Chapter 8: Reaction Stoichiometry

Major Concept Area: The Atomic/Molecular View of Matter

• Chemical reactions involve rearrangements of atoms.
• The mole allows the chemist to measure a known number of atoms in the laboratory.
• Chemical equations are recipes for the quantities of materials involved in chemical reactions.
• Electron transfer reactions can be partitioned into half reactions, one producing and one consuming electrons.

The quantitative aspects of chemical equations are crucially important. How much oil will you buy to produce enough heat for the winter? How much gas is needed to get you to New York City? How much sulfur does Union Carbide have to mine to produce 1 million tons of sulfuric acid? How many Calories do you need for lunch in order to run 5 miles? Chemical reactions are everywhere. We now take some time to develop methods for answering questions of this type. In the course of this we will develop the full power of the mole concept. In addition, we will review and extend our knowledge of the three reaction classifications introduced in Chapter 1: electron transfer reactions, proton transfer reactions, and double displacement reactions.

8-1 Reaction Stoichiometry as a Recipe. In Chapter 1 we stressed the idea that a chemical equation for a reaction is essentially a recipe. Our purpose in doing that was to ground the rather abstract notions of chemical formulas and equations to something concrete, with which almost everyone has experience. All of the important quantitative aspects of chemical equations--collectively known as reaction stoichiometry--can be understood in terms of the recipe idea. It is instructive to return to the fruit salad recipe for our first example.

Example 8-1. You intend to make 20 fruit salads for a large dinner party according to the following recipe:

You go to the store to purchase starting materials, and come home with the 10 pounds of apples, 15 pounds of oranges, and 1 pound of grapes. Are these materials sufficient to produce 20 fruit salads?

Solution. We have a problem at the outset: the equation is expressed in terms of numbers of apples, oranges, and grapes; however, we know only the total amounts of each fruit by mass. Clearly we need more information, specifically, the average mass of an apple, an orange, and a grape. Suppose that we can weigh a typical apple, orange, and grape on a simple kitchen scale. Suppose further that the apples, oranges, and grapes are of unusually uniform size, so that the weight of one apple faithfully represents the weight of any apple. Our weighings give the following results:

These unit weights enable us to convert grocery store masses into numbers of apples, oranges, and grapes:

Now we are in a position to apply the recipe, which tells us that for each apple used, we must use 5 oranges and 10 grapes. To use 40 apples would require 200 oranges and 400 grapes. We have more than enough grapes to use up all of the apples, but we do not have enough oranges. We can conclude that the number of fruit salads will be limited by the number of oranges that we have. Because we need 5 oranges per fruit salad, we can make only 6 fruit salads from 30 oranges. This will require 6 apples and 6 * 10 = 60 grapes. We will need to return to the store to get quite a few more oranges. Specifically, we will require enough for 14 more fruit salads, or a total of 14 * 5 = 70 oranges. Since each orange weighs a third of a pound, we need somewhat over 20 pounds more of oranges.

From this example we learn a few important ideas. First, the recipe is expressed in numbers of units of each ingredient, not in terms of their masses. Similarly a chemical equation is expressed in terms of numbers of atoms, molecules, or formula units, not in terms of their masses. Second, the amounts of our fruit salad ingredients are expressed as masses when we buy them at the store. Similarly, when we buy chemical substances, the amounts are expressed as masses. Third, to apply the recipe, we must convert the masses to numbers of things using the mass per thing. Having done this, we figure out how many fruit salads it is possible to make with the given amounts of ingredients. We must do the same thing with a chemical equation. Finally, one of the ingredients is found to limit the number of fruit salads that can be made. We will find that in many cases, one of the reactants limits the amount of product that can be formed in a chemical reaction. This will be the reactant that is used up first; it is called the limiting reactant. Let's now see how the recipe analogy helps us in doing stoichiometric calculations for chemical reactions.

Example 8-2. What mass of pure silicon can be obtained by reduction of SiCl₄ with 10.0 g of magnesium metal at elevated temperature? (This reaction is used in the process of extracting silicon from naturally occurring ores. Silicon is used to make semiconductors.) Assume that a large supply of SiCl₄ is on hand.

Solution. We require first a balanced chemical equation (recipe), which follows from the description of the process. We are told that silicon is produced by reaction of Mg and SiCl₄. We put Mg and SiCl₄ as reactants, and Si as a product. The reaction is carried out at elevated temperature, so magnesium is present as a liquid.

The equation is not and indeed can not be balanced as it is, because there are no magnesium and chlorine atoms in the products. We must first propose at least one other product for the reaction in order to balance it. The simplest choice is magnesium chloride, which can be shown experimentally to be correct. Thus

Next we must balance the equation to conform with the law of mass conservation. This is accomplished by putting coefficient 2 before MgCl₂ to balance chlorine; then coefficient 2 before Mg. Balancing is simple in this case.

At this point, we are ready to approach the problem.

The equation gives us relationships between the numbers of moles of reactants and products. Keeping in mind the recipe analogy, the following roadmap results:

With the exception of the last step, this is exactly what we did in the recipe calculation in Example 8-1.

The stoichiometric factor 1 mole of Si per 2 moles of Mg follows directly from the chemical equation, which tells us that 2 moles of magnesium is required to produce, and is therefore equivalent to, 1 mole of Si. Finally, we convert to mass of Si:

We can therefore expect to obtain 5.78 g of pure silicon from 10 g of magnesium. Note that the amount of SiCl₄ does not play a role in the problem because we are told to assume that we have more than enough of it. This is equivalent to saying that Mg is the limiting reactant.

The situation can be made more realistic by specifying starting amounts for both SiCl₄ and Mg. This is done in the following example.

Example 8-3. How much pure silicon can be produced by reaction of 38.0 g of Mg and 26.7 g of SiCl₄(l)?

Solution. In order to use the chemical equation developed in the preceding example, we must determine the number of moles of each reactant:

The chemical equation indicates that we require 2 moles of Mg for each mole of SiCl₄ that we desire to react; i.e., at the least, we must have twice as much Mg in moles as we have SiCl₄. The numbers above show that we have more than enough Mg to react with all of the SiCl₄ available. We conclude that SiCl₄ is the limiting reagent. The amount of Si product will be limited by the amount of SiCl₄ available.

Now convert back to mass:

8-2 Electron Transfer Reactions. Before considering additional examples of reaction stoichiometry, we will take some time to extend our knowledge of electron transfer reactions. The reaction between Mg and SiCl₄ used in the preceding two examples is an electron transfer reaction. Magnesium begins as an element with uncharged atoms, but finishes as an ionic compound in which it has a charge of 2+. It has lost electrons, and has thus been oxidized. Silicon begins as a component of SiCl₄, in which it has a charge of 4+; and ends as elemental silicon, with zero charge. Si has been reduced because it has gained electrons. For this reaction, the charge changes are fairly easy to see. In the reaction below, however, it may not be obvious whether or not electrons are transferred, and it is certainly not obvious how many have been transferred:

There is a simple set of rules that can be used to readily determine the charge on an atom in any compound. The charge determined according to these rules is called the oxidation state of the atom. This is generally different from the formal charge, introduced in Chapter 3, because the rules for assigning electrons are different. Oxidation state, like formal charge, is an electron bookkeeping procedure. Also like formal charge, it is very useful. The oxidation state of an element in a compound can be determined using the following simple rules for oxidation state.

The Assignment of Oxidation State

1. An atom of an element in its uncombined form has oxidation state zero. Thus, each oxygen atom in O₂ has oxidation state (O.S.) = 0. Each atom in a sample of iron metal, Fe(s), has O.S. = 0.
2. For simple monatomic ions the oxidation state is the charge on the ion. Thus sodium is present in NaCl as Na+, and has O.S. = 1+; chlorine is present as Cl- and has O.S. = -1.
3. In compounds and polyatomic ions, oxygen has oxidation state -2. Because oxygen is the second-most electronegative element, it either gains two electrons to form the O^2- ion, or has the lion's share of electrons in a covalent compound. In Na₂O, the O.S. of oxygen is -2, and that of sodium is 1+. In SO₃, the oxidation state of oxygen is -2, and that of sulfur is 6+.

   There are two exceptions to this rule. The first is hydrogen peroxide, in which the two oxygen atoms are bonded together and must be considered to equally share the electrons of that bond. The O.S. of oxygen is -1 in peroxides. The other is the compound OF₂, in which the O.S. of oxygen is 2+.

4. In compounds hydrogen has oxidation state 1+. The most common ion of hydrogen is the proton, H⁺. With only a few exceptions, hydrogen may be considered to have O.S. = 1+ in its compounds.
5. In most compounds, the oxidation states of elements not mentioned explicitly in these rules can be calculated from the overall charge on the molecule or ion, and the known oxidation states of the remaining elements.

Let's examine a few examples.

Example 8-4. Determine the oxidation state of each element in each molecule or ion:

Solution.

NaBr. This is an ionic compound. The ions have their normal charges. Consequently, O.S.(Na) = 1+; O.S. (Br) = 1-.

H₂SO₄. This is a covalent compound. We can use the rules for the oxidation states of oxygen and hydrogen to determine the oxidation state of sulfur. Each oxygen atom has an oxidation state of 2-. The total for four oxygens is then 8-. Each of the two hydrogens has O.S. 1+, for a total of 2+. The compound is electrically neutral overall, so the oxidation state of sulfur must be such that its sum with the total oxidation states of hydrogen and oxygen is zero:

This gives O.S.(S) = 6+. We can use the following schematic approach to determine unknown oxidation states.

• Write the formula: H₂SO₄
• Under the formula, write the oxidation state of each atom for which the oxidation state is known:
  

  H2	S	O4
  1+		2-

• to determine the total charge of each element, multiply the per-atom oxidation state by the number of atoms, and write the results above the formula:

  2+		8-
  H2	S	O4
  1+		2-

• Determine the total charge of the element of unknown oxidation state using the known element charges and the known overall charge on the molecule or ion, and write the result above the element.

  2+	6+	8-
  H2	S	O4
  1+		2-

• Divide the total charge of the element of unknown O.S. by the number of atoms of the element, and write the result under the formula. This is the oxidation state of the element.

  2+	6+	8-
  H2	S	O4
  1+	6+	2-

The oxidation state of sulfur in H₂SO₄ is 6+.

Using a similar approach gives 6+ for Cr in Na₂Cr₂O₇ and 7+ for Cl in HClO₄. We work through C₂H₆O using the shorthand system developed above to obtain 2- as the oxidation state for carbon:

We are now equipped to examine reaction 8-2-1a for changes in oxidation states. Using our newly developed methods, we conclude that the oxidation state of chromium decreases from 6+ to 3+ as a result of the reaction, while that of carbon increases from 2- to 0. The reaction is clearly electron transfer.

To develop a method for balancing the equations for redox reactions like that in 8-2-1a, we take advantage of the fact that in all electron transfer processes, the total number of electrons lost by the oxidized species is exactly the same as the total number of electrons gained by the reduced species. This is seen NOT to be true in 8-2-1a, because the two chromium atoms of Cr₂O₇^2- gain a total of six electrons, whereas the two carbon atoms of C₂H₆O lose a total of only four. We have some balancing to do.

Balancing Electron Transfer Equations by the Method of Half Reactions. The method of half reactions is a systematic approach to the balancing of electron transfer reactions. We treat it as a series of steps.

Step 1: Divide the reaction into two half-reactions, one involving the oxidized element, the other involving the reduced element:

Step 2: Balance each half reaction separately, first balancing atoms of the element that loses or gains electrons; then balancing oxygen and hydrogen; and finally, balancing electrical charge.

Step 3. Equalize electrons in the two half reactions. It is at this point that we make use of the key fact about redox reactions: that every electron produced in oxidation must be consumed in reduction. We equalize the electrons transferred if we multiply the chromium half reaction by two, and the carbon half reaction by 3:

Step 4. Recombine the balanced half reactions to give a balanced overall electron transfer reaction.

Step 5. Simplify and check. Here we cancel common species, reduce coefficients to lowest whole numbers, and check to make sure that the same numbers of atoms of all types, and charges, occur on both sides. We obtain

This checks for balance, with 4 Cr atoms, 17 O atoms, 34 H atoms, and 12+ charge on each side.

Example 8-5. What mass of acetic acid, C₂H₄O₂, is obtained from reaction of 146.2 g of potassium dichromate, K₂Cr₂O₇ with 84.6 g of ethyl alcohol, C₂H₆O? Discuss changes in molecular stereochemistry when C₂H₆O is converted to C₂H₄O₂.

Solution. The balanced equation is available in 8-2-1b. First, we calculate moles of each reactant:

The equation tells us that we need 3 moles of ethyl alcohol for each 2 moles of dichromate anion reacted. Because 1 mole of potassium dichromate contains one mole of dichromate anion, moles Cr₂O₇^2- = 0.497. Using the 3 to 2 ratio

This is all we need to use up all of the dichromate. Clearly we have available much more ethanol than we need. Dichromate is the limiting reactant and determines the quantity of acetic acid obtained:

The NMR spectrum of ethanol shows an upfield triplet (intensity 3), a midfield quartet (intensity 2) and a midfield singlet (intensity 1), consistent with the structural formula in Figure 8-1a. The spectrum for acetic acid shows an upfield singlet of intensity 3, and a far downfield singlet of intensity 1, consistent with the structural formula in Figure 8-1b. In ethanol, both carbon atoms are surrounded by four electron groups, tetrahedrally distributed, and each group is used to attach an atom or group of atoms to carbon. The stereochemistry is therefore tetrahedral at both carbon atoms. The oxygen atom also has four electron groups distributed tetrahedrally, but only two are used to attach atoms. The stereochemisty at oxygen is bent. Figure 8-1c shows a wedge-hash drawing of the structure. The CH₃ carbon of acetic acid has the same stereochemistry as the CH₃ carbon of ethanol: tetrahedral. However, the second carbon atom is surrounded by only three electron groups, distributed in a trigonal planar arrangement. Since all three groups are used to attach atoms, the shape at this carbon is also trigonal planar. The stereochemistry at the singly bonded oxygen is bent. A wedge-hash representation is shown in Figure 8-1d.

8-3 Additional Aspects of Reaction Stoichiometry . It is usually helpful in learning new concepts to have a visual representation of the concept. Example 8-6 provides such a representation for a double displacement reaction.

Example 8-6. AgNO₃ and NaCl react in solution via double displacement according to

We react a fixed mass (1.500g) of AgNO₃ with varying masses of NaCl and weigh the amount of AgCl obtained as a solid in each reaction. We then plot mass AgCl obtained versus mass NaCl used, as shown in Figure 8-2a.

1. Why does the plot increase linearly in region 1?
2. Why is the plot horizontal in region 2?
3. What is the significance of the point where the region 1 and 2 lines meet (the breakpoint)?
4. Suppose we were to plot mass NaCl/mass AgNO₃ on the x axis. At what value of this quantity would the breakpoint occur?
5. Suppose we were to plot moles NaCl/moles AgNO₃ on the x-axis. At what value of this quantity would the breakpoint occur?
6. If we were to use KCl rather than NaCl, what would the various plots look like?

Solution. 1) In this region, the amount of product goes up linearly with the amount of NaCl added, because there is sufficient AgNO₃ in solution to react with all of the added NaCl. Reaction runs until the limiting reagent, NaCl, is gone. Region 1 defines the range in which NaCl is the limiting reagent.

2) In region 2, the same amount of product is obtained no matter how much NaCl is added. Now the amount of product is determined by the fixed amount of AgNO₃ present in the solution. The same amount of product is always obtained because the amount of AgNO₃ is always the same. In region 2, AgNO₃ is the limiting reagent.

3) The two lines intersect at the point where stoichiometrically equivalent amounts of NaCl and AgNO₃ are present. This gives the amount of NaCl that will exactly react with 1.500 g AgNO₃. When reaction is finished, both reactants will be gone. Neither reactant is limiting.

4) What mass of NaCl is required to just use up 1.500 g AgNO₃? The two substances react in a 1:1 mole ratio. We convert 1.500 g AgNO₃ to moles, then calculate what mass of NaCl gives this same number of moles of NaCl:

The breakpoint would occur at a value of 0.516/1.500 = 0.344.

5) The breakpoint would occur at a mole ratio of 1, based on the stoichiometry of the chemical reaction.

6) It would be necessary to add a larger mass of KCl to reach the breakpoint because it has a larger formula weight. The mass AgCl versus mass KCl plot would have a region 1 with smaller slope than region 1 for NaCl (see Figure 8-2b). However, the same limiting mass of AgCl would be reached. The plot of moles KCl/mole AgNO₃ would coincide with the mole/mole plot for NaCl, because the stoichiometry is the same for the two reactions (Figure 8-2c).

The power of the mole concept is illustrated in the next example.

Example 7. In order to obtain copper metal, a sample of the ore, CuFeS₂, is taken through a process consisting of several steps, as detailed below:

The copper is then solidified and refined by electrochemical methods. What mass of copper can be obtained from 150.0 g of ore, CuFeS₂, by the above sequence of reactions, assuming each reaction runs at 100% efficiency?

Solution. It might at first seem necessary to work through the reactions one at a time to finally arrive at a mass of Cu. This is unnecessary, however. Seemingly complex problems of this type are really quite simple, if we take advantage of the conservation of mass. If we assume that all of the copper containing material produced in one step is used as input for the next, we can expect by the law of conservation of mass that all of the copper contained in the ore ends up as pure copper metal after the 3-step process. Thus

The calculation is simple:

It is interesting to note that the detailed stoichiometries of the three reactions are not used at all in solving this problem. You might want to try to classify each of the reactions as electron transfer, proton transfer, or double displacement.

Stoichiometric calculations are crucially important in the methods used by chemists to determine the relative amounts of elements in a new compound. Example 8-8 introduces the method of combustion analysis. Elemental analysis is the chemist’s name for the process by which the identities and amounts of the elements constituting a particular compound are determined. For the numerous compounds containing carbon and hydrogen, combustion analysis is useful. In this process, a precisely known mass of compound is burned in an excess of oxygen. The carbon in the compound is completely converted to carbon dioxide, CO₂, and the hydrogen to water, H₂O. The CO₂ and H₂O are collected separately and weighed. From the masses of CO₂ and H₂O produced, the masses of carbon and hydrogen in the original mass of compound can be obtained. From this information, their relative numbers of moles can be calculated. Masses of other elements present can be determined by separate methods, leading to an empirical formula.

Example 8-8. When 0.617 grams of a compound of carbon, hydrogen, and nitrogen is burned in excess oxygen, 1.716 g CO₂ and 0.3513 g H₂O are collected. What is the empirical formula of the compound?

Solution. The masses of carbon dioxide and water can be converted to moles. From this information, the moles and consequently the masses of carbon and hydrogen in the original mass of compound can be obtained. The mass of nitrogen is then obtained by difference. Once the masses of all elements in the original mass of compound are known, conversion to the relative numbers of moles gives the formula.

Now the mass of nitrogen in the original mass of compound can be obtained:

We now have

Dividing all 3 by moles N gives the formula: C₅H₅N

Example 8-9. A 32.1 mg sample of a compound containing carbon, hydrogen, and oxygen was burned in oxygen to produce 78.4 mg of CO₂ and 32.0 mg of H₂O. ¹H-NMR and IR data for the compound are given below:

What is the structural formula for the compound?

Solution. The combustion analysis data allow us to determine empirical formula. Then the NMR and IR data can be used to deduce the arrangement of atoms--the structural formula.

(Here we have used the fact that the molar mass is the same in units of mg/mmole as in g/mole.)

Convert oxygen mass to mmoles:

Divide mmoles of each element by the smallest value, 0.445:

The empirical formula is C₄H₈O. The infrared data suggest that a C=O group is present, so we use the NMR data to build a structure based around this group. There are two 3-proton signals in the NMR spectrum, suggestive of CH₃ groups. One CH₃ group produces a singlet signal, so there are no protons on the atom attached to the CH₃ carbon atom. This CH₃ group is probably attached to the carbon atom of the C=O group. So far, we have the following arrangement:

The second CH₃ signal is split to a triplet, suggesting that there are two protons on the neighboring atom. Consistent with this, the two proton signal is split to a quartet, suggesting coupling to a CH₃ group. It appears that the structure is as shown below:

This compound is called methyl-ethyl ketone.

8-4 Reactions Involving Gases . In Chapter 5 we discussed the properties of gases extensively, learning that a very simple relationship describes the interrelationship of volume, pressure, temperature, and moles of gas. Gases are frequently reactants, products, or both in chemical reactions, and often it is convenient to determine the extent of reaction by measuring the amount of gas produced or consumed. The following example illustrates this application.

Example 8-10. 17.24 g of liquid C₆H₁₄ (hexane) is enclosed with 3.80 moles of O₂(g) in a cylinder fitted with a piston. The initial temperature of the mixture is 27 ^oC,and the external (outside) pressure on the piston is 1.00 atm. The hexane and oxygen are then caused to react according to the following equation:

All of the hexane is used up. The heat produced by the reaction causes the temperature to rise to 77 ^oC. The external pressure is maintained at 1.00 atm.

Solution. This problem will give us a workout on gas and stoichiometry concepts.

a. The equation is balanced by placing 6 before CO₂, 7 before H₂O, and 19/2 before O₂. Fractions are then cleared via multiplication by 2. The result is

b. Initial volume of gaseous reactants in the cylinder. We initially have only liquid hexane and gaseous O₂ in the cylinder. We can ignore the very small volume occupied by the hexane. We have 3.80 moles O₂(g) at 27 ^oC and a pressure of 1.00 atm. The ideal gas law gives us the volume:

c. Initial amounts of both reactants are specified; we must determine which is limiting. Oxygen is already expressed in moles, so all we need do is compute moles hexane:

The balanced equation indicates that 19 moles of oxygen are needed for each 2 moles of hexane. For 0.200 moles hexane,

Much more than this is available; hexane is limiting. Conclusions thus far:

To calculate the moles of CO₂ formed is simple:

The total amount of gas at the end of reaction is 1.200 moles CO₂ + 1.900 moles O₂ = 3.100 moles gas. The volume can be calculated from the ideal gas law, using T = 77 + 273, and P = 1 atm:

The piston moves slightly in during reaction.

8-5 Solutions and Concentration. A solution is a mixture of pure substances that is homogeneous at the molecular level. That is, the molecules of the substances are intimately intermixed. Usually, one of the substances is a liquid, called the solvent. A relatively small amount of a second substance, usually a solid, is then dissolved in the liquid. The substance that dissolves is called the solute. A solution is obtained when table sugar is dissolved in water. Water is the solvent, and sugar is the solute. Solutions are extremely important in chemistry. Most chemical reactions are carried out by dissolving the reactants in a solvent, and some substances are sold commercially as solutions because they are unstable in pure form. The amount of solute per unit amount of solvent is called the concentration of the solution. There are several ways of expressing this, the most common of which is called the molarity. Molarity, symbol M, is defined as the number of moles of solute per liter of solution. You will be expected to use and understand molarity in the laboratory; we introduce it here to initiate this process.

Example 8-11. A solution of K₂SO₄ (potassium sulfate) in water is prepared by the following stepwise procedure:

What is the molarity of the solution?

Solution. We calculate the moles of K₂SO₄ and divide by the total volume of solution prepared:

Molarity is a useful concept for dealing with the stoichiometry of reactions in solution. We consider two examples here, one involving double displacement and the other involving proton transfer.

Example 8-12. 34.6 mL of the 0.3004 M K₂SO₄ solution from Example 8-11 is mixed with 52.3 mL of a 0.0954 M solution of BaCl₂. How much BaSO₄(s) forms?

Solution. Aqueous potassium sulfate and barium chloride undergo double displacement in aqueous solution to produce barium sulfate as a solid and leaving potassium chloride dissolved in the water. The reaction is 8-5-1.

This equation relates moles of reactants and products, which can be obtained easily from molarities.

Because 1 mole of K₂SO₄ is required per mole of BaCl₂, BaCl₂ is the limiting reagent. It follows that 4.99 * 10^-3 moles BaSO₄ are produced. Converting to grams,

Example 8-13. 31.05 mL of a 0.151 M solution of the acid, H₂SeO₄, is mixed with 58.74 mL of a 0.201 M solution of the base, NaOH. The acid and base react via proton transfer. Will the final solution be acidic (i.e., have an excess of H⁺) or basic (i.e., have an excess of OH-)?

Solution. The proton transfer reaction involving H₂SeO₄ and NaOH is given in 8-5-2.

As in any reaction stoichiometry calculation, we must determine the numbers of moles of reactants that are initially present.

The equation indicates that 2 mmoles NaOH are required to react with each mmole H₂SeO₄. There is more than enough NaOH available. We conclude that H₂SeO₄ is the limiting reactant, and there will be some NaOH left over when reaction is complete. The final solution will contain an excess of OH-; it will be basic.

Although we have answered the question at this point, we can take things one step further by determining the amount of NaOH remaining after reaction. This is

This quantity of NaOH is present in a total solution volume of 31.05 + 58.74 = 89.79 mL. The molarity of NaOH remaining is thus

8-6 Stoichiometry of Incomplete Reactions . We have made an implicit and drastic assumption about all of the chemical reactions that we have worked with thus far in the text: that they proceed until the limiting reactant is completely used up. When this happens, chemists say that the reaction has "proceeded to completion." In practice, many reactions do not proceed to completion. In most cases this is because the reaction reaches a state of chemical equilibrium in which both reactants and products are present. We deal with chemical equilibrium in Chapter 12 and 13. We want to anticipate those chapters now by developing methods to handle the stoichiometric changes in incomplete reactions. The best approach is by example.

Example 8-14. 5.25 moles of nitrogen and 12.00 moles of hydrogen are mixed and allowed to react according to 8-6-1. Calculate the amount of ammonia produced, assuming that the reaction proceeds only 20% of the way to completion, based on the limiting reactant.

Solution. We must first determine the limiting reactant, based on the given starting amounts and the reaction stoichiometry. The extent of reaction does not enter in to this determination, which is done in the usual way. In this problem, determination of limiting reagent is simple because the amounts are given in moles. The equation requires 3 moles of H₂ for each mole of N₂ used, but only 12.00/5.25 = 2.29 moles of H₂ are available per mole of N₂. We conclude that H₂ is limiting. At this point we bring in the extent of reaction. Based on hydrogen, reaction proceeds only 20%. This means that 20% of the hydrogen initially present will be converted to ammonia, leaving 80% unconverted.

The equation dictates that for each 3 moles of H₂ reacted, 2 moles NH₃ must form:

Incomplete reactions can be nicely handled in terms of a formalism that we describe as follows.

Step 1. Under the chemical equation, write the initial amounts of reactants and products present:

Zero is written under ammonia, because there is no ammonia initially present.

Step 2. Under the initial amounts, write the amounts of substances that are used (reactants) and formed (products). These numbers should be negative for reactants (because their amounts decrease during reaction), positive for products (because their amounts increase during reaction), and should be in accord with the stoichiometry of the reaction.

The number, -2.40, under H₂ is 20% of the initial amount; this is the amount of hydrogen that is used up. The number is negative to indicate this. What numbers should be placed under N₂ and NH₃? This is determined from reaction stoichiometry:

Enter these numbers in the "reacts/forms" line:

Step 3. Calculate the final amounts of the three substances by adding the amount reacted or formed to the initial amount:

This gives us a straightforward system for dealing with incomplete reactions.

8-7 Oxidation States from Lewis Structures . In an earlier section we presented some simple rules for the determination of oxidation state. In this section we show that it is possible to determine oxidation states of the atoms in covalent species (molecules and ions) directly from the Lewis structure of the species, using a procedure very similar to that used to obtain formal charge in Chapter 3. Again, the procedure is systematic. It is carried through using the thiosulfate anion as an example.

Step 1. Develop an acceptable Lewis structure for the species, S₂O₃^2-, using the methods of Chapter 3. For S₂O₃^2-, the structure in Figure 8-3 is obtained.

Step 2. Assign electrons to atoms according to these rules:

• Each atom is assigned all electrons of its lone pairs
• The bond electrons are assigned to the more electronegative atom of the bond. If both atoms are of the same element, the bond electrons are divided in half.

Step 3. Compare the number of electrons assigned with the number of valence electrons that the atom normally has. The oxidation state is given as the difference between the normal number and the number assigned in the Lewis structure.

Note that these differ from formal charges, which would be 1- for each oxygen; 1- for the terminal sulfur; and 2+ for the central sulfur.

Applying the rules in section 8-2 to the thiosulfate anion gives oxidation states 2- for oxygen and 2+ for sulfur. The sulfur value is the average of the values obtained from the Lewis structure. The advantage of the Lewis structure method is that it produces individual atom oxidation states, rather than averages. You might try assigning oxidation states to the carbon atoms of acetic acid, C₂H₄O, using the Lewis structure approach. You should find that the central carbon has an oxidation state of 3+, and the other carbon has oxidation state 3-. These average to zero.

Supplement: Job's Method for Determination of Stoichiometry

Job's Method, also called the Method of Continuous Variation, is a simple and effective approach to the determination of chemical reaction stoichiometry. We will discuss it in the context of generic reaction (8S-1),

which can be rewritten in the form of (8S-2) by dividing all coefficients by "a".

where k = b/a and m = d/a. Job's method is based on the following fact: if a series of solutions is prepared, each containing the same total number of moles of A and B, but a different ratio, R, of moles B to moles A, the maximum amount of product, D, is obtained in the solution in which R = k (the stoichiometric ratio). To implement Job's Method experimentally, one prepares a series of solutions containing a fixed total number of moles of A and B, but in which the R is systematically varied from large to small, and measures the amount of product obtained in each solution. One then plots amount of product versus R, and obtains a maximum at the initially-unknown value of k.

That the maximum amount of product should occur at the stoichiometric ratio can be justified both intuitively and mathematically. The intuitive justification runs as follows. When R is greater than k, there is an excess of reagent B, so reagent A is the limiting reagent. As R is systematically decreased toward k (i.e., as moles A increases and moles B decreases such that moles A + moles B stays constant) the amount of product increases with the amount of limiting reagent, A, until R becomes equal to k. In contrast, when R is less than k, there is an excess of reagent A, and B is limiting. As R is systematically increased toward k (i.e., as moles B increases and moles A decreases), the amount of product increases with the amount of limiting reagent B, until R becomes = k. Putting this all together, we see that as R is varied over the range from zero to the maximum value investigated, the amount of product obtained increases until R = k, then decreases as R becomes larger than k. This demonstrates Job's Method intuitively.

The mathematical justification is also quite simple. We use variable "x" to represent the moles of A in a particular solution, and assume that the total moles of A and B is to be kept at 1.0 throughout the series of solutions. Then in each solution it will be true that

Our goal is to show that the maximum amount of product is obtained when R = moles B/moles A = (1-x)/x is equal to k. We approach this by finding the value of x that maximizes product.

According to equation (8S-2), if x is less than the stoichiometrically correct amount of A, then A is limiting and moles product = mx. A plot of moles product versus x over a series of solutions should be linear, with slope m. Similarly, if x exceeds the stoichiometrically correct amount of A, then B is limiting and moles product = m(1-x)/k. A plot of moles product versus x over a series of solutions should also be linear, with slope = -m/k. The first plot will proceed up to the right as x increases. The second plot will proceed down to the right as x increases. At some point then, the two straight lines will intersect. At the intersection, they have a point in common. The value of x corresponding to this point is obtained by equating the ordinate values and solving for x:

Because the amount of product increases as k is approached from either direction, the point of intersection of the lines occurs at the maximum amount of product obtainable. We have therefore shown that maximum product is obtained when R = k. This is what we set out to demonstrate.

Example. A and B are known to react to form D, but the stoichiometry is uncertain. A Job's Method study yields the following data. Plot quantity of product versus moles A to determine the stoichiometry.

Solution. The plot is shown in Figure 8S-1. The value of k is clearly 4, because at the maximum, moles B/moles A = 1.6/0.4 = 4.

A Detailed Example: Stoichiometry of Copper Amino Acid Complexes by Job's Method. We now describe the application of Job's method to the determination of the stoichiometry of amino acid complexes with the Cu²⁺ ion. Job's Method may be applied to stoichiometric detemination in either of 2 ways. For clarity, we discuss it in terms of the general reaction, (8S-3):

The Standard Method. A series of solutions is prepared in which the sum of the number of moles of A and the number of moles of B is kept constant, but in which the relative amounts of A and B are systematically varied. Thus

For example, the following series of solutions might be prepared from two stock solutions, one 0.1 M in A and the other 0.1 M in B:

Once the solutions are prepared, the amount of product formed is measured in each solution, and a plot of the amount of product formed versus moles of B in the solution is constructed. The crux of Job's Method is that the maximum amount of product will form when A and B are present in the correct stoichiometric ratio. For the data in the table, this plot results. Normally the plot looks like an inverted "V", with the vertex skewed toward one side or the other. Straight lines are drawn through the points on both sides of the vertex, and the moles of B corresponding to their intersection point allows the stoichiometry to be determined. The value of x is then calculated from equation (8S-4):

Because the intersection in the plot occurs at moles B = 0.072/(0.09 - 0.072) = 4, the coefficient of reactant B in equation (1) is 4.

The challenging aspect of the application of Job's Method is in finding a way to measure the amount of product formed in each solution. If the product precipitates, the product in each solution may be recovered by filtration and weighed. If the product absorbs light at a known wavelength, then each solution can be placed in a spectrometer and the absorbance at that wavelength determined. There are many other possibilities. The primary restraint, particularly when using a spectrometric method, is that ONLY THE AMOUNT OF PRODUCT is measured. If, for example, one of the reactants also has significant absorbance at the wavelength being used to monitor the amount of product, the shape of the Job's method plot will be affected, sometimes to an extent that prevents determination of the stoichiometry in a straightforward way. It is important to be alert for this problem.

The Limiting Reagent Method. In this modification of Job's Method, a series of solutions is prepared containing a fixed number of moles of reactant A and varying amounts of reactant B, such that the ratio, moles B/moles A runs the gamut from 0 to a value known to be larger than x. The amount of product in each solution is then measured. Once the amount of B exceeds the stoichiometrically required amount, A is the limiting reagent, and the amount of product formed will remain constant. Thus by noting the ratio [B]/[A] at which the amount of product formed levels off, the stoichiometry can be determined. A plot of amount of product versus [B]/[A] will reveal this point visually. Table 2 provides data for a series of solutions. Look here for a plot of the data.

8-1. Reaction of calcium hydride with water is used to prepare small amounts of hydrogen gas; balanced, it is

8-2. Each year the US Chemical industry produces more sulfuric acid than any other chemical, using the following series of reactions:

8-3. Reaction of carbon with calcium oxide produces carbon monoxide and calcium carbide, CaC₂. 2.45 g CaC₂ is isolated from reaction of 5.00 g calcium oxide with 2.50 g carbon. What is the percent yield of CaC₂?

8-4. A 0.537 g sample of an organic compound containing carbon, hydrogen and oxygen was burned in air to give 1.030 g CO₂ and 0.632 g H₂O. What is the empirical formula of the compound?

8-5. White phosphorus (P₄) is used in military incendiary devices because it ignites spontaneously in air. The product of its reaction with oxygen is P₄O₁₀. How many grams of P₄ will produce 50.0 g of P₄O₁₀?

8-6. Titanium tetrachloride was used during World War II to produce smoke screens for naval vessels, according to the reaction

8-7. Dimethylhydrazine, (CH₃)₂NNH₂, was used as a fuel in the Apollo lunar descent module, with liquid N₂O₄ as the oxidizer. Products of reaction of these two species are H₂O, CO₂, and N₂.

8-8. Fermentation of sugar to ethyl alcohol follows the equation

What is the maximum mass of alcohol that can be obtained from 5.00 * 10² g sugar?

8-9. Freon-12, a refrigerant, is prepared by the reaction

If 1.50 * 10² g CCl₄ is mixed with 1.00 * 10² g SbF₃,

8-10. AgNO₃ and CaCl₂ react in solution according to

A fixed mass, 2.130 g, of AgNO₃ is reacted with varying masses of CaCl₂. The amount of AgNO₃ left unreacted in each reaction is plotted against the mass of CaCl₂ used. The plot is shown in Figure 8-4.

8-11. Consider the following combining weight data:

Deduce the relative atomic weights of hydrogen, oxygen, and nitrogen; deduce the formula for ammonia. Use only the given information please.

8-12. Ammonia is arguably the most important commercially-produced substance because it is essential as a fertilizer in agriculture. It is produced on a huge scale by the reaction below:

What mass of H₂(g) will exactly react with 10.0 g of N₂(g)?

8-13. What mass of iron metal, Fe(s), can be dissolved by reaction with 2.54 x 10⁶ g of sulfuric acid, H₂SO₄? You must write a balanced equation to proceed.

8-14. Sulfuric acid is made commercially by the following reactions, run in sequence:

How much sulfur (S(s)) must be mined to make 1.00 ton (2.00 * 10³ lb) of sulfuric acid by this process?

8-15. A mixture of pure AgCl and pure AgBr is found to contain 60.94% Ag by mass. What are the mass percentages of Cl and Br in the mixture?

8-16. An element X forms an iodide XI₃ and a chloride XCl₃. The iodide is quantitatively converted to the chloride when it is heated in a stream of chlorine:

If 0.5000 g of XI₃ is treated, 0.2360 g of XCl₃ is obtained. Calculate the atomic weight of the element X and identify it.

8-17. Coal contains on average about 5% sulfur by weight. How much SO₂ is produced in one 24-hour day by a coal-fired power plant that burns coal at the rate of 2.0 tons/hour?

8-18. Ca(OH)₂ is commonly used to "scrub" sulfur dioxide from stack gases in power plants, so that it does not escape to pollute the atmosphere. How much CaSO₄ is produced during 30 days of operation of a power plant that burns 5% sulfur coal at the rate of 2.0 tons/hour? The chemical equations are given below.

8-19. A tanker truck spills 120 m³ of concentrated sulfuric acid, which has a density of 1.841 g/mL and is 97.0% sulfuric acid by weight. What mass of sodium carbonate should be brought to the spill site to neutralize the acid?

8-20. 1.000 g of a mixture contains an unknown amount of Ag₂S. Ag₂S is converted to AgCl via the reaction sequence below. If 0.112 g AgCl is obtained, what fraction of the mixture was Ag₂S?

8-21. Which answer is correct? A mole of H₂O and a mole of O₂

8-22. One mole of oxygen molecules contain more independent units (O₂) than one mole of oxygen atoms (O).

8-23. Circle the letter of the "best" answer and give reasons for your choice. One molecule of sulfur contains 8 S atoms. Then one mole of sulfur molecules contains

8-24. Each carbon atom has 6 electrons. How many carbon atoms contain 1 mole of electrons?

8-25. What is incorrect with defining the limiting reagent of a reaction simply as the reagent present in the smallest amount?

8-26. 5 molecules of H₂ and 2 molecules of O₂ in a container react according to the equation

Which diagram best represents the system after reaction is complete?

8-27. 1.429 g of cysteine, an amino acid containing sulfur, is burned in excess oxygen to produce 1.761 g CO₂, 0.841 g H₂O, and 0.855 g of SO₂ (from sulfur combustion). What is the empirical formula for cysteine?

8-28. In the gravimetric determination of phosphorus, an aqueous solution of the dihydrogenphosphate ion (H₂PO₄^-) is treated with a mixture of ammonium (NH₄⁺) and magnesium (Mg²⁺) ions to precipitate magnesium ammonium phosphate (MgNH₄PO₄·6H₂O). This is heated and decomposed into magnesium pyrophosphate (Mg₂P₂O₇), which is weighed. The reactions are

A solution of H₂PO₄^- yields 1.054 g Mg₂P₂O₇. What weight of NaH₂PO₄ would be required to give the original amount of H₂PO₄^-?

8-29. When 3.10 g of a compound containing only carbon, hydrogen, and oxygen was completely burned in oxygen, 4.40 g CO₂ and 2.70 g H₂O were produced. What is the empirical formula of the compound? What is the molecular formula if the molecular weight is 62.1 g/mole?

8-30. A siliceous rock contains the mineral ZnS. To analyze for Zn, a sample of the rock is pulverized and treated with HCl to dissolve the ZnS. Zinc is precipitated from soluton by the addition of potassium ferrocyanide, K₄Fe(CN)₆. After filtering, the precipitate is dried and weighed. The reactions that occur are

If a 2.00 gram sample of rock yields 0.969 g of Zn₂Fe(CN)₆, what is the percentage of Zn in the sample?

8-31. The arsenic in an 8.67-g sample of pesticide was converted to arsenate and precipitated as Ag₃AsO₄ with 50.0 mL of 0.02504 M silver nitrate (AgNO₃). The excess Ag⁺ was then titrated with 3.64 mL of 0.05441 M potassium thiocyanate (KCNS, consisting of the K⁺ and CNS^- ions) to form a precipitate of AgCNS. Calculate the percentage of As₂O₃ in the sample.