). Figure 2-7 refers to given and target pure materials, i.e., elements or compounds. This reflects the fact that calculations such as these are limited to pure materials, as opposed to mixtures, because only with pure materials can you find the number of moles using the corresponding molar mass; the scheme shown above for the glucose problem is essentially the same mass as Figure 2-7 but is written explicitly for that problem. In the more generalized form, the "given" material referred to in the top box of Figure 2-7 is the one that you have the necessary mass information that you need to calculate moles, and the "target" is the compound you are calculating for. The arrows between each box in Figure 2-7 represent conversion factors between the terms. For example, the molar mass allows for the conversion of mass to moles or vice versa, and the stoichiometric factor allows for the conversion of moles of the given material to moles of the target one. 

Figure 2-7. The general pathway followed for stoichiometry calculations mass measurements. There are three essential steps: 1) using the molar mass of the pure substance for which you have a given mass, convert the mass to moles; 2) from the balanced equation, use the corresponding stoichiometric factor to find the number of moles of the pure substance you are calculating for (the target material); 3) using the molar mass of the target compound, calculate the mass of number of moles from step 2.

 

Example 

Problem 2-18. If 500. g of oxygen (O₂) is generated by photosynthesis according to Equation 1 above: a) what mass of glucose is generated along with it? b) What mass of CO₂ was consumed? c) What mass of water? 

Solution

a) 469 g glucose (full solution shown below)

\[ \text{mass}\  \ce{C6H12O6} = (500.\ \cancel{g}\ \ce{O2})(\dfrac{1\ \cancel{mol\ \ce{O2}}}{32.00\   \cancel{g\  \ce{O2}}})(\dfrac{1\  \cancel{mol\ \ce{C6H12O6}}}{6\ \cancel{mol\ \ce{O2}}}) (\dfrac{180.16\ g\ \ce{C6H12O6}}{1\cancel{mol\ \ce{C6H12O6}}}) = 469\  g\ \ce{C6H12O6}  \nonumber \]

b) 688 CO₂

c) 281 g H₂O

 

Exercise 

Problem 2-19. The combustion of elemental sulfur in the presence of moisture yields sulfuric acid according to the unbalanced equation below:

\[ \ce{S + O2 + H2O -> H2SO4} \nonumber \]

a) Balance the chemical equation above.

b) What mass of H₂SO₄ can be generated by the complete reaction of 2.5 kg of pure sulfur?

c) If 500 g of H₂SO₄ is desired, what mass of sulfur is required?

 

The logic of reaction stoichiometry forms the basis for a range of analytical techniques, as described in the following sections.

 

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Additional Problems.

2-20. Calcium carbonate (CaCO₃), the main component of chalk as well as many types of seashells, reacts at high temperatures to form calcium oxide (CaO) according to the following equation:

\[ \ce{CaCO3 ->[heat] CaO + CO2} \nonumber \]

What mass of CaO can be generated if 50.0 g of CaCO₃ reacts according to the above equation?

 

 

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