MathConnections
CH 3 CHCH 2 CH3CH2CH2CH2
4.2 Fundamentals of Stoichiometry
Stoichiometry is a term used to describe quantitative relationships in chemistry.
Any chemistry question that asks “How much?” of a particular substance will be consumed or formed in a chemical reaction is a stoichiometry problem. And at the heart of every such stoichiometry problem, you’ll always fi nd a balanced chemical equation.
We have seen already that chemical equations are always written in terms of the numbers of particles involved. Whether we interpret them in terms of individual mol- ecules or moles of molecules, the stoichiometric coeffi cients that balance a chemical equation refer to numbers of particles and not to masses. Usually, we can’t measure the number of particles directly in the laboratory; masses and volume of liquids are the quantities that are more likely to be measurable. Thus if we want to make quantitative calculations for a chemical reaction, frequently we need to convert between the mea- sured value of a mass or volume and the desired value of a number of moles. Because
Figure 4.2 ❚ Here we show the same sort of molecular picture as in Figure 4.1 but this time for the incomplete combustion of octane to give carbon monoxide and water.
Note that both equations balance, but they clearly show different stoichiometry.
2 C8H18 ⫹ 17 O2 16 CO ⫹ 18 H2O
such calculations are common and important, chemists have developed a standard ap- proach to overcome this mismatch of units. Although you might think of this approach as an algorithm for solving a particular class of chemistry problems, it is instructive to understand its conceptual underpinnings. The key concept is the use of the bal- anced chemical equation to establish ratios among moles of the various materials in the reaction.
Obtaining Ratios from a Balanced Chemical Equation
The simplest combustible hydrocarbon—and therefore one of the simplest of all fuels—is methane, CH4. So the combustion of methane will provide a good place for us to begin our exploration of reaction stoichiometry. We start by writing a balanced chemical equation:
CH4(g) + 2O2(g) CO2(g) + 2H2O(ᐉ)
In Chapter 3, we have already considered how we might read this chemical equation.
We could think of the equation as representing one molecule of methane gas react- ing with two molecules of oxygen or a mole of methane molecules reacting with two moles of oxygen molecules. In either case, we must have a 1:2 ratio of methane mol- ecules to oxygen molecules, a 1:1 ratio of methane molecules to carbon dioxide mol- ecules, a 1:2 ratio of methane molecules to water molecules, and a 2:2 (or 1:1) ratio of oxygen molecules to water molecules.
As we discussed in Section 3.4, the coeffi cients in a chemical equation can be interpreted as molar relationships as well as molecular relationships. So we can say that the equation shows that one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water. From the chemical equa- tion, we can write the following set of mole ratios:
1 mol CH4 : 2 mol O2
1 mol CH4 : 1 mol CO2
1 mol CH4 : 2 mol H2O 2 mol O2 : 2 mol H2O We could also write the last one as 1:1.
When doing stoichiometric calculations, mole ratios are very often written to look like fractions. They are used analogously to unit conversion factors, relating the amount of one substance to that of another:
1 mol CH4
——————
2 mol O2 1 mol CH4
——————
1 mol CO2 1 mol CH4
——————
2 mol H2O 2 mol O2
——————
2 mol H2O qor 1 mol O2
——————
1 mol H2O r Just like unit conversion factors, these mole ratios can be inverted as needed to carry out a particular calculation. Example Problem 4.1 shows how to use mole ratios in a stoichiometry problem.
E X A M P L E P RO B L E M 4 .1
In the combustion of methane, how many moles of O2 are required if 6.75 mol of CH4 is to be completely consumed?
The physical state of the water formed in combustion will depend on the reaction conditions.
The physical state of the water formed in combustion will depend on the reaction conditions.
Strategy We start with the balanced chemical equation and use the stoichiometric co- effi cients to establish the mole ratio between methane and oxygen. Then, we can use that ratio to relate the amount of methane in the reaction to the amount of oxygen needed.
