A. Simple Problems
A balanced equation is a quantitative statement of a reaction. It relates amounts of reactants to amounts of products. Let us see what this statement means in terms of a particular reaction.

Pentane, C₅H₁₂, burns in oxygen to form carbon dioxide and water. The balanced equation for the combustion of pentane is:

C₅H₁₂ + 8 O₂ 5 CO₂ + 6 H₂O

In qualitative terms, this equation shows that pentane reacts with oxygen to form carbon dioxide and water. In quantitative terms, the equation states that 1 molecule of pentane reacts with 8 molecules of oxygen to form 5 molecules of carbon dioxide and 6 molecules of water. For 15 molecules of pentane, (8 x 15) or 120 molecules of oxygen would be needed for complete reaction; (5 x 15) molecules of carbon dioxide and (6 x 15) molecules of water would be formed. Whatever number (n) of molecules of pentane were used in the reaction, 5n molecules of carbon dioxide and 6n molecules of water would be formed.

If 6.02 x 10²³ molecules (1 mole) of pentane are burned, [8(6.02 x 10²³)] molecules (8 moles) of oxygen are needed. The reaction would form [5(6.02 x 10²³)] molecules (5 moles) of carbon dioxide and [6(6.02 x 10²³] molecules (6 moles) of water. These quantitative relationships are summarized in Table 8.5.

Any balanced equation gives the ratio between moles of reactants and moles of products. Given the number of moles of one component used or produced in a reaction, the number of moles or grams of any other component used or produced can be calculated. Such calculations are called stoichiometry.

A stoichiometric problem can be stated in many ways, but it always contains the following information:

1. The reaction involved.

2. A stated amount of one component of the reaction.

3. A question asking "how much" of another substance is needed or formed in the reaction.

The quantitative problems in previous chapters were solved by answering a series of questions:

1. What is wanted?

2. What is given?

3. What conversion factors are needed to go from "given" to "wanted"?

4. How should the arithmetic equation be set up so that the units of the "given" are converted to the units of the "wanted"?

Stoichiometric problems can be solved by answering the same set of questions. The only difference is that some of the conversion factors are derived from the balanced chemical equation for the reaction involved. The steps to follow in solving a stoichiometric problem are the following:

We will now do some stoichiometry problems. Before we begin, several points should be emphasized. First, to prevent confusion and errors, the name and units of each item in the equation are always shown. Second, no arithmetic is done until the whole equation is written out. Third, all units in the final equation must cancel except for those required in the answer.

Example How many grams of carbon dioxide are formed when 61.5 g penane are burned in oxygen? Equation Wanted ? g CO2 Given 61.5 g C5H12 Conversion factors C5H12, mass to moles: 72.2g C5H12 = 1 mol C5H12 mol C5H12 to mol CO2: 1 mol C5H12 yields 5 mol CO2 CO2, moles to mass: 1 mol CO2 = 44.0 g CO2 Arithmetic equation Answer 187 g CO2

Example What mass of iron(III) oxide (rust) is formed when 6.23 g iron react completely with oxygen in the air to form this product? Equation The name iron(III) oxide shows that iron is prsent in the product as Fe3+, the oxide ion is O-2. Iron(III) oxidethe is Fe2O3. The equation for the reaction is: Wanted ? g Fe2O3 Given 6.23 g Fe Conversion factors Fe, mass to moles: 55.85 g Fe = 1 mol Fe mol Fe to mol Fe2O3: 4 mol Fe yield 2 mol Fe2O3 Fe2O3, moles to mass: 1 mol Fe2O3 = 2(55.85)+ 3(16.0) = 159.7 g Fe2O3 Arithmetic equation Answer 8.91 g Fe2O3

 

Example How many molecules of hydrogen will be formed by the reactions of 2.65 x 10-3 g zinc with hydrochloric acid? Solution Equation Wanted ? molecules H2 Conversion factors Zn, mass to moles: 65.4 g Zn = 1 mol Zn mol Zn to mol H2: 1 mol Zn yields 1 mol H2 H2, moles to number of molecules: 1 mol H2 = 6.02x1023molecules H2 Arithmetic equation Answer 2.44x1019 molecules H2

B. Percent Yield
In Examples 8.11, 8.12 and 8.13 we have calculated the theoretical yield of the reaction--that is, the amount of product that would be obtained if the reaction proceeded completely and only as stated in the equation. Very often--in fact, in most cases--the actual yield of a reaction does not equal the theoretical yield but is a lesser amount. The reasons for such discrepancies are varied: The reactant may have been impure, some of the product may have been lost, or small amounts of substances other than those shown in the equation may have been formed. In these cases the actual yield is less than the calculated theoretical yield. The percent yield is calculated by comparing the actual yield with the theoretical yield.

The percent yield of a reaction depends on results obtained in the laboratory. If we know the amount of reactants used, we can calculate the theoretical yield. We then measure the amount of obtained product, the actual yield. The ratio between the two gives the percent yield.

Percent yield =

actual yield theoretical yield

X 100%

As an example of this process, suppose that, for the reaction in Example 8.11, the actual yield is only 165 g CO₂. The theoretical yield, the one calculated, is 175 g CO₂. The percent yield would be:

165 g 187 g

X 100% = 88.2% yield

Example 8.14 illustrates how to calculate percent yield.

Example The reaction of ethane with chlorine yields ethyl chloride and hydrogen chloride: When 5.6 g ethane react with chlorine, 8.2 g ethyl chloride are obtained. Calculate the percent yield of ethyl chloride. Solution Because this problem involves percent yield, the first step is to calculate the theoretical yield. Equation Wanted ? g C2H5Cl (theoretical yield) Given 5.6 g C2H6, 8.2 g C2H5Cl (actual yield) Conversion factors C2H6, mass to moles: 30.1 g C2H6 = 1 mol C2H6 mol C2H6 to mol C2H5Cl: 1 mol C2H6 yields 1 mol C2H5Cl C2H5Cl, moles to mass: 1 mol C2H5Cl = 64.5 g C2H5Cl Arithmetic equation Answer

The ratio of actual yield to theoretical yield can be used to measure the purity of a sample. Such a determination uses a reaction in which the active ingredient of the sample reacts quantitatively--that is, the reaction gives 100% yield. The percent yield of the reaction measures the percent of active ingredient in the sample. Example 8.15 illustrates this process. The problem investigates the purity of a zinc sample and utilizes the reaction of zinc with hydrochloric acid, according to the equation:

Zn + 2 HCl ZnCl₂ + H₂

This reaction is quantitative. A weighed sample of the impure zinc is immersed in excess hydrochloric acid. An excess of the acid assures that more than enough is present to react with all the zinc in the sample.

Example When a sample of impure zinc weighting 7.45 g reacts with an excess of hydrchloric acid, 0.214 g hydrogen is released. Calculate the percent zinc in the sample. Analysis of the problem 1. The zinc sample is impure. The part that is zinc reacts to form hydrogen; the rest does not. 2. We can calculate the amount of zinc needed to form 0.214 g hydrogen. This amount is the actual amount of zinc in the sample; it must be less than the actual weight of the sample. 3. We can calculate the ratio between the amount of zinc needed to form 0.214 g H2 and the actual weight of the sample. This ratio multiplied by 100% is the percent purity of the sample. Solution 1. Calculate the amount of zinc that would yield 0.214 g hydrogen. Equation Wanted ? g Zn Given 0.214 g H2, 7.45 g sample Conversion factors H2, mass to moles: 2.016 g H2 = 1 mol H2 mol H2 to mol Zn: 1 mol H2 requires 1 mol Zn (from the equation) Zn, moles to mass: 1 mol Zn = 65.4 g Zn Arithmetic equation This amount is the actual amount of zinc in the sample. 2. Calculate the percent ratio, which will equal the percentage of zinc in the sample.

C. Problems Involving a Limiting Reactant
All the stoichiometry problems encountered thus far have had two reactants, but the amount was given for only one of them. In calculating the solutions, we assumed that enough of the second reactant was present to allow complete reaction of the first. This is not always the case. We will encounter many problems in which we know the amounts of all reactants and must calculate which reactant is present in excess before we can calculate a theoretical yield.

A problem of this type might be encountered if you owned a bicycle shop. Among other parts, each bicycle requires two wheels and one set of handlebars. Suppose that, after a busy season, your stockroom contains only 14 wheels and 6 sets of handlebars, although it contains huge quantities of all the other necessary parts. How many bicycles can you build before a new shipment of wheels and handlebars arrives? You have enough wheels for seven bicycles (14 ÷ 2). You have enough handlebars for six bicycles (6 ÷ 1). Clearly, you can put together only six bicycles and will have two unused wheels. No matter how you juggle the parts, you have only enough handlebars for six bicycles. Even with 100 wheels, you could make only six bicycles. The number of handlebars is the limiting factor in the number of bicycles you can make.

Similarly, in a chemical reaction in which the amount of each reactant is known, one reactant is usually present in excess; the amount of the other limits the amount of product that can be formed. The problem becomes how to decide which is the limiting reactant<.

Consider the reaction of hydrogen with chlorine to form hydrogen chloride. The balanced equation for this reaction is:

H₂ + Cl₂ 2 HCl

According to the equation, each mole of hydrogen that reacts requires one mole of chlorine to form two moles of hydrogen chloride (Figure 8.7a). If only 0.5 mol hydrogen is present, only 0.5 mol chlorine is needed, and 1.0 mol hydrogen chloride is formed (Figure 8.7b). If 2.0 mol hydrogen are to react, then 2.0 mol chlorine are necessary, and 4.0 mol hydrogen chloride are formed. Suppose you have 2.0 mol hydrogen but only 1.0 mol chlorine (Figure 8.7d). Only 1.0 mol hydrogen can react, because you have only 1.0 mol chlorine. In this instance, chlorine is the limiting reactant; only 2.0 mol hydrogen chloride are formed, and the second mole of hydrogen remains unreacted.

A limiting reactant problem, then, consists of two simpler problems. Suppose we have the reaction

A + B C

and we know now much A and B are available. To determine how much C will be formed, we must calculate (1) how much C can be prepared from the given amount of A and (2) how much C can be prepared from the given amount of B. The smaller of these two amounts is the theoretical yield of C.

FIGURE 8.7 The amount of product formed by a chemical reaction is limited by the amounts of reactants. In (a) and (b), the reactants are present in the ratio of the balanced equation, H2 + Cl2 2 HCl, so they react completely. In (c) and (d), one of the reactants is present in excess; some of this reactant is left unreacted after the reaction is complete.

Example What mass of lithium chloride can be formed by the reaction of 5.00 g lithium with 5.00 g chlorine? Analysis of the problem We know how much of each reactant is present. The theoretical yield of product might be calculated from either: The true or limiting theoretical yield is the smaller of the two. Solution Equation Wanted ? g LiCl Given Amounts of two reactants: 5.00 g Li and 5.00 g Cl2 Conversion factors For lithium: 6.94 g Li = 1 mol Li 1 mol Li yields 1 mol LiCl For chlorine: 70.9 g Cl2 = 1 mol Cl2 1 mol Cl2 yields 2 mol LiCl For both: 1 mol LiCl = 42.4 g LiCl Arithmetic equation There are two equations: Equation A is based on the amount of lithium present. Equation B is based on the amount of chlorine present. Equation A is:   Equation B is: We now ask ourselves "Will this reaction yield 30.6 g or 5.98 g of lithium chloride?" We must choose the smaller amount, 5.98 g, because only enough chlorine is available to prepare 5.98 g lighium chloride; we cannot get 30.6 g. Thus, chlorine is the limiting reagent, and lithium is present in excess.