Solution The balanced equation was given earlier in the chapter:
CH4(g) + 2 O2(g) : CO2(g) + 2 H2O(𝓵)
The coeffi cients from this equation give us the ratio between CH4 and O2, which can be expressed in either of the following forms:
1 mol CH4
——————
2 mol O2 or 2 mol O2
——————
1 mol CH4
To calculate the desired amount of O2 from the known amount of CH4, we should use the second form. This gives us the result needed:
6.75 mol CH4× 2 mol O2
——————
1 mol CH4 = 13.5 mol O2
Discussion The answer that we obtained is a direct result of the 2:1 mole ratio and so is reasonably intuitive. You might think of the choice of the form of mole ratio as an application of dimensional analysis. Many stoichiometry problems seem more complicated than this because the known quantity is usually not in moles. But the simple use of the mole ratio illustrated here is always the pivotal step in any reaction stoichiometry calculation.
Check Your Understanding How many moles of H2O are formed in the reaction above?
The balanced chemical equation provides all the mole ratios needed to relate the amounts of the compounds in a reaction. To carry out the reaction in the laboratory, however, we often use a balance to measure out the materials needed, giving us a mea- surement of mass in grams. The ratios from the chemical equation are in numbers of molecules or moles. So we need to be able to convert between grams and moles. As we have seen, the molar mass provides the way to do this.
The use of ratios to convert from one unit to another is no more complicated in stoichiometry than in simple physical measurements. The important difference, how- ever, is that in many stoichiometry problems, ratios are used three or more times.
Each individual use of the ratio follows principles you already know, and if you keep that in mind, you should be able to fi nd your way through even the most complex stoichiometry problems. The main challenge is to determine which ratio to use and when to use it. One way to keep track of the steps needed is to create a fl ow chart using blocks to indicate when information is derived from a relationship, such as the balanced chemical equation or the molar mass.
In Figure 4.3, we can see that if we begin with the mass of one species in the reac- tion and we want to know the mass of some other species consumed or produced, we have to use three ratios: a ratio that uses molar mass of the substance given, a mole ratio from the balanced chemical equation, and a ratio using the molar mass of the
Figure 4.3 ❚ This fl ow diagram illustrates the various steps involved in solving a typical reaction stoichiometry problem.
Mass given
(grams) Molar mass RATIO
Molar mass RATIO Balanced
chemical
equation RATIO Moles
in
Moles out
Mass desired (grams)
substance whose mass we wish to fi nd. The simplest and most reliable way to identify which ratio to use for each step is to write all quantities with their complete units attached. In setting up the simple calculation in Example Problem 4.1, we wrote the initial quantity with units of “mol CH4,” and this helped us choose the correct form of the mole ratio for the calculation. Of course, keeping proper units attached to all the quantities in a calculation is always a good idea, whether it is a chemical stoichiometry problem or an engineering calculation. The following examples apply this idea to slightly more common stoichiometry calculations, where amounts are measured in mass rather than numbers of moles.
E X A M P L E P RO B L E M 4 . 2
In Section 3.2, we considered the reaction between hydrogen and oxygen to form water. How many grams of water can be produced if suffi cient hydrogen reacts with 26.0 g of oxygen?
Strategy First, it will help if we recognize that this is a reaction stoichiometry prob- lem. Two signs of this are that it asks “How much . . .” and that there is obviously a chemical reaction involved. Once we see that, we should immediately realize that we need a balanced chemical equation to give us the mole ratio between water produced and oxygen reacted. We will convert the mass of oxygen given to moles of oxygen, use the mole ratio between oxygen and water to provide moles of water, and then use the molar mass of water to fi nd the mass of water produced.
Solution Recall that both hydrogen and oxygen are diatomic gases. So the balanced equation is
2 H2(g) + O2(g) : 2 H2O(g)
The molar mass of oxygen is 32.0 g/mol, and the molar mass of water is 18.0 g/mol.
Using the diagram in Figure 4.3, we write sequential ratios, starting with the mass of oxygen given and keeping in mind that the answer should be in grams of water:
Mass given
Molar mass ratio for oxygen
Mole ratio from equation
Molar mass ratio for water
26.0 g O2 ⫻ ⫻ 2 mol H2O ⫻ 18.0 g H2O ⫽ 29.3 g H2O 32.0 g O2 1 mol O2
1 mol O2
1 mol H2O
Analyze Your Answer These two masses—for the O2 reactant and the H2O product—are similar. Does this make sense? Looking at the equation for the reac- tion, we can see that all of the oxygen atoms end up in water molecules. Because hydrogen is such a light element, the mass of water formed will not be too much greater than the mass of oxygen reacting. Our answer of 29.3 g of H2O is slightly larger than the 26.0 g of oxygen we started with, so it seems reasonable. (Note that this similarity between the reactant and product masses will not always be the case.)
Discussion Note that we start with the substance whose quantity was given (26.0 g O2). The fi rst and last steps are just conversions between mass and moles, with which you should be comfortable. The use of the mole ratio from the balanced equation is the only new concept here.
This calculation can also be carried out in a more stepwise fashion, in which you might fi rst convert the 26.0 g O2 to 0.813 mol O2 and so forth. The result will be the same as long as all of the calculations are carried out correctly.
Check Your Understanding Calculate the mass of hydrogen needed to produce 29.2 grams of water.
E X A M P L E P RO B L E M 4 . 3
Tetraphosphorus trisulfi de, P4S3, is used in the manufacture of “strike anywhere”
matches. Elemental phosphorus and sulfur react directly to form P4S3: 8 P4+ 3 S8: 8 P4S3
If we have 153 g of S8 and an excess of phosphorus, what mass of P4S3 can be produced by this reaction?
Strategy The heart of any stoichiometry problem is the balanced chemical equa- tion that provides the mole ratio we need. We must convert between masses and the number of moles in order to use this ratio, fi rst for the reactant, S8, and at the end of the problem for the product, P4S3. Molar mass is used to provide the needed con- version factors. The phrase, “excess of phosphorus,” tells us that we have more than enough P4 to consume 153 g of S8 completely.
Solution The molar mass of S8 is 256.6 g/mol and that of P4S3 is 220.1 g/mol. Along with the balanced equation, those masses allow us to set up the needed relationships:
153 g S8× 1 mol S8
—————
256.6 g S8 × 8 mol P4S3
——————
3 mol S8 × 220.1 g P4S3
——————
1 mol P4S3 = 3.50 3 102 g P4S3 Check Your Understanding The smell of a burning match is due to the forma- tion of sulfur oxides. We can represent the relevant reaction as S + O2: SO2. If 4.8 g of sulfur is burned in this manner, what mass of sulfur dioxide is produced?
Now, we can relate the amount of carbon dioxide emitted by an automobile engine and the amount of gasoline burned. What mass of CO2 will be produced for every gallon of gasoline used? While this may sound much different from a typical chemistry book stoichiometry problem, that’s really all it is. Start by assuming that our simple model using the complete combustion of octane to represent the burning of gasoline is accurate. The balanced chemical equation is one we saw earlier:
2 C8H18+ 25 O2: 16 CO2+ 18 H2O
We’ll need to know the mass of a gallon of gasoline. As you might guess, this depends on the exact composition of the gasoline in question. But because we are basing our calcula- tion on octane, it might be reasonable to use the density of octane, which is 0.7025 g/mL.
Use that and some unit conversion factors to fi nd the mass in grams of a gallon of octane:
1 gallon × ————— 3.7854 L
1 gallon × ————— 1000 mL
1 L × 0.7025 g octane
————————
mL = 2659 g
Now we’ve reduced our original question to something that looks a lot more like the previous examples: What mass of CO2 is produced by the complete combustion of 2659 g of C8H18? Use molar masses and the mole ratio from the balanced equation to set up the calculation:
2659 g C8H18× 1 mol C8H18
————————
114.23 g C8H18 × 16 mol CO2
——————
2 mol C8H18 × 44.010 g
——————
1 mol CO2 = 8196 g CO2
To put that in more familiar units, use the fact that a pound is approximately 454 g.
Our calculation then says that burning a gallon of gasoline will produce 18 pounds of carbon dioxide. Although we made a few simplifying assumptions, our result should certainly be a reasonable estimate